2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS131

2.2

Di¤erentiation and Integration of Vector-Valued Functions

Simply put, we di¤erentiate and integrate vector functions by di¤erentiating and integrating their component functions. Since the component functions are realvalued functions of one variable, we can use the techniques studied in calculus I and II.

2.2.1

Di¤erentiation

De…nition 123 Let ! r (t) be a vector function. The derivative of ! r with re! d r is de…ned to be spect to t, denoted ! r 0 (t) or dt ! r (t + h) ! r (t) ! r 0 (t) = lim h!0 h ! 0 Geometrically, r (a) is the vector tangent to the curve at t = a. De…nition 124 The line tangent to a curve C with position vector ! r (t) at t = a is the line through ! r (a) in the direction of ! r 0 (a). De…nition 125 (Unit Tangent Vector) The unit tangent vector, denoted ! T (t) is de…ned to be ! ! r 0 (t) T (t) = !0 (2.4) k r (t)k ! Remark 126 Of course, the above de…nition makes sense only if ! r 0 (t) 6= 0 .

The derivative is de…ned in terms of limits. Taking the limit of a vector function amounts to taking the limits of the component functions. Thus, we have the following theorem: Theorem 127 If ! r (t) = hf (t) ; g (t) ; h (t)i then ! r 0 (t) = hf 0 (t) ; g 0 (t) ; h0 (t)i. There is a similar result for plane curves. Since the component functions are real-valued functions of one variable, all the properties of the derivative will hold. We have the following theorem: Theorem 128 Suppose that ! u and ! v are di¤ erentiable vector functions, c is a scalar and f is a real-valued function. Then: 1. (! u (t)

0 ! v (t)) = ! u 0 (t)

! v 0 (t)

0 2. (c! u (t)) = c! u 0 (t)

u (t) + f (t) ! u 0 (t) 3. (f (t) ! u (t)) = f 0 (t) ! 0

0 4. (! u (t) ! v (t)) = ! u 0 (t) ! v (t) + ! u (t) ! v 0 (t)

132

CHAPTER 2. VECTOR FUNCTIONS

0 5. (! u (t) ! v (t)) = ! u 0 (t) ! v (t) + ! u (t) 0 6. (! u (f (t))) = f 0 (t) ! u 0 (f (t))

! v 0 (t)

D E 2 Example 129 Let ! r (t) = t; et ; sin 2t . Find ! r 0 (t) and the unit tangent vector at t = 0. Then, …nd the equation of the tangent at t = 0. 1. Computation of ! r 0 (t). D E 2 ! r 0 (t) = 1; 2tet ; 2 cos 2t

2. Computation of the unit tangent at t = 0. First, we must …nd the tangent vector at t = 0 . This vector is ! r 0 (0). From our computation above, we see that ! r 0 (0) = h1; 0; 2i ! The unit tangent vector at t = 0 is T (0). ! ! r 0 (0) T (0) = k! r 0 (0)k = =

h1; 0; 2i p 5 1 2 p ; 0; p 5 5

3. Computation of the tangent line. The parametric equations of the line through ! r (0) = h0; 1; 0i with direction vector h1; 0; 2i is 8 < x=t y=1 : z = 2t

De…nition 130 (Smooth Curve) A curve C given by a position vector ! r (t) on an interval I is said to be smooth if the conditions below are satis…ed: 1. ! r 0 (t) is continuous. ! 2. ! r 0 (t) 6= 0 except possibly at the endpoints of I.

Smooth curves will play an important role in the next sections. Geometrically, a curve is not smooth at points where there is a corner also called a cusp. Example 131 Consider the curve given by ! r (t) = 1 + t2 ; t3 . Find if it is 1

smooth on R. What about on (0; 1)? We start by computing ! r 0 (t). ! r 0 (t) = 2t; 3t2 . Wee that ! r 0 (t) is always ! ! 0 continuous. However, r (t) = 0 when t = 0. Hence, this curve is not smooth on R. It is smooth on (0; 1) since ! r 0 (t) is continuous there and never equal ! to 0 .

2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS133 We …nish with the proof of a well known result which we state sition. Proposition 132 Let C be a curve given by a position vector ! r (t). c (a constant) then ! r 0 (t) ? ! r (t) for all t. 2 Proof. We know that k! r (t)k = ! r (t) ! r (t). So, we have ! r (t)

as a propoIf k! r (t)k =

! r (t) = c2 . If we di¤ erentiate each side and use our rules of di¤ erentiation, we get ! r 0 (t) ! r (t) + ! r (t) ! r 0 (t) = 0 2! r 0 (t) ! r (t) = 0 ! r 0 (t) ! r (t)

=

0

This says that ! r 0 (t) ? ! r (t). You may not recognize this result the way it is stated. Think of a circle. The position vector of a circle is its radius. The theorem stated in the case of a circle says that the radius of a circle is perpendicular to the tangent to the circle.

2.2.2

Integration

De…nition 133 If ! r (t) = hf (t) ; g (t) ; h (t)i then *Z Z b Z b Z b ! r (t) dt = f (t) dt; g (t) dt; a

and

a

Z

! r (t) dt =

Z

a

f (t) dt;

Z

b

h (t) dt

a

g (t) dt;

We have similar de…nitions for plane curves.

Z

+

h (t) dt

R ! 1 . Find R (t) = ! r (t) dt Example 134 Let ! r (t) = cos 2t; 2 sin t; 2 1+t ! which satis…es R (0) = h3; 2; 1i. Z ! ! R (t) = r (t) dt Z Z Z 1 = cos 2tdt; 2 sin tdt; dt 1 + t2 1 = sin 2t + C1 ; 2 cos t + C2 ; tan 1 t + C3 2 ! Since we want R (0) = h3; 2; 1i, we must have h3; 2; 1i = h0 + C1 ; 2 + C2 ; 0 + C3 i Thus, C1 = 3, C2 =

4 and C3 = 1. It follows that

! R (t) =

1 sin 2t + 3; 2 cos t 2

4; tan

1

t+1

134

CHAPTER 2. VECTOR FUNCTIONS

2.2.3

Velocity and Acceleration

In this section, we look at direct applications of the derivative and the integral of a vector function. De…nition 135 (Velocity and Acceleration) Consider an object moving along C, a smooth curve, twice di¤ erentiable, with position vector ! r (t). 1. The velocity of the object, denoted ! v (t) is de…ned to be ! v (t) = ! r 0 (t)

(2.5)

2. The acceleration of the object, denoted ! a (t) is de…ned to be ! a (t) = ! v 0 (t) = ! r 00 (t)

(2.6)

3. The speed of the object is k! v (t)k. Example 136 (Finding the velocity and acceleration of a moving object) An object is moving along the curve ! r (t) = t; t3 ; 3t for t 0. Find ! v (t), ! a (t) and sketch the trajectory of the object as well as the velocity and acceleration when t = 1. Velocity:

Acceleration:

! v (t) = 1; 3t2 ; 3 ! a (t) = h0; 6t; 0i

Sketch: When t = 1, we have ! v (1) = h1; 3; 3i and

! a (1) = h0; 6; 0i

The trajectory of the object as well as the velocity and acceleration at t = 1 are shown in …gure 2.2. In many applications, we do not know the position function. Instead, we know the acceleration and we must …nd the velocity and position function. The next example illustrates this. Example 137 (Finding a Position Function by Integration) A moving particle starts at position ! r (0) = h1; 0; 0i with initial velocity ! v (0) = h1; 1; 1i. ! Its acceleration is a (t) = h4t; 6t; 1i. Find its velocity and position function at time t.

2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS135

Figure 2.2: Motion of an object along ! r (t) = t; t3 ; 3t Velocity: Since ! a (t) = ! v 0 (t), it follows that Z ! v (t) = ! a (t) dt Thus

! v (t) = 2t2 + C1 ; 3t2 + C2 ; t + C3

We …nd the constants by using the initial condition. ! v (0) = h1; 1; 1i = hC1 ; C2 ; C3 i Thus, C1 = 1, C2 =

1 and C3 = 1. It follows that ! v (t) = 2t2 + 1; 3t2

1; t + 1

Position Function: Since ! v (t) = ! r 0 (t), it follows that Z ! ! r (t) = v (t) dt =

2t3 + t + C1 ; t3 3

t + C2 ;

t2 + t + C3 2

We …nd the constants by using the initial condition. ! r (0) = h1; 0; 0i = hC1 ; C2 ; C3 i

136

CHAPTER 2. VECTOR FUNCTIONS Thus, C1 = 1 and C2 = C3 = 0. It follows that ! r (t) =

2t3 + t + 1; t3 3

t;

t2 +t 2

Remark 138 The above problem can be set up as a di¤ erential equation, that is an equation which involves the derivatives of an unknown function. Solving the equation amounts to …nding the unknown function. We were given the acceleration ! a (t) and we had to …nd the position function ! r (t). Since ! ! 00 a (t) = r (t), the above problem could have been stated as: Find ! r (t) given ! ! ! 00 that r (t) = h4t; 6t; 1i, r (0) = h1; 0; 0i and v (0) = h1; 1; 1i. The last two conditions are called the initial conditions because they are the conditions when t = 0. Of course, we would solve the problem the same way in this particular example. However, many di¤ erential equations which appear in applied mathematics are far more challenging to solve. Remark 139 In general, we can recover velocity by integration when acceleration is known. Similarly, we can recover position by integration when velocity is known. More precisely, if ! a (t0 ) and ! v (t0 ) are known, then ! v (t) = ! v (t0 ) +

Z

t

t0

! a (u) du

This can be derived from the de…nition of acceleration. Since a (t) = v 0 (t), Rt Rt integrating from t0 to t gives t0 a (u) du = t0 v 0 (u) du. Using the fundamental Rt theorem of calculus, we get t0 a (u) du = v (t) v (t0 ) hence the result. Similarly, if ! v (t) and ! r (t0 ) are known, then ! r (t) = ! r (t0 ) +

Z

t

t0

! v (u) du

This can be proven the same way as the formula for velocity. Remark 140 When studying the motion of an object, the time at which we start tracking the object is usually set to 0. The acceleration at time t = 0 is called initial acceleration and denoted ! a 0 . Similarly, the velocity at time t = 0 is called initial velocity and usually denoted ! v 0 . The position at time t = 0 is called initial position and usually denoted ! r 0 . The term initial means at time t = 0.

2.2.4

Projectile Motion

If you wondered how we can know the acceleration and not the other quantities, this section will hopefully answer your questions. We begin with a little bit of ! physics. Newton’s second law of motion states that if at time t a force F (t)

2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS137 acts on an object of mass m, this action will produce an acceleration ! a (t) of the object satisfying ! F (t) = m! a (t) Thus, if we know the force acting on the object, we can …nd the acceleration. We will then be in the situation of the previous example. We illustrate this with a classical problem in physics, the problem of …nding the trajectory of an object being thrown in the air and subject to the laws of physics. Example 141 A projectile is …red with an angle of elevation and initial velocity ! v 0 as shown on the next slideIf we ignore air resistance, the only force acting on the object is gravity. 1. Find the position of the object ! r (t) in terms of 2. Express the range d in terms of 3. Find the value of

and ! v 0.

.

which maximizes the range d.

4. What is the maximum height reached by the object? Answer to 1. When translating a real world problem into mathematics so we can solve it, we will have to introduce a coordinate system so we can measure distances and position objects. This step is crucial. We want to do it in a way that makes all the equations we will derive as simple as possible. We begin by setting the axes so that y is the vertical direction and x the horizontal direction. In addition, the motion of the projectile begins at the origin. The only force the object is subject to is gravity. Since the force of gravity acts downward, we have ! a (t) = h0; gi where g = 9:81m=s2 . We can now compute ! v (t). Let us introduce some notation, let ! v (0) = hv0x ; v0y i let ! v (t) = hvx ; vy i and let v0 = k! v 0 k. Now, by de…nition, Z ! ! v (t) = a (t) dt =

Then, we have Thus

hC1 ; gt + C2 i

! v (0) = hv0x ; v0y i = hC1 ; C2 i ! v (t) = hv0x ; gt + v0y i

Elementary trigonometry tells us that v0x

= v0 cos

v0y

= v0 sin

138

CHAPTER 2. VECTOR FUNCTIONS Thus

! v (t) = hv0 cos ; gt + v0 sin i Now, we can compute ! r (t). Z ! ! r (t) = v (t) dt =

1 2 gt + (v0 sin ) t + C2 2

(v0 cos ) t + C1 ;

Since ! r (0) = h0; 0i, both C1 and C2 are 0 hence ! r (t) =

(v0 cos ) t;

1 2 gt + (v0 sin ) t 2

So, we can see that the motion in the x-direction and in the y-direction follow di¤ erent laws. This can be made more obvious by writing the parametric equations of the trajectory of the object. (

x = (v0 cos ) t 1 2 gt + (v0 sin ) t y= 2

Answer to 2. d, the range, is the horizontal distance corresponding to y = 0. We can …nd d by …nding the value of t for which y = 0 and plugging this value in the equation for x. y

=

0 ()

()

t

1 2 gt + (v0 sin ) t = 0 2

1 gt + v0 sin 2

=0

2v0 sin . t = 0 corresponds to the g 2v0 sin initial position. The value of t we want is t = . This value gives g us This happens when either t = 0 or t =

d =

(v0 cos )

d =

v02 sin 2 g

2v0 sin g

Answer to 3. We see that d is maximum when sin 2 = 1 that is when

=

4

.

Answer to 4. The object reaches its maximum altitude when its vertical velocity is 0 that is when vy = gt + v0 sin = 0 or when t = v0 sin . At that g

2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS139

Figure 2.3: Trajectory of a Projectile time, the altitude is y

v0 sin g

2

v0 sin g

=

1 g 2

=

(v0 sin ) 1 (v0 sin ) + 2 g g

+ (v0 sin ) 2

v0 sin g

2

2

=

2.2.5

(v0 sin ) 2g

Things to Know

Be able to tell if and where a curve is smooth. Be able to compute the derivative of a space curve. Know what the unit tangent vector is and be able to …nd it. Be able to compute the integral of a space curve. Know the relationship between the position vector, the velocity and acceleration of a particle in motion. Be able to …nd the velocity and acceleration of a particle given its position vector. Be able to …nd the velocity and position of a particle given its acceleration and initial velocity and position. Be able to do problems similar to the example on projectile motion.

140

2.2.6

CHAPTER 2. VECTOR FUNCTIONS

Problems

1. Evaluate 2. Evaluate 3. Evaluate

R1

t3 ; 7; t + 1 dt

0

R

sin t; 1 + cos t; sec2 t dt

4 4

E 1 1 ; ; t 5 t 2t dt

R 4 D1 1

4. Determine if the curves below are smooth on R. (a) ! r (t) = t3 ; t4 ; t5 (b) ! r (t) = t3 + t; t4 ; t5 5. ! r (t) = t + 1; t2 + 1 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 …nd the particle’s velocity and acceleration when t = 1. 6. ! r (t) = et ; 92 e2t 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 …nd the particle’s velocity and acceleration when t = ln 3. 7. Motion on a circle x2 + y 2 = 1. ! r (t) = hsin t; cos ti is the position of a particle in the xy-plane at time t. Find the particle’s velocity and acceleration when t = 4 and t = 2 and plot them as vectors along with the curve. 8. Motion on a cycloid. ! r (t) = ht sin t; 1 cos ti is the position of a particle in the xy-plane at time t. Find the particle’s velocity and acceleration when t = and t = 32 and plot them as vectors along with the curve. 9. ! r (t) = t + 1; t2 1; 2t is the position of a particle in space at time t. Find the particle’s velocity, acceleration and speed when t = 1. 10. ! r (t) = h2 cos t; 3 sin t; 4ti is the position of a particle in space at time t. Find the particle’s velocity, acceleration and speed when t = 2 . E D 2 11. ! r (t) = 2 ln (t + 1) ; t2 ; t2 is the position of a particle in space at time t. Find the particle’s velocity, acceleration and speed when t = 1. p 12. Find the angle between the velocity and acceleration if ! r (t) = 3t + 1; 3t; t2 at t = 0. 13. Find the angle between the velocity and acceleration if ! r (t) = ln t2 + 1 ; tan at t = 0. 14. Find the equation of the line tangent to ! r (t) = sin t; t2 t = 0.

cos t; et at

1

p t; t2 + 1

2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS141 15. Find the equation of the line tangent to ! r (t) = ha sin t; a cos t; bti at t=2 . 16. Consider particles moving along the unit circle x2 + y 2 = 1. Answer the questions below for each of the particle a e. i. Does the particle have constant speed? If so, what is it? ii. Is the particles’s acceleration always orthogonal to its velocity vector? iii. Does the particle move clockwise or counterclockwise around the circle? iv. Does the particle begin at the point (1; 0)? (a) ! r (t) = hcos t; sin ti, t 0. ! (b) r (t) = hcos 2t; sin 2ti, t 0. ; sin t (c) ! r (t) = cos t 2

(d) ! r (t) = hcos t; sin ti, t ! (e) r (t) = cos t2 ; sin t2 , t

2

,t

0.

0. 0.

17. A particle moves along the top of the parabola y 2 = 2x from left to right at constant speed 5 units per second. Find the velocity of the particle as it moves through (2; 2). hint: You …rst need to …nd the parametric equations of the curve. Recall, if y = f (x) then the parametric equations x = at are where a is a constant. In this case, …rst write x = f (y) y = f (at) then use a similar trick for the parametric equations. 18. Solve for ! r (t) in

d! r dt

= h t; t; ti . ! r (0) = h1; 2; 3i

19. 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 h3; 1; 1i. Find the position vector ! r (t). 20. A projectile is …red at a speed of 840m=s at an angle of 60 . How long will it take to get 21km downrange? 21. A projectile is …red with an initial speed of 500m=s at an angle of elevation of 45 . (a) When and how far away will the projectile strike? (b) How high overhead will the projectile be when it is 5km downrange? (c) What is the greatest height reached by the projectile?

142

CHAPTER 2. VECTOR FUNCTIONS

2.2.7

Answers

1. Evaluate

R1

t3 ; 7; t + 1 dt

0

Z

1

t3 ; 7; t + 1 dt =

0

2. Evaluate

3. Evaluate

R

1 3 ; 7; 4 2

sin t; 1 + cos t; sec2 t dt

4 4

Z

R 4 D1 1

D sin t; 1 + cos t; sec2 t dt = 0;

4

4

2

+

p

2; 2

E

E 1 1 ; ; t 5 t 2t dt Z

1

4

1 1 1 ; ; t 5 t 2t

dt = hln 4; ln 4; ln 2i

4. Determine if the curves below are smooth on R. (a) ! r (t) = t3 ; t4 ; t5 Not smooth. ! (b) r (t) = t3 + t; t4 ; t5 Smooth.

5. ! r (t) = t + 1; t2 + 1 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 …nd the particle’s velocity and acceleration when t = 1. 2

y = (x

1) + 1

! v (t) = h1; 2ti Thus

! v (1) = h1; 2i

and ! a (t)

=

h0; 2i ! = a (1)

6. ! r (t) = et ; 92 e2t 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 …nd the particle’s velocity and acceleration when t = ln 3. y=

2 2 x 9

2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS143 ! v (t) =

4 et ; e2t 9

Thus ! v (ln 3) = h3; 4i and ! a (t) =

8 et ; e2t 9

So ! a (ln 3) = h3; 8i 7. Motion on a circle x2 + y 2 = 1. ! r (t) = hsin t; cos ti is the position of a particle in the xy-plane at time t. Find the particle’s velocity and acceleration when t = 4 and t = 2 and plot them as vectors along with the curve. *p p + 2 2 ! r = ; (blue) 4 2 2 ! r

= h1; 0i

(red)

! v (t) = hcos t;

sin ti

So ! v

=

4 ! v

2

*p

p + 2 (light blue) 2

2 ; 2

= h0; 1i (light red)

2

! a (t) = h sin t; ! a

4 ! a

=

2

* p 2

2

;

cos ti

p + 2 (cyan) 2

= h 1; 0i (pink)

144

CHAPTER 2. VECTOR FUNCTIONS

1.0

y

0.8 0.6 0.4 0.2

-1.0 -0.8 -0.6 -0.4 -0.2 -0.2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

x

-0.4 -0.6 -0.8 -1.0

8. Motion on a cycloid. ! r (t) = ht sin t; 1 cos ti is the position of a particle in the xy-plane at time t. Find the particle’s velocity and acceleration when t = and t = 32 and plot them as vectors along with the curve. ! r ( ) = h ; 2i (blue) 3 2

! r

3 + 1; 1 2

=

! v (t) = h1

(red)

cos t; sin ti

So ! v ( ) = h2; 0i (light blue) 3 2

! v

= h1; 1i (light red)

! a (t) = hsin t; cos ti ! a ( ) = h0; 1i (cyan) ! a

3 2

= h 1; 0i (pink)

2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS145

y

2

1

0 1

2

3

4

5

6

x

-1

9. ! r (t) = t + 1; t2 1; 2t is the position of a particle in space at time t. Find the particle’s velocity, acceleration and speed when t = 1. ! v (t) = h1; 2t; 2i

! a (t) = h0; 2; 0i p v (t) = 5 + 4t2 ! v (1) = h1; 2; 2i ! a (1) = h0; 2; 0i v (1) = 3

10. ! r (t) = h2 cos t; 3 sin t; 4ti is the position of a particle in space at time t. Find the particle’s velocity, acceleration and speed when t = 2 . ! v (t) = h 2 sin t; 3 cos t; 4i

! a (t) = h 2 cos t; 3 sin t; 0i p v (t) = 4 sin2 t + 9 cos2 t + 16 ! v

2

= h 2; 0; 4i

! a

2

= h0; 3; 0i

v

2

=

p

20

146

CHAPTER 2. VECTOR FUNCTIONS

E D 2 11. ! r (t) = 2 ln (t + 1) ; t2 ; t2 is the position of a particle in space at time t. Find the particle’s velocity, acceleration and speed when t = 1. 2 ; 2t; t t+1

! v (t) = ! a (t) = v (t) =

*

s

2 (t + 1)

5t2 +

+

2 ; 2; 1

4 (t + 1)

2

! v (1) = h1; 2; 1i ! a (1) = h 1; 2; 1i p v (1) = 7 p 12. Find the angle between the velocity and acceleration if ! r (t) = 3t + 1; 3t; t2 at t = 0. Let be the angle in question. Then ! v (0) ! a (0) = cos 1 ! k v (0)k k! a (0)k =

2

13. Find the angle between the velocity and acceleration if ! r (t) = ln t2 + 1 ; tan at t = 0. Let be the angle in question then ! v (0) ! a (0) = cos 1 ! ! k v (0)k k a (0)k =

2

14. Find the equation of the line tangent to ! r (t) = sin t; t2 t = 0. The equation of the tangent is 8 < x=t y= 1 : z =1+t

cos t; et at

15. Find the equation of the line tangent to ! r (t) = ha sin t; a cos t; bti at t=2 . The equation of the tangent is 8 x = at < y=a : z = 2 b + bt

1

p t; t2 + 1

2.2. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS147 16. Consider particles moving along the unit circle x2 + y 2 = 1. Answer the questions below for each of the particle a e. i. Does the particle have constant speed? If so, what is it? ii. Is the particles’s acceleration always orthogonal to its velocity vector? iii. Does the particle move clockwise or counterclockwise around the circle? iv. Does the particle begin at the point (1; 0)? (a) ! r (t) = hcos t; sin ti, t 0. i. yes, 1 ii. yes iii. counterclockwise iv. yes ! (b) r (t) = hcos 2t; sin 2ti, t 0. i. yes, 2 ii. yes iii. counterclockwise iv. yes ! (c) r (t) = cos t 2 ; sin t i. yes, 1 ii. yes iii. counterclockwise iv. no, (0; 1) ! (d) r (t) = hcos t; sin ti, t 0. i. yes, 1 ii. yes iii. clockwise iv. yes (e) ! r (t) = cos t2 ; sin t2 , t 0. i. no ii. no iii. counterclockwise iv. yes

2

,t

0.

17. A particle moves along the top of the parabola y 2 = 2x from left to right at constant speed 5 units per second. Find the velocity of the particle as it moves through (2; 2). hint: You …rst need to …nd the parametric equations of the curve. Recall, if y = f (x) then the parametric equations x = at are where a is a constant. In this case, …rst write x = f (y) y = f (at) then use a similarptrick p for the parametric equations. The velocity is 2 5; 5 .

148

CHAPTER 2. VECTOR FUNCTIONS

18. Solve for ! r (t) in

d! r dt

= h t; t; ti . ! r (0) = h1; 2; 3i

! r (t) =

t2 t2 t2 + 1; + 2; +3 2 2 2

19. 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 h3; 1; 1i. Find the position vector ! r (t). This amounts to …nding ! r (t) given ! a (t) = h3; 1; 1i, v (0) = 2 and ! r (0) = h1; 2; 3i. ! v (t) = ! r (t) =

2 2 p ;t + p 11 11

6 3t + p ; t 11

3t2 6 t2 + p t + 1; 2 2 11

2 t2 2 p t + 2; + p t + 3 2 11 11

20. A projectile is …red at a speed of 840m=s at an angle of 60 . How long will it take to get 21km downrange? t = 50s 21. A projectile is …red with an initial speed of 500m=s at an angle of elevation of 45 . (a) When and how far away will the projectile strike? t x (72:1)

72:1 25484m

(b) How high overhead will the projectile be when it is 5km downrange? First, we …nd t1 such that x (t1 ) = 5000, then compute y (t1 ). p t = 10 2 14:14 Hence, the height is y (14:14)

4018m

(c) What is the greatest height reached by the projectile? The maximum height is given by ymax

6371m