## VECTOR CALCULUS. B.Sc. Mathematics (V SEMESTER) CORE COURSE (2011 ADMISSION ONWARDS) UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION

VECTOR CALCULUS B.Sc. Mathematics (V SEMESTER) CORE COURSE (2011 ADMISSION ONWARDS) UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION Calicut Univer...
VECTOR CALCULUS B.Sc. Mathematics (V SEMESTER) CORE COURSE (2011 ADMISSION ONWARDS)

UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION Calicut University, P.O. Malappuram, Kerala, India-673 635

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UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION

B.SC. MATHEMATICS (2011 A DMISSION ONWARDS )

V SEMESTER CORE COURSE:

VECTOR CALCULUS Prepared by: Sri. Nandakumar. M. Assistant Professor, NAM College, Kallikkandi, Kannur. Scrutinized by: Dr. Anil Kumar. V Head of the Dept. Dept. of Maths, University of Calicut.

Layout & Settings: Computer Section, SDE © Reserved

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CONTENTS

PAGES

MODULE - I

5  ‐   70

MODULE - II

71  ‐  137

MODULE - III

138  ‐  166

MODULE - IV

167  ‐  222

SYLLABUS

223

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Vector Calculus

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MODULE - 1 ANALYTIC GEOMETRY IN SPACE

VECTORS 1. A vector is a quantity that is determined by both its magnitude and its direction; thus it is an arrow or a directed line segment. For example force is a vector. A velocity is a vector giving the speed and direction of motion. We denote vectors by lowercase boldface letter a,b,v, etc. 2. A scalar is a quantity that is determined by its magnitude, its number of units measured on a scale. For Problem, length temperature, and voltage are scalars. 3. A vector has a tail, called its initial point, and a tip, called its terminal point. 4. The length

of a vector a is the distance between its initial point and terminal point.

5. The length (or magnitude) of a vector a is also called the norm (or Euclidean norm) of a and is denoted by . 6. A vector of length 1 is called a unit vector. 7. Two vector

and

are equal, written, a=b, if they have the same length and the

same direction. Hence a vector can be arbitrarily translated, that is, its initial point can be chosen arbitrarily. COMPONENTS OF A VECTOR

We consider a

Cartesian coordinate system in space, that is, a usual rectangular

coordinate system with the same scale of measurement on the three mutually perpendicular and terminal point coordinate axes. Then if a given vector has initial point , then the three numbers ………..(1)

Are called the components of the vector a with respect to that coordinated system, and we write simply In terms of components, length of a is given by …….(2) Problem

Find the components and length of the vector a with initial point

and terminal point

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Solution.

The components of a are Hence

. Using (2), the length of the vector is

POSITION VECTOR

A Cartesian coordinate system being given, the position vector r of a point is the vector with the origin as the initial point and as the terminal point. From (1), the components are given by So that Theorem I

(Vectors as ordered triples of real numbers)

A fixed Cartesian coordinate system being given each vector is uniquely determined by its ordered triple of corresponding components. Conversely, to each ordered triple of real numbers

there corresponds precisely one vector

, with

corresponding to the zero vector 0, which has length 0 and no direction. VECTOR ADDITION Definition (Addition of Vectors)

The sum

of two vectors

and

the corresponding components. ....(3) Basic Properties of Vector Addition a)

(commutativity)

b)

(associativity)

c) d) where

denotes the vector having the length

and the direction opposite to that of

SCALAR MULTIPLICATION Definition (Scalar Multiplication by a Number) The product ca of any vector and any scalar c (real number c) is the vector obtained by multiplying each component of a by c. That is,  Vector Calculus

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…...(4)

Geometrically, if , then with has the direction of and with the direction opposite to . In any case, the length of is and if

or

has

(or both)

Basic Properties of Scalar Multiplication (a) (b) (written cka)

(c)

…(5)

(d) Remarks (4) and (5) imply for any vector a (a) (b) Instead of

we simply write and

Problem Given the vectors

Find

and

Solution: ; ; ;

Unit Vectors i,j,k A vector

can also be represented as a=a1i+a2j+a3k

…… (6)

In this representation i,j,k are the unit vectors in the positive directions of the axes of a Cartesian coordinate system . Hence …..(7) Problem The vectors

and

can also be written as

and Inner Product Definition (Inner Product (Dot Product) of vectors) The inner product or dot product

dot b”) of two vectors a and b is the product of their lengths times the cosine of their angle.  Vector Calculus

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…(1)

if

The angle between a and b in measured when the vectors have their initial points coinciding. In components, …...(2) Definition

A vector a is called orthogonal to a vector b if

Then b is also orthogonal to a and we call these vectors orthogonal vectors. 1. The zero vector is orthogonal to every vector. For nonzero vectors if and only if thus

2.

Theorem 1 (Orthogonality) The inner product of two nonzero vector is zero if and only if these vectors are perpendicular. Length and Angle in terms of inner Product we get . Hence From (1), with …(3) From (3) and (1) we obtain for the angle

between two nonzero vectors ……(4)

Problem Find the inner product and the lengths of

and

as well as

the angle between these vectors Solution and (4) gives the angle radians. Properties of Inner Products For any vectors

and scalars

(a)

(Linearity)

(b)

(Symmetry)

(c)

if and only if

(Positive definiteness)

Hence dot multiplication is commutative and is distributive with respect to vector and we have addition, in fact from the above with (Distributivity) Furthermore, from (1) and

we see that (Schwarz inequality)

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Result: Prove the following triangle inequality:

Proof using (3)

since using (3) and Schwarz inequality

Taking square roots on both sides, we obtain

Result: Prove the Parallelogram equality (parallelogram identity)

Proof

, using (3) Derivation of (2) from (1) We can write the given vectors a and b in components as and

Since i,j and k are unit vectors, we have from (3)

Since they are orthogonal (because the coordinate axes perpendicular) Orthogonality Theorem gives

Hence if we substitute those representations of a and b into Symmetry, we first have a sum of nine inner products.

and use Distributivity and

Since six of these products are zero, we obtain (2)

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APPLICATIONS OF INNER PRODUCTS Work done by a force as inner product Consider a body which a constant force p acts. Let the body be given a displacement d. Then the work done by p in the displacement d is defined as

that is, magnitude of the force times length of the displacement times the cosine of the , then . If p and d are orthogonal, then the work is angle between p and d. If zero. If , then which means that in the displacement one has to do work against the force. PROJECTION OF A VECTOR IN THE DIRECTION OF ANOTHER NON ZERO VECTOR

Components or projection of a vector a in the direction of a vector

is defined by

……..(5) where is the angle between a and b. Thus is the length of the orthogonal projection o a on a straight line l parallel to b, taken with the plus sign if b has the direction of b and with the minus sign if has the direction opposite to b Multiplying (5) by

, we have

ie

(b

……(6)

if b is a unit vector,as it is often used for fixing a direction then (6) simply gives p=a.b

(|b|=1)

Definition An orthonormal basis I,j k associated with a Cartesian coordinate system. Then {i, j, k} form an orthonormal basis, called standard basis An orthonormal basis has the advantage that the determination of the coefficients in representation v=l1a+l2b+l3c

(v a given vector)

is very simple. This is illustrated in the following Problem. Problem

Vector Calculus

When

(v a given vector), show that

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Solution , since

and and

Similarly, it can be shown that Normal Vector to a given line •

in the plane are perpendicular (or

i,e, if

orthogonal) if •

and

Two non-zero vectors Consider a line

The line

though the origin and parallel to

when can also be written Now line

. Hence a is perpendicular to the line

and also to

is called a normal vector to

is another normal vector to

(and to

Two straight lines

.

vectors of

and

(and to

and respectively

and

because

and

).

). are perpendicular if their

normal vectors are perpendicular. Since

Problem

and

implies that a is perpendicular to position vector of each points on the

are parallel •

where

is

and

are normal

are perpendicular if

Find a normal vector to the line

Solution Let given line be . Then the through the origin and parallel to is which can also be written as where and Hence by the discussion above a normal vector to the line is Problem Find the straight line perpendicular to the straight line

through the point

Also find the point of intersection of the lines

in the xy-plane and

and

Solution Suppose the required straight line be . Then vector to and is perpendicular to the normal vector . That is

i.e,

is a normal of the line

….(8)

Now, if we take and we have is a normal vector to and hence Since it passes through by substituting in the equation of we have or Hence the equation of the required line is  Vector Calculus

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or

Now the point of intersection of of obtained by solving the following systems of equations.

and

Solving, we obtain is

and

and

can be

Hence the point of intersection of the lines

NORMAL VECTOR TO A PLANE

Let

be a plane in space. It can also be writer as ….(9) and

where

Dividing (9) by

The unit vector in the direction of a is

, we get …..(10)

where

Representation (10) is called Hesse’s normal form of a plane.

In (10), p is the projection of r in the direction of n. Note that the projection has the same constant value

for the position vector r of any point in the plane.

(10) holds if and only if n is perpendicular to the plane. n is called unit normal vector to the plane (the other being –n)

Remark origin.

From the above discussion it follows that

is the distance of the plne from the

Problem 6 Find a unit vector perpendicular to the plane distance of the plane from the origin.

. Also find the

Solution A normal vector to the given line is

Hence

Vector Calculus

and the unit normal vector is given by

12

and since

we have

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and the plane has distance

from the origin

Assignments ,

For the vectors 1.

,

and 2

,

,

4.

find 3.

5

Find the work done by the force p acting on a body if the body displaced from a point A to a point B along the straight segment AB. Sketch p and AB. 6 7 8

Can work be zero or negative? In what cases? Let

. Find the angle between 10

9 11

Find the angle between the straight lines

and

12

Find the angle between the planes

13

Find the angles of the triangle with vertices

14.

Find the angles of the parallelogram with vertices

and

Find the component of a in the direction of b: 15 16. 17 Vector Product Definition

The vector product (cross product)

and

of two vector

is a vector

as follows: •

If a and b have the same direction,

If a and b have the opposite direction,

In any, other case,

has the length …..(1)

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where is the angle between a and b. This is the area of the parallelogram with a and b as adjacent sides The direction of is perpendicular to both a and b and such that a,b,v in this order, form a right handed triple.

CROSED PRODUCT IN COMPONENTS In components, the cross product

is given by ….(2)

Notice that in (2)

Hence

is the expansion of the determinant

by the first row. Definition A Cartesian coordinate system is called right-handed if the corresponding unit vectors I,j,k in the positive directions of the axes form a right handed triple. The system is called left-handed if the sense of k is reversed. Problem Find the vector product coordinates.

of

and

in right-handed

Solution

or Problem

With respect to a right-handed Cartesian coordinate system, let

and

Then

Vectors Product and Standard basis vectors Since i,j,k are orthogonal (mutually perpendicular) unit vectors, the definition of vector product gives some useful formulas for simplifying vector products; in right-handed  Vector Calculus

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coordinates these are …… (3)

For left-handed coordinates, replace k by –k , Thus

GENERAL PROPERTIES OF VECTOR PRODUCT Cross multiplication has the following properties: •

For every scalar l …(4)

It is distributive with respect to vector addition, that is, (a) (b)

….(5)

Vector product is not commutative but anticommutative, that is, …(6)

It is not associative, that is in general

so that the parentheses cannot be omitted. Proof. (4) follows directly from the definition.

…(8) The sum of the two determinants in (8) is the first component of the right side of (5a). For the other components in (5a) and in (5b), equality follows by the same idea. Now to get (6), note that , using (2**)

, as the interchange of rows 2

And 3 multiples the determinants by -1 again using (2**)  Vector Calculus

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We can confirm this geometrically if we set and (11) and for b, a,w to from a right handed triple, we must have

then

by

For the proof of (7), note that , where as

SCALAR TRIPLE PRODUCT Definition (Scalar Triple Product) The scalar product of three vectors a,b and c, denoted by , is defined as :

The scalar triple product of the three vectors

is also denoted by

and is also called the box product

Remarks •

The scalar triple product is a scalar quantity.

Since the scalar triple product involves both the signs of ‘cross’ and ‘dot’ it is some times called the mixed product.

Geometrical meaning of Scalar triple product The scalar triple product has a geometrical interpretation. Consider the parallelepiped with a, b and c as co terminus edges. Its height is the length of the component of a on . To be precise, we should say that this height is the magnitude of , where is the angle between a and . Now,

Where the sign or depends on which is positive or negative according obtuse that is according as a, b, c is right handed or left handed.

is acute or

Hence the volume of the parallelepiped with co terminal edges a, b and c is up to sign, the scalar triple product.

,

Expression for the scalar triple product as a determinant Let

and

Then the scalar triple product can be easily evaluated using the following formula:  Vector Calculus

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Problem

Compute

if

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and

Solution Problem

Find the volume of the parallelepiped whose co-terminal edges are arrows

representing the vectors

and

Solution

Problem

Find the volume of the tetrahedron with co terminal edges representing the vectors and

Solution The volume of the tetrahedron

Hence, the volume of the tetrahedron is Theorem (Linear independence of three vectors) Three vectors form a linearly independent set if and only if their scalar triple product is not zero.

The following is restatement of the above Theorem. Theorem

Scalar triple product of three coplanar vectors is zero.

Proof. Let be three coplanar vectors. Now represent a vector which is perpendicular to the plane containing b and c in which also lies the vector a and hence is perpendicular to a. Therefore Thus when three vectors are coplanar.

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Conversely, suppose that . That is , which shows that is perpendicular to a. But is vector perpendicular to the plane containing b and c hence a should also lie in the plane of b and c. That is a, b, c are coplanar vectors. Problem 11 Show independent.

that

the

are

vectors

linearly

Solution By the Theorem, it is enough to show that the scalar triple product of the given vectors is not zero. It can be seen that the scalar triple product is not zero. Hence the given vectors are linearly independent. Problem

Find the constant

so that the vectors are coplanar.

Solution i.e,

Three vectors a, b, c are coplanar if

if

Problem

or

if

or if and

Prove that the points

are

coplanar. Solution Let the given points be respectively and if these four points are coplanar then the vectors are coplanar, so that their scalar triple product is zero. i.e Now

sition vector of

position vector of

Similarly Now Hence the given points are coplanar.

Equation of a Plane with three points Let and c=x3i+y3j+z3k be the position vectors of three points and Let us assume that the three points Hence they determine a plane. Let  Vector Calculus

and

do not lie in the same straight line. be the position vector of any point  18

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in the plane. Consider the vectors which all lie in the plane. That is and condition for coplanar vectors.

are coplanar vectors. Now we apply the

or or

Problem Find the equation for the plane determine by the points

and

Solution The equation of a plane with three points given by

and

Hence here, the equation of the plane with the points is given by

is

and

or i.e, the equation of the plane is

or Assignments In Assignments

1-9 with respect to a right-handed Cartesian coordinate system, let Find the following expressions.

1.

2.

3

4

5

6

7

8

.

9  Vector Calculus

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10

What properties of cross multiplication do Assignments 1,4 and 5 illustrate ?

11

A wheel is rotating about the x-axis with angular speed 3 The rotation appears clockwise if one looks from the origin in the positive x-direction. Find the velocity . and the speed at the point

12

What are the velocity and speed in Exercise 11 at the point about the y-axis and ?

if the wheel rotates

A force p acts on a line through a point A. Find the moment vector m of p about a point

are

13

14 15

Find the area of the parallelogram if the vertices are

16

Find the area of the triangle in space if the vertices are

17

Find the plane through

18

Find

19

Find the volume of the tetrahedron with the vertices

20

Are the vectors

the

volume

of

the

parallelepiped

if

the

edge

vectors

are

linearly independent ?

LINES AND PLANES IN SPACE In this chapter we show how to use scalar and vector products to write equations for lines, line segments, and planes in space. Lines and Line Segments in Space is line in space passing through a point and is parallel to a vector Then is the set of all points for which is parallel to v. That lies on if and only if is a scalar multiple of .

Suppose is,

Vector equation for the line through

and parallel to v is given by …(1)

Expanding Eq. (1), we obtain

Equating the corresponding components of the two sides gives three scalar equations involving the parameter t:  Vector Calculus

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When rearranged, these equations give us the standard parameterization of the line for the interval as follows : ……(2) Standard parameterization of the line through is given by above. Problem

and parallel to

Find parametric equations for the line through

and parallel to

.

Solution With become

equal to

and

Problem Find parametric equations for the line

equal to

through

Eq (2)

and

.

Solution The vector

is parallel to the line , and Eg.(2) with “base point”

give …..(3)

If we choose

as the “base point” we obtain …..(4)

The equations in (4) serve as well as the equations in (3); they simply place you at a different point for a given value of . Line Segment Joining Two Points To parameterize a line segment joining two points, we first parameterize the line through the points. We then find the values for the end points and restrict to lie in the closed interval bounded by these values. The line equations together with this added restriction parameterize the segment. Problem

Parameterize the line segment joining the points

and

.

Solution We begin with equations for the line through

and , which obtained in Problem 2: …..(5)

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We observe that the point

at and Passes through to Eq (3) to parameterize the line segment:

at

The Distance from a Point to a Line in Space To find the distance from a point to to a line that passes through a point parallel to a vector v, we find the length of the component of normal to the line . In the notation which is

of the figure, the length is Distance from a Point

to a Line Through

parallel to

is given by

......(7) Problem

Find the distance from the point

to the line ......(8)

Solution Putting

we see from the equations for

that

passes through

and is parallel

(v is obtained by comparing (8) with (2) to get

to

With

and

Eq. (7) gives

Equations for planes in Space Suppose plane to the nonzero vector is orthogonal to That is,

lies on

passes through a point and is normal (perpendicular) Then is the set of all points for which if and only if

This equation is equivalent to

or  Vector Calculus

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i.e. Plane Through following equivalent equatations:

normal to

is given by the

Vector equation :

….(9)

Component equatation : ….(10) Problem

Find an equatation for the plane through

perpendicular to

.

Solution Using Eq. (10),

Problem Find the plane through Solution We find a vector normal to the plane and use it with one of the points (it does not matter which) to write an equatation for the plane. The

cross

Product

is

normal

to

the

plane.

Note

that

Similiarly, Hence

We substitute the components of this normal vector and the co ordinates of the point (0,0,1) into Eq. (10) to get

i.e Problem Find the point where the line

Intersects the plane Solution The point  Vector Calculus

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…(12) Lies in the plane if its coordinates satisfy the equatation of the plane; that is, if

the point of intersection is

Putting

The Distance from a Point to a Plane ProblemFind the distance from

to the plane

Solution We follow the above algorithm. Using (11) vector normal; to the given plane is given by

The poins on the plane easiest to find from the plane’s equation are the intercepts .If we take to be the y-intercept, then putting Hence

is

The distance from

and

in the equation of the plane

or

, and then

to the plane is

Angles Between Planes; Lines of Intersection The angle between two intersecting planes is defined to be the ?(acute) angle determined by their normalvectors  Vector Calculus

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Problem Find the angle between the planes

and

Solution Using (11), it can be seen that the vectors

are normals to the given planes and between them (using the definition of dot product) is

respectively. The angle

Problem Find the vector parallel to the line of intersection of the planes

and

. Solution

The line of intersection of two planes is perpendicular to the plane’s normal vector’s and , and therefore parallel to .In particular is a vector parallel to the plane’s line of intersection. In our case, ,

We note that any nonzero scalar multiple of line of intersection of the planes and

is also a vector parallel to the .

Problem Find parametric equations for the line in which the planes intersect.

and

Solution v =14i+2j+15k as a vector parallel to the line. To find a point on the line, we can take any point common to the two planes. Substituting z=0 in the plane equations we obtain

and solving for the x and y simultaneously gives x=3, y= -1. Hence one of the point common to the plane is (3,-1,10) The line is [Using eq.(2)]

Assignments Find the parametric equations for the lines in Exercise 1-6  Vector Calculus

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1. The line through the point 2. The line through

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parallel to the vector i+j+k and

3. The line through the origin parallel to the vector 2j+k 4. The line through (1,1,1) parallel to the z-axis 5. The line through perpendicular to the plane 6. The x-axis Find the parametrizations for the line segments joining the points in Assignments 7-10. Draw coordinate axes anD sketch each segments indicating the direction of increasing t for your parametrization. 7.

,

9

,

.

8.

,

10

,

Find equations for the planes in Assignments 11-13. 11. The plane through

normal to

12 The plane through

and

13 The plane through

perpendicular to the line

14 Find the point of intersection of the lines and 15 Find the plane determined by the intersection of the lines:

16 Find a plane through planes

and perpendicular to the line of intersection of the

In Assignments 17-19, find the distance from the point to the line. 17 18 19 In Exercise 20-22 , find the distance from the point to the plane. 20 21  Vector Calculus

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22 23 Find the distance from the plane

to the plane,

24 Find the angles between the following planes:

Use a calculator to find the acute angles between planes in Assignments 25-26 to the nearest hundredth of a radial. 25. 26 In Exercise 27-28 , find the point in whixh the line meets the given planes 27 28.

Find parametrizations for the lines in which the planes in Assignments 29 intersect. 29. 30.

Given two lines in space, either they are parallel, or they intersect or they are skew (imagine, for Problem the flight paths of two planes in the sky). Exercise 31 give three lines. Determine whether the lines, taken two at a time, are parallel, intersect or are skew. If they interesect, find the point of intersection. 31

CYLINDERS,SPHERE,CONE AND QUADRIC SURFACES Definition A cylinder is a surface generated by a line which is always parallel to a fixed line passes through (intersects) a given curve. The fixed line is called the axis of the cylinder and the given curve is called a guiding curve or generating curve Remark

If the guiding curve is a circle, the cylinder is called a right circular cylinder.

• Since the generator is a straight line, it extends on either side infinitely. As such , a cylinder is an infinite surface.

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• The degree of the equation of a cylinder depends on the degree of the equation of the guiding curve. •

A cylinder, whose equation is of second degree, is called a quadric cylinder.

When graphing a cylinder or other surface by hand or analyzing one generated by a computer, it helps to look at the curves formed by intersecting the surface with planes parallel to the coordinate planes. These curves are called cross sections or traces.

We now consider a cylinder generated by a parabola. Problem Find an equation for the cylinder made by the lines parallel to the z-axis that pass through the parabola Solution Suppose that the point lies on the parabola in the plane. Then, for any value of z, the point will lie on the cylinder because it lies on the line through parallel to the z-axis. Conversely any point whose ycoordinate is the squre of it x-coordinate lie on the cylinder because it lies on the line through parallel to the z-axis Remark Regardless of the value of z, therefore, the points on the surface are the points whose coordinates satisfy the equation . This makes an equation for the cylinder. Because of this, we call the cylinder “the cylinder ’’. As Problem 1 or the Remark follows it suggests, any curve defines a cylinder parallel to the z-axis whose equation is also

in the

- plane

defines the circular cylinder made by the lines Problem The equation parallel to the z-axis that pass through the circle in the xy-plane.  Vector Calculus

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Problem

The

equation

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defines

the

elliptical cylinder made by the lines parallel to the z-axis that passes through the ellipse in the In the similar way, we have the following: •

Any curve

in the

space equation is also •

Any curve also

defines a cylinder parallel to the y-axis whose .

defines a cylinder parallel to the x-axis whose space equation is .

We summarize the above as follows : Problem The equation defines surface made by the lines parallel to the xaxis that passes through the hyperbola in the plane Quadric Surfaces A quadric surface is the graph in space of a second-degree equation in the The most general form is

and z.

Where A,B,C and so on are constants, but the equation can be simplified y translation and rotation, as in the two-dimensional case. We will study only the simpler equations. Although the definition did not require it, the cylinders considered so far in this chapter were also Problem of quadric surfaces. We now examine ellipsoids (these include spheres as a special case), paraboloids, cone, and hyperboloids. Problem

The ellipsoid ……(1)

cuts the coordinate axes at and . It lies within the rectangular box defined by the inequalities The surface is symmetric with respect to each of the coordinate planes because the variables in the defining equation are squared. The curves in which the three coordinate planes cut the surface are ellipses. They are

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The section cut from the surface by the plane

is the ellipse …..(2)

Special Cases: If any two of the semi axes a, b and c are equal, the surface is an ellipsoid of revolution. If all three are equal, the surface is sphere. Problem

The elliptic paraboloid …..(3)

and as the variables x and y in the is symmetric with respect to the planes defining equation are squared. The only intercept on the axes is the origin (0,0,0). Except for this point, the surface lies above or entirely below the xy-plane, depending on the sign of c. The sections cut by the coordinate planes are

…..(4) Each plane

Problem

above the xy-plane cuts the surface in the ellipse

The circular paraboloid or paraboloid of revolution …..(5)

is obtained by taking by b=a in Eg. (3) for the elliptic paraboloid. The cross sections of the surface by planes perpendicular to the z-axis are circles centered on the z-axis. The cross sections by planes containing the z-axis are congruent parabolas with a common focus at the . point (0,0, Application : Shapes cut from circular paraboloids are used for antennas in radio telescopes, satellite trackers, and microwave radio links. Definition A cone is a surface generated by lines all of which pass through a fixed point (called vertex) and (i) all the lines intersect a given curve (called guiding curve) or (ii)

all the lines touch a given surface

Vector Calculus

30

or (iii)

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all the lines are equally inclined to a fixed line through the fixed point.

The moving lines which generate a cone are known as its generators. When the moving lines satisfy condition (ii) in the definition of a cone, we term the cone as enveloping cone. Problem The elliptic cone …..(6)

is symmetric with respect to the three coordinate planes Fig (7). The sections cut by coordinate planes are …(7) ……(8)

The secions cut by planes above and below the xy-plane are ellipse whose enters lie on the z-axis and whose vertices lie on the lines in Eq.(7) and (8). If a=b, the cone is a right circular cone.

Problem 9

The hyperboloid of one sheet ….(9)

is symmetric with respect to each of the three coordinate planes . The sections cut out by the coordinate planes are

…..(10)

The plane cuts the surface in an ellipse with center on the z-axis and vertices on one of the hyperbolas in (10) If a=b, the hyperboloid is a surface of revolution Remark : The surface in Problem 9 is connected, meaning that it is possible to travel from one point on it to any other without leaving the surface. For this reason it is said to have one sheet, in contrast to the hyperboloid in the next Problem, which as two sheets. Problem

The hyperboloid of two sheets ….(11)

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is symmetric with respect to the three coordinate planes. The plane z=0 does not intersect the surface; in fact, for a horizontal plane to intersect the surface, we must have The hyperbolic sections

have their vertices and foci on the . The surface is separated into two portions, one above the plane and the other below the plane . This accounts for the name, hyperboloid of two sheets. Eq.(9) and (11) have different numbers of negative terms. The number in each case is the same as the number of sheets of the hyperboloid. If we replace the I on the right side of either Eq.(9) or Eq.(11) by 0, we obtain the equation

for an elliptic cone (Eq.6). The hyperboloids are asymptotic to this cone in the same way that the hyperbolas

are asymptotic to the lines

In the xy-plane.

Problem The hyperbolic paraboloid …..(12) Has symmetry with respect to the planes

and

. The sections in these planes are ………. (13) ….(14)

In the plane x=0, the parabola opens upward from the origin. The parabola in the plane y=0 opens downward. If we cut the surfaces by a plane

, the section is a hyperbola. ……(15)

With its focal axis parallel to the y-axis and its vertices on the parabola in (13).

Vector Calculus

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If is negative, the focal axis is parallel to the x-axis and the vertices lie on the parabola in (14). Near the origin, the surface is shaped like a saddle. To a person travelling along the surface in the yz-plane, the origin looks like a minimum. To a person travelling along the surface in the xz-plane, the origin looks like a maximum. Such a point is called a minimax or saddle point of surface. Assignments Sketch the surfaces in Assignments 1-32 1. 3

2. .

4

5.

6

7.

8.

9. 10.

11.

12.

13.

14.

15.

16. 17

18.

19

20.

21

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32

CYLINDRICAL AND SPHERICAL COORDINATES Cylindrical and Spherical Coordinates This section introduces two new coordinate systems for space: the coordinate system and the spherical coordinate system. Cylindrical  Vector Calculus

cylindrical coordinates  33

simplify the equations of cylinders. Spherical coordinates spheres and cones.

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simplify

the

equations

of

Cylindrical Coordinates We obtain cylindrical coordinates for space by combining polar coordinates in the xyplane with the usual z-axis. This assigns to every point in space one or more coordinates triples of the form (r, θ, z), as shown in Fig 1. Definition Cylindrical coordinates represent a point P in space by ordered triples (r, θ , z) in which 1. r and θ are polar coordinates for the vertical projection of P on the xy - plane, 2. z is the rectangular vertical coordinate. The values of x, y, r , and θ in rectangular and cylindrical coordinates are related usual equations.

by

the

Equations Relating Rectangular (x,y,z) and Cylindrical (r, θ, z) Co ordinates x = r cos θ, y = r sin θ, z = z

;

r2 = x2 + y2 tan θ =

In cylindrical coordinates , the equation r = a describes not just a (1)  circle in the xyplane but an entire cylinder about the z-axis . The z-axis is given by r = 0. The equation θ= θ0 describes the plane that contains the zaxis and makes an angle θ0 with the positive x-axis. And, just as in rectangular coordinates, the equation z= z0 describes a plane perpendicular to the z-axis. Problem

What points satisfy the equations r = 2, θ =

Solution These points make up the line in which the cylinder r = 2 cuts the portion of the plane θ = where r is positive. This is the line through the point (2, , 0)parallel to the z -axis. Along this line, z varies while r and θ have the constant values r = 2 and θ = Problem Sketch the surface r = 1 + cos θ Solution

Vector Calculus

The equation involves only r and θ; the coordinate variable z is missing . Therefore, the surface is a cylinder of lines that pass through the cardioid r = 1 + cos θ in the r θ-plane and lie parallel to the z-axis. The rules for sketching the cylinder are the same as always: sketch the x-, y-, and z-axes, draw a few

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perpendicular cross sections, connect the cross sections with parallel lines , and darken the exposed parts. Problem Find a Cartesian equation for the surface z = r2 and identity the surface. Solution From Eqs. (1) we have z=r2 = x2 + y2. The surface is the circular paraboloid x2 + y2 = z Problem Find an equation for the circular cylinder 4x2 + 4y2 = 9 in cylindrical coordinates.

Solution

The cylinder consists of the points whose distance from the z-axis is =

The corresponding equation in cylindrical coordinates is

r= Problems Find an equation for the cylinder x2 + (y−3)2 = 9 in cylindrical coordinates Solution

The equation for the cylinder in cylindrical coordinates is the same as the polar equation for the cylinder's base in the xy- plane:

Spherical Coordinates

Definition Spherical coordinates represent a point P in space by ordered triples ( ,

in which

1.

is the distance from P to the origin.

2.

is the angle

3.

is the angle from cylindrical coordinates.

makes with the positive z-axis (0

),

The equation = a describes the sphere of radius a centered at the origin. The equation = 0 describes a single cone whose vertex lies at the origin and whose axis lies along the z-axis. (We broaden our interpretation to include the xy-plane as the cone ) If

0 is

greater than

, the cone

=

0

opens downward.

Equations Relating Spherical Coordinates to Cartesian and Cylindrical Coordinates r= r=

Problem

Solution  Vector Calculus

, x = r cos , y = r sin

=

=

, ,

(2)

Find a spherical coordinate equation for the sphere

We use Eqs. (2) to substitute for x, y and z:  35

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1

1

Problem

Find a spherical coordinate equation for the cone z =

Solution Use geometry. The cone is symmetric with respect to the z-axis and cuts the first quadrant of the yz-plane along the line z = y. The angle between the cone and the positive z-axis is therefore radians. The cone consists of the points whose spherical coordinates have

equal to

so its equations

is

Assignments

In Assignments 1-26, translate the equations and inequalities from the given coordinates system (rectangular, cylindrical, spherical ) into equations and inequalities in the other two systems. Also , identify the figure being defined. 1. r = 0

2.

3. z = 0

4. z = −2

5. z =

6. z =

7.

8.

9.

10.

11.

= 5 cos

12.

=1

= −6 cos

13. r = csc θ

14. r = −3 sec θ

15.

16.

17. 18. 19. 20.

Vector Calculus

36

21. z = 4−4

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,0

22. z = 4−r, 0 23. 24.

,0 ,0

25. z +

=0

26. 27.

Find

the

rectangular

coordinates

of

the

center

of

the

sphere

___________________________________

VECTOR- VALUED FUNCTIONS AND SPACE CURVES Space Curves

In this chapter, we shall consider the equations of the form                                 …..(1)

where,

and

are real valued functions of the scalar variable t. As t increases from its

initial value   to the value   the point

trace out some geometric object in space;

it  may  be  straight  line  or  curve.  This  geometric  object  is  called  space  curve  or  arc.  Simply,  the  equations    represent a curve  in space. A space curve is the locus of the point  functions of a single variable

whose co‐ordinates are

Definitions  When a particle moves through space during a time interval I, then the particle’s coordinates  can be considered as functions defined on I.    The points

make up the curve in space that we call the particle’s

path. The equations and interval in (1) parameterize the curve.  The vector                                                Vector Calculus

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from  the  origin  to  the  particle’s  position

at  time    is  the  particle’s  position

vector. The functions   and h are the component functions (components) of the position vector.  We think of the particle’s path as the curve traced by   during the time interval                  Equation  (1)  defines    as  a  vector  function  of  the  real  variable    on  the  interval    More  generally,  a  vector  function  or  vector‐valued  function  on  a  domain  set  D  is  a  rule  that  assigns  a  vector in space to each element in D. For now the domains will be intervals of real numbers. There  are situations when the domains will be regions in the plane or in space. Vector functions will then  be called “vector fields”. A detailed study on this will be done in later chapter.  We  refer  to  real‐valued  functions  as  scalar  functions  to  distinguish  them  from  vector  functions. The components of r are scalar functions of t. When we define a vector‐valued function  by  giving  its  component  functions,  we  assume  the  vector  function’s  domain  to  be  the  common  domain of the components.  It  is  most  convenient  to  use  the  Problem  Consider  the  circle  trigonometric  functions  with  t  interpreted  as  the  angle  that  varies  from .  Then  we  have

In  this  chapter  we  show  that  curves  in  space  constitute  a  major  field  of  applications  of  vector  calculus. To track a particle moving in space, we run a vector r from the origin to the particle and  study the changes in r.  Problem  A  straight  line  L  through  a  point  A  with  position  vector    direction of constant vector   can be represented in the form

in  the

……(2)

If  b  is  a  unit  vector,  its  components  are  the  direction  cosines  of  L.  In  this  case  distance of the points of L from A.

measures  the

Problem The vector function

is defined for all real values of t. The curve traced by r is a helix (from an old Greek word for “spiral”)  that winds around the circular cylinder  . The curve lies on the cylinder because  the i‐ and j‐ components of r, being the x‐ and y‐ coordinates of the tip of r, satisfy the cylinder’s  equation:    The  curve  rises  as  the  k‐component  increase.  Each  time    increases  by  completes one turn around the cylinder. The equations

the  curve

Parameterize the helix, the interval   Vector Calculus

being understood.   38

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Limits  The  way  we  define  limits  of  vector‐valued  functions  is  similar  to  the  way  we  define  limits  of  real‐ valued functions.  Definition   be a vector function and   a vector. We say that   has limit   as

Let  approaches   and write    If, for every number

, there exists a corresponding number

such that for all

If

, then

precisely when

The equation    provides a practical way to calculate limits of vector functions.    Problem

If

then

and

We define continuity for vector functions the same way we define continuity for scalar functions.  Continuity  Definition

is  continuous  at  a  point    in  its  domain  if  A  vector  function  . The function is continuous if it is continuous at every point in its domain.

Component Test for Continuity at a Point  Since limits can be expressed in terms of components, we can test vector functions for continuity by  examining  their  components.  The  vector  function    is  continuous  at   if and only if the component functions  are continuous at

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Problem  The  function

is  contitinuous  because  the  components

,cost,sint and t are continuous  Consider the function and

is discontinuous at every integer. We note that the components everywhere. But the function Hence

are continuous

is discontinuous at every integer

is discontinuous at every integer.

Derivatives and Motion

Suppose that   is the position vetor of a particles moving along a curve  in  space  and  that    are  differentiable  functions  of    Then  the  difference  between  the  particle’s positions at time   and time   is    In terms of components,         As

approaches zero, three things seem to happen simultaneously.

First,

Second, the secant line

Third, the quotient

approaches

along the curve. seems to approach a limiting position tangent to the curve at approaches the following limit

We are therefore led by past experience to the following definitions.  Definitions

The

vector

function

is  differentiable  at  .Also,   is said to be differentiable if it is differentiable at  every point of its domain. At any point   at which   is differentiable, its derivative is the vector     Vector Calculus

40

Problem

If

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find

Solution  Given

Hence

Definition  The curve traced by   is smooth if   is continuous and never 0, i.e., if  have continuous first derivatives that are not simultaneously 0.  Remark

The vector

Definition

The tangent line to the curve at a point

through the point parallel to

when different from0, is also a vector tangent to the curve.

at

is defined to be the line

Remark  We require   for a smooth curve to make sure the curve has a continuously  turning tangent at each point. On a smooth curve there are no sharp corners of cusps.  Definition  A  curve  that  is  made  up  of  a  finite  number  of  smooth  curves  pieced  together  in  a  continuous fashion is called piecewise smooth                                         Definitions

If r is the position vector of a particle moving along smooth curve in space, then

is  the  particle’s  velocity  vector,  tangent  to  the  curve.  At  any  time    the  direction  of    is  the  direction of motion, the magnitude of v is the particle’s speed, and the derivative   when  it exists, is the particle’s acceleration vector. In short,  1. Velocity is the derivative of position : 2  Speed is the magnitude of velocity   :

Speed=

3  Acceleration is the derivative of velocity

4  The vector

is the direction of motion at time  .

We can express the velocity of a moving particle as the product of its speed and direction.    Problem

The vector

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gives the position of a moving body at time . Find the body’s speed and acceleration when  times, if any, are the body’s velocity and acceleration orthogonal ?

At

Solution        At

, the body’s speed and direction are

Speed :

Direction

To find the times when   and   are orthogonal, we look for values of   for which    The only value is  Problem 8

A particle moves along the curve

Find the velocity and acceleration at

Solution  Here the position vector of the particle at time   is given by    Then the velocity   is given by    and the  acceleration a is given by    When

Problem

and

Show that if b, c, d are constant  vectors, then

is a path of a point moving with constant acceleration.  Solution  The velocity v is given by   Vector Calculus

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since the derivative of  the constant vector d is 0  The acceleration a is given by    The above is a constant vector, being a scalar multiple of the constant vector b. Hence the result.  Problem

A particle moves so that its position vector is given by

where   is a constant. Show that  (i) the velocity of the particle is perpendicular  to r  (ii) the  acceleration  is  directed  towards  the  origin  and  has  magnitude  proportional  to  the  distance  from the origin.  (iii)

is a constant vector.

Solution  (i) The velocity v is given by    Now      Hence velocity of the particles is perpendicular to r.

(ii) The acceleration a is given by

Thus direction of acceleration is opposite to that vector r and as such   it is directed towards the origin and the magnitude is proportional   to

. i.e., the acceleration is directed towards the origin and has magnitude

proportional to the distance from the origin   (iii)

a constant vector.  Vector Calculus

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Problem  (Centripetal acceleration) suppose a particle moves along a circle C having radius R in  the counter clock wise sense. Then its motion is given by the vector function

….  (1)

The velocity vector    Is a tangent at each point   to the circle C and it magnitude    is constant, Hence

so that   is the called the angular speed.  The acceleration vector is    with magnitude  . Since   are constants this implies that there is an  acceleration of constant magnitude   towards the origin.( due to negative sign). This acceleration  is  called  centripetal  acceleration.  It  results  from  the  fact  that  the  velocity  vector  is  changing  direction at a constant rate. The centripetal force is   Where   is the mass of  . The opposite  vector  motion.

is called centrifugal force, and the two forces are in equilibrium at each instant of the

Differentiation Rules  Because  the  derivatives  of  vector  functions  may  be  computed  component  by  component,  the  rules  for  differentiating  vector  functions  have  the  same  form  as  the  rules  for  differentiating  scalar functions They are :  Constant Function Rule :

(any constant vector C)

If   and   are differentiable vector functions of t, then  Scalar Multiple Rules :

(any number C)

(any differentiable scalar function f(t))   Sum Rule :  Difference Rule :  Dot  Product Rule :   Vector Calculus

44

Cross product rule

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Chain Rule (Short Form) : If r is a differentiable function of t and t is a differentiable function of s, then

We will prove the dot and cross product rules and Chain Rule and leaving the others as Assignments .  Proof of the Dot Product Rule Suppose that    and    Then

u'.v                                     u.v’  Proof of the Cross Product Rule   According to the definition of derivative,

To change this fraction into an equivalent one that contains the difference quotients for the derivates of u   in the numerator. Then

and v, we subtract and add

The last of these equalities holds because the limit of the cross product of two vector functions is  the cross product of their limits if the latter exist. As   approaches zero,         approaches    because v, being differentiable at   is continuous at   The two fractions approach the values of   and   at   In short     Vector Calculus

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Proof the Chain Rule   Suppose that

is a differentiable vector function of

and  that  t  is  a  differentiable  scalar  functions  of  some  other  variable  .  Then

are  differentiable

functions of   and the Chain Rule for differentiable real‐valued functions gives

Problem Show that

Solution   then

Let

and

Also

or

Constant Vectors  A  vector  changes  if  either  its  magnitude  changes  or  its  direction  changes  or  both  direction  and  magnitude change.  Theorem A  that

The necessary and sufficient condition for the vector function

to be constant is

the zero vector,

Solution   Necessary Part

If

is constant, then

Hence     Vector Calculus

46

Sufficiency Part

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Conversely, suppose that

Let

Then     we have

Hence using the assumption

Equating the coefficients, we get    and this implies that   and

and

are  constants, (in other words this means that

are  independent of

Therefore

is a constant vector.

Theorem B

The necessary and sufficient condition for the vector function

magnitude is that  Solution

to have constant

Let F be a vector function of the scalar variable

Suppose

to have constant magnitude, say F, so that

Then

Therefore,

or

or

and this implies that  magnitude.  Theorem C

Let

or

or

is a constant or   is a constant. i.e. the vector function

have constant   to have constant

be a vector function of the scalar variable   and n be a unit vector in the direction of

the magnitude of   then   Vector Calculus

or

The necessary and sufficient condition for the vector function

direction is that  Solution

. Then

Conversely, suppose that

or

If   be

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(since   by theorem B)

Necessary Part

…….(1)

Suppose that F has constant direction. Then n is a constant vector. Therefore,

. Hence by (1) above we have

Sufficiency Part  Conversely, suppose that

. Then from (1), we get   or

….(2)

Since n is of constant length, by the last theorem, we have                                               From  (2)  and  (3),  we  obtain

…(3)

.  Hence  n  is  a  constant  vector.  Therefore  the  direction  of  F  is

constant.  Vector Functions of a Constant Length  When  we  track  a  particle  moving  on  a  sphere  centered  at  the  origin  the  position  vector  has  a  constant length equal to the radius of the sphere. The velocity vector , tangent to the path of  motion,  is  tangent  to  the  sphere  and  hence  perpendicular  to  r,  This  is  always  the  case  for  a  differentiable vector function of constant length (as seen in Theorem above): The vector and its first  derivative  are  orthogonal.  With  the  length  constant,  the  change  in  the  function  is  a  change  in  direction only, and direct changes take place at the right angles.                                     If   is differentiable vector functions of   of constant length, then

…..(3)

For the proof of (3) see Theorem above.  Problem

Show that

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has constant length and is orthogonal to its derivative.  Solution

Integrals of Vector Functions  A differentiable vector function   is an anti derivative of a vector function   on an interval   if    at  each  point  of    If    is  an  anti  derivative  of  on  it  can  be  shown,  working  one   for some constant  component at a time, that every anti derivative of   on   has the form of   vector  . The set of all anti derivates of   on   is the indefinite integral   on    Definition  denoted by

The  indefinite  integral  or    with  respect  to    is  the  set  of  all  anti  derivatives  of  ,   If   is any anti derivative of   then

The usual arithmetic rules for indefinite integrals apply.  Problem

……  (4)

……(5)   with

As in the integration of scalar functions, it is recommended that you skip the steps in (4) and (5) and  go directly to the final form. Find an anti derivative for each component and add a constant vector  at the end.  Problem 15

Evaluate

Solution

We know that

where A is a vector function in the variable

Hence   Vector Calculus

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where   is an arbitrary constant vector.  Problem Prove that  Solution

We know that

Differentiating with respect to  on both sides, we get    On integration, we get    Definite integrals  Definite integrals of vector functions are defined in terms components.  Definition  If  the  component  of  definite integral of   from   to   is

are  integral  over

then  so  is  r,  and  the

The usual arithmetic rules for definite integrals apply.  Problem

find

If

Solution         Vector Calculus

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Where   is an arbitrary constant vector  (ii)

Problem

Problem The velocity of a particle moving in space is    Find the particle’s position as a function of   if  Solution

when

Our goal is to solve the initial value problem that consists of

The differential equation :  The initial condition :

Integrating both sides of the differential equation with respect to   gives

We then use the initial condition to find the right value for C :     implies

The particle’s position as a function of t is     To check (always a good idea), we can see this formula that            and   Vector Calculus

=2i+k   51

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Problem  The  acceleration  of  a  particle  at  time    is  given  by  the velocity    and the displacement   be zero at   Find  Solution

If

and

at time

Given

On integration,         when

Putting

we get

so that  (ii)

hence

Since

we have

Integrating we get    Putting

when

we get

So that

hence

Assignments   Assignments  1‐2, r(t) is the position of a particle in the xy‐plane at time t. Find an equation in x and  y whose graph is the path of the particle. Then find the particle’s velocity and acceleration vectors at  the given value of t.  1. 2. Assignments  3‐4 give the position vectors of particles moving along various curves in the xy‐plane.  In each case, find the particles velocity and acceleration vectors at the stated times and sketch them  as vectors in the curve

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3.

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(Motion on the circle) and

(Motion on the cycloid)

and

In Assignments  5‐7, r(t) is the position of particles in space at time r(t) Find the particle’s velocity  and acceleration vectors. Then find the particles speed and direction of motion at the given value of  t. Write the particles velocity at the time as the product of its speed a nd direction.  5.

In Assignments  8‐10 r(t) is the position of a particle in space at time t. and the angle between the  velocity  and acceleration vectors at time t=0.

8  9  10

The position vector of a particle in space at time   is given by  .  Find  the  time  or  times  in  the  given  time  interval when the velocity and acceleration vectors are orthogonal.  Evaluate the integrals in Assignments  11‐13.

11  12

13

Solve the initial value problems in Assignments  14‐16 for r a vector function of t.  14

Differential equation :

Initial condition :

15

Differential equation :

Initial condition :

16

Differential equation :

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Initial condition :

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and

The tangent line to smooth curve   at   is the line that passes through the  point   parallel to  , the curve’s velocity vector at  . IN Assignments  17‐18,  find  parametric  equations  for  the  line  that  is  tangent  to  the  given  curve  at  the  given  parameter  value  .  17

18  19

Each  of  the  following  equations  (a)‐(e)  describes  the  motion  of  a  particle  having  the  same  path, namely the unit circle  . Although the path of each particle in (a)‐(e) is the  same, the behavior, or “dynamics”, of each particle is different. For each particle, answer the  following questions.  i)

Does the particle have constant speed? If so, what is its constant speed?

ii)

Is the particle’s acceleration vector always orthogonal to its velocity vector ?

iii)

Does the particle move clockwise or counter clockwise around the circle ?

iv)

Does the particle begin at the point (1,0) ?

a)

b)  c)  d)  e)

.  .

20  At time t=0, a particle is located at the point (1,2,3). It travels in a straight line to the point  (4,1,4),  has  speed  2  at  (1,2,3)  and  constant  acceleration  3i‐j+k.  Find  an  equation  for  the  position  vector r(t) of the particle at time t  ARC LENGTH AND   THE UNIT TANGENT VECTOR T  Arc Length Along a Curve    One of the special features of smooth space curves is that they have a measurable length.  This enable us to locate points along these curves by giving their directed distance   along the curve  from some base point, the way we locate points on coordinate axes by giving their directed distance  from  the  origin    Time  is  the  natural  parameter  for  describing  a  moving  body’s  velocity  and  acceleration, but   is the natural parameter for studying a curve’s shape. Both parameters appear in  analyses of space flight.                                       Vector Calculus

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To measure distance along a smooth curve in space, we add a z‐term to the formula we use  for curves in the plane.  Definition

The  length  of  a  smooth  curve

once as   increases from

,  that  is  traced  exactly    is

……….(1)

We usually take

then

…..(2)

Just  as  for  plane  curves,  we  can  calculate  the  length  of  a  curve  in  space  from  any  convenient  parameterization that meets the stated conditions. The square root in either or both of Eqs.(1) and  (2) is   the length of the velocity vector  . Hence we have the Length Formula (Short Form)

Problem

Find the length of one turn of the helix

Solution

The helix makes one full turn as   runs from

Using the length formula (short form), the length of this portion of this curve is

This is

times the length of the circle in the

If  we  choose  a  base  point  determines a point

plane over which the helix stands.

on  a  smooth  curve    parameterized  by

each  value  of

on   and a “directed distance”.

…(4)

measured  along    from  the  base  point  .  If  ,    is  the  distance  from    to  .    If    is  the  negative  of  the  distance.  Each  value  of    determines  a  point  on    and  this

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parameterizes   with respect to  . We call   an arc length parameter for the curve,. The parameter‘s  value increase in the direction of increasing .                                          Arc length parameter with base point                         Problem

is given by       ……(5)

If

, the arc length parameter along the helix

from   to   is

Using Esq.(4)

Thus,

and so on

Problem Show that if

is a unit vector, then the directed distance along the line

From the point

when

is   itself.

Solution    So (nothing that, being unit vector,

Speed on a Smooth Curve    Since the derivatives beneath the radical in Esq.(5) are continuous  (the curve is smooth), the  Fundamental Theorem of Calculus tells us that  is a differentiable function of   with derivative

……....(6)

As we except, the speed with which the particle moves along its path is the magnitude of

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Notice that while the base point   plays a role in defining  in Esq. (5) it plays no role in Esq.(6).  The rate at which a moving particle covers distance along its path has nothing to do with how far  away the base point is.  Notice also that   since, by definition,  again that   is an increasing function of

is never zero for a smooth curve. We see once

The Unit Tangent Vector T  Since   for the curves we are considering,   is one‐to‐one and has an inverse that gives    as a differentiable function of   The derivative of the inverse is

……(7)

This makes   a differentiable function of  whose derivative can be calculated with the Chain Rule to  be                 ………………..(8)

Equation (8) says that   is a unit vector in the direction of v. We call  vector of the curve traces by   and denote it by T.   Definition

The unit tangent vector of a differentiable curve

the unit tangent

is

…….(9)

The unit tangent vector   is a differentiable function of   whenever   is a differentiable function of   As we will see in the next chapter   is one of three unit vectors in a travelling reference frame that  is used to describe the motion of space vehicle and other bodies moving in three dimensions.  Problem

Find the unit tangent vector of the helix

Solution

Problem

Find the unit tangent vector to the curve

At the point   Vector Calculus

.   57

Solution

Then

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The position vector of a point on the curve is given by

Hence

Therefore,

Therefore at t=2, the unit tangent vector is    Problem

Find the unit tangent vector at a point t to the curve

Solution

Problem

Find the unit tangent vector of the curve

Solution

Definition  The  involute  of  a  circle  is  the  path  traced  by  the  endpoint    of  a  string  unwinding  from a circle. In the above Problem  it is the unit circle in the  plane.  Problem

For the counterclockwise motion

around the unit circle,    is already a unit vector, so   Vector Calculus

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Assignments   In  Assignments    1‐4,  find  the  curve’s  unit  tangent  vector.  Also,  find  the  length  of  the  indicated  portion of the curve.  1. 2. 3. 4. 5.

Find the point on the curve

at a distance

units along the curve from the origin in the direction of increasing arc length.

In  Assignments    6‐7,  find  the  arc  length  parameter  along  the  curve  from  the  point  where   by evaluating the integral

From eq.(3). Then find the length of the indicated portion of the curve.

6  7

8  Find the length of the curve

From

to

a)  Show  that  the  curve    is  an  ellipse by showing that it is the intersection of a right circular cylinder and a plane. Find  equations for the cylinder and plane.

(b)  Write  an  integral  for  the  length  of  the  ellipse  (Evaluation  of  the  integral  is  not  required, as it is no elementary)

CURVATURE, TORSION AND TNB FRAME  In  this  chapter  we  define  a  frame  of  mutually  orthogonal  unit  vectors  that  always  travels  with  a  body  moving  along  a  curve  in  space  .  The  frame  has  three  vectors.  The  first  is  T,  the  unit  tangent  vector.  The  second  is  N,  the  unit  vector  that  gives  the  direction  of .  The  third  . These vectors and their derivatives, when available, give useful information about a  is vehicle’s orientation in space and about how the vehicle’s path turns and twists.

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For Problem,   tells how much a  vehicle’s path turn to the  left or right as it moves  along;  it  is called  the  curvature  of  the  vehicle’s  path.  The number .N  tells  how  much  a  vehicle’s path rotates or twists out of its plane of motion as the vehicle moves along; it is called the  torsion  of  the  vehicle’s  path.  If    is  a  train  climbing  up  a  curved  track,  the  rate  at  which  the  headlight turns from side to side per unit distance is the curvature of the track. The rate at which  the engine tends to twist out of the plane formed by T and N is the torsion.                  Every moving body travels with a TNB frame that characterizes the geometry of its path  of motion.  The Curvature of a Plane Curve  As a particle moves along a smooth curve in the plane,   turns as the curve bends. Since  T is an unit vector, its length remains constant and only its direction changes as the particle moves  along the curve. The rate at which T turns per unit of length along the curve is called the curvature .  The traditional symbol for the curvature function is the Greek letter   (‘’Kappa”).  Definition

If T is the unit tangent vector of smooth curve, the curvature function of the curve is    If   is large, T turns sharply as the particle passes through   and the curvature at   is large.  If    is  close  to  zero,  T  turns  more  slowly  and  the  curvature  at    is  smaller.  Testing  the  definition, we see in the following Problems 1and 2 that the curvature is constant for straight lines  and circles.  Problem

(The curvature of a straight line is zero)

On a straight line, the unit tangent vector T always points in the same direction, so its components  are constants. Therefore .                               Problem The parameterization for a circle having radius  is    and substitute   to parameterize in terms of arc length s. (Note that if the radius of the circle    and   is the angle between two rays emanating from the centre, then the length of the arc of the  circle included between the rays is given by      Then     Vector Calculus

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and    Hence, for any value of

The Principal Unit Normal Vector for Plane Curves

Since T has constant length, the vector

is orthogonal to   This conclusion is using the

result “If   is a differentiable vector function of   of constant length, then

’’

Therefore, if we divide  is given in the following definition.

by the length   we obtain a unit vector orthogonal to T  and

Definition

the principal unit normal vector for a curve in the plane is

At a point where

The vector  points in the direction in which T turns as the curve bends. Therefore, if   points toward the right if T turns  we face in the direction of increasing arc length, the vector  clockwise  and  toward  the  left  if  T  turns  counter  clockwise.  In  other  words,  the  principal  normal  vector N will point toward the concave side of the curve   Because the arc length parameter for a smooth curve

is defined with

and the Chain Rule gives

positive,

This formula enables us to find N without having to find   and   first.  Problem   Vector Calculus

Find T and N for the circular motion   61

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Solution

We first find T:

From this we find      and

Circle of Curvature and Radius of Curvature  The circle of curvature or osculating circle at a point the plane of the curve that

on a plane curve where

1.

is tangent to the  curve at   ( has the same tangent line the curve has);

2.

has the same curvature the curve has at   and

lines toward the concave or inner side of the curve

is the circle in

The  radius  of  curvature  of  the  curve  at    is  the  radius  of  the  circle  of  curvature,  which  according to Problem 2 is                               Radius of curvature=

To find  , we  find   and take the reciprocal. The center of curvature of the curve at   is the  center of the circle of curvature.  Curvature and Normal Vectors for Space Curves

Just as it for a curve in the plane, the arc length parameter   gives the unit tangent vector   for a smooth curve in space. We again define the curvature to be                                             The vector

Problem   Vector Calculus

…..(3)

is  orthogonal to T and we define the principal unit normal to be                            …….(4)

Find the curvature for the helix    62

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Solution

We calculate T from the velocity vector v:                                            as

Then, using the Chain Rule, we find

Chain Rule

nothing that

so

Therefore

……(5)

From Eq. (5) we see that increasing   for a fixed   decreases the curvature. Decreasing   for a fixed    eventually decreases the curvature as well .Stretching a spring teds to straighten it  if b=0 the Helix reduces to   again as it should (we have seen earlier that the curvature of a straight  line is  .  Problem Find   for the helix in the previous Problem.  Solution  We have (using the previous Problem)

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Torsion and the Binormal Vector  The Binormal Vector of a curve in space is   a unit vector orthogonal to both   and   . Together   and   define a moving right‐handed vector frame that plays a significant role in  calculating the flight paths of space vehicles.

How does  product, we have

behave in relation to

and   ? From the rule for differentiating a cross

Since   is the direction of

and

From  this  we  see  that  factors.

……(6)    is  orthogonal  to    since  a  cross  product  is  orthogonal  to  its

Since    is also orthogonal to   (the latter has constant length), it follows that   is  orthogonal  to  the  plane  of  .  In  other  words,    is  parallel  to  ,so    is  scalar  multiple of   In symbols,    The  minus  sign  in  this  equation  is  traditional.  The  scalar    is  called  the  torsion  along  the  curve. Notice that.    So that

Definition  Let

. The torsion function of a smooth curve is

Unlike the curvature   which is never negative, the torsion   may be positive, negative or zero.  The curvature   can be thought of as the rate at which the normal planes turns as the    is  the  rate  at  which  the  point    moves  along  the  curve.  Similarly,  the  torsion  osculating plane turns about   as    moves along the curve. Torsion measures how the curve twists.   Vector Calculus

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The Tangential and Normal Components of Acceleration  When  a  body  is  accelerated  by  gravity,  brakes,  a  combination  of  rocket  motors,  or  whatever,  we  usually  want  to  know  how  much  of  the  acceleration  acts  to  move  the  body  straight  ahead  in  the  direction of motion, in the tangential direction T. We can find out if we use the Chain Rule to rewrite  v as    and differentiable  both ends of this string of equalities to get                                                                 Where

……(7)

……(8)

are the tangential and normal scalar components of acceleration.  Equation (7) is remarkable in that B does not appear. No matter how the path of the moving body  we are watching may appear to twist and turn in space, the acceleration a always lies in the plane of  T and N orthogonal to B. The equation also tells us exactly how much of the acceleration takes place  tangent to the motion  To  calculate

and how much takes place normal to the motion

we  usually  use  the  formula   for

equation  first.

Problem

,  which  comes  from  solving  the

. With this formula we can find

without having to calculate

…..(9)

Without finding T and N, write the acceleration of the motion

in the form  Solution

We use the first of Eqs. (8) to find

:

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Knowing

we use Eq.(9) to find

We then use Eq.(7) to find a:

Formulas for computing Curvature and Torsion  We  now  give  some  easy‐to‐use  formulas  for  computing  the  curvature  and  torsion  of  a  smooth  curve. From Eq.(7), we have

it follows that

A Vector Formula for Curvature  Solving for   the discussion above gives the following vector formula for curvature

……(10)

Equation  (10)  calculates  the  curvature,  a  geometric  property  of  the  curve,  from  the  velocity  and  acceleration  of  any  vector  representation  of  the  curve  in  which    is  different  from  zero.  Take  a  moment to  think about how remarkable  this really is : From any  formula for  motion along a curve,  no matter how variable the motion may be (as long as v is never zero), we can calculate a physical  property of the curve that seems to have nothing to do with the way the curve is traversed.    The most widely used formula for torsion is traversed.   Vector Calculus

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…..(11)

Where

and so on.

This  formula  calculate  s  the  torsion  directly  from  the  derivates  of  the  component  functions    that  make  up  The  determinant’s  first  row  comes  from    the  second row comes from   and the third row come from  .  Problem

Use Eqs (10) and (11) to find  and   for the helix

Solution  We calculate the curvature with Eq.(10):

…….(12)

Notice  that  Eq.(12)  agrees  with  Eq.(5)  in  an  earlier  Problem,  where  we  calculate  the  curvature  directly from this its definition.  To evaluate Eq.(11) for the torsion, we find the entries in the determinant by differentiating   with  respect  to  We already have   and   and    Hence,

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………….(13)

*Formula for radius of curvature at a point on a curve in Cartesian co‐ordinates  The formula for radius of curvature at a point?  ordinates is

on the curve.

in the Cartesian co‐

where

and

Problem

Find the radius of curvature of

at

Solution  Differentiating both sides of the given equation with respect to x,

……(1)

Differentiating (1), with respect to x, we obtain

at (3,4)

The centre of curvature

is given by the formula

Problem  Solution   Vector Calculus

Find the centre of curvature at the point

to the curve

68

Given

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which can be written  as

we obtain

Diffrentiating  both sides with respect to    and

Hence at

and

i.e., the centre of curvature at

is

Assignments   Find T,N and   for the plane curves in Exercise 1‐2  1. 2. In Exercise 3, write a in the form

without finding T and N

3. 4.

(A formula for the curvature of the graph of a function in the xy-plane) in the xy plane automatically has the parameterization (a) The graph and the vector formula differentiable function of

(b) Use the formula for

Use this formula to show that if

, is a twice-

then

in (a) to find the curvature of

.

Compare your answer with the answer in Exercise 1. 5.  (Normal’s to plane curves)  (a)  Show  that curve

Vector Calculus

and  , at the point

,  are  both  normal  to  the  .

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To obtain N for a particular plane curve, we can choose the one of   or   from part (a) that  points  toward  the  concave  side  of  the  curve,  and  make  it  into  a  unit  vector.  Apply  this  method to find N for the following curves.  b)

c)

Find T, N, B,  , and   for the space curves in Exercise 6‐9.  6

7

8    9    10

Write a in the form

without finding T and N, where

In Exercise 11‐12, write a in the form  T and N  11.

at the given value of t without finding

12.

In  Exercise  13,  find  r,  T,  N  and  B  at  the  given  value  of  t,  Then  find  equations  for  the  osculating, normal, and rectifying planes at that value of t.  13

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MODULE - II MULTIVARIABLE FUNCTIONS AND PARTIAL DERIVATIVES FUNCTIONS OF SEVERAL VARIABLES Introduction In this chapter and coming chapters we discuss the domain, graph, limit and continuity of functions of two or more (independent) variables. Also some theorems of elementary type are discussed. Definitions Suppose D is a set of n- tuples of real numbers function on is a rule that assigns a real number

A real valued

to each element in . The set is the functions domain. The set of w- values taken on by is the function’s range. The symbol is the dependent variables of , and is said to be a function of the independent variables . We also call the the function’s input variables and call the function’s output variables. Remark

If If

Problem

is a function of two independent variables, the domain is a region in the is a function of three independent variables, the domain is a region in space. at the point

The value of

is

Remark In defining functions of more than one variable, we follow the usual practice of excluding inputs that lead to complex number or division by zero. This is illustrated in the following Problems. Problem Function

Domain

Range

Entire Plane Entire Space

Half-space

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Definitions

In plane : a closed disk consists of the region of points inside a circle together with the circle. an open disk consists of the region of points inside a circle without the circle. In space : a closed ball consists of the region of points inside a sphere together with the sphere. an open ball consists of the region of points inside a sphere without the bounding sphere. Definitions

In plane : A point

in a region

center of a disk that lies entirely in centered at

in the

plane is an interior point of

. A point

is a boundary point of

contains points that lie outside

boundary point itself need not belong to interior points of

if every disk

as well as points that lie in

) . The interior of the region

The boundary of the region

if it is the (The

is the set of

is the set of boundary points of

A

region is open if it consists entirely of interior points. A region is closed if it contains all of its boundary points. in a region

In space :A point of a ball that lies entirely in

in space is an interior point of

.A point

is a boundary point of

encloses points that lie outside

centered at

is the set of boundary points of

entirely of interior points.

if every sphere

as well as points that lie inside .

(The boundary point itself need not belong to ). The interior of . The boundary of

if it is the center

. A region

is the interior points of is open if it consists

A region is closed if it contains all of its entire boundary points.

Definitions A region in the plane is bounded if it lies inside a disk of fixed radius. A region is unbounded if it is not bounded. Problem

The parabola

is the boundary domain of

and is

and hence the domain is closed. The points above the contained in the domain of parabola make up the domains interior. Also note that the domain is unbounded. Graphs and Level curves of Functions of Two Variables Definitions •

The set of points

in the plane where a function

has a constant value

is called a level curve of •

The set of all points

in space, for

in the domain of

is called the

graph of •

The graph of

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is called the surface

.

72

Problem

Consider

the

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function

.

and

Plot

the

in the domain of

level

curves

in the plane.

Solution The domain of is the entire plane, and the range of than or equal to 50. The graph is the paraboloid The level curve

is the set of points in the

is the set of real numbers less

plane at which

or centered at the origin. Similarly, the level curves which is the circle of radius and are the circles

The level curve

consists of the origin alone.

Definition The curve in space in which the plane up of the points that represent the functions value Remark Note that some times counter lines. Problem

cuts a surface is made . It is called the counter line

represents the level curves and sometimes

Examine that the graph of the function …..(1)

is the hemisphere above the xy plane Solution such that i.e., It can be seen that the domain of is the set of all the domain is the set of all points in the xy plane which lie or within the circle whose center is at the origin and radius 4. Now the range of

is

From Eq.(1) being positive square root, the possible values of z are always positive real numbers, so that

Also

Hence the range of is (1). Squaring both sides of (1), we get

Vector Calculus

. Now the graph of

has the equation given in

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which is the equation of the sphere with centre at the origin and radius 4. But, since the graph of is just the top half of this sphere. i.e., the graph of is the hemisphere above plane. the Assignments Give the domains of the following functions in Assignments 1-15 1.

2.

3.

4.

5 Sketch the graph of the functions in Assignments 6-8 6 7 8

In Assignments 9-14, (a) find the function’s domain, (b) find the function’s range, (c) describe the function’s level curves, (d) find the boundary of the functions domain, (e) determine if the domain is an open region, a closed region, or neither, and (f) describe if the domain is bounded or unbounded. 9

10

11

12

13

14

Display the values of the functions in Assignments 15- 19 into ways: (a) by sketching and (b) by drawing an assortment of level curves in the function’s the surface domain. Label each level curve with its functions value. 15

16

17

18

19 In Assignments 20-21, find an equation for the level curve of the function through the given point.

that passes

20. 21  Vector Calculus

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LIMITS AND CONTINUITY OF FUNCTIONS OF SEVERAL VARIABLES Limit of a Function of Two Variables Definition

(Limit of a function of two independent variables)

We say that a function

if, for every number domain of .

Theorem 1

approaches the limit as

approaches

there exists a corresponding number

and write

such that for all

in the

: Properties of Limits

The following rules hold if

and

(L and M real

numbers) Sum Rule : i.e., the limit of the sum of two functions is the sum of their limits. 2.

Difference Rule

:

i.e., the limit of the difference of two functions is the difference of their limits. 3

Product Rule :

i.e., the limit of the product of two functions is the product of their limits. (any number k)

Constant Multiple Rule :

i.e., the limit of the constant times a function is that constant times the limit of the function. 5

Quotient Rule :

,

i.e., the limit of the quotient of two functions is the quotient of their limits, provided the limit of the denominator is not zero. 6

Power Rule : If m and n are integers, then is a real number.

i.e., the limit of any rational power of a function is that power of the limit of the function, provided the latter is a real number. Problem

1 .

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75

Problem

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Find

Solution Since the denominator

approaches 0 as

However, if we multiply numerator and denominator by

we cannot use the Quotient Rule. , we obtain

Hence

The Two-Path Test for the Non existence of a Limit If a function

has different limits along two different paths as does not exist.

then Problem 3

approaches

Using two path test, show that

has no limit as Solution

approaches Along the curve

, the function has the constant value given by

Hence We consider two paths : For

(i.e., the path is the curve

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i.e., the x- axis)

76

For

(i.e., the path is the curve

Hence by the two-path test,

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)

has no limit as

approaches

Assignments Find the limits in Assignments 1-6 2. 3

4

5

6

Find the limits in Assignments 7-10 by rewriting the fractions first. 7.

8

9

10

By considering different paths of approach, show that the functions in Assignments 11-14 have no limit as . 11.

12

13

14

In Assignments 15-18, evaluate the limits, if they exit. 15

16

17

18

CONTINUITY OF A FUNCTION OF TWO VARIABLES Definition (Continuity at a point) point if

A function

of two variables is said to be continuous at the

is defined at exists, and

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77

Definition (Continuous function) at every point in the domain of .

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A function

is said to be continuous if it is continuous

Theorem 2 If and are two functions which are continuous at the point the following functions are also continuous at the point :

, then

: ; ; : where c is a constant; , provided Problem

The function

Problem

Being the quotient of two continuous functions, the function

is continuous everywhere in the plane.

continuous everywhere in the plane except at the points . the points on the set Also, the domain of

i.e.,

is continuous on

is given by

Hence we conclude that Problem

where

is

is continuous at every point where it is defined.

The function

is continuous everywhere in the plane. Composite of continuous functions is continuous : If and , and is a continuous function of z, then the composite Problem composite Problem

and

The rational function

functions. Hence the composite Problem

is continuous function of is continuous.

are continuous functions. Hence the is continuous. and

are continuous is continuous.

Show that

is continuous at every point except the origin.  Vector Calculus

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Solution The functions is continuous at any point a rational function of x and y.

because its valued are then given by

At the value of is defined, but we show that has no limit as test for the non-existence of a limit’. the function

For every value of because

Therefore,

has a constant value on the “punctured” line

has this number as its limit as

The limit changes with The line with

using ‘two path

approaches

along the line

We consider two paths :

i.e., the line

There is therefore no single number we may call the limit of limit fails to exist, and the function is not continuous.

as

approaches the origin. The

Assignments At what points

in the plane are the functions in Assignments 1-8 continuous?

1

2.

3

4

5

6

7

8

9 10

Examine that Show that

is continuous at is continuous at every point in the plane except

the origin. 11

Examine that the function

12

Examine that function

Vector Calculus

where is continuous at

is discontinuous at

.

.  79

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HINTS to selected Assignments : 10

Verify that

11

Check that at least pone of conditions (i), (ii), (iii) in the Definition fails.

doesn’t exist.

Functions of More Than Tow Variables The definitions of limit and continuity for functions of two variables and the conclusions about limits and continuity for sums, products, quotients, powers, and composites all extend to functions of three or more variables. Problems 10

and

are continuous functions throughout their domains.

Problem 11

Evaluate the limit

Solution

By direct substitution, we have

Assignments Find the limits I n Assignments 1-6 (P denote the point 1.

2.

3

4.

5

6.

At what points

.

in space are the functions in Assignments 7-14 continuous ?

7

8.

9

10

11

12

13.

14

ARTIAL DERIVATIVES Definitions (Partial Derivatives of f) The derivative of constant is called the partial derivative of

with respect to

. Similarly, the derivative of constant is called the partial derivative of  Vector Calculus

with respect to

with respect to

keeping

as

and is denoted by

with respect to

keeping

and is denoted by

.

as and  80

are sometimes denoted by derivatives of

and

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respectively and are called first order partial

The definition is explained in the following equations.

Problem

Find the valued of

at the point

, If

=

Solution Treating as a constant and functions of the variable we get

as a product of

and

, and using product rule of

treating as constant

At the point

Problem

and

and hence

Find the first order partial derivatives of

when

Solution

Problem

If

find

Solution We regard a quotient. Treating of function of one variable, we obtain  Vector Calculus

as a constant, and using the quotient rule of differentiation

81

Problem The plane the tangent to the parabola at

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intersects the paraboloid

in a parabola. Find the slope of

Solution The slope is the value of the partial derivative

at

Verification :We

can treat the given parabola as the graph of the single-variable function (obtained by putting ) and ask for the slop at The slope, calculated now as an ordinary derivative, is

Problem 9

Find

independent variables

if the equation

defines

as a function of the two

and

Solution We treat as a constant and as a differentiable function of equation with respect to , we obtain

Differentiating both sides of the given

…….(3) Now (treating y as a constant)

and Hence, (3) gives i.e.,

Assignments

Vector Calculus

82

In Assignments 1-11 , find

and

.

1

2.

3

4

5

6

7

8

9

10

11

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( g continuous for all t)

Compute all the first derivatives of the functions in Assignments 12-19 : 12.

13

14

15

16

17

18

19

PARTIAL DERIVATIVES OF HIGHER ORDERS We have seen that and

and

are the first order partial derivatives of

. Partially differentiating

, we obtain partial derivatives of second order. viz.,

Similarly higher order partial derivatives can be defined. Notations We sometimes denote ; by by

or by ;

Problem

;

by

or by

by Compute all the first and second partial derivatives of the function

.

Solution Here is a function of two independent variables

and

We have first to calculate the first order

partial derivatives and then the second order partial derivatives  Vector Calculus

,

.  83

Now

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and

Also

and

Theorem (Euler’s Theorem- The Mixed Derivate Theorem) If a function

and its partial derivatives

and

are defined and continuous at a point

and in some open region containing it, then i.e; Problem Find

, where

Solution The symbol

tells us to differentiate first with respect to

and then with respect to

However if

we postpone the differentiation with respect to and differentiate first with respect to more answer more quickly (This is possible by Euler’s Theorem). Now,

we get the

Hence Problem If z = ex(xcosy-ysiny),prove that Solution Here

and Also and Adding the second partial derivatives, we obtain  Vector Calculus

84

Problem

If

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, prove that

Solution is a function of the independent variables Here is a dependent variable depending on the independent variables and

and . In other words

….(4)

………..(5) Adding (4) and (5), we get Problem

Find

all first .

and

second

order

partial

derivatives

of

the

function

Solution Treating

as constant, we obtain

Treating

as constant, we obtain

Treating

as constant, we obtain

Similarly,

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Assignments Find all the first and second order partial derivatives of the functions I n Assignments 1-8 2. 3

4

5

6

7

8

Show that the functions in Assignments 9-15 are all solutions of the wave equation 9

10

11

12

13

14

15

, where

is a differentiable function of

and

, where

is a constant.

In Assignments 16-17, verify that

16

17

In Assignments 18-24, verify that 18

19

20

21

22

23

24

25

26 27

If

, verify that and

28

If

29

If

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, show that , prove that

86

30

Given

31

Find the value of

32

Find the value of

33

If

34

If

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, show that when when

.

, show that , show that

is a constant. (HINT: Here v is function of the independent variables 35

If

, where

36

If

, where

37

If

38

If

39

If

40

If

prove that

, prove that

42

If

43

If

. . .

, prove that

.

, prove that , show that ,

If

and )

prove that

(a)

41

(b)

, prove that , prove that where

.

show that .

44

If

show that

45

If

prove that

46

If

find the value of n which make

In Assignments 47-52, which order of differentiation will calculate

.

. faster: x first, or y first ? Try

to answer without within anything down.  Vector Calculus

87

47

48

49

50

51

52

53 Show that dimensional Laplace equation

54

If

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and

find

and

are solutions of two-

at (1,2).

FUNCTIONS OF MORE THANTWO VARIABLES The definitions of the partial derivatives of functions of more than two independent variables are like the definitions for functions of two variables. They are ordinary derivatives with respect to one variable taken, while the other independent variables are kept constant. Problem

If

then prove that

Solution Since

, we have

…..(6) Similarly, or by symmetry, we have ….(7) and …….(8) Adding (6), (7) and (8), we obtain

Problem If  Vector Calculus

show that  88

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Solution of this Problem is done in the coming chapter “Chain Rule”. Problem

If

with

Show that Solution Here

….(9)

and . In other words is a function three independent variables depending on the independent variables and .

is a dependent variable

…..(10) Proceeding similarly, or by symmetry of (10), we can find that …..(11) ….(12) Adding (10), (11) and (12) we get the desired result. Problem

If resistors

and

ohms are connected the parallel to make an R-ohm resistor,

the value of R can be found from the equation and Solution

To find

when

ohms. , we regard

given equation with respect to

i.e.,

. Find the value of

and

as constants and differentiate both sides of the

:

-

When

Vector Calculus

89

Hence

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and

Existence of Partial Derivatives even at a point of discontinuity In the case of functions of a single variable, the existence of a derivative implied continuity But, the following Problem illustrate that a function to both and at a point without being continuous there.

can have partial derivatives with respect

Problem Show that function

is not continuous at

find the first order partial derivatives if they exist.

Solution The limit of as approaches along the line Definition of continuity , is not continuous at Note that the graph of

is

but

. Hence, by the

is the surface in the space consists of

the horizontal line

parallel to the

coordinate axis and passing through

the horizontal line

parallel to the

coordinate axis and passing through

the four open quadrants of the Now, the partial derivates

and

plane

and

are the slopes of the horizontal line

and

and both exist at

. Assignments In Assignments 1-6, find 1.

and 2.

3

4

5

6

In Assignments 7-9 , find the partial derivative of the function with respect to each variable. 7.

8

9

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Show that each function in Assignments 10-13 satisfies the three dimensional Laplace equation

11

10.

13

12 14

Compute

if

.

DIFFERENTIALS AND LINEARIZATION Differentiability If from changing

is differentiable at , then the change in the value of from to is given by an equation of the for

that results

(1) in which as We now discuss the analogous property for functions of two variables in the following theorem. Theorem 1 (The increment Theorem for Functions of Two Variables) Suppose that the first derivatives of containing the point

and that

and

are defined throughout an open region are continuous at

in the value of that results from moving from satisfies an equation of the form

. Then the change

to another point

in

…….(2) as

in which Definition

.

A function

Eq.(2) holds for domain.

at

is differentiable at . We call

Corollary to Theorem 1

then

and

exist and

differentiable if it is differentiable at every point in its

If the partial derivatives

throughout an open region

if

and

of a function

are continuous

is differentiable at every point of

Remark continuous.

The Corollary says that a function is differentiable if its partial derivatives are

Theorem 2

If a function

Proof

If we replace

is differentiable at

in Eq. (2) by the expression

, then

is continuous at and rewrite the equation as ….(3)

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we see that the right-hand side of the new equation approaches

as

and

. Thus

Hence the theorem. LINEARIZATION OF FUNCTION OF TWO VARIABLES . We want the effective approximation of this function

Consider the function near a point Since

at which the values of

is differentiable, Eq. (3) holds for by increments

point

and at

, the new value of

as . If the increments where will eventually be smaller still and we will have

the linear function follows: Definitions

and

which we call linear function

Suppose the function

linearization of

at

is differentiable.

. Therefore, if we move from

and

In other words, for small values of

are known and at which

and

to any

obtained using (3), is

are small, the products

and

will have approximately the same value as which we call linearization of . The Definition

is differentiable at a point

. Then the

is the function …….(4)

The approximation

is the standard approximation of Problem

at at the point (3,2)

Find the linearization of

Solution Here

. To use (4), we first evaluate the following:

and

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Substituting these values in Eq.(4)

That is, the linearization of

at

is

Total differential – Predicting Change with Differentials Definition

to a point

If we move from

nearby, the resulting differential in

is (6)

This change in the linearization of Remark

is called the total differential of

The total differential gives a good approximation of the resulting change in

Problem A company manufactures right circular cylindrical molasses storage tanks that are 25 ft high with a radius of 5 ft. How sensitive are the tanks’ volumes to small variations in height and radius? What happens if the values of and are reversed? Solution

As a function of radius and height , the typical tank’s volume is

Using Eq.(6), The change in volume caused by small changes height is approximately

and

Thus, a 1-unit change in will change by about units. A 1-unit change in will units. The tank’s volume is 10 times more sensitive to a small change change by about in that it is to small change to equal size in . A quality control engineer, who is concerned with being sure the tanks have the correct volume, would want to pay special attention to their radii. If the values of and differential in becomes

are reversed to make

Now the volume is more sensitive to change in

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and

then the total

than to changes in

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Absolute, Relative, and Percentage Change If we move from value of a function

to point nearby, we can describe the corresponding change in the in three different ways. True

Estimate

Absolute Change: Relative Change: Percentage Change: Problem

Estimate the resulting absolute; relative and percentage changes in the values of the , when the variables

function

by the amounts

and

change from the initial values of

and

Solution An estimate of the absolute change is given by

An estimate of the relative change is given by

Finally, an estimate of percentage change is given by

Problem The volume of a right circular cylinder is to be calculated from measured values of and Suppose that is measured with an error of no more than 2% and with an error of no more than 0.5%. Estimate the resulting possible percentage error in the calculation of Solution It is given that the percentage error in more than 0.05%. i.e.

is no more than 2% and that of

is no

and

Now using (6), we have

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So that

We estimate the error in the volume calculate is at most Problem

.

Find a reasonable square about the point will not vary by more than .

Solution

We approximate the variation

in which the value of

by the differential

Since the region to which we are restricting our attention is a square , we may set

to get

We have to take

i.e;

in such a way that

is no larger than 0.1. In that case we have

, expressing

With

As long as and we may expect

in terms of

the square we want is described by the inequalities

stays in this square, we may expect to be approximately the same size.

Problem The size and angles of a triangle radius is constant,

to be less than or equal to 0.1

vary in such a way that its circum-

Solution If A,B,C are the angles and a, b, c are the corresponding opposite sides, then by the law of sins, we have  Vector Calculus

95

where

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Taking differentials, …..(7) …….(8) …….(9) Being the sum of the angles of a triangle,

, and hence ….(10)

since differentia of the constant 180 is 0. From (7), (8) and (9) we have

Substituting in (10), we obtain

Problem If the kinetic energy is given by

, find approximately the change in kinetic

energy as changes from 49 to 49.5 units and v changes from 1600 to 1590 units (g is a constant given by g=32 units) Solution

Since

is a function of

and

by (6) ……(11)

As

, (11) becomes …….(12) It is given that

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Substituting these values in (12), we obtain

Hence the change in kinetic energy

units

Problem Find the percentage error in the area of an ellipse when an error of 1 percent is made in measuring the major and minor axes. Solution

The area

of the ellipse

is given by …..(14)

where

are the semi-major and semi-minor axes. Taking logarithms on both sides of (14), we obtain

and then taking differentials,

Hence …..(15) It is given that percentage error in computation of major axis is 1. Hence Similarly, Substituting these values in (15) we obtain Hence the percentage error in the area is 2%. That is an error of two percentages occurs in the computation of the area of the ellipse. Problem Find the percentage error in the area of an ellipse if an error of is made while measuring the major axis and an error of is made while measuring the minor axis. Solution Proceeding as the previous Problem, we obtain

Hence percentage error in the area i.e; an error of  Vector Calculus

percentage occurs in the computation of the area of the ellipse.  97

Problem

By a measurement the angle

area is calculated by the formula

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of a triangle

is found to be

and the

. Find the percentage error in the calculated value

of the area due to an error of 15 in the measured value of Solution

Let

denote the area of the triangle. Then

Taking logarithms, we obtain Taking differentials on both sides,

Since all other differentials are 0. and the

Hence Percentage error in

is given by

Given that

Percentage error in area = is calculated from the lengths of the sides . If Problem The area of a triangle is diminished and is increased by the same small amount , prove that the change in the area is given by

Solution

The are of the triangle is given by

where On squaring, Taking logarithms on both sides,

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Taking differentials, we obtain

As becomes

and

and

(as there is no change in the length of side c), the above

Hence i.e; i.e.,

Assignments In Assignments 1-5, find the linearization 1.

at

2.

at

of the function at each point.

at

3.

In Assignments 4-6, find the linearization of the function at . Then use inequality (5) to find an upper bound for the magnitude of the error in the approximation over the rectangle R. 4

at

5

,

at

,

(Use 6

in estimate E). at

(Use

, in estimating E).

7

You plan to calculate the area of a long, thin rectangle from measurements of its length and width. Which dimension should you measure more carefully? Give reason for your answer.

8

Suppose is to be found from the formula be In 2with maximum possible errors of maximum possible error in the computed value of

Vector Calculus

where and

and

are found to . Estimate the

99

9

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If and to the nearest millimeter, what should we expect the maximum percentage error in calculating to be ? will not

10. Give reasonable square centered at (1,1) over which the value of very more than 11 The resistance produced by writing resistors of calculated from the formula

and

ohms in parallel can be

Show that FUNCTIONS OF MORE THAN TWO VARIABLES So far we have discussed linearization and total differentials of functions of two variables. They can naturally be extended to functions of more than two variables. Definitions Suppose the function at linearization of

is differentiable at a point is the function

. Then the

……(19) The approximation

of

by

is the standard linear approximation of

at

.

The error in the standard Linear Approximation If has continuous first and second partial derivatives throughout an open set centered at and if is any upper bound for the values containing a rectangle of

on on

by its linearization

, then the error

incurred replacing

satisfies the inequality …..(20)

Total Differential Definition

If we move from

resulting differential in

to a point

nearby, the

is …..(21)

This change in the linearization of

is called the total differential of

. This gives a good

approximation of the resulting change in . Problem  Vector Calculus

Find the linearization

of  100

Find an upper bound for the error incurred in replacing

at the point on the rectangle

Solution

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by

On evaluation, we get

With these values, Eq. (19) gives

Eq.(20) gives an upper bound for the error incurred by replacing by

we make

to be

on

Since

Hence

The error will not be greater than 0.0024. Assignments Find the linearization’s 1.

of the functions in Assignments 1-3 at the given points. at

a)

(1,1,1)

2.

(b)

(1,0,0)

(c)

(0,0,0)

(1,1,0)

(c)

(1,2,2)

(0, ,0)

(c)

(0, , )

at a)

(1,0,0)

(b) at

3. a)

(0,0,0)

(b)

In Assignments 4-5, find the linearization

of the function

upper bound for the magnitude of the error E in the approximation

at

. Then find an over the

region 4.

5

Vector Calculus

at

at

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THE CHAIN RULE The Chain Rule 1 for Functions of One Variable Let

be a differentiable function of

then

and

is a differentiable composite functions of

be a differentiable function of and the derivative

could be

calculated using the Chain Rule 1 given by ….(1)

Functions of Two Variables – Composite Functions Composite function of a single variable Suppose

…(2)

where

…(3)

and

…(4)

Here is a function of and where and are themselves functions of another variable The Eqs (2),(3) and (4) are said define as a composite function of . For Problem, the system of equations composite function of

define

a

Chain Rule 2 for functions of Two Independent Variables

The following theorem is the Chain Rule 2 for one independent variable intermediate variables and .

and two

is Theorem 1 : Chain Rule 2 (Chain rule for composite function of single variable) If and are differentiable, then is differentiable function differentiable and of and ...(5) Problem and

Find

using chain rule, when and

What is the derivative’s value at Solution By Chain Rule 2,

Vector Calculus

102

Now

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and

since

Also,

at

Problem

and

is

Use the chain rule to find the derivative of

with respect to

along the path

What is the derivative’s value at Solution Using the chain rule

and proceeding as in Problem 2, we obtain

. ,

At

Chain Rule 3 for Functions of Three Variables The following is the Chain Rule 3 for one independent variable intermediate variables and . is differentiable and Theorem 2 : Chain Rule 3. If functions of then is a differentiable function of and

and

and three

are differentiable

…..(6) Problem at

Find

if

. What is the derivative’s value

.

Solution Here is a function of three independent variables and where and are functions of another variable . Hence is a composite function of and by Chain Rule 3, we obtain

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Hence Assignments by chain rule and (b) verify the result using direct substitution and

In Assignments 1-3 find differentiation. 1. 2. 3

find

If

In Assignments 4-6 (a) express in terms of

as a function of , both by using the Chain Rule and by expressing

and differentiating directly with respect to

Then (b) evaluate

at the given value

of 4. 5 6 In Assignments 7-8, draw tree diagram and write a Chain Rule formula for each derivative. 7

for

8

for

CHAIN RULES FOR TWO INDEPENDENT VARIABLES AND THREE INTERMEDIATE VARIABLES

Theorem 3 (Chain Rule) Suppose that are differentiable, then

and has partial derivative with respect to

and

. If all four functions given by the formulas

….(7) …(8) Problem

Express

and

in terms of

and

if

Solution Using Chain Rule 4, we obtain

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and

possess continuous first order partial derivatives

Corollary 1 to Chain Rule 4 Suppose with respect to Then

and

let

and

possess continuous first order derivatives.

and and

in terms of

and

if

Problem

Express

Solution

Using Corollary to Chain Rule 4, we obtain

and

Problem

If

is a function of

and

and

then prove that Solution Here is a function of two independent variables and , where and are functions of two other variables and Hence is a composite function of and . By Corollary 1 to Chain Rule 4.

Vector Calculus

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and and

Hence

Corollary 2 to Chain Rule 4

Suppose . Then

possess continuous first order derivatives with respect to

and let

and

is the ordinary (single- variable) derivative.

where

Problem

Show that any differentiable function of the form , where

is a solution of the partial differential equation ( is a non zero constant)

Solution

Using the chain rule it can be seen that since since

Hence Showing that

is a solution of the partial differential equation

The Chain Rule 5 for Functions of Many Variables Suppose and the

is a differentiable function of finite number of variables are differentiable function finite number of variables

differentiable function of the variables through these variables are given by equations of the form

and the partial derivatives of

. Then

is a

with respect to

…(9)  Vector Calculus

106

The other equations are obtained by replacing

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by

, one at a time.

Assignments In assignments (1), (a) express Rule and by expressing

and

directly in terms of

as functions of and

and

both by using the Chain

before differentiating. Then (b) evaluate

and

at the given point 1. In Assignments 2, (a) express by expressing

and

directly in terms of

as a functions of and

and

both by using the chain Rule and

before differentiating. Then (b) evaluate

and

at

the given point 2. In Assignments 3, (a) express

directly in terms of

and by expressing and

and

as functions of and

and

both by using the Chain Rule

before differentiating. Then (b) evaluate

at the given point

3.

In Assignments 4-8 draw a tree diagram and write a chain Rule formula for each derivative. 4.

for

5.

for

6.

for

7.

for

8.

for

IMPLICIT DIFFERENTIATION Consider the functions

(i)

(ii)

(iv)

and

Vector Calculus

(iii) (v)

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It can be see that the functions defined in (i) to (iii) gives the value of directly when the value of is given, because is explicitly given. A function in which the value of is explicitly given is called explicit function. Now the functions defined in (iv) and (v) given the value of , that cannot be readily obtained as is not explicitly given. Such functions defined are called implicit functions. Definition

The equation ……(10)

defined

as an implicit function of , since solving (10) we obtain

Problems

In the equation

as a function of .

is an implicit function of .

Implicit Differentiation Suppose : 1. The function

is differentiable, and defines

2. The equation

Also, take

. Then

function of so that we can regard Tree Diagram below)

implicitly as a differentiable function of

is a function of two variables

and

and

say

is again a

as a composite function of . So by the Chain Rule (See

…(11) Now,

and by (10),

or

so that (11) becomes

provided

Differentiating again with respect to , regarding

….(12)

and

composite functions, we

obtain …(13) Provided Problem By implicit differentiation, find

Vector Calculus

if

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Solution Take

. Then

Assignments In Assignments 1-2 find the value of

at the given point. Assume that the equations define

as

a differentiable function of

1. 2. 3. 4. 5. If equation where

determined

as a differentiable function of

and , then, at points

,

Use these equations to find the values of

and

at the points in Assignments 6-9

6. 7. 8. 9. 10. Find

when

11. Find

and

when

12. Find

and

when

Assignments 13-17, find

if

if and

.

in the following cases

13.

14

15

16

16

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PARTIAL DERIVATIVES WITH CONSTRAINED VARIABLES Problem

Find

if

and

Solution

We are given two equations in the four unknowns and As we are asked to find , we have to take two of and as dependent variables and an one of the variables and as independent variables. The possible choices for the variables come down to Dependent

Independent

Case 1) Case 2) In either case, we can express explicitly in terms of the selected independent variables. We do this by using the second equation eliminates the remaining dependent variable in the first equation. In the first case, the remaining dependent variable is . We eliminate it from the first . The resulting expression for is equation by replacing it by

and

….(1) when

This is the formula for

and

are the independent variables.

In the second case, where the independent variables are and and the remaining dependent variable is we eliminate the dependent variable in the expression for by by This gives replacing

and

….(2) This is formula for

Problem Find

and

and

when

at the point

and

are the independent variables. if

are independent variables.

Solution  Vector Calculus

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It is not convenient to eliminate in the expression for We therefore differentiate both equations implicitly with respect to , treating and as independent variables and and as dependent variables. This gives

and

…….(4)

These equations may now be combined to express Esq.(4) for to get

in terms of x,y, and z. We solve

and substitute into Eqs.(3) to get

is

The value of this derivative at

Notation To show what variables are assumed to be independent in calculating a derivative, we can use the following notations: with

and

with Problem Find Solution

if

independent.

and independent. and are independent, and hence we have

From the notation, we have

Assignments In assignments variables. 1.

If

Vector Calculus

1-2, begin by drawing a diagram that shows the relations among the and

, find  111

a) 2.

b).

Let

c)

b). b).

Find a)

if

at the point 4.

be the internal energy of a gas that obeys the ideal gas law (n and R constant). Find

a) 3.

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Suppose that

and

and

as in polar coordinates.

and

Find

DIRECTIONAL DERIVATIVES, GRADIENT VECTORS AND TANGENT PLANE Directional Derivatives in the Plane Consider a function

that is defined through a region

is a unit vector, and that of the line through

in the direction of

is a point in

in the

plane. Suppose

Then the parametric representation

is given by

, where the parameter

measures the arc length from

in the direction of

The next definition describes that the rate of change of found out by calculating Definition the number

at

in the direction of

can be

at

The derivative of

at

in the direction of the unit vector

is

…(1) provided the limit exists. Notation

The directional derivative of

Problem

Find the derivative of

at

in the direction of

is also is denoted by

in the direction of the unit vector

at Solution

Here  Vector Calculus

and  112

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Hence

at

Hence, the rate of change of

in the direction

RELATION BETWEEN DERIVATIVE AND GRADIENT VECTOR Definition The greatest vector (gradient) of at a point

Theorem 1

are defined at

If the partial derivatives of

is is the vector.

then

…(4) the scalar product of the gradient

at

and the unit vector

Proof Consider the line …(2) through unit vector

parameterized with the arc length parameter

increasing in the direction of the

. Then , by the Chain Rule since

and …(3)

u  Vector Calculus

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Hence the Theorem, Remark.

in the direction of

Eq.(4) says that the derivative of

at

at

is the dot product of

.

Problem Find the derivative of

at the point

in the direction of

Solution The unit vector in the direction of

is obtained by dividing

The partial derivatives of

are

at

at

by its length:

is

Hence, by theorem 1, the derivative of f at

in the direction of A is

Properties of Directional Derivatives Evaluating the dot product in the formula

reveals the following properties : 1.

When

i.e., when u is in the direction of

this direction is

then

That is, at each point

most rapidly in the direction of the gradient vector 2.

Similarly, When

Vector Calculus

in its domain,

increases

at .

decreases most rapidly in the direction of

derivatives in the direction is 3

and hence the derivative in

. (i.e., when

).The

.

(i.e., when u is orthogonal to the gradient), then

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Hence any direction u orthogonal to the gradient is a direction of zero change in Remark

The above properties hold in three dimensions also.

Problem Consider the function (a)

(b)

Find the directions in which (i)

increases most rapidly at the point

and

(ii)

decreases most rapidly at the point

.

What are the directions of zero change in

at

?

Solution The gradient is given by

a) (i) The function increases most rapidly in the direction of

at

. The

gradient at Its unit direction is

(ii) The function decreases most rapidly in the direction of

b) The directions

and

of zero change at

at

which is

are the directions orthogonal to

. Let

. Then, as we suppose that

is orthogonal to ,

i.e.,

implies we take and Gradients and Tangents to level Curves If a differentiable function

has a constant value

along a smooth curve

(making the curve a level curve of ), then

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Differentiating both sides of this equation with respect to

obtain

using Chain Rule ….. (5) .

Equation for Tangent Line to Level Curves Recall that the line through a point equation

normal to a vector

has the

….(6) By the Remark above tangent line is the line normal to the gradient. Hence, if

Eq.(6) becomes …(7) Problem 4

Find an equation for the tangent to the ellipse at the point

Solution

The ellipse is a level curve of the function

at

is

Hence, by Eq. (7), the equation of the tangent line is (as Functions of Three Variables We obtain three- variable formulas by adding the a differential function

and a unit vector

terms to the two variable formulas. For in space, we have

and  Vector Calculus

116

Problem

Find the derivative

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in the direction of the vector 4i+4j-2k at the point

if

Solution

Let u be the unit vector in the direction of the given vector. Then

Note to Problem 5: The above means that the value of the function

is increasing 3 unit distance if

in the direction of the vector

we proceed from Problem

What is the maximum possible

Solution

By the Remark, we know that

, if

at the point

gives the maximum possible value of

Now

Problem

. A mosquito

The temperature of points in space is given by

desires to fly in such a direction that he will get cool as soon as possible. In what located at direction should he move ? Solution By Problem 5,

By Remark above,

is decreasing most rapidly in the

. Hence, the mosquito should move in the direction opposite to

direction of

ie in the direction of Problem

If

find

at the point

Solution

Vector Calculus

117

Problem

Find unit normal to the surface

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at the point

Solution This is a level surface for the function

. We know that

is

normal to the given urface at i.e.,

Hence a unit normal to the surface at

Note to Problem 9

is

Another unit normal is

having direction opposite to the unit

normal vector in the Problem. Problem a)

Find the derivative of

b)

In what directions does directions ?

at change most rapidly at

in the direction of

and what are the rates of change in these

Solution a)

The unit vector u in the derivative of A obtained by dividing A by its length:

i.e.,

The partial derivatives of at

Vector Calculus

are

118

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at

is

Thus, the derivative of

at

in the direction of A is

b)

By the Remark, the function increases most rapidly in the direction of and decreases most rapidly in the direction of the directions are, respectively,

The rates of change in

and

Tangent Plane and Normal Plane Definition

The tangent plane at the point

plane through

normal to

on the level surface

is the

and have the equation ….(8)

Definition through

The normal line of the surface at parallel to

on the level surface

is the line

and have the parametric equations …..(9)

Problem

Find

the

tangent

plane

and

normal

line

of

the

surface

at the point Solution Here

.

Using (8), the tangent plane is ,or Using (9), the line normal to the surface at

Vector Calculus

is

119

Problem

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Find an equation for the tangent plane to the surface

at the

point Solution Here, Also,

Hence, using (8) the equation of the tangent plane is

and

Problem Find the angle between the surfaces

at the point

Solution We know that the angle between the surfaces at the given point is the angle between the normal to the surfaces at that point. at

Now, a normal to

at

and a normal to Let

is

is

be the required angle between the surfaces at the point. Then

is obtained from

which is

or

i.e Problem

Consider

the

cylinder

and

the

plane

that meet in an ellipse E . Find parametric equations for the tangent to E at the point Solution

The tangent line is orthogonal to both

and

. The components of v and the coordinates of

at

, and therefore parallel to

gives us equations for the line. We

have

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Using Eq.(9) and the idea in that Definition, the normal line is given by the parametric equations.

Planes Tangent to a Surface at a point

To find an equation for the plane tangent to a surface , we first observe that the equation

where

. The surface

is equivalent to

is therefore the zero level surface of the function

The partial derivatives of

are

The formula given by (8) for the plane tangent to the level surface at

therefore reduces to

….. (10) Problem

Find the plane tangent to the surface

Solution

We first calculate the partial derivatives of

at and use Eq.(10)

The tangent plane is therefore

or

.

Increments and Distance: The directional derivative plays the role of an ordinary derivative when we want to estimate how much a function to another point nearby.

If

is ordinary derivative

increment)

For a function of two or more variables, we use the formula (i.e.,

is directional derivative

increment)

Where u is the direction of the motion away from  Vector Calculus

from a point

is a function of a single variable, then

(i.e., •

changes if we move a small distance

.  121

Estimating the Change in

in a Direction

The formula to estimate the change in in a particular direction

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when we move a small distance

from a point

is

directional derivative

distance increment

Problem Estimate how much the value of

moves 0.1 unit from

will change if the point

straight toward

Solution We first find the derivative of

at

in the direction of the vector

The direction of this vector is given by the unit vector

at

is

Therefore, The change approximately

in that result from moving

unit away from

in the direction of

is

Constant Multiple Rule :

2.

Sum Rule :

3

Different Rule :

4

Product Rule :

5

Quotient Rule :

(any number k)

Problem For the functions evaluate the following:  Vector Calculus

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and Solution

and

Note that

Using Algebraic Rules for Gradients, we obtain

1. 2. 3. 4. 5.

Note: Directly determining the gradients, we obtain 1. 2. 3 4 5

since

Assignments 1.

If

2.

If

find find

and

at the point

In Assignments 3-4 find the gradient of the function at the given point. Then sketch the gradient together, with the level curve that passes through the point. 3. 4 In Assignments 5-6 find

at the point at the point at the point given point

5  Vector Calculus

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6 7

at the point

Find the directional derivative of (b)

direction of (a) 8

Find the directional derivative at

9

in the

in the direction of

(a)

(b)

(c)

(d)

if

in the directional of

Find the directional derivative of

at

. 10

Find the directional derivatives of

at the point

in the direction

. 11

If

find the directional derivatives of

at

in the direction of

EXTREME VALUES AND SADDLE POINTS OF FUNCTIONS OF TWO VARIABLES Definitions Let

be defined on a region R containing the point

is a local maximum value of

Then

if

for all domain points

if

for all domain points

in an open disk centered at •

is a local minimum value of in an open disk centered at

As an Problem, local maxima correspondence to mountain peaks on the surface and local minima correspond to valley bottoms. At such points the tangent planes, when they exist, are horizontal. Local extreme are also called relative extreme. Like functions of single variable, the key to identifying the local extreme is a first derivative test. Theorem (First Derivative Test for Local Extreme Values If

has a local maximum or minimum value at an interior point

and if the first partial derivatives exist there, then Proof Then

1.

Suppose that

and

has a local maximum value at an interior point

is an interior point of the domain of the curve

of its domain, . of its domain.

in which the plane

cuts the surface

Vector Calculus

124

2. The

function

is

a

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differentiable

function

of

at

has a local maximum value at

3. The function

at

4. The value of the derivative of derivative is

is therefore zero

Since this

, we conclude that

A similar argument with the function

show that

This proves the theorem for local maximum values. Similarly the result for local minimum can be proved and is left as an assignments . This completes the proof of the theorem. and

If we substitute the values

for the argument plane to the surface

into the equation

at

, the equation reduces to

or Thus, Theorem I says that the surface does indeed have a horizontal tangent plane at a local extremum, provided there is a tangent plane there. can ever have an

As in the single variable case, Theorem I says that the only place a function extreme value are

1. Interior points where 2. Interior points where one or both of

and

do n ot exist,

3. Boundary points of the function’s domain Definition An interior point of the domain of a function and

one or both of

where both

and

are zero or where

do not exist is a critical point of

Thus, the only points where the function boundary points.

can assume extreme values are critical points and

As with differentiable functions of a single variable not every critical point gives rise to a local extremem. A differentiable function of a single variable might have a point of inflection. A differentiable function of two variables might have a saddle point. Definition

A differentiable function

every open disk centered at domain points surface Problem  Vector Calculus

has a saddle point at a critical point

there are domain points

where

if in

where

. and

. The corresponding point

on the

is called a saddle point of the surface. (In , origin is a saddle point). Find the local extreme value of  125

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Solution The domain of

is the entire plane (so there are no boundary points) and the partial

and

derivatives

exist everywhere. Therefore, local extreme values can occur only

where and The only possibility is the origin, where the value of see that the origin gives a local minimum. Problem

Find the local extreme values (if any) of

Solution

The domain of

derivatives

is zero. Since

is never negative, we

is the entire plane (so there are no boundary points) and the partial

and

exist everywhere.

Therefore, local extreme values can occur only where and The only possibility is the origin, where the value of

is zero.

Therefore local extrema can occur only at the origin axis

has the value of

:along the

. However along the positive

positive

axis

has the value of

. Therefore every open disk in the -plane centered at contains points where the function is positive and points where it is negative. The function has a saddle point at the origin instead of a local extreme value. We conclude that the function has no local extreme values. at an interior point

The fact that sure

has a local extreme value there. However, if

continuous on

of

does not tell us enough to be the

and its first and second partial derivatives are

we may be able to learn the rest from the following theorem.

Theorem 2 (Second derivative Test for Local Extreme Values) Suppose centered at

and its first and second partial derivatives are continuous throughout a disk and that

. Then

1.

has a local maximum at

if

and

at

2.

has a local tminimum at

if

and

at

3.

4.

if

and

The test is inconclusive at

if

some other way to determine the behavior of The expression

at at

. In this case, we must find

at

is called the discriminant of

. It is sometimes easier to

remember the determinant form.

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126

Problem

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Find the local extreme values of the function

Solution The function is defined and differentiable for all

and

and its domain has no boundary

points. The function therefore has extreme values only at the points where

and

are

simultaneously. This leads to or Therefore, the point does so, we calculate

is the only point where

may take on an extreme value. To see if it

, The discriminant of

at

is

The combination and Tell us that

has a local maximum at

Problem

. The value of

at this point is

Find the local extreme values of

Solution Since

is differentiable everywhere, it can assume extreme values only where and

Thus, the origin is the only point where there, we calculate

might have an extreme value. To see what happens

The discriminant,

is negative, therefore the function has a saddle point at local extreme values.

. We conclude that

has no

Absolute Maxima and Minima on Closed Bounded Regions Problem

Vector Calculus

Find the absolute maximum and minimum values of

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on the triangular plate in the quadrant bounded by the lines Solution Since

is differentiable, the only places where and point the boundary.

triangle where Interior Points

For these we have

The value of

Yielding the single point Boundary Points 1.

can assume values are points inside the

there is

We take the triangle one side at a time.

On the segment

The function

may now be regarded as a function of may occur at the endpoints.

defined on the closed interval

Its extreme values

where where is

and at the interior points where

2.On the segment OB,

and

We know from the symmetry of in the candidates on this segment are

We have already accounted for the values of interior points of

where

With

and

and from the analysis we just carried out that

at the endpoints of

so we need only look at the

, we have

gives

Setting

At this value of and We list all the candidates: The minimum is  Vector Calculus

which

. The maximum is 4, which assumes at

and

assumes at

.  128

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Assignments Find all the local maxima, local minima and saddle points of the functions in Assignments 1-15 1. 2. 3 4. 5 6 7. 8. 9 10 11. 12. 13 14 15 In Assignments 16-19, find the absolute maxima and minima of the functions on the given domain. on the closed triangular plate bounded by the lines

16

on the closed triangular plate in the first quadrant bounded by the lines

18

on the rectangular plate

19 20

on the rectangular plate Find two numbers a and b with

such that

has its largest value. 22

Find the maxima, minima and saddle points of a)

Vector Calculus

if any, given that

and  129

b)

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and

c)

and

Describe your reasoning in each case 23

Show that

is a critical point of

constant k has. (Hit: consider two cases: 24

a)

If

no matter what value the and

).

must f have a local maximum or minimum value at

Among all the points on the graph of

find the point farthest from the plane. 26

The function

fails to have an absolute maximum value in the closed first

and . Does this contradict the discussion of finding absolute extrema quadrant given in the text? Give reason for your answer. To find the extreme values of a function we treat find where

on a parameterized curve

as a function of the single variable t and use the Chain Rule to

is zero. As in any other single variable case.

a) Critical points (points where

is zero or fails to exist), and

b) Endpoint of the parameter domain Find the absolute maximum and minimum values of the following functions on the given curves. 27.

Functions: b)

a)

c)

curves:

i)

The semicircle

ii)

The quarter circle

use the parametric equations 28

Functions: Curves:

i)

The line

ii)

The line segment

iii)

The line segment

Vector Calculus

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LAGRANGE MULTIPLIERS In this chapter we consider the method of Lagrange multipliers to solve max-min problems with constrained variables. Constrained Maxima and Minima Problem

Find the point

closest to the origin on the plane

Solution The problem asks us to find the minimum value of the function

Subject to the constraint that

Since

has a minimum value wherever the function

has a minimum value, we solve the problem by finding the minimum value of . If we regard

the constraint equation and write

and

subject to

as the independent variables in this

as

our problem reduces to one of finding the points

at which the function

has its minimum value or values. Since the domain of test of the previous chapter tells us that any minima that

is the entire

- plane, the first derivative

might have must occur at points where

This leads to and the solution We may apply a geometric argument together with the second derivative test to show that these values minimize . The z coordinate of the corresponding point on the plane

is

Therefore, the closest point we seek is The distance from  Vector Calculus

to the origin is  131

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Remark Attempts to solve a constrained maximum or minimum problem by substitution, as we might call the method of Problem 1, do not always go smoothly. Hence the need of the method of Lagrange multipliers. Before formally give the Lagrange method we consider as Problem. Problem Find the points closest to the origin on the hyperbolic cylinder Solution To find the points on the hyperbolic cylinder closest to the origin we imagine a small sphere centered at the origin expanding like a soap bubble until it just touches the hyperbolic cylinder At each point of contact, the hyperbolic cylinder and sphere have the same tangent plane and normal line. Therefore, if the sphere and cylinder are represented as the level surfaces obtained by setting. and and

equal to 0, then the gradient vectors

(being normal’s to the surfaces at the point of contact) will be parallel where the surfaces

touch. At any point of contact we should therefore be able to find a scalar

such that

or Thus, the coordinates equations.

and

of any point of tangency will have to satisfy the three scalar ……(1)

For what values of lie on the surface

will a point

whose coordinates satisfy the equations in (1) also

? To answer this equation, we use the fact that no point on the in the first equation in (1). This means that

surface has a zero x- coordinates to conclude that only if or For

, the equation

becomes

. If this equation is to be satisfies as well as z

also (from the equation must be zero. Since have coordinates of the form

), we conclude that the points we seek all

have coordinates of this form? The points

What points on the surface ,

for which

,

The points on the cylinder closest to the origin are the points

.

The Method of Lagrange Multipliers Theorem 1

Suppose that  Vector Calculus

is differentiable in a region whose interior contains a smooth curve  132

If

is a point on

where

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has a local maximum or minimum relative to its values on

then

is

on

are

at

orthogonal to

Proof We show that

is orthogonal to the curve’s velocity vector at

, whose derivative with respect to is

given by the composite

At any point

The values of

where

has a local maximum or minimum relative to its values on the curve,

so

By dropping the z- terms in Theorem I, we obtain a similar result for functions of two variables. Corollary of Theorem 1

where a

At the points on a smooth curve

differentiable function

takes on its local maxima and minima relative to its values on the

curve, Theorem 1 is the key to the method of Lagrange multipliers. Suppose that are differentiable and that

is a point on the surface

where

minimum value relative to its other vales on the surface. Then minimum at

has a local maximum or

takes on a local maximum or

relative to its values on every differentiable curve through . Therefore,

through

and

But so is

on the surface

is orthogonal to the velocity vector of every such differentiable curve (because

is orthogonal to the level surface

‘Directional Derivatives’). Therefore, at Multiplier.

.

as seen in the chapter

is some scalar multiple

is called Lagrange

The Method of Lagrange Multipliers Suppose that

and

are differentiable. To find the local maximum and minimum

values of subject to the constraint satisfy the equations

find the values of

and

that simultaneously

and For functions of two independent variables, the appropriate equations are and Problem

Find the greatest and smallest values that the function

takes on the ellipse  Vector Calculus

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Solution subject to the constraint

We want the extreme values of

and

To do so, we first find the values of

for which

and Now

and

from which we find

so that

or

Case 1 : If

. We now consider these two cases. , then

But

is not on the ellipse.

.

Hence, Case 2: If

, then

and

Substituting this in the equation

gives

and therefore takes on its extreme values on the ellipse at the four points

The function , Problem

. The extreme values are

and

Find the maximum and minimum values of the function

on the

circle Solution

We model this as a Lagrange multiplier problem with ,

and look for the values of

Vector Calculus

and

that satisfy the equations

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and gives

These equations tell us, among other things, that and

and

have the same sign. With these values for

the equations

gives so

and

Thus,

, has extreme values at

and

at the points

By calculating the value of

, we see that its maximum and minimum

are

values on the circle

and Lagrange Multipliers with Two Constraints Many problems require us to find the extreme value of a differentiable function variables are subject to two constraints. If the constraints are

whose

and and

and

are differentiable, with

maxima and minima of points

where

not parallel to

, we find the constrained local

by introducing two Lagrange multipliers

and . That is, we locate the

takes on its constrained extreme values by finding the values of

and that simultaneously satisfy the equations ….(2) Problem The plane cuts the cylinder on the ellipse that lie closest to and farthest from the origin.

in an ellipse Find the point

Solution We find the extreme values of

[The square of the distance from (x,y,z) to the origin] subject to the constraints …(3)  Vector Calculus

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……(4) The gradient equation in (2) then gives

or

….(5)

The scalar equations in (5) yield …..(6)

Equations (6) are satisfied simultaneously if either If

and

or

and

, then solving Eqs. (6) are satisfied simultaneously if either

and

or

. If

then solving Eqs (3) and (4) simultaneously to find the corresponding points on the ellipse

gives the two points If

and

.

he Eqs (3) and (4) give

The corresponding points on the ellipse are and But here we need to be careful. While farther from the origin than .

and

both give local maxima of

The points on the ellipse closest to the origin are from the origin is .

and

on the ellipse,

is

. The point on the ellipse farthest

Assignments where

1.

Find the points on the ellipse

2.

Find the maximum value of

3

Find the points on the curve

4

Use the method of Lagrange multipliers to find a)

the minimum value of

b)

the maximum value of

Vector Calculus

has its extreme values on the line

nearest the origin. subject to the constraints

;

subject to the constraint

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5

Find the dimensions of the closed right circular cylindrical can of smallest surface area whose volume is .

6

Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest with sides parallel to the area that can be inscribed in the ellipse coordinate axes.

7

Find the maximum and minimum values of .

8

on a metal plate is . An ant on The temperature at a point the plate walks around the circle of radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant?

9

Find the point on the plane

10

Find the minimum distance from the surface

11

Find the point on the surface

12

Find the maximum and minimum values of .

13

Find three real numbers whose sum is 9 and the sum of whose squares is as small as possible.

14

Find the dimensions of the closed rectangular box with maximum volume that can be included in the unit sphere.

15

enters the earth’s A space probe in the shape of the ellipsoid atmosphere and its surface begins to heat. After one hour, the temperature at the point on the probe’s surface is Find the hottest point on the probe’s surface.

16

(Application to economics) In economics, the usefulness or utility of amounts and of two capital goods and is sometimes measured by a function . For Problem, and might to be two chemicals a pharmaceutical company needs to have on hand and the gain from manufacturing a product whose synthesis requires different amounts of the chemicals depending on the process used. If costs dollars per kilogram, costs dollars per kilogram, and the total amount allocated for the purchase of and together is given that . dollars, then the company’s managers want to maximize Thus, they need to solve a typical Lagrange multiplier problem. Suppose that . Find the maximum value of constraint.

subject to the constraint

closest to the point

.

to the origin.

closest to the origin. on the sphere

and that the equation and the corresponding values of

simplifies to and

subject to this latter

17

Maximize the function

18

Find the point closets to the origin on the line of intersection of the planes and

19

Find the extreme values of the sphere

20

Find the extreme values of the function intersects the sphere

subject to the constraints

on the intersection of the plane

and

with

on the circle in which the plane

Vector Calculus

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MODULE ‐III  MULTIPLE INTEGRALS DOUBLE INTEGRALS: INTEGRTION OF FUNCTIONS OF TWO INDEPENDENT VARIABLES

Integration of functions of two Variables – Double integration Let

be a continuous function of two variables

defined on a region

bounded by a closed curve be divided, in any manner, into

Let the

and

be any point in the sub region of area

sub-regions of areas

. Let

and consider the sum

over the region R, written

Limit of theabove sum is defined as the double ntegral of mathematically as follows : ….(1) The region

is called the region of integration (this region in the case of definite integral of the form

integration

corresponds to the interval of of a function of single

variable). In order to simplify the evaluation of the double integral we often consider sub regions of

and one common choice is the rectangular sub regions or rectangular grids, obtained by by lines parallel to the coordinate axes. Since the area of a typical rectangular grid is

subdividing

, it follows from (1) that

Properties of Double Integrals (1)

for any number k y)dA

(2) (3)

if f(x,y)

(4)

on R if

on

=

(5) Where

and

Property 3 above is called domain in additivity property.

Vector Calculus

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Fubini’s Theorem (First Form) We now describe Fubini’s Theorem for calculating double integrals over rectangular regions. If

Problem

is continuous on the rectangular region

Evaluate

for

then

. Verify that

the change in the order of integration doesn’t effect the result. Solution

Reversing the order of integration, we have

Hence the order of integration doesn’t effect the result. Problem

Evaluate

Solution

Problem

Vector Calculus

Evaluate

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Solution

Fubini’s Theorem (Stronger form) We now describe Fubini’s Theorem for calculating double integrals over bounded nonrectangular regions. is continuous on a (rectangular or non-rectangular) region R.

Let 1.

If R is defined by

with

2.

and

continuous on

then

continuous on

then

If R is defined by

with

Remark respect to .

and

In I above integration is first with respect to

while in 2 integration is first with

Problem Evaluate

Problem Calculate line

and the line

Vector Calculus

where

is the triangle in the

plane bounded by the

axis, the

whether changing the order of integration works.  140

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Solution The region of integration is determined in Problem 2. The region is given by the following system of inequalities:

If we integrate first with respect to

and then with respect to , we find

By Problem 2, The above system of inequalities representing the triangle is equivalent to the following:

If we consider this, then

cannot be expressed in terms of elementary functions.

and we are stopped by the fact that

Hence in this Problem changing the order of integration doesn’t work. Problem Evaluate

over the first quadrant of the circle

.

Solution Here the circle meets the quadrant the circle meets the line from

to the curve

when

and this implies

Consider strips parallel to the

Hence in the first axis. Each strip varies

To cover the entire region each strip move from

to

Hence the region enclosed in the first quadrant of the circle is given by the set of inequalities:

Problem Evaluate Solution

over the region

in which

and

Region of integration is (Ref. Problem 5)

Vector Calculus

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Problem

Evaluate

Solution

Proceeding as in Problem 14, the region of integration is

where

is the domain in the first quadrant of the circle

Hence the required integral I is given by

……(1) (Put

in (1), then

ProblemEvaluate Solution

or

when

when

by changing the order of integration.

Here the region of integration

of the double integral

is given by

means of the following system of inequalities: and

….(1)

Alternatively, the region of integration is (Ref. Problem 6)  Vector Calculus

142

and

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….(2)

To change the order of integration (i.e., to evaluate the integral by integration first with respect to x) we have to consider the alternative form (2)

Problem

Evaluate

Solution

=

Problem Evaluate Solution

Vector Calculus

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Assignments 1.

3.

2.

4.

5

6

7

8

.

9 10

Evaluate

11

Evaluate

12

Evaluate

over the quadrant of the xydxdy where

is the region of the circle

where

is

(i)

the region in the first quadrant of the circle

(ii)

the region in the first quadrant of the circle

14.

Evaluate

15

Evaluate

16

Evaluate

17

Evaluate

over the region

Evaluate

and

over the region in which each of where

where

and

is the triangle with vertices

(Exercise 22 is a special case, with 18

in which each of

.

of this exercise)

is the region bounded by the

dinate

and the parabola taken over the postive quadrant of the ellipse

19.

Find the value of the integral

20

Evaluate

21

Evaluate

22

Change the order of integration in

where

axis, or

is the region bounded by the ellipse

by changing the order of integration. and hence evaluate the given

integral. 23.

Change the order of integration in

24

Change the order of integration in

Vector Calculus

and hence evaluate the given integral. and hence evaluate the given integral.  144

25

Change the order of integration in

26

Change the order of integration in

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and hence evaluate the given integral. and hence evaluate the given

integral. 27

Evaluate

28.

Evaluate

29.

Evaluate

30.

Evaluate

31.

Evaluate

32.

Evaluate

by changing the order of integration. by changing the order of integration. by changing the order of integration. by changing the order of integration.

over the positive quadrant of the circle and supposing

33.

Evaluate

34.

Show that

35.

Show that

36.

Show that

.

over the region in the positive quadrant for which

Hint: 37.

Evaluate

38

Evaluate

over the area bounded by the ellipse .

Applications of Double Integral Area by double Integral Suppose it is required to find the area A enclosed between Let the area be divided into rectangle elements of the type and

and of

area

. If this elements is moved along the vertical strip from

,

where to

we obtain the area of that strip as  Vector Calculus

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which represents the area of the vertical strip. Adding up all such strips between we get the required are

and

,

as

and

Problem Find the area enclosed between

and

.

Solution Here

and

. So using the same idea just given above, we have

Problem Find the area bounded between the curve

above the

axisand below the line

Problem Find thearea enclosed by the ellipse Solution

Required area is

i.e., the region of intregration

where

is the region enclosed by the ellipse

.

is

Required area is given by

Assignments 1.

Find the area enclosed by the lines

2

Find the area enclosed by the parabola

Vector Calculus

and the

axis.  146

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3

Find the area in the first quadrant bounded by the

4

Find the area of the cardioid

5

Find by double integration, the area which lies inside the cardiod outside the circle

axis and thecurves

and

and

. , using double integration.

6

Find the area of the ellipse

7

Using double integration, find the area of the region enclosed by the parabola

and the

line

Volume by Double integral Volume of Solid of Revolution the lines and the x- axis Let this We consider the area boudned by area revolve about the x-axis. To find this volume of the solid of revolution, which is generated, we consider, an elementary area , where . The area of the rectangle is when revolve about the axis, the elementary area generates a hollow circular disk whose and thickness is so that its volume is internal radius is and external radius is i.e.,

, as we can neglect the term . Hencethe total volume is given by

Problem Find the volume of a spherical segment of height

Deduce the

volume of the solid enclosed by the sphere of radius Solution The equation of the generating circle is , the centre being the origin and the x- axis being into plane, which cuts of the segment. The required solid is generated by rotating the region

Volume of the segmet is given by

The volume of the solid enclosed by the sphere is

Vector Calculus

147

obtained by taking

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so that its volume is

Assignments 1.

Find the colume of the ellipsoid

2.

Find the volume of the paraboloid of revolution

3

Find the volume of the solid enclosed by the sphere

cut off by the plane

Volume of Solid as Double Integral Suppose

is defind over a rectangular or non-rectangular region

interpert the double integral of

over

above by the surface

Each term

. Then we can

as the volume of the solid prism bounded below by

and

in the sum

is the volume of a vertical rectangular prism that approximates the volume of the portion of the solid that stands directly above the base solid. That is

. The limit of the sum

is defined as the volume of the

Volume = Problem Find the volume enclosed by the co-ordinate planes and the portion of the plane in the first octant (Fig. 28) Solution In the first quadrant the values of

are always greater than or equal to 0. The required volume

is given by zdxdy Where

Vector Calculus

and

is region bounded by

and

i.e.,

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DOUBLE INTEGRALS IN POLAR FORM Integrals are sometimes easier to evaluate if we change to polar coordinates. The chapter shows how to accomplish this. Problem

Find the limits of integration for integrating

the cardioids

and outside the circle

over the region

that lies inside

Also find the integral value when

Solution: Step 1: (A sketch) We sketch the region and label the bounding curves (Fig.1) Step 2: (The

limits of integration) .A typical ray from the origin enters the region

where

.

Step 3: (The

limits of integration). The rays from the origin that intersect the region

run from

to The integral is

If

is the constant function whose value is I, then the integral of

The area of a closed and bounded region

over

is the area of

in the polar coordinate plane is ….(1)

Changing Cartesian Integrals into Polar Integrals The procedure for changing a Cartesian integral  Vector Calculus

into a polar integral has two steps.  149

Step 1: Substitute

and

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and replace

by

in the Cartesian integral.

Step 2: Supply polar limits of integration for the boundary of The Cartesian integral then becomes ….(2) where G denotes the region of integration in polar coordinates. Attention :

Notice that

Problem

Find the polar moment of inertia about the origin of a thin place of density

is replaced by

bounded by the quarter circle

Solution We sketch the plate to determine the limits of integration (fig.3) In Cartesian coordinates, the polar moment is the value of the integral …..(3) and replacing

Substituting

Problem where

by

, we get

Evaluate is the semicircular region bounded by the

axis and the curve

Solution In Cartesian coordinates, the integral in question is a non elementary integral and there is no direct way to integrate

with respect to either and replacing

Vector Calculus

by

or

Polar coordinates help us. Substituting

enables to evaluate the integral as

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Assignments In Assignments 1-16, change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. 1.

2.

3

4.

5

6.

7

8

9

10.

11

12

13

14

15

16

17

Find the area of the region cut from the first quadrant by the curve . .

18

Find the area enclosed by one leaf of the rose

19

Find the area of the region cut from the first quadrant by the cardioids

20

Find the area of the region that lies inside the cardioids

21

Find

the

area

of

the .

region The

enclosed

region

by

common

the to

. and outside the circle

positive the

x-

interiors

axis of

and

the

spiral

cardioids

and

TRIPLE INTEGRALS IN RECTANGULAR COORDINATES The Triple Integral of a function of three independent variables, over a surface defined in the same manner as a double integral.

Vector Calculus

is

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Properties of Triple Integrals Triple integrals have the same algebraic properties as double and the single integrals. If and

are continuous, then =

1.

(any number k )

=

2. 3.

0

if

on if

4

on

Triple integrals also have an additivity property. If the domain

of a continuous function

partiotioned by smooth surfaces into a finite number of non-overlapping cells =

+

is

then

+…….+

Application of Triple Integrals: Volume by Triple Integrals Definition

The volume of a closed bounded region

in space is

…..(1) We also note that

Where

is the region given by the system of inequalities:

Problem

Evaluate

Solution

Problem

Find the volume of the region

enclosed by the surfaces

Solution  Vector Calculus

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Using (1), the volume is

To find the limits of integration for evaluating the integral, we take the following steps.

Step 1 :

(A

sketch)

The

or

surfaces

(fig.

1)

intersect

Step 3:

at

onto

The “upper” boundary of

is the

and leaves at

Step 4:

(The x- limits of integration). As

we obtain

and

The volume of

Vector Calculus

.

and leaves at through

(The y- limits of integration). The line

at

cylinder

passing through a typical point

(The z- limits of integration). The line

parallel to the z-axis enters

elliptical

, the projection of

. The lower boundary is the curve

Step 2:

the

. The boundary of the region

the xy plane, is an ellipse with the same equation: curve

on

) the value of

parallel to the y-axis enters

. sweeps across varies from

(putting at

in to

at

is

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(now put

)

. Problem

where

Evaluate

is the volume bounded by the planes

and Solution

(Fig.2)

to

Problem

Here

the

region

of and

integration

is

bounded

by

the

planes

i.e.,

Using the idea of triple integral, find the volume of the solid enclosed by the sphere .

Solution Because of symmetry, we need to complete the volume in the first octant only. Hence if the total volume, then  Vector Calculus

is

154

Where

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is the region of the sphere in the first octant

Now the region of integration varies from

to

varies from

to

is such that

varies from

to

Assignments Evaluate the integrals in Exercise 1-14 1. 2. 3. 4. 5. 6. 7. 8.  Vector Calculus

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9. 10. 11. 12. 13. 14. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES When a calculation in physics, engineering, or geometry involves a cylinder, cone or sphere, we can often simplify our work by using cylindrical or spherical coordinates.

Triple Integrals in Cylindrical Coordinates In Cylindrical coordinates (Fig.1) the surfaces like the following have equations of constant coordinate values:

Cylinder, radius 4, axis the z-axis Plane containing the z-axis  Vector Calculus

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Plane perpendicular to the z-axis The volume element (Fig.2) for subdividing a region in space with cylindrical coordinates is

Triple integrals in cylindrical coordinates are then evaluated as integrated integrals, as in the following Problem. Problem 1

Find the limits of integration in cylindrical coordinates for integrating a function

over the region D bounded below by the plane , and above by the paraboloid

laterally by the circular cylinder .

Solution Step 1 : (A sketch) (Fig.3). The base of

is also the region’s projection

is the circle

boundary of

on the

plane. The

. It polar coordinate equation is obtained as follows:

r =2sin Step 2: (The z- limits of integration). A line at

axis enters

and leaves at

and Step 3: (The leaves at  Vector Calculus

through a typical point

in

parallel to the z-

(Here we have used the fact, with

, limits of integration). A ray

through

from the origin enters

at

and

.  157

Step 4: (The

limits of integration). As

axis runs from

to

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sweeps across

, the angle

it makes with the positive

.

The integral is …(2) Problem 2

) enclosed by the cylinder

Find the centroid of the solid (with density given by bounded above by the paraboloid

Solution

First

we

note

that

and below by the xy-plane. with

and and below by the plane

We sketch the solid, bounded above by the paraboloid (Fig. 4). Its base

is the disk

in the xy-plane. (Details:

corresponds to

and hence the base

). The solid’s centeroid To find

,

corresponds to the xy-plane. such that

, that is

lies on its axis of symmetry, here the z-axis. This makes

we divide the first moment

by the mass

.

To find the limits of integration for the mass and moment integrals, we continue with the four basic steps. We completed step 1 with our initial sketch. The remaining steps give the limits of integration. Step 2: (The z- limits). A line the solids at

Step 4 : (The - limits). As positive x- axis runs from

in the base parallel to the z- axis enters

and leaves

Step 3 : ( The r-limits ). A ray

Vector Calculus

through a typical point through

from the origin enters

at

sweeps over the base like a clock hand, the angle to

. The value of

and leaves at it makes with the

is

158

The value of

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is

Therefore,

and the centroid is

. Notice that the centroid lies outside the solid.

Triple Integrals in Spherical Coordinates Spherical coordinates (Fig.5) are good for describing spheres centered at the origin, half-planes hinged along the z-axis, and single napped cones whose vertices lie at the origin and whose axes lie along the z- axis. Surfaces like these have equations of constant coordinate value: Sphere, radius 4, center at the origin Cone opening up from the origin, Sphere, radius 4, center at the origin making an angle of radians with the positive z- axis. Half- plane, hinged along, the z- axis, making an angle of

Vector Calculus

radians with the positive x- axis.

159

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The volume element in spherical coordinates is the volume of a spherical wedge defined by the differentials

and

(Fig.6) The wedge is approximately a rectangular box with one side a

, another side a circular arc of length circular arc of length Therefore the volume element in sphere in spherical coordinates is

, and thickness

.

….(3) and triple integrals take the form …(4) In particular, the volume of a region

in space is given by (with …(5)

V=

To evaluate these integrals, we usually integrate first with respect to . The procedure for finding the limits of integration is illustrated in the following Problem. We restrict our attention to integrating over domains that are solids of revolution about the z- axis (or portions thereof) and for which the limits for Problem 3

and

are constant.

Find the volume of the upper region

cut from the solid sphere

by the cone

Solution The volume is To find the limits of integration for evaluating the integral, we take the following steps. Step 1:

(A sketch) We sketch

Step 2 :

(The

and its projection

on the

limits of integration) We draw a ray

with the positive z- axis. We also draw , the projection of that

makes with the positive x- axis. Ray

Vector Calculus

enters

at

plane (Fig.7) from the origin making an angle

on the xy-plane, along with the angle and leaves at

160

Step 3:

(The

limits of integration). The cone

positive z- axis. For any given , the angle Step 4: volume is

makes an angle of

can run from

(The - limits of integration).The ray

Problem 4 A solid of constant density moment of inertia about the z- axis. Solution

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to

sweeps over

occupies the region

with the

. as

runs from

to

. The

in Problem 3. Find the solid’s

In rectangular coordinates, the moment is dv , Hence,

In spherical coordinates,

For the region in Problem 3, this become

Assignments Evaluate the cylinder coordinated integrals in Assignments 1-5. 1.

Vector Calculus

2

161

3.

.

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4

5 6.

Let

be the region bounded below by the plane

. Set up the triple integrals in cylindrical

and on the sides by the cylinder coordinates that give the volume of (a) 7

, above by the sphere

using the following orders of integration.

(b)

(c)

Give the limits of integration for evaluating the integral

on the side by

as an iterated integral over the region that is bounded below by the plane , and on top by the paraboloid

the cylinder

Evaluate the spherical coordinate integrals in Assignments 8-12. 8 9 .

10 11 12 13. Let

be the region bounded below by the plane

above by the sphere

and on the sides by the cylinder give the volume of a)d

using the following orders of integration.

b)d

SUBSTITUTIONS IN MULTIPLE INTEGRALS In this chapter we show how to evaluate multiple integrals by substitution. Substitutions in Double Integrals Suppose that a region in the plane by equations of the form

plane is transformed one-to-one into the region

in the xy-

. We call defined on

the image of

under the transformation, and

can be thought of as a function

the pre image of defined on

Any function as well. Then

….(1)

Vector Calculus

162

The factor

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is the Jacobian determinant or Jacobian of the coordinate transformation defined by ….(2)

Notation

The Jacobian is also denoted by

to help remember how the determinant in Eq(2) is constructed from the partial derivatives of

and

Problem Evaluate

by applying the transformation …(3) plane

and integrating over an appropriate region in the Solution ,

(3) gives

.

equations for the  Boundary of

Corresponding

…..(4)

equations

Simplified

for the boundary of G

equations

+1

‐

=  Vector Calculus

163

Problem

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Evaluate

Solution The integrand suggests the transformation

and

. That is, ….(5)

equations for the  boundary of

Corresponding

equations

Simplified

for the boundary of G

equations

‐

Substitutions in Triple Integrals Suppose that region

in

space is transformed one-to-one into the region

in

space by differentiable equations of the form

as in Fig.4 on the next page Then any function function

Vector Calculus

defined on

can be thought of as a

164

defined on

. If

and

over

by the equation

= where

have continuous first partial derivatives, then the integral of

is related to the integral of

over

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dudvdw

…..(6)

is the Jacobian determinant defined by

….(7)

Assignments 1.

(a)

Solve the system

for

and

in terms of

and . Then find the value of the Jacobian

b) Find the image under the transformation with vertices

and

in the

. of the triangular region

-plane. Sketch the transformed region in the

plane 2.

(a)

Solve the system

for

and

in terms of

and . Then find the value of the Jacobian

of the triangular region

b) Find the image under the transformation in the

-plane bounded by the

transformed region in the 3.

4.

Find the Jacobian

axis, the

axis, and line

Sketch the

plane. for the transformation

a)

,

b)

,

Evaluate the Integral

from Problem 1 directly by integration with respect to 5

.

and

to confirm that its value is 2.

Use the transformation in Exercise 2 to evaluate the integral for the region

in the first quadrant bounded by the lines and

Vector Calculus

165

6

Let

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be the region in the first quadrant of the and the lines and

plane bounded by the hyperbolas

Use the transformation

with

to rewrite

as an integral over an appropriate region

in the

plane. Then evaluate the

integral over 7

A

thin

plate

of

constant

density in the

the origin. (Hint: use the transformation

covers

the

region

bounded

by

the

ellipse

plane. Find the first moment of the plate about ).

Vector Calculus

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MODULE –IV  INTEGRATION IN VECTOR FIELDS LINE INTEGRALS Introduction The concept of a line integral is a simple and natural generalization of a definite integral …(1) known from calculus. In (1) we integrate the integrand

from

along the

axis to

a line integral we shall integrate a given function, called the integrand, along a curve in the plane). Hence curve integral would be a better term, but line integral is standard.

In

in space (or

We represent the curve C by a parametric representation …(2) We call C the path of integration, now oriented. The direction from

its initial point, and

to

in which

its terminal point. C is

increases is called the positive direction on C.

We can indicate the direction by an arrow Fig. 1. The points called a closed path.

and

may coincide (Fig.2). Then C is

We call C smooth curve if C has a unique tangent at each of its points whose direction varies continuously as we move along C. Mathematically this is equivalent to saying that C has a representation (2) such that is differentiable and the derivative and different from the zero vector at every point of C.

is continuous

Definitions and Notation Suppose

that

is

a

function

whose

domain

contains

the

curve

We partition the curve into a finite number of sub arcs (Fig.3). The typical sub arc has length the sum

. In each sub arc we choose appoint

and form …(2)

Vector Calculus

167

If

is continuous and the functions

approach a limit as

and

have continuous first derivatives, then the sums in (2)

increases, and lengths

over the curve from the integral is

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approach zero. We call this limit the integral of

to . If the curve is denoted by a single letter, C for Problem, the notation for ….(3)

over C”

Evaluation for Smooth Curves If

is smooth for

(i.e., if

is continuous and never 0), we can use the

equation (obtained from Eq. (5) of chapter “Arc Length and the Unit Tangent Vector T” with

in Eq(3) as

to express

evaluate the integral of

. A theorem from advanced calculus says that we can then

over C as ……(4)

This formula will evaluate the integral correctly no matter what parameterization we use, as long as the parameterization is smooth. Problem point

Integrate

over the line segment C joining the origin and the

. (Fig.4)

Solution We choose the simplest parameterization we can think of :

The components have continuous first derivatives and so the parameterization is smooth. The integral of

Vector Calculus

is never 0,

over C is

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Additivity Line integrals have the useful property that if a curve

is made by joining a finite number of

end to end, then the integral of a function over curves the curves that make it up =

+

+……..+

is the sum of the integrals over

…..(6)

...(6) Mass and Moment Calculations Mass and Moment formulas for coil springs, thin rods and wires lying along a smooth curve in space are given below:

Mass : First moments about the coordinate planes: , Coordinates of the center of mass: ,

,

Moments of Inertia:

distance from the point

to line

Radius of gyration about a line Problem A coil spring lies along the helix

The spring’s density is a constant,

Find the spring’s mass and center of mass, and its

axis.

Solution Because of the symmetries involved, the center of mass lies at the point  Vector Calculus

on the z- axis.  169

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For the remaining calculations, we first find

Then

Notice that the radius of gyration about the z- axis is the radius of the cylinder around which the helix winds. Problem

A slender metal arch, denser at the bottom than top, lies along the semicircle in the yz-plane . Find the center of the arch’s mass if the density at the point

on the arch is Solution We know that

and

because the arch lies in the

-plane with its mass

distributed symmetrically about the z-axis. To find , we parameterize the circle as

For this parameterization

Then, we obtain =

Vector Calculus

=

=2

170

With

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to the nearest hundredth, the center of mass is

Assignments 1

where C is the straight line segment

Evaluate to

2

from

. along the curve

Evaluate

. 3

over the straight line segment from

Find the line integral of

to

. 4

Integrate

over the path from

and

given by

5

over the path

Integrate

In the following assignemnts, integrate f over the given curve 6 7

8

to

Find the mass of a wire that lies along the curve density is

9

, if the

.

Find the mass of a thin wire lying along the curve if the density is

10

A circular wire hoop of constant density lies along the circle Find the hoop’s moment of inertia and radius of gyration about the z- axis.

in the xy-plane.

VECTOR FIELDS, WORK, CIRCULATION AND FLUX Scalar Field Definition point scalar field. If

(Scalar fields) A scalar field is a scalar valued function of three variables. i.e. if to each of a region in space there corresponds a unique scalar

is a scalar field, any surface defined by

isotimic surface or a level surface. For Problem, in Physics, if

Vector Calculus

. We say that

is a

, where c is a constant, is called an denotes either or gravitational field

171

potential, such surfaces are called equipotential surfaces. If isothermal surfaces.

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denotes temperature, they are called

Vector fields Definition

(Vector field) A vector field is a vector valued function of three variables. i.e. vector

field F is a rule associating with each point of a region in space, unique vector field three –dimensional vectors might have a formula like

The field is continuous if the components functions if like

and

and

.A

are continuous, differentiable

are differentiable, and so on. A field of two-dimensional vectors might have a formula

Definition vectors

The gradient field of differentiable scalar functions

Problem

is the

Problem Find the unit normal to the surface Solution This is level surface for the function the given surface.

is normal to

Now

at the point Hence a unit normal to the surface is

Note: Another unit normal is

having direction opposite to the unit normal vector in the

Problem

Problem If …(1)  Vector Calculus

172

Find

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, such that

Solution

….(2)

We have

From (1) and (2)

….(3) …(4) …(5)

Integrating (3) with respect for x, we get ….(6) where

is free from x, arbitrary constant.

Differentiating (6) partially with respect to y and using (4), we get . Then ….(7) Now integrating (7) with respect to y, we get

. So (6) is ..(8)

Differentiating (8) partially with respect to z and using (5), we get

Hence

or

a constant.

Therefore (8) becomes, Given

Here

or

or

Hence

The Work Done by a Force over a Curve in Space Suppose that the vector field represents a force throughout a region in space (it might be the force of gravity or an electromagnetic force of some kind) and that

is a smooth curve in the region. Then the integral of of the curve’s unit tangent vector, over the curve is called the work done by Definition  Vector Calculus

, the scalar component of

over the curve from

in the direction

to .

The work done by a force  173

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from

over a smooth curve

to

is

….(1) Derivation of (I): The work done by

along the curve segment

along the curve

The work done by

to

(Fig.2) will be approximately

will be approximately

As the norm of the partition of approaches zero, the norm of the induced partition of the curve approaches zero and these sums approach the line integral

The sign of the number we calculated with this integral depends on the direction in which the curve traversed as

increases. If we reverse the direction of motion, we reverse direction of T and its integral.

and change the sign of Problem

Find the work done by from

over the curve to

(Fig.3)

Solution Step 1: (Evaluate F on the curve)

. Step 2: (Find

)

Step 3: (Dot F with

Step 4: (Integrate from

Vector Calculus

)

to

)

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Problem Find the work done in moving a particle once round a circle circle has centre at the origin at radius 3 and the force field is given by

Solution

in the xy-plane: the

In the xy plane , and Now

Hence we can choose the parametric equations of the circle radius 3 as:

with center at the origin and

varies from 0 to

Where

Similarly

Also

Hence

Problem

Find the total work done in moving a particle in a force field given by along the curve

from

to

Solution Let

denote the arc of the given curve from

Vector Calculus

to  175

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Work done is given by Given

is in the parametric form

,y=2t2, z=t3 etc. in the above we get

substituting

In the line integral varies from

to

unit of work Problem

and

Find the value of (3) when

is the helix

... (4) Solution From (4) we have

Thus

Hence (3) gives

Problem If

evaluate

(i)

is a curve from

to

(ii)

is the straight line joining

where

is:

with parametric form and

Solution  Vector Calculus

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Here

...(5)

(i) C is in the parametric form

. At the point

. Substituting

Moving from the point

in (5) we get

to

(ii) The straight line joining and

and at

means from

to varies from

to and

Therefore

is given in the parametric form by to

Substituting

in (5) we

get in

varies from

to .

Problem Evaluate the line integral (3) with

along two different paths with the same initial point

and the same terminal point

, namely

(a)

the straight line segment

(b)

the parabolic arc

By substituting

into F we obtain

We also need

Vector Calculus

and

Hence the integral over

is

177

b)

Similarly, by substituting

into F and calculating

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we obtain for the integral over the path

.

Flow Integrals and Circulation Definition

If

is a smooth curve in the domain of a the flow along the curve from

continuous velocity field to

is the integral of

over the curve from

to . …(2)

The integral in this case is called a flow integral. If the curve is a closed loop, the flow is called the circulation around the curve. Problem

Find the flow along the helix

A fluid’s velocity field is .

Solution Step 1: (Evaluate F on the curve)

Step 2:

(Find

)

Step 3:

Step 4:

Vector Calculus

178

Problem 14

Find

the

circulation

of

the

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around

field

the

circle

Solution

1. On the circle, 2. 3. 1

4. Circulation=

Flux Across a Plane Curve Definition

If C is a smooth closed curve in the domain of a continuous vector field

in the plane, and if n is the outward-pointing unit normal vector C, the flux of F across C is given by the following line integral: ….(3) The Formula for calculating Flux Across a Smooth Closed Plane Curve …(4) The

integral

can

be

evaluated

from

any

smooth

parameterization

that traces C counter clockwise exactly once. Problem

Find the flux of

across the circle

in the

plane.

Solution The parameterization , traces the circle clockwise exactly once. We can therefore use this parameterization in Eq(4). With

We find  Vector Calculus

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Since the answer is positive, the net flow across the curve is The flux of F across the circle is outward. A net inward flow would have given a negative flux. Assignments Find the gradient fields of the functions in Exercise 1-2 1. 2. 3. Give a formula for the vector field in the plane that has the property that F points towards the origin with magnitude inversely proportional to the square of the distance from

to the origin.

In Assignments 4-7find the work done by force F from paths.

to

over each of the following

a) The straight line path b) The curved path c) The path

consisting of the segment from

segment from

to

followed by the

to

4.

5.

6 In Assignments 7-8 find the work done by F over the curve in the direction of increasing t. 7 8 9

Evaluate

10

Evaluate

along the curve for the vector field

from

to

along the curve

from

to 11

Find the work done by the force

12

Find the circulation and flux of the fields

over the straight line from

to

and  Vector Calculus

180

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around and across each of the following curves.

(a) The circle (b) The ellipse In Assignments 13-14 find the circulation and flux of the field F around and across the closed semicircular path that consists of the line segment , 13 15

followed

by

the

line

segment

14 Evaluate the flow integral of the velocity field following paths from

to

along each of the

in the xy- plane.

a)

The upper half of the circle

b)

The line segment from

c)

The line segment from

to to

followed by the line segment from

to

In Assignments 16-17, F is the velocity field of a fluid flowing through a region in space. Find the flow along the given curve in the direction of increasing t 16 17 18

around the closed path consisting of the Find the circulation of following three curves traversed in the direction of increasing t:

19.

The field to (Hint: Use

is the velocity field of a flow in space. Find the flow from along the curve of intersection of the cylinder

and the plane

.

as the parameter).

PATH INDEPENDENCE, POTENTIAL FUNCTIONS AND CONVERSATIVE FIELDS In gravitational and electric fields, the amount of work it takes to move a mass or a charge from one point two another depends only on the object’s initial and final positions and not on the path taken in between. This chapter discusses the notion of path independence of work integrals and describes the remarkable properties of fields in which work integrals are path independence,

Vector Calculus

181

Definitions

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Let F be a field defined on an open region D in space, and suppose that for any two

points A and B in the D the work

done in moving from A to B is the same over all paths is path independent in D and the field F is conservative on D.

from A to B. Then the integral

for some scalar function

Definition If F is a field defined on D and called a potential function for F.

on D, then

is

Theorem 1, (The Fundamental Theorem of Line Integrals)

1. Let

be a vector field whose components are continuous throughout

an open connected region D in space. Then there exists a differentiable function such that

If and only if all points A and B in D the value of

is independent of the path

joining A to B in D. 2. If the integral is independent of the path from A to B, its value is Prove that Proof. Suppose curve,

that

A

and

B

implies path independent of the integral.

are two points in D and that C: r(t)= is a smooth curve in D joining A and B. Along the

is a differentiable function of and using chain rule

, because Therefore,

Thus, the value of the work integral depends only on the values of

at A and B and

not on the path in between. Remark In between proving the result, we have proved Part 2 as well as the forward implication in Part 1. The more technical proof of the reverse implication is omitted. Problem

Vector Calculus

Find the work done by the conservative field

182

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along any smooth curve C joining the point

to

Solution With

we have as by Part 2 of Fundamental Theorem

Problem Evaluate the integral

to

from

by showing that F has a potential and applying (3)

Solution Recall that the vector field F has a potential

if

grad i.e., (here) if i.e, if i.e., (here) if We show that we can satisfy these conditions. Integrating

with respect to

Where

is an arbitrary function of

respect to

we obtain

and

(and not involving

Now differentiating

with

we obtain

But it is given that Hence Integrating

Where

is a function involving

Vector Calculus

with respect to y, we obtain

alone.  183

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Differentiating with respect to z

Hence

Hence h is a constant;

say

This gives and by (3) Theorem 2 The following statements are equivalent: 1. 2.

around every closed loop in D. The field F is conservative on D

Proof. We want to show that for any two points A and B in D the integral of F.dr has the same value over any two paths make a path

Thus the integrals over

and

from A to B. We reverse the direction on

from B to A. (Fig.1) Together,

and

and

is zero over any post loop C. We pick two

points A and B on C and use them to break C in to two pieces back to A. (Fig.2). Then

to make a closed loop C and

give the same value.

We want to show that the integral of

Vector Calculus

to

from A and B followed by

from B

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Finding Potentials for Conservative Fields Now we give a test for being conservative. The Component Test for Conservative fields Let be a field whose component functions have continuous first partial derivative. Then, Fis conservative if and only if and

…(1)

Proof. We show that Eqs. (1) must hold if F is conservative. There is a potential function

such that

as continuity implies that the mixed partial derivatives are equal

The other two equations in (1) are proved similarly. The second half of the proof, that Eq.(1) imply that F is conservative, is a consequence of Stokes’s Theorem and is omitted. is conservative and

Problem Show that find a potential function for it. Solution We apply the test in Eqs. (1) to

and calculate

Together, these equalities tell us that there is a function

with

We find by integrating the equations …..(2) We integrate the first equation with respect to  Vector Calculus

holding

and

fixed, to get  185

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We write the constant of integration as a function of and

change. We then calculate

and

because its value may change if

from this equation and match it with the expression for z

in Eq.(2). This gives

So

Therefore, g is a function of z alone and

We now calculate This gives

from this equation and match it to the formula for

in Eq.(2).

or Integration yields, Hence We have found infinitely many potential functions for F, one for each value of C. Problem

Show that

is not conservative.

Solution We apply the component test in Eqs (1) and final that

The two are unequal, so F is not conservative. No further testing required. Exact Differential Forms Definitions The form differential form is exact on a domain D in space if

For some (scalar) function

throughout D

Notice that if on

. Conversely, if

on D, then then the form

exact is therefore the same as the test for Problem

is called a differential form. A

is gradient field of is exact. The test for the form’s being

being conservative.

Show that the differential form under the integral sign of I=

Vector Calculus

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is exact, so that we have independence of path in any domain, and find the value of I from to Solution Here Exactness follows from

To find compare with

we integrate

, which gives

(which is “long”, so that we save work) and then differentiate to

and

where

Integrating with respect to , hence

is a function of z alone (Already

is not involved in the function ). and hence

constant. Taking

so that,

is a

we have

From this and (5) we get

Problem

is exact, and evaluate the integral

Show that

over the line segment from

to

Solution We let

These equalities tell us that

Vector Calculus

and apply the test of Eq.(3):

is exact, so

187

for some function We find

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and the integral’s value is

.

up to a constant by integrating the equations …(4)

From the first equation we get The second equation tells us that or Hence, g is a function of z alone, and

The third of Eqs. (4) tells us that or

.

Therefore, The value of the integral is

Assignments Which fields in Assignments 1-3 are conservative, and which are not ? 1.

2

3 In Exercise 4-6 find a potential function 4.

for the field F. 5

6 In Exercise 7-11, show that the differential forms in the integrals are exact. Then evaluate the integrals. 7 8 9 10

Vector Calculus

11

188

12

Evaluate the integral

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from Problem 4 by finding parametric for the

to and evaluating the line integral of line segment from along the segment. Since F is conservative, the integral is independent of the path. GREEN’S THEOREM IN THE PLANE We need to ideas for Green’s theorem. The first is the flux density of a vector field at a point which in mathematics is called the divergence of the vector field. The second one is the idea of circulation density of a vector field, which in mathematics is called the curl of the vector field. Definition

The flux density or divergence of a vector field

at the point

is

at the point

is

...(1) Problem

Find the divergence of

Solution

We use the formula in Eq.(1) :

Definition

The circulation density or curl of a vector field ……(2)

Problem

Find the curl of the vector field

Solution

We use the formula (2) :

Green’s Theorem in the Plane Theorem 1: field enclosed by

Green’s Theorem (Flux-Divergence or Normal Form) across a simple closed curve

The outward flux of a

equals the double integral of div F over the region

That is, …(3)

Theorem 2:

Green’s Theorem (Circulation-Curl Tangential Form) The counter clockwise

circulation of a field integral of curl

over the region

around a simple closed curve enclosed by

in the plane equals the double

That is, ….(4)

Problem  Vector Calculus

Verify both forms of Green’s theorem for the field  189

and the region

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bounded by the unit circle

Solution We first express all functions, derivatives, and differentials in terms of

L.H.S of Eq.(3) :

R.H.S of Eq.(3) :

Hence Theorem 1 is verified. L.H.S of Eq.(4) :

R.H.S of Eq.(4) :

Hence Theorem 2 is verified

Using Green’s Therom to evaluate line integrals Problem Evaluate the line integral Where C is the square cut from the first quadrant by the lines x=1 and y=1  Vector Calculus

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Solution We can use either from of greens theorem to change the line integral in to a double integral over the square.

1.

With eq. (3) : Taking

and

interior gives (nothing that the region

2.

With Eq.(4): Taking

and

and

as the square’s boundary and

is given by the system of inequalities

gives the same result:

Extending Green’s Theorem to other Regions Green’s theorem can be extended to regions of the form annular disk etc. The method is illustrated through Problems. Problem

Verify

the

circulation

form

of

Green’s

theorem

on

the

annular

ring

(Fig. 2), if

Solution The boundary of

consists of the circle

traversed counter clockwise as increases, and the circle

Vector Calculus

191

Traversed clockwise as

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increases. The functions

and

and their parial derivatives are

continuous throughout , Moreover,

The integral of

over the boundary of

is

d   Calculating Area with Green’s Theorem If a simple closed curve Green’s theorem, the area of

in the plane and the region

it encloses satisfy the hypotheses of

is given by the following Green’s Theorem Area formula.

Green’s Theorem Area Formula Area of

….(5)

The reason is that by Eq.(3) run backward, Area of

Area of a plane region as a line integral over the boundary and obtain

In the equation (1) of Green’s theorem choose

Also, setting

Now the double integral

we get

is the area of

of

where we integrate as indicated in Green’s theorem. This interesting formula expresses the in terms of a line integral over the boundary. It has various application; for instance, the area of theory of certain planimeters (instruments for measuring area) is based on it.  Vector Calculus

192

Problem

For an ellipse

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or

we get

thus from (4) we obtain the familiar result

Assignments In Assignments 1-2 verify Green’s theorem by evaluating both sides of Eqs. (3) and (4) for the field Take the domains of integration in each case to be the disk

and its

bounding circle 1.

2.

In Assignments 3-5, use Green’s theorem to find the counterclockwise circulation and outward flux for the field

F and curve

3. The square bounded by 4 The Triangle bounded by 5 The right hand loop of the lemniscates 6

Find the counter clockwise circulation and outward flux of the field and over the boundary of the region enclosed by the curves quadrant.

7.

around and

in the first

Find the outward flux of the field

Across the cardioids 8.

Find the work done by around the curve.

in moving a particle once counterclockwise

The boundary of the “triangle” region in the first quadrant enclosed by the x-axis, the line , and the curve Apply Green’s theorem to evaluate the integrals in Assignments 9-10 9. The triangle bounded by

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10 The circle Use the Green’s theorem area formula (5) to find the area of the regions enclosed by the curves Assignments 11-12 11

The circle

12

The asteroid SURFACE AREA AND SURFACE INTEGRALS

In this chapter our objective is the integration of function defined over a curved surface. For that purpose we first consider the surface area. The Formula for Surface Area over a closed and bounded plane region

The area of the surface

…(1)

Surface area = where p is a unit vector normal to

is

and

That is, the surface area of the surface

is the double integral over

divided by the magnitude of the scalar component of

of the magnitude

normal to

Attention! Eq.(1) is obtained under the assumption that through out that is continuous. Whenever the integral exists, however, we define its value to be the area of the portion of the surface

lies over

Problem 1 Find the area of the surface cut from the bottom of the paraboloid

by

the plane

Vector Calculus

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Solution We sketch the surfaces

and the region

below it in the , and

part of the level surface

plane (fig.1). The surface is the disk

is

in the

plane. To get a unit vector normal to the plane of , we can taken At any point

on the surface, we have

Therefore,

In the region Surface area

dxdy by changing to polar coordinates

Problem 2

by the

Find the area of the cap cut from the hemisphere

cylinder

(Fig.2)

Solution The cap

is part of the level surface

onto the disk

in the

plane. The vector

It projects one-to-one is normal to the plane of

At any point on the surface,  Vector Calculus

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as

at points of

Therefore, Surface area =

=

….(2)

, since z is the z- coordinates of a point on the sphere, we can express it in terms of

and

as

changing to polar coordinates with

Surface integrals We show how to integrate a function over a surface. Definitions

If

is the shadow region of a surface

is a continuous function defined at the points

where p is a unit vector normal to

and

defined by the equation

then the integral of g over

and g

is the integral

. The integral itself is called a surface integral.

The Surface Area Differential and the Differential Form for Surface Integrals Surface area differential is

and differential formula for surface integral is

Surface integrals behave like other double integrals, the integrals of the sum of two functions being the sum of their integrals and so on. The domain additivity property takes the form

The idea is that if

is partitioned by smooth curves into a finite number of non overlapping

smooth patches (i.e.,if is piecewise smooth), then the integral over is the sum of the integrals over the patches. Thus, the integral of a function over the surface of a cube is the sum of the integrals over the faces of the cube. We integrate over a turtle shell of welded plates by integrating one plate at a time and adding the results.  Vector Calculus

196

Problem 3

Integrate and

planes

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over the surface of the cube cut from the first octant by the (Fig.3)

Solution We integrate over each of the six sides and add the results. Since that lie in the coordinate planes, the integral over the surface of the cube reduces to = Side

on the sides

+

is the surface

over the square region

in the xy-

plane. For this surface and region,

and +

=

dy=

Symmetry tells us that the integrals of

over sided

and

are also

Hence

Orientation We call a smooth surface

orient able or two-sided if it is possible to define a field n of unit

that varies continuously with position. Any patch or sub portion of an orientable. vectors on Spheres and other smooth closed surfaces in space (smooth surfaces that enclose solids) are orientable. By convention, we choose n on a closed surface to point outward. Once n has been chosen, we say that we have oriented the surface, and we call the surface together with its normal field an oriented surface. The vector n at any point is called the positive direction at that point (fig.5)  Vector Calculus

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There are non orientable surfaces also. A well-known Problem of such a surface is the show in Fig. 3 When a normal vector, which is given at

is displaced continuously

in Fig. 3, the resulting normal vector upon returning to

along the curve original vector at

. A model of a

is opposite to the

can be made by taking a long rectangular piece of

paper, making a half- twist and sticking the shorter sides together so that two points

and the two

in Fig. 3 coincide.

points

Definition The flux of three-dimensional vector field F across an oriented surface direction of n is given by the formula

in the

….(3) Problem

outward through the surface

Find the flux of , by the planes

cut from the cylinder

and

Solution The

outward

normal

field

on

may

be

calculated

from

the

of

to be

With

we also have

We can drop the absolute value bars because The value of

on the surface is given by the formula

as

on

Therefore, the flux of F outward through

Vector Calculus

on

is

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Moments and Masses of Thin Shells Thin shells of materials like bowls, metal drums, and domes are modeled with surfaces. Their moments and masses are calculated with the following formulas.

Masses and Moment Formulas for very Thin Shells

First moments about the coordinate planes:                          Coordinate of Center of mass:                           Moments of inertia:

Find the center of mass of a thin hemispherical shell of radius a and constant density

Solution We model the shell with the hemisphere

The symmetry of the surface about the z-axis tells us that It remains only to find

.

from the formula

The mass of the shell is

To evaluate the integral for

Vector Calculus

we take

and calculate

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Then =

The shell’s center of mass is the point

Assignments 1.

Find the area of the surface cut from the paraboloid

2.

Find the area of the region cut from the plane

that lies above the triangle bounded by the

Find the area of the surface and

lines

in the

plane.

4

Find the area of the ellipse cut from the plane

5

Find the area of portion of the paraboloid in the

6

by the cylinder whose walls

and

are 3.

by the plane

by the cylinder that lies above the ring

plane.

Find the area of the surface

above the square

, in the xy-plane 7

Integrate

over the surface of the cube cut from the first octant by the

planes 8

Integrate

over the surface of the rectangular solid cut from the first octant by

the planes 9

Integrate the first octant.

10

Find the flux of the field by

rectangular surface

over the portion of the plane

that lies in

across the portion of the surface

given

in the direction k.

In Assignments 11-13, find the flux of the field F across the portion of the sphere in the first octant in the direction away from the origin. 11 12 13

Vector Calculus

200

14.

Find the flux of the field Let

and

be the portion of the cylinder

in the first octant the projects parallel to the x-axis in the

onto the rectangle vector normal to

plane (Fig.8). Let n be the unit

that points away from the yz-plane. Find the flux of the field across

16

upward through the surface cut from

thye parabolic cylinder 15

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in the direction of n

Find the outward flux of the field

across the surface of the cube cut

from first octant by the planes 17

Find the centroid of the portion of the sphere

that lies in the first octant.

PARAMETRIZED SURFACES We know that explicit form of a surface in space is

and implicit from is

. In this chapter we discuss the parametrization of surface. let

……..(1)

be a continuous vector function that is defined on a region interior of domain

. We call the range of

the surface

in the

plane and one-to-one on the

defined or traced by , and Eq.(1) together with the

constitute a prametrization of the surface. The variables

and

is the parameter domain. To simplify our discussion, we will take

to be rectangle defined by

The requirement that

inequalities of the form

are the parameters, and

be one-to-one on the interior

ensures that does not cross itself. Notice that Eq.(1) is the vector equivalent of three of parametric equations:

Problem

Find a paramtrization of the cone .

Solution Here, cylindrical coordinates provide everything we need. A typical point (Fig.2) has and

Problem

and

, with 0

on the cone . Taking

in Eq.(1) gives the parametrization

Find a parametrization of the sphere

.

Solution Spherical coordinates provide what we need. A typical point

on the sphere (Fig.3) has . Taking

and

in Eq.(1) gives the parametrization  Vector Calculus

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, Problem 3

Find a parametrization of the cylinder

Solution has

In cylindrical coordinates, a point the cylinder

and

For points on

,

We have implies implies A typical point on the cylinder therefore has

Taking

and

in Eq.(1) gives the parameterization ,

.

Surface Area Definition

A parameterized surface

is smooth if the first order partial derivatives the parameter domain, where

and

are continuous and

is never zero on

Parametric Formula for the Area of a Smooth Surface The area of the smooth surface

…(2) We can abbreviate the integral in (1) by writing

Vector Calculus

for

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Surface Area Differential and the Differential Formula for Surface Area Surface area differential is

and differential formula for surface area is

……………….(3) Problem

Find the surface area of the cone in Problem 1 (Fig.2)

Solution

In Problem 1 we found the parameterization

To apply Eq.(2) we first find

Noting that

and we have

r Thus,

The area of the cone is using Eq.(1) with

Problem

Find the area of a sphere of radius

Solution

We use the parameterization from Problem 2:

Determination of

Thus,  Vector Calculus

203

for

since

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. Therefore the area of the sphere is

Surface Integrals Having found the formula for calculating the area of a parameterized surface, we can now integrate a function over the surface using parameterized form

is smooth surface defined parametrically as

Definition

If

and

is a continuous function defined on , then the integral of

Problem

Integrate

Solution

Continuing the work in Problem 1 and 4, we have

over

is

over the cone and with

, we have

Problem

Find the flux of

outward through the parabolic cylinder

Solution The formula for a flux is

Vector Calculus

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On the surface of the given parabolic cylinder we have

and

so we

automatically have the parameterization The cross product of tangent vectors is

The unit normal pointing outward from the surface is

On the surface,

so the vector field is

Thus,

The flux of F outward through the surface is

Assignments In Assignments 1-8, find a parameterization of the surface. 1.

The paraboloid

2

The first octant portion of the cone

3

The cap cut from the sphere

Vector Calculus

between the planes

and

by the cone  205

4

The portion of the sphere

5

The surface cut from the parabolic cylinder

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. Between the planes

and

by the planes

. 6

The portion of the cylinder

7

The portion of the plane

8

a)

inside the cylinder

b)

inside the cylinder

between the planes

The portion of the cylinder

and

between the planes inside the cylinder

and Exercise 9-13; Use the parameterization to express the area of the surface as a double integral. Then evaluate the integral. inside the cylinder

9

The portion of the plane

10

The portion of the cone

between the planes

and

11

The portion of the cylinder

between the planes

and

12

The cap cut from the paraboloid

13

The lower portion cut from the sphere

by the cone

.

by the cone

.

Exercise 14-17, Integrate the given function over the given surface 14

over the parabolic cylinder

15

, over the unit sphere

16

, over the portion of the plane above the square

in the

17

that lies above the square plane

, over the parabolic dome

In Assignments 18-22 use a parameterization to find the flux

across the surface in the

given direction. outward (normal away from the x- axis) through the surface cut from

18

the parabolic cylinder 19

across the portion of the sphere away from the origin.

and

above the square

Vector Calculus

.

in the first octant in the direction

upward across the portion of the plane

20

by the planes

that lies

in the xy-plane

206

21

outward

(normal

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away

from

the

z-

axis)

through

the

cone

outward (normal away from the z- axis) through the portion of the cone

22

between the planes

and

. that lies in the first octant.

23

Find the centroid of the portion of the sphere

24

Find the moment of inertia about the z- axis of a thin spherical shell

of

constant density .

STOKES’S THEOREM Definition curl

or

For the vector field

the curl of

is the vector field, denoted by

defined by

is pronounced as del cross F. Definition by

(Here we must assume that

of a given scalar function

is the vector function given

is differentiable). We introduce the differential operator ….(1)

and write ….(2) is read as Theorem 1

” as well as (Stokes’s Theorem)

The circulation of around the boundary of an oriented surface in the direction counterclockwise with respect to the surface’s unit normal vector n equals the integrals of over ….(1) Problem

Evaluate Eq.(1) for the hemisphere , and the field

Vector Calculus

its bounding circle

.  207

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Solution We calculate the counterclockwise circulation around

(as viewed from above) using the

parameterization (of the given circle)

F.dr=-9sin2 d —9cos2 d =-9d

Also,

is the outer unit normal

(See Problem in the chapter “Surface

Area and Surface Integrals”)

Hence the curl integrals of F is = The circulation around the circle equals the integrals of the curl over the hemisphere, as it should. Problem

around the curve

Find the circulation of the field

which the plane

meets the cone

in

counterclockwise as viewed from above

Solution Stoke’s theorem enables us to find the circulation by integrating over the surface of the cone. Traversing C in the counterclockwise direction viewed from above corresponds to taking the inner normal n to the cone (which has a positive z- component). We parameterize the cone as

We then have ,

Vector Calculus

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as Hence,

and the circulation is ,using Stokes’s theorem

Problem 6

A fluid of constant density where

, find

rotates around the z-axis with velocity

is a positive constant called the angular velocity of the rotation. If

and relate it to the circulation density.

Solution With

By Stokes’s theorem, the circulation of F around a circle plane normal to

Thus

Problem

in a

say the xy-plane, is 2

( F.dr

Eq.(1) relates

….(1)

with the circulation density

Use stokes’s theorem to evaluate

boundary of the portion of the viewed as viewed above (Fig.3)

if

and C is the

in the first octant traversed counterclockwise

Solution The plane is the level surface normal vector

of the function

The unit

is a consistent with the counterclockwise motion around C. To apply stokes’s theorem, we find  Vector Calculus

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so

On the plane,

and The surface area element is

Using Stokes’s theorem, the circulation is

Definition A region D is simply connected if every closed path in D can be contracted to a point in D without ever leaving D. at every point of a simply connected open region D in space, then on any Theorem 2 If piecewise smooth closed path C in D. Problem

Prove that if

is a scalar function.

Solution We have

Hence

as by Euler’s theorem

Vector Calculus

and so on.

210

Problem

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Solution curl grad since by the previous Problem curl Problem

If

and

for any scalar

and in particular for the scalar

then show that

(i) (ii) (iii) Solution

(i) , Since

and so on

(ii)

(iii)

Taking

Problem If

in (ii) above, we obtain

, prove that

Solution We first note that

Vector Calculus

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Hence

=

+

-

Assignments Assignments 1-3, use the surface integral in Stokes’s theorem to calculate the circulation of the field F around the curve C in the indicated direction.

1. The ellipse

in the xy-plane, counterclockwise when viewed from above

2. The boundary of the triangle cut from the plane wise when viewed from above.

by the first octant, counter clock

3 The square bounded by the lines viewed from above 4

and

in the xy-plane, counterclockwise when

Let n be the outer unit normal of the elliptical shell

and let

Vector Calculus

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Find the value of

(Hint: 5

one Let

parameterization

of

the

ellipse

be the cylinder

at

the

base

of

the

shell

is

together with its Top,

, Let

. Use Stokes’s theorem to calculate the flux of

outward

through 6.

Show

that

has the same value of all oriented surfaces direction on C.

that span C and that induce the same positive

In Exercise 7-9, use the surface integral in Stokes’s theorem to calculate the flux of the curl of the field F across the surfaces

direction of the outward unit normal n.

7

8

DIVERGENCE THEOREM Definition by

The divergence of the vector field

as well as

Problem

denoted

, is the scalar function

If

, find

Solution

Definition

A vector is solenoidal if its divergence is zero.

Problem

Show that

Vector Calculus

is solenoidal  213

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Solution Since Problem

, the vector F is solenoidal Prove that

.

Solution Let

Problem

, then

Prove that

Also verify the same for

, where

is a scalar function

and

Solution

The rest of the work is left to the Assignments We now consider Divergence Theorem, which transform surface integrals into triple integrals. Theorem 1

(The Divergence Theorem)

(Transformation between surface integrals and volume integrals) The flux of a vector field

across a closed oriented surface

of the surface’s outward unit normal field n equals the integral of by the surface:

in the direction

over the region

enclosed

…(1) (i.e., outward flux is equal to divergence integral) Problem

(Verification of the Divergence Theorem)

Divergence Theorem for the field  Vector Calculus

Evaluate both sides of Eq(1) in the

over the sphere  214

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Solution The outer unit normal n to follows:

, calculated from the gradient of

, as

Hence

on the surface. Therefore,

because

The divergence of F is

so Hence L.H.S and R.H.S of Eq.(1) are the same and the Divergence Theorem is verified. Problem

Evaluate

over the sphere

by the Divergence

Theorem Solution

Hence =

= , since the volume of the sphere is

Problem

from the first octant by the planes

Vector Calculus

outward through the surface of the cube cut

Find the flux of and

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Solution Instead of calculating the flux as a sum of six separating integrals, one for each face of cube, we can calculate the flux by integrating the divergence

over the cube’s interior. That is,

Where

is the cube surface and

is the cube interior. In the ‘given cube’ interior

so that

The Divergence Theorem for the Other Regions The Divergence Theorem can be extended to regions that can be partitioned into a finite number of simple regions and to regions that can be defined as limits of simpler regions in certain ways. For Problem, suppose that D is the region between two concentric spheres and that F has continuously differentiable components throughout D and on the bounding surfaces. Split D by an equatorial plane and apply the Divergence Theorem to each half separately. The bottom half,

is

consist of an outer hemisphere, a plane washer shaped shown in Fig.1 The surface that bounds base, and an inner hemisphere. The Divergence Theorem says that

…. ..(1) The unit normal that points outward from points away from the origin along the outer surface, equals k along the flat base, and points toward the origin along the inner surface. Next apply the Divergence Theorem to

as shown in Fig. 2

….(2) As we follows over pointing outward from we see that equals –k along the washer-shaped base in the xy-plane, points away from the origin on the outer sphere, and points toward the origin on the inner sphere. When we add Eqs. (1) and (2) the integrals over the flat base cancel because of the opposite signs of

Vector Calculus

and

. We thus arrive at the result\

216

with

the region between the spheres,

unit normal to Problem 8

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the boundary of

consisting of two spheres, and n the

directed outward from Find the net outward flux of the field.

across the boundary of the region Solution The flux can be calculated by integrating

over

. We have

and

Similarly, and Hence =0

and

Hence the net outward flux across the boundary of D is zero Problem

Find the net outward flux of the field

across the boundary of the region Solution The outward unit normal on the sphere of radius is

(with

as sphere Hence on the sphere,

and  Vector Calculus

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Hence the outward flux of F across any sphere centered at the origin is

.

Gauss’ Law In this section we consider Gauss’ Law, one of the four great Laws of Electromagnetic Theory. For that purpose we need the Problem just above. In electromagnetic theory, the electric field created by a point charge

where

located at the origin is the inverse square field

is physical constant, r is the position vector of the point

and

. In the notation of the previous Problem,

The calculations in the previous Problem show that the outard flux of E across any sphere centered at the origin is E across any closed surface

. But this result is not confined to spheres. The outward flux of that encloses the origin (and to which the Divergence Theorem applies)

is also This statement, called Gauss’s law, applies to charge distributions that are more general than the one assumed here. Unifying the Integral Theorems If we think of a two-dimensional field whose k- component is zero, then

as a three-dimensional field and the normal form of Green’s Theorem can be

written as

similarly,

=

so the tangential form of Green’s Theorem can be written as

With the equations of Green’s theorem now in del notation, we can see their relationships to the equations in Stoke’s theorem and the Divergence Theorem. Green’s Theorem and its Generalization to Three Dimensions Normal form of Green’s Theorem: Divegence Theorem : Tangential form of Green’s Theorem :  Vector Calculus

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Stokes’s Theorem : Problem Prove that

, where

is the unit vector in the direction of r and

Solution Note that Hence

Problem 11

Prove that

Solution

, using Problem 8 since

and

since =  Vector Calculus

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Formulae of expansion If

are differentiable vector functions and

is a differentiable scalar function. Then

(i) (ii) (iii) Proof

(i)

(ii)

(iii)  Vector Calculus

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= Problem

If F is a vector having fixed direction, show that curl F is perpendicular to F.

Solution Since F has a fixed direction, we can write

is a scalar function. Then

where a is a constant vector and

, using formula (i) of expansion

is perpendicular to both

Now Problem

and a. Hence, in particular,

is perpendicular of F.

is solenoidal.

If a and b are irrational, prove that

Solution Given a and b are irrational, so that and Now

is a solenoidal.

Hence

Assignments In Assignments 1-4 find the total mass of a mass distribution of density 1.

Vector Calculus

T the box

in a region

in space.

,

221

2.

T the box

3.

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,

the tetrahedron with vertices (0,0,0) (1,0,0), (0,1,0) (0,0,1)

4. of a mass of density 1 in T about the x- axis,

Find the moment of inertia where T is 5. The cube 6. The cylinder

by the divergence theorem for the following data.

Evaluate the surface integral

the surface of the box given by the inequalities

7. 8.

the surface of

9.

the surface of the tetrahedron with vertices (0,0,0),(1,0,0), (0,1,0), (0,0,1)

10.

is the sphere

SYLLABUS FOR CALICUT UNIVERSITY  Vector Calculus

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FIFTH SEMESTER B.Sc. DEGREE PROGRAMME Mathematics - Core Course MM 5B 05 – VECTOR CALCULUS 4 Modules

30 weightage

Module I A quick review of Vector Algebra - Lines and planes in space - Cylinders and Quadric surfaces Cylindrical and spherical coordinates - Vector valued functions and space curves - Arc length and Unit tangent vector - Curvature, torsion and TNB frame Module II – Multivariable functions and Partial Derivatives Functions of several variables - Limits and Continuity - Partial derivatives -Differentiability linearization and differentials - Chain rule - Partial derivatives with constrained variables Directional derivatives, gradient vectors and tangent planes -

Extreme value and saddle points -

Lagrange multipliers - Taylor's formula Module III Double Integrals - Double integrals in polar form - Triple integrals in Rectangular Coordinates Triple integrals in cylindrical and spherical co-ordinates - Substitutions in multiple integrals. Module IV – Integration in Vector Fields Line integrals -Vector fields, work circulation and flux-Path independence, potential functions and conservative fields-

Green's theorem in the plane-Surface area and surface integrals -

Parametrized surfaces-Stokes' theorem (statement only)-Divergence theorem and unified theory (no proof).

Vector Calculus

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Vector Calculus

224