Ventsi Rumchev, Engineering Mathematics 233, Notes for Lecture 1

VECTORS AND VECTOR FUNCTIONS Overview: In this lecture we revisit the dot and cross products of two vectors. We introduce the concept of vector functions to model and analyse the motion of objects in space. Motivation: The notion of vectors is invaluable in mathematics, science and engineering. Vectors can represent, for example, positions in space, forces, velocities, accelerations, etc. The use of vectors leads to a greater understanding to a variety of problems in mathematics, science and engineering. Key concepts in this lecture:  Vectors (revisited): dot (scalar) product and cross (vector) product.  Vector functions and space curves.  Derivatives and integrals of vector functions.  Motion in space: position, displacement, velocity and acceleration. Contents:  Dot (scalar) product (revisited): definition, properties, angle between two nonzero vectors, orthogonality, scalar and vector projections, and work as a dot product.  Cross (vector) product (revisited): definition, determinant formula, geometric interpretation, properties.  Vector functions: definition, parametric and vector equations of space curves, limit and continuity of vector functions.  Derivatives of vector functions: definition, evaluation, smooth and piecewise smooth vector functions, differentiation rules.  Integrals of vector functions: indefinite and definite integrals, evaluation.  Motion in space: trajectory, velocity vector, speed, direction of the motion, acceleration, Newton’s Second Law of Motion. Learning Outcomes:  Ability to calculate dot products and apply them.  Ability to calculate cross products and apply them.  Differentiate vector functions.  Integrate vector functions  Model and analyse elementary motions of objects in space.

VECTORS (revisited). 1. Dot product The dot (scalar) product of vectors a = a1i + a2j + a3k and b = b1i + b2j + b3k is the scalar denoted by a.b given by a.b = a1 b1 + a2 b2 + a3 b3.

Ventsi Rumchev, Engineering Mathematics 233, Notes for Lecture 1

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Example 1. Find the dot product of a = 3i  j + 2k and b =  i + 4j + k. Solution: a.b = 3(-1) + (-1)4 + 2(1) = - 5. □ Example 2. (Dot product of vectors in component form). If a = (1, − 2, 2) and b = (−2, 3, 1) find the dot product a.b. Solution: a.b = 1(-2) + (-2)3 + 2(1) = - 6. □ Properties of the dot product. If a and b are vectors in R3 and k is a scalar, then 1. Magnitude of a vector: a.a = |a|2 2. Zero product: 0.a = 0 3. Commutativity: a.b = b.a 4. Multiple of a dot product: k(a.b) = (ka).b = a.(kb) 5. Distributivity: a.(b + c) = a.b + a.c

Angle between two vectors. If  is the angle between two nonzero vectors a and b, then a.b cos  = . | a || b | Geometric formula for the dot product. a . b = |a| |b| cos Orthogonal vector theorem. Nonzero vectors a and b are orthogonal if and only if a . b = 0. Example 3. Consider the triangle  ABC with vertices A(1, 1, 8), B(4, −3, − 4) and C(− 3, 1, 5). Find the angle formed at A. Solution: AB = (4 – 1) i + (-3 – 1) j + (-4 – 8) k = 3 i - 4 j – 12 k and AC = - 4 i – 3 k. Then, cos  =

AB  .AC = AB.AC

and the required angle is

3.( 4)  (4).0  (12).(3) 3 2  (4) 2  (12) 2 (4) 2  (3) 2

=

24 , 65

Ventsi Rumchev, Engineering Mathematics 233, Notes for Lecture 1

 = cos –1(

3

24 )  1.19. 65

□ 2

Example 4. For what values of x are the vectors ( 6, x, 2) and (x, x , x) orthogonal? Solution: For orthogonality, the dot product must be equal to zero, that is ( 6, x, 2) . (x, x2, x) = -6x + x3 + 2x = 0  x(x2 – 4) = 0  x =  2 or x = 0. The value x = 0 leads to a zero vector so the given vectors are orthogonal if and only if x =  2. □

Scalar projection of b onto a: Vector projection of b onto a:

a.b |a| a.b a a.b projab = ( ) = a | a | | a | | a |2 compab =

Example 5. Find the scalar and vector projections of b = i + 6j  2k onto a = 2i  3j + k. Solution: The scalar projection of b onto a is 2.1  (3).6  1.( 2) a.b 18 9 compab = = =   14 . 7 |a| 14 2 2  (3) 2  12 The vector projection is the scalar projection multiplied by the unit vector in the direction of a, i.e. a 9 (2,3,1) 18 27 9 projab = compab = =  i + j  k. 14 7 7 7 |a| 7 14 □

Work as a dot product. The work W done by a constant force F on an object moving along the line from point P to point Q is given by W = F . PQ where PQ is the displacement vector of the object’s motion. Example 6. A boat sails north aided by a wind blowing in a direction of N30oE (30o east of north) with magnitude 500 lb. How much work is performed by the wind as the boat moves 100 ft.

Ventsi Rumchev, Engineering Mathematics 233, Notes for Lecture 1

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Solution: The wind force F acts in direction  = 30o, so, since its magnitude is |F| = 500 lb, we have F = 500cos60o i + 500sin30oj = 250 i + 250 3 j Since the boat sails north, the displacement vector is PQ = 100 j. Therefore, the work performed by the wind is W = F . PQ = ( 250 i + 250 3 j) . 100 j = 25 000

3 lb.ft.

□

2. Cross product If a = (a1, a2, a2) and b = (b1, b2, b2), the cross (vector) product of a and b is the vector a  b = (a2b3  a3b2, a3b1  a1b3, a1b2  a2b1).

The form of the cross product may seem strange but we will soon discover that cross products have a variety of useful properties and applications. In particular, we will show that the vector a  b is orthogonal (perpendicular) to both vectors a and b, that is a  b is orthogonal to the plane defined by the vectors a and b. The cross product is interpreted in mechanics as a torque vector, which is defined as a cross product of the position and the force vectors. A simple way of remembering (and evaluating) a cross product is by the following determinant form.

Determinant formula for the cross product.

i

j

k

a  b = a1 b1

a2

a3

b2

b3

Example 7. Find a  b if a = (2, -4, 6) and b = (5, 1, 3). Solution: i j

k a  b = 2  4 6 = (-4.3 –6.1)i – (2.3 – 6.5)j + (2.1 – (-4).5)k = - 8i + 24j + 22k. 5 1 3 □

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It is not difficult to see that a  b is orthogonal to the vectors a and b by calculating the dot products: (a  b) . a = (-18, 24, 22) . (2, -4, 6) = (-18).2 + 24.(-4) + 22.6 = -36 –96 + 132 = 0 and (a  b) . b =(-18, 24, 22) . (5, 1, 3) = -90 + 24 + 66 = 0. This is a general property of the cross product.

Geometric interpretation of the cross product: Orthogonality property. The vector a  b is orthogonal to both vectors a and b. Length (magnitude) of the cross product. If  is the angle between two nonzero vectors a and b (0    ), then | a  b | = |a| |b| sin . Parallel vectors. Two nonzero vectors a and b are parallel if and only if a  b = 0. The area of the parallelogram determined by the vectors a and b is equal to the length of the cross product a  b. Example 8. Find the area of the triangle with vertices P(−2, 4, 5), Q(0, 7, − 4) and R(−1, 5, 0). Solution: The triangle  PQR is half the area of the parallelogram determined by the vectors PQ = (0 – (−2), 7 – 4, − 4 – 5) = (2, 3, −9) and PR = (− 1 – (− 2), 5 – 4, 0 – 5) = (1, 1, − 5); that is Area( PQR) = (1/2) | PQ PR |. Now, i j k PQ PR = 2 3  9 = (-15 + 9) i – (-10 + 9) j + (2 – 3) k = – 6i + j – k, 1 1 5 and | PQ PR | = (6) 2  12  (1) 2  38 . Thus, Area( PQR) = (1/2) 38 [unit2]. □

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Properties of the cross product. If a, b and c are vectors and t is a scalar, then 1. Anticommutativity: a  b = – b  a. 2. Scalar distributivity: (ta)  b = t (a  b) = a  (tb). 3. Distributivity of the cross product over addition: a  (b + c) = a  b + a  c (a + b)  c = a  c + b  c 4. Product of a multiple: a  (ta) = 0; in particular, a  a = 0. 5. Zero product: a  0 = 0  a = 0 6. Scalar triple product: a . (b  c) = (a  b) . c = b . (c  a) 7. Vector triple formula: a  (b  c) = (a.c)b – (a.b)c.

VECTOR FUNCTIONS

A vector-valued function, or vector function, is a function with a domain T, a set of real numbers, and a range R, a set of vectors R. In other words, for every number t in the domain T of the function there is a unique vector r(t) = (f(t), g(t), h(t)) = f(t)i + g(t)j + h(t)k that is an element of R. The real-valued functions f(t), g(t) and h(t) are called component functions of r(t).

When a particle P moves through space during time interval T, we think of particle’s coordinates as (real-valued) functions defined on T x = x(t) y = y(t) z = z(t)

, t  T.

(1)

The points P (x, y, z) = (x(t), y(t), z(t)) and t  T, make up a curve C in space (a space curve) called particle’s path. The equations (1) are termed parametric equations of C and t is called a parameter. The position vector of the particle defines the vector equation of C: r(t) = OP = (x(t), y(t), z(t)) = x(t)i + y(t)j + z(t)k. Example 9. Line Segment. Find the vector and parametric equations for the line segment that joins the point A(x1, y1, z1) to the point B(x2, y2, z2). Solution:

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Let rA= OA = (x1, y1, z1) be the position vector of the point A and rB = OB =(x2, y2, z2) be the position vector of the point B. Then, AB = OB – OA. Now, take any point P(x, y, z) on the line segment AB. Then, the position vector OP = r(t) of the point P can be written as OP = OA + AP. On the other hand we have AP = t AB for 0  t  1 and OP = OA + AP = OA + t AB = OA + t (OB – OA) = (1 – t) OA + t OB for t  [0, 1]. Thus, the vector equation of the line segment that joins the point A(x1, y1, z1) to the point B(x2, y2, z2) is r(t) = (1 – t) rA + t rB

for t  [0, 1],

and the corresponding parametric equations are x = (1 - t) x1 + t x2, y = (1 - t) y1 + t y2 and z = (1 - t) z1 + t z2 for t  [0, 1].

□

The limit of vector function r(t) = (f(t), g(t), h(t)) as t  to is the vector

lim r(t) = ( lim f(t), lim g(t), lim h(t)), t t 0

t t 0

t t 0

t t 0

provided the limits of the component functions exist. A vector function r(t) is said to be continuous at t = to if to is in the domain of r(t) and lim r(t) = r(to), t t 0

that is r(t) is continuous at to if and only if its component functions f(t), g(t) and h(t) are continuous at to.

Example 10. If r(t) = (cos t)i + (sin t)j + t k, then lim r(t) = ( lim cos t)i + ( lim sin t)j +( lim t)k = 0 i +1 j +( / 2) k = j +( / 2)+k.

t  / 2

t  / 2

t  / 2

t  / 2

□ The derivative of a vector function r(t) is defined as r =

dr r (t  h)  r (t ) = lim . h 0 dt h

The function r(t) is said to be differentiable at t = to if r  exists at to. It can be shown that if r(t) is differentiable at to and r   0 then the derivative r  ( to) is a tangent vector to the curve traced by the function r(t) at the point t = to and points in the direction of increasing t.

Ventsi Rumchev, Engineering Mathematics 233, Notes for Lecture 1

Evaluation of the derivative. If r(t) = (f(t), g(t), h(t)) = f(t)i + g(t)j + h(t)k, where f(t), g(t) and h(t) are differentiable functions, then r  = ( f (t ) , g (t ) , h(t ) ) = f (t ) i + g (t ) j + h(t ) k.

The curve traced by r(t) is smooth on any interval of t if r  (t) is continuous and never 0; that is, if f(t), g(t) and h(t) have continuous first-order derivatives which are not simultaneously 0. The curve is piecewise smooth on an interval that can be subdivided into a finite number of subintervals on which r(t) is smooth.

Example 11. (Determining whether a curve is smooth) Determine whether the graph of the vector function r(t) = (t2 – 1, cos t, et + e-t) is smooth for all t. Solution: The derivative

r  (t) = (2t, – sin t, et – e-t)

is continuous for all t but r  (0) = (0, 0, 0) = 0, so the curve traced by the function r(t) is not smooth but it is smooth on any interval not containing t = 0. □

Differentiation Rules for Vector Functions. Let u and v be differentiable vector functions of t, k being a scalar and f(t) a realvalued function, then d [ u(t)  v(t)] = u (t)  v (t). dt d 2. Scalar Multiple Rules: [k u(t)] = k u (t). dt d [ f(t) u(t)] = f (t ) u(t) + f(t) u (t). dt d 3. Dot Product Rule: [ u(t) . v(t)] = u (t) . v(t) + u(t) . v (t). dt d 4. Cross Product Rule: [u(t)  v(t)] = u (t)  v(t) + u(t)  v (t). dt

1. Sum/Difference Rule:

5. Chain Rule:

d [ u(f(t)] = f (t ) u (f(t). dt

6. Vector Functions of Constant Length: | u(t) | = const. d u(t) . u(t) = 0, that is the vectors u(t) and u (t) are orthogonal. dt

8

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Example 12. Let u(t) = (cos t)i + j + t k, v(t) = i + e-2t j + 2t3 k and f(t) = ln t. d d Find [ f(t) u(t)] and [ u(t) . v(t)]. dt dt Solution: Let us first evaluate the derivatives u (t), v (t) and f (t ) . They are d d d (cos t))i + ( (1))j + ( (t))k = (-sin t) i + 0j + k = (-sin t) i + k, dt dt dt v (t) = -2e-2t j + 6t2 k

u (t) = ( and

f (t ) = 1/t. Now, d [ f(t) u(t)] = f (t ) u(t) + f(t) u (t) = (1/t) [(cos t)i + j + t k] + (ln t)[ (-sin t) i + k] = dt = [(1/t)cos t  (ln t)(-sin t)]i + j + (1 + ln t) k. (Scalar Multiple Rule) and

d [ u(t) . v(t)] = u (t) . v(t) + u(t) . v (t) = [(-sin t) i + k] . [i + e-2t j + 2t3 k] + dt + [(cos t)i + j + t k] . [-2e-2t j + 6t2 k] = (-sin t)(1) + (0)( e-2t) + (1)( 2t3)+ (cos t)(0) +

+ (1)( 2e-2t) + (t)( 6t2) = -sin t + 8 t3 2e-2t. (Dot Product Rule) □

The indefinite integral of r(t) with respect to t is the set of all the antiderivatives of r(t), denoted by  r (t)dt . If R(t) is any antiderivative of r(t), then

 r(t)dt

= R(t) + c,

where c is a constant vector.

Example 13. (Finding Antiderivatives). Find the indefinite integral of r(t) = (cos t)i + j + t k. Solution:

 r(t)dt = (  cos tdt ) i + ( 1dt ) j + (  tdt ) k = (sin t + c1) i + (t + c2) j + ((1/2)t 2

2

+

2

c3)k = (sin t) i + t j + (1/2)t k + c1i + c2 j + c3 k = (sin t) i + t j + (1/2)t k + c.

□

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If the component functions of the vector function r(t) = f(t)i + g(t)j + h(t)k are integrable over [a, b], then the same follows for r(t), and the definite integral of r(t) from a to b is b

b

b

b

a

a

a

a

 r(t)dt = (  f (t )dt )i + (  g (t )dt )j + (  h(t )dt )k. Example 14. (Evaluating Definite Integrals). If r(t) = (cos t)i + j + t k, then the integral of r(t) over [0, /2] is  /2

 /2

0

0

 r(t )dt = (

 /2

 /2

0

0

2  costdt ) i + ( 1dt ) j + (  tdt ) k = [(sin t) i + t j + (1/2)t k]

t  /2 = t0

= i + (/2) j + ( /8) k. 2

 □

Motion in Space. Consider a particle (rigid body) moving in space. The particle’s path (a space curve) is traced by its position vector r(t). The vector function r(t) is named also the trajectory of the particle. Then, the following functions characterise the particle’s motion: velocity vector at time t is the derivative of the position vector:

v(t) = r  (t);

speed at time t is the magnitude of the velocity:

| v(t) |;

direction of the motion at time t is given by the unit vector:

v(t) / |v(t)|;

acceleration is the derivative of the velocity vector:

a(t) = v (t) = r (t ) .

Example 15. (Motion in Space). Find the velocity, speed and acceleration of a particle with position function r(t) = (sin t)i + (2cos t)j . Sketch the particle’s path and draw the velocity and acceleration vectors for t = /6. Solution: The motion is on the plane. The trajectory is r(t) = (sin t)i + (2cos t)j, so the velocity vector at time t is v(t) = r  (t) = (cos t)i + (- 2sin t)j. The speed | v(t) | is the magnitude of the velocity - | v(t) | = (cost ) 2  (2 sin t ) 2  1  3 sin 2 t , and the acceleration is the derivative of the velocity - a(t) = v (t) = r (t ) = (- sin t)i +

Ventsi Rumchev, Engineering Mathematics 233, Notes for Lecture 1 (-2cos t)j. At t = /6 we have: r(/6) = (sin /6)i + (2cos /6 )j = (1/2) i +

11 3 j, v(/6)

3 i – j and a(/6) = (-1/2)i - 3 j. 2 The parametric equations of the particle’s path are x = sin t and y = 2cos t. To sketch the particle’s path let us find its Cartesian equation. Since

=

y x 2  ( ) 2 = sin2 t + cos2 t = 1, 2

the particle’s path is an ellipse centred at the origin with focal axis along the y – axis. □ Example 16. (Motion in Space). Find the velocity and position vector of a particle with acceleration a(t) = – 10 k, initial velocity v(0) = i + j – k and position r(0) = 2 i + 3 j. Solution.: a(t) = – 10 k  v(t) =  a(t)dt = – 10 t k + c1 and v(0) = 0 + c1 = i + j – k, so c1 = = i + j – k and v(t) = = i + j – (10 t + 1) k. Now, r(t) =  v(t)dt =  [ i + j – (10 t + 1) k]dt = t i + t j - (5t2 + t) k + c2. Since the initial position is r(0) = 2 i + 3 j, we have r(0) = 2 i + 3 j = 0 i + 0 j + 0 k + c2, so c2 = 2 i + 3 j and the position vector is r(t) = (t + 2) i + (t + 3)j + (5t2 + t) k. □ The Newton’s Second Law of Motion states that the rate at which a body’s momentum changes is equal to the net force acting on the body (Philosophiae Naturalis Principia Mathematica, first published in 1687). Newton expresses his laws in terms of mass, momentum, and force. The momentum p of a body is defined as the product of its mass m and velocity v, i.e. p = m v. If the net force acting on a body is F, then the Newton’s second law becomes F=

d d p= ( mv). dt dt

If the mass m of the body is constant, Newton’s Second law can be rewritten as F = m a, d where the acceleration a = v. dt

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Example 17. (Motion in Space). A projectile is fired with an angle of elevation  and initial velocity vo. Assuming that air resistance is negligible and the only external force is gravity, find the position function r(t) of the projectile. What value of  maximises the range (the horizontal distance travelled)? Solution: The axis can be set up so that the projectile starts at the origin. y

V0



x

d

The force due to gravity acts downwards, so it is equal to  mg j, where the (freefall) acceleration due to gravity is g = 9.8 m/s2, and it is the only force acting on the projectile, since air resistance is negligible. According to Newton’s second law of motion, the sum of all forces acting on the projectile must equals ma, so  mg j = F = ma. Thus, a =  g j, and, since v (t) = a, we have v(t) =  gt j + C1, where C1 = v(0) = vo is the initial velocity, which is given. Therefore, r  (t) =  gt j + vo,

and after integrating again we find the position vector of the projectile r(t) =  (1/2)gt2 j + t vo + C2,

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At t = 0 the projectile is at the origin, that is r(0) = 0, and, clearly, C2 = 0, so that the position vector of the projectile is r(t) =  (1/2)gt2 j + t vo. The initial velocity vo of the projectile can be written as vo = vo cos i+ vo sin  j, The expression for the position vector then becomes r(t) = (vo cos)t i + ((vo sin )t  (1/2)gt2) j , Consequently, the parametric equations of the trajectory are x(t) = (vo cos)t and y(t) = (vo sin )t  (1/2)gt2. The range d is defined as the total horizontal distance travelled by the projectile during its flight. Thus the range is the value of x when y = 0. The solution of (vo sin )t  (1/2)gt2 = 0 yield t = 0 and t = (2 vo sin ) / g. The latter value of t only defines the range, namely d = xmax = (vo cos)

2vo sin  vo sin 2 = . g g

The range d is maximal when sin 2 = 1, so that the maximal range is achieved for  =  / 4. □