1. Vector-Valued Functions Example. Consider the function r(t)=(t + 1)i ...

CHAPTER 11

Vector-Valued Functions

1. Vector-Valued Functions Example. Consider the function r(t) = (t + 1)i + (t2

2)j = ht + 1, t2

2i.

Each value of t yields a particular vector — think of t as time. What does the curve generated by this function look like? We have the following parametric equations: x = t + 1 and y = t2

1 and t2 = y + 2 =)

t=x t2 = x2

2 =)

2x + 1 and t2 = y + 2 =)

y = x2

2x

1,

a parabola.

We have r( 2) = h 1, 2i, r(0) = h1, 2i and r(2) = h3, 2i.

The arrows on the graph indicate the orientation, the direction with increasing values of t. 35

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11. VECTOR-VALUED FUNCTIONS

Definition. A vector-valued function r(t) is a mapping from D 2 R to R ✓ V3, so that for each t 2 D, r(t) = v for exactly one vector v 2 V3. We write r(t) = f (t)i + g(t)j + h(t)k = hf (t), g(t), h(t)i for some scalar functions f , g, and h, the component functions or r. In 3 dimensions, we get a space curve, and in 2 dimensions, a plane curve. p Problem (Page 863 #48). Find the intersection of z = x2 + y 2 and y + 2z = 2. First eliminate z:

p 1 z = x2 + y 2 = (2 y) =) 2 2 2 2 4(x + y ) = (2 y) = y 2 4y + 4 =) 3y 2 + 4y + (4x2

4) = 0 =) p p 4 ± 16 48x2 + 48 4 ± 64 48x2 y= = = 6 6 2 2p y= ± 4 3x2 3 3 Parametric equations for the intersection:

Thus f1(t) = f2(t) =

*

x=t 2 2p y= ± 4 3t2 3 3 ⇣ ⌘ 4 1p p 1 2 2 2 z = 2+ ⌥ 4 3t = ⌥ 4 2 3 3 3 3

t,

*

t,

2 2p + 4 3 3 2 3

2p 4 3

3t2,

4 3

1p 4 3

4 1p 2 3t , + 4 3 3

p 4±4 4 6

3x2

3t2

+

2 2 p tp 3 3

3t2 ,

2 2 p tp 3 3

3t2 , +

1. VECTOR-VALUED FUNCTIONS

37

are vector-valued functions describing the intersection.

The intersection is an ellipse, with each of the two vector-valued functions describing half of it. Example. Find a vector-valued function whose graph is the ellipse of major diameter 10 parallel to the y-axis and minor diameter 4 parallel to the z-axis, centered at (5, 2, 1). The ellipse is in the plane x = 5, parallel to the yz-plane. The equation for the ellipse in the yz-plane is (y

2)2 (z 1)2 + = 1. 25 4

Set x = 5 and find y and z in terms of t. x = 5,

y = 2 + 5 cos t,

z = 1 + 2 sin t =) r(t) = h5, 2 + 5 cos t, 1 + 2 sin ti.

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11. VECTOR-VALUED FUNCTIONS

Example. Match (1)f1(t) = hcos t, ln t, sin ti (3)f3(t) = h3 sin 2t, t, ti

with the graphs below.

(2)f2(t) = ht cos t, t sin t, ti

(4)f4(t) = h5 sin3 t, 5 cos3 t, ti

(1) x2 + z 2 = 1 =) a cylinder along the y-axis, so graph B. (2) x2 + y 2 = t2 = z 2 =) rising circular motion, so graph C (3) y = z =) a plane. Letting t = y, x = 3sin2y =) a sin curve in the plane y = z, so graph D (4) graph A

1. VECTOR-VALUED FUNCTIONS

39

Arc Length in R3. Suppose a curve r(t) = hf (t), g(t), h(t)i is traced once from t = a to t = b.

Partition [a, b] into n subintervals of equal size a = t0 < t1 < · · · tn = b where ti

ti

1

=

t=

b

a n

for i = 1, . . . , n.

Let si = arclength from r(ti 1) to r(ti).

For

si ⇡ d(r(ti 1), r(ti)) p = [f (ti) f (ti 1)]2 + [g(ti) g(ti 1)]2 + [h(ti) h(ti 1)]2 p = [f 0(ci) t]2 + [g 0(di) t]2 + [h0(ei) t]2 (by MVT) p = [f 0(ci)]2 + [g 0(di)]2 + [h0(ei)]2 t

t small, ci ⇡ di ⇡ ei =) n p X s⇡ [f 0(ci)]2 + [g 0(ci)]2 + [h0(ci)]2 t =) i=1

n p X s = lim [f 0(ci)]2 + [g 0(ci)]2 + [h0(ci)]2 t n!1

i=1

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11. VECTOR-VALUED FUNCTIONS

Thus the arclength is

Z bp s= [f 0(t)]2 + [g 0(t)]2 + [h0(t)]2 dt. a

Example. Find the length of r(t) = h4 cos t, 4 sin t, 3ti, 0  t  s=

Z

0

⇡ 2

p [ 4 sin t]2 + [4 cos t]2 + [3]2 dt = Z

⇡ 2

p 16 + 9 dt =

0

Z

⇡ 2

⇡ . 2

⇡ 2

5 dt = 5t

0

= 0

5⇡ . 2

Note. Most integrals for arclength need to be evaluated by numerical methods. Maple. See vectorvalued(11.1).mw or vectorvalued(11.1).pdf 2. The Calculus of Vector-Valued Functions Note. We will work in 3 dimensions. Just drop o↵ a component for two dimensions. Overall, we just import what we know from Calculus I into the components. Definition. For r(t) = hf (t), g(t), h(t)i, the limit of r(t) as t ! a is ⌦ ↵ lim r(t) = limhf (t), g(t), h(t)i = lim f (t), lim g(t), lim h(t) t!a

t!a

t!a

t!a

t!a

provided all the indicated limits exist. If any of the limits on the RHS fail to exist, then lim r(t) does not exist. t!a

2. THE CALCULUS OF VECTOR-VALUED FUNCTIONS

41

Example. lim⇡ hsin t cos t, cos2 t, sin ti =

t! 4

D

E lim sin t cos t, lim⇡ cos t, lim⇡ sin t = t! ⇡4 t! 4 t! 4 D E D 1 1 p2 E ⇡ ⇡ ⇡ ⇡ sin cos , cos2 , sin = , , 4 4 4 4 2 2 2 Definition. The function f : R ! R is continuous at a if 2

lim f (t) = f (a)

t!a

The function above is not continuous at t = 2 since lim f (t) does not exist since t!2

1 = lim f (t) 6= lim f (t) = 4. t!2

t!2+

At every other point, lim f (t) does exist. But f is not continuous at t = 1 since t!a

lim f (t) = 2 6= 3 = f (1)

t!1

and f is not continuous at t = 3 since f (3) is not defined. These last two cases are called removable discontinuities since they can be removed by either defining a value at a point (f (3) = 1) or redefining a value (f (1) = 2). At every other point, f is continuous. For example, at t = 5, lim f (t) = 1 = f (5).

t!5

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11. VECTOR-VALUED FUNCTIONS

Definition. r(t) = hf (t), g(t), h(t)i is continuous at t = a whenever lim r(t) = r(a).

t!a

Theorem (2.1). r(t) = hf (t), g(t), h(t)i is continuous at t = a if and only if all of f , g, h are continuous at t = a. Example. For which values of t is continuous?

r(t) = hsin t,

csc t, tan ti

f (t) = sin t is continuous everywhere. 1 g(t) = csc t = is discontinuous for sin t = 0, i.e., t = n⇡, n = 0, ±1, ±2, . . . . sin t sin t ⇡ h(t) = tan t = is discontinuous for cos t = 0, i.e., t = + n⇡, n = cos t 2 0, ±1, ±2, . . . . n⇡ Thus r(t) is continuous everywhere except at , n = 0, ±1, ±2, . . . . 2 Definition. The derivative r0(t) of r(t) is defined by r(t + t) r(t) r0(t) = lim t!0 t for all values of t for which the limit exists. When the limit exists for t = a, we say r is di↵erentiable at t = a.

2. THE CALCULUS OF VECTOR-VALUED FUNCTIONS

43

Theorem (2.2). Let r(t) = hf (t), g(t), h(t)i and suppose the components are all di↵erentiable for some value of t. Then r is also dei↵erentiable at that value of t and r0(t) = hf 0(t), g 0(t), h0(t)i. Proof. r(t + t) r(t) t!0 t h ↵ ⌦ ↵i 1 ⌦ = lim f (t + t), g(t + t), h(t + t) f (t), g(t), h(t) t!0 t ↵i 1 h⌦ = lim f (t + t) f (t), g(t + t) g(t), h(t + t) h(t) t!0 t D f (t + t) f (t) g(t + t) g(t) h(t + t) h(t) E = lim , , t!0 t t t D f (t + t) f (t) g(t + t) g(t) h(t + t) h(t) E = lim , lim , lim t!0 t!0 t!0 t t t 0 0 0 = hf (t), g (t), h (t)i.

r0(t) = lim

Example. If r(t) = hsin t cos t, cos2 t, sin ti, Dd⇥ ⇤ d⇥ 2 ⇤ d⇥ ⇤E 0 r (t) = sin t cos t , cos t , sin t dt dt dt ⌦ ↵ = cos2 t sin2 t, 2 sin t cos t, cos t ⌦ ↵ = cos 2t, sin 2t, cos t .

44

11. VECTOR-VALUED FUNCTIONS

Theorem (2.3). Suppose r(t) and s(t) are di↵erentiable, f (t) is di↵erentiable, and c 2 R. Then ⇤ d⇥ (1) r(t) + s(t) = r0(t) + s0(t) dt ⇤ d⇥ (2) cr(t) = cr0(t) dt ⇤ d⇥ (3) f (t)r(t) = f 0(t)r(t) + f (t)r0(t) dt ⇤ d⇥ (4) r(t) · s(t) = r0(t) · s(t) + r(t) · s0(t) dt ⇤ d⇥ (5) r(t) ⇥ s(t) = r0(t) ⇥ s(t) + r(t) ⇥ s0(t) dt ⌦ ↵ Proof. Let r(t) = a(t), b(t), c(t) . ⇤ ⌦ ↵⇤ d⇥ d⇥ f (t)r(t) = f (t) a(t), b(t), c(t) dt dt ↵⇤ d ⇥⌦ = f (t)a(t), f (t)b(t), f (t)c(t) dt ⌦ 0 ↵ 0 0 0 0 0 = f (t)a(t) + f (t)a (t), f (t)b(t) + f (t)b (t), f (t)c(t) + f (t)c (t) ⌦ 0 ↵ ⌦ ↵ 0 0 0 0 0 = f (t)a(t), f (t)b(t), f (t)c(t) + f (t)a (t), f (t)b (t), f (t)c (t) ⌦ ↵ ⌦ ↵ = f 0(t) a(t), b(t), c(t) + f (t) a0(t), b0(t), c0(t) = f 0(t)r(t) + f (t)r0(t)

Definition. r(t) = hf (t), g(t), h(t)i is smooth on an interval I if r0(t) is continuous on I and r0(t) 6= 0, except possibly at endpoints of I.

Corollary. r(t) is smooth on I if f 0, g 0, and h0 are all continuous on I and are not all 0 at the same time.

2. THE CALCULUS OF VECTOR-VALUED FUNCTIONS

D Example. Where is r(t) = t3

⇣ 2 2 ⌘E t, cos t smooth? 3 3 D ⇣ 2 2 ⌘E 0 2 r (t) = 3t 2t, t , sin t 3 3 D ⇣ 2 2 ⌘E = t(3t 2), t , sin t 3 3 2 is smooth except at t = . 3

45

t2 t, 2 2

The Tangent Vector

We refer to r0(a) as the tangent vector to the curve C at t = a. In the figures above, we visualize the tangent vector as the limit as t ! 0.

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11. VECTOR-VALUED FUNCTIONS

Example (Uniform Circular Motion). Let r(t) = hR cos !t, R sin !ti, R > 0, ! > 0.

2⇡ . ! r0(t) = h R! sin !t, R! cos !ti is the tangent vector. The motion is in the circle of radius R with period

r0(t) · r(t) = R2! sin !t cos !t + R2! sin !t cos !t = 0 =) r0(t) ? r(t) =) r0(t) is tangent to the circle. p 0 speed = kr (t)k = R2! 2 sin2 !t + R2! 2 cos2 !t p = R! sin2 !t + cos2 !t = R!, a constant. Note. r0(t) points in the direction of the motion. Theorem (2.4). kr(t)k = constant if and only if r(t) and r0(t) are orthogonal for all t. Definition. R(t) is an antiderivative of r(t) whenever R0(t) = r(t).

2. THE CALCULUS OF VECTOR-VALUED FUNCTIONS

47

Definition. If R(t) is any antiderivative of r(t), the indefinite integral of r(t) is Z r(t) dt = R(t) + c

where c is an arbitrary constant vector. This means Z Z Z Z DZ E ⌦ ↵ r(t) dt = f (t), g(t), h(t) dt = f (t) dt, g(t) dt, h(t) dt . Example.

DZ

Z

3t

e

dt,

Z

2

⌦ e

3t

↵ , t2 cos t3, t cos t dt =

t

3

t cos t dt,

Z

cos t +

E t cos t dt =

&

1

& +

0 D

sin t cos t

E 1 3t 1 3 e + c1, sin t + c2, t sin t + cos t + c3 = 3 3 D 1 E 3t 1 3 e , sin t , t sin t + cos t + hc1, c2, c3i . | {z } 3 3 c

Definition. For r(t) = hf (t), g(t), h(t)i, the definite integral of r(t) on the interval [a, b] is Z b Z b Z b Z b DZ b E ⌦ ↵ r(t) dt = f (t), g(t), h(t) dt = f (t) dt, g(t) dt, h(t) dt . a

a

a

a

a

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11. VECTOR-VALUED FUNCTIONS

Theorem (2.5 — Fundamental Theorem of Calculus). Suppose R(t) is an antiderivative of r(t) on [a, b]. Then Z b r(t) dt = R(b) R(a). a

Example.

Z

⇡/3 ⌦

↵ sec t tan t, tan t, 2 sin t cos t dt = 0 Z ⇡/3 Z ⇡/3 D Z ⇡/3 E sec t tan t dt, tan t dt, 2 sin t cos t dt = 0

D

0

sec t

⇡/3

2

⇡/3 E

ln|cos t| , sin t = 0 0 D E ⇡ ⇡ 2⇡ 2 sec sec 0, ln|cos | + ln|cos 0|, sin sin 0 = 3 3 3 D E 1 3 2 1, ln + 0, 0 = 2 4 D 3E 1, ln 2, 4 Maple. See vectorcalculus(11.2).mw or vectorcalculus(11.2).pdf 0

,

0

⇡/3

3. MOTION IN SPACE

3. Motion in Space Consider the curve traced out by r(t) = hf (t), g(t), h(t)i, t 2 [a, b].

r0(t) = hf 0(t), g 0(t), h0(t)i. p 0 kr (t)k = [f 0(t)]2 + [g 0(t)]2 + [h0(t)]2.

For t0 2 [a, b], the arclength from u = t0 to u = t is Z tp s(t) = [f 0(u)]2 + [g 0(u)]2 + [h0(u)]2 du. t0

By the FTC,

p s (t) = [f 0(t)]2 + [g 0(t)]2 + [h0(t)]2 = kr0(t)k. 0

s0(t) is the instantaneous change of arclength with respect to time = speed. Thus r0(t) = velocity vector = v(t) and r00(t) = v 0(t) = accelleration vector = a(t).

49

50

11. VECTOR-VALUED FUNCTIONS

Problem ⌦ 3t (Page↵ 785 # 14). Find the position function r(t) given that a(t) = e , t, sin t , v(0) = h4, 2, 4i, and r(0) = h0, 4, 2i. Z D 1 E 2 ⌦ 3t ↵ 3t t v(t) = e , t, sin t dt = e , , cos t + c1 3 2 D 1 E D 13 E v(0) = , 0, 1 + c1 = h4, 2, 4i =) c1 = , 2, 5 3 3 D 1 E 13 t2 3t Thus v(t) = e + , 2, cos t + 5 . 3 3 2 Z D E 1 3t 13 t2 r(t) = e + , 2, cos t + 5 dt = 3 3 2 D1 E 13 t3 3t e + t, 2t, sin t + 5t + c2 9 3 6 D1 E D 1 E r(0) = , 0, 0 + c2 = h0, 4, 2i =) c2 = , 4, 2 9 9 D1 E 13 1 t3 3t Thus r(t) = e + t , 2t + 4, sin t + 5t 2 . 9 3 9 6

3. MOTION IN SPACE

51

Centripetal Force Newton’s Second law is F = ma where F is the net force on an object. We return to uniform circular motion.

d✓ = ! (a constant) =) ✓ = !t + c. dt ⌦ ↵ r(t) = hb cos ✓, b sin ✓i = b cos(!t + c), b sin(!t + c) .

Since the path is the same for all c, take ✓ = 0 for t = 0, i.e., take c = 0. Thus ⌦ ↵ r(t) = b cos !t, b sin !t =) ⌦ ↵ v(t) = r0(t) = b! sin !t, b! cos !t =) kv(t)k = !b. Now

a(t) = v0(t) = r00(t) =

⌦

↵ b! 2 cos !t, b! 2 sin !t = ⌦ ↵ 2 ! b cos !t, b sin !t =

! 2r(t).

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11. VECTOR-VALUED FUNCTIONS

Then

F(t) = ma(t) = m! 2r(t), a centripetal force (a force pointing to the center).

a constant.

kF(t)k = k m! 2r(t)k = m! 2kr(t)k = m! 2b,

Note. Force increases as the rotation rate ! increases.

3. MOTION IN SPACE

53

Problem (Page 786 #51(c)). A baseball is hit from a height of 3 feet with initial speed 120 feet per second and at an angle of 31 degrees above the horizontal. Find a vector-valued function describing the position of the ball t seconds after it is hit. To be a home run, the ball must clear a wall that is 385 feet away and 6 feet tall. Determine whether this is a home run. Solution

g ⇡ 32 ft/sec. Newton’s 2nd Law =) v0(t) = a(t) = gj = 32j =) Z Z v(t) = a(t) dt = ( 32j) dt =

mgj = F(t) = ma(t).

32tj + c1.

v(0) = h0, 0i+c1 = h120 cos 31 , 120 sin 31 i =) c1 = h120 cos 31 , 120 sin 31 i. Thus v(t) = h120 cos 31 , 120 sin 31 32ti. R r(t) = v(t) dt = h120 cos 31 t, 120 sin 31 t r(0) = h0, 0i + c2 = h0, 3i =) c2 = h0, 3i. Thus r(t) = h120 cos 31 t, 120 sin 31 t

16t2i + c2.

16t2 + 3i.

When first component = 120(cos 31 )t = 385, t = 3.75 =) second component = 120(sin 31 )(3.75)

16(3.75)2 + 3 = 9.77 =) a home run.

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11. VECTOR-VALUED FUNCTIONS

Problem (Page 785 #34). A merry-go-round of radius 5 feet and moment of inertia I = 10 rotates at 4 rad/s. Find the constant force needed to stop the merry-go-round in 2 seconds. Solution Rotational version of Newton’s Second Law: For an object rotating in two dimensions, with ⌧ = torque, T = k⌧ k = (force in direction of motion F)(distance from axis of rotation r). If ✓(t) is measured from some fixed ray, !(t) = ✓0(t) = angular velocity, ↵(t) = ! 0(t) = ✓00(t) = angular accelleration. I = moment of inertia = how much force must be applied to start object rotating. Equation of Rotational Motion: T = I↵ Here, since F = constant and distance = 5 = r, T is constant =) ↵ is constant since I = 10. We find ↵: We have !(0) = 4 and want !(2) = 0, so Z 2 2 !(2) !(0) = ↵ dt = ↵t = 2↵ = 0 0

0

4 =) ↵ =

Since I↵ = F r,

(10)( 2) = F (5) =) F =

4 ft-pounds.

2 rad/sec.

3. MOTION IN SPACE

Rotational Force in 3 Dimensions Recall ⌧ = r ⇥ F.

linear momentum = p = mv. angular momentum = L(t) = r(t) ⇥ mv(t). i dh 0 L (t) = r(t) ⇥ mv(t) dt = r0(t) ⇥ mv(t) + r(t) ⇥ mv0(t) = v(t) ⇥ mv(t) +r(t) ⇥ ma(t) | {z } =0 since vectors are k

= r ⇥ F = ⌧.

This implies the conservation of angular momentum: In the absence of torque (⌧ = 0), angular momentum remains constant.

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