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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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
is obtained by adding
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
(read “a
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
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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
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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
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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
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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
.
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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
We add the restriction …..(6)
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
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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
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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).
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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.
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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
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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
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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
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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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(Motion on the circle) and
4
(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.
6
7
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
9
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
3
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
:
Vector Calculus
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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
Vector Calculus
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
73
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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 .
Vector Calculus
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
Vector Calculus
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
Vector Calculus
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
Vector Calculus
, 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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90
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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
in radius 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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is the circum radius. Hence
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
radian. Hence
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
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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,
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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.
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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
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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
with the gradient 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
The gradient of
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
is the gradient
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
The gradient of
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
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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
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, Hence, the gradient of
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
The gradient of
is
Therefore, The change approximately
in that result from moving
unit away from
in the direction of
is
Algebra Rules for Gradients 1.
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
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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.
has a saddle poin at
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.
Vector Calculus
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
127
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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
in the first quadrant. 17
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
? Give reasons for your answer. 25
Among all the points on the graph of
that lie about the plane
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
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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
(The Orthogonal Gradient Theorem)
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
133
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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
Hence the gradient equation gives
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
134
The gradient equation implies that
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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
135
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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
136
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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
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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.
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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
139
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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
141
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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
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143
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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
over the quadrant bonded by
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
145
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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
of a sphere of radius
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
about the x-axis
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
in the first quadrant.
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
151
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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
152
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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
153
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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
155
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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
156
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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.
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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)
and read “the integral of
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
moment of inertia and radius of gyration about the
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
in the first quadrant from
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
Find the gradient field of
is the field gradient
Solution The gradient field of
is the
Problem Find the unit normal to the surface Solution This is level surface for the function the given surface.
at the point grad
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
176
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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=
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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
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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
By addition, we have …(4)
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
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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
gradient
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:
Radius of gyration about a line Problem
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
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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
The gradient grad
(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 ,
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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
of radius bounding a disk
….(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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Prove that curl grad
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
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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
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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
, since grad and in particular, grad
, 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.
,
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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
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