May 16, 2011

13 CALCULUS OFVECTOR-VALUEDFUNCTIONS

13.1 Vector-Valued Functions (LT Section 14.1)

Preliminary Questions1. Which one of the following does not parametrize a line?

(a) r1(t) = 8 t, 2t, 3t(b) r2(t) = t3i 7t3j + t3k(c) r3(t) =

8 4t3, 2 + 5t2, 9t3

solution(a) This is a parametrization of the line passing through the point (8, 0, 0) in the direction parallel to the vector 1, 2, 3,since:

8 t, 2t, 3t = 8, 0, 0 + t 1, 2, 3(b) Using the parameter s = t3 we get:

t3, 7t3, t3 = s, 7s, s = s 1, 7, 1This is a parametrization of the line through the origin, with the direction vector v = 1, 7, 1.(c) The parametrization

8 4t3, 2 + 5t2, 9t3 does not parametrize a line. In particular, the points (8, 2, 0) (at t = 0),

(4, 7, 9) (at t = 1), and (24, 22, 72) (at t = 2) are not collinear.2. What is the projection of r(t) = t i + t4j + etk onto the xz-plane?

solution The projection of the path onto the xz-plane is the curve traced by t i + etk = t, 0, et . This is the curvez = ex in the xz-plane.

3. Which projection of cos t, cos 2t, sin t is a circle?solution The parametric equations are

x = cos t, y = cos 2t, z = sin tThe projection onto the xz-plane is cos t, 0, sin t. Since x2 + z2 = cos2 t + sin2 t = 1, the projection is a circle in thexz-plane. The projection onto the xy-plane is traced by the curve cos t, cos 2t, 0. Therefore, x = cos t and y = cos 2t .We express y in terms of x:

y = cos 2t = 2 cos2 t 1 = 2x2 1The projection onto the xy-plane is a parabola. The projection onto the yz-plane is the curve 0, cos 2t, sin t. Hencey = cos 2t and z = sin t . We find y as a function of z:

y = cos 2t = 1 2 sin2 t = 1 2z2

The projection onto the yz-plane is again a parabola.

4. What is the center of the circle with parametrization

r(t) = (2 + cos t)i + 2j + (3 sin t)k?solution The parametric equations are

x = 2 + cos t, y = 2, z = 3 sin tTherefore, the curve is contained in the plane y = 2, and the following holds:

(x + 2)2 + (z 3)2 = cos2 t + sin2 t = 1We conclude that the curve r(t) is the circle of radius 1 in the plane y = 2 centered at the point (2, 2, 3).

250

May 16, 2011

S E C T I O N 13.1 Vector-Valued Functions (LT SECTION 14.1) 251

5. How do the paths r1(t) = cos t, sin t and r2(t) = sin t, cos t around the unit circle differ?solution The two paths describe the unit circle. However, as t increases from 0 to 2 , the point on the path sin t i + cos tjmoves in a clockwise direction, whereas the point on the path cos t i + sin tj moves in a counterclockwise direction.

6. Which three of the following vector-valued functions parametrize the same space curve?

(a) (2 + cos t)i + 9j + (3 sin t)k (b) (2 + cos t)i 9j + (3 sin t)k(c) (2 + cos 3t)i + 9j + (3 sin 3t)k (d) (2 cos t)i + 9j + (3 + sin t)k(e) (2 + cos t)i + 9j + (3 + sin t)ksolution All the curves except for (b) lie in the vertical plane y = 9. We identify each one of the curves (a), (c), (d)and (e).

(a) The parametric equations are:

x = 2 + cos t, y = 9, z = 3 sin tHence,

(x + 2)2 + (z 3)2 = (cos t)2 + ( sin t)2 = 1This is the circle of radius 1 in the plane y = 9, centered at (2, 9, 3).(c) The parametric equations are:

x = 2 + cos 3t, y = 9, z = 3 sin 3tHence,

(x + 2)2 + (z 3)2 = (cos 3t)2 + ( sin 3t)2 = 1This is the circle of radius 1 in the plane y = 9, centered at (2, 9, 3).(d) In this curve we have:

x = 2 cos t, y = 9, z = 3 + sin tHence,

(x + 2)2 + (z 3)2 = ( cos t)2 + (sin t)2 = 1Again, the circle of radius 1 in the plane y = 9, centered at (2, 9, 3).(e) In this parametrization we have:

x = 2 + cos t, y = 9, z = 3 + sin tHence,

(x 2)2 + (z 3)2 = (cos t)2 + (sin t)2 = 1This is the circle of radius 1 in the plane y = 9, centered at (2, 9, 3).We conclude that (a), (c) and (d) parametrize the same circle whereas (b) and (e) are different curves.

Exercises1. What is the domain of r(t) = et i + 1

tj + (t + 1)3k?

solution r(t) is defined for t = 0 and t = 1, hence the domain of r(t) is:D = {t R : t = 0, t = 1}

What is the domain of r(s) = es i + sj + cos sk?3. Evaluate r(2) and r(1) for r(t) =sin 2 t, t

2, (t2 + 1)1.

solution Since r(t) =sin 2 t, t

2, (t2 + 1)1, then

r(2) =sin , 4, 51

=

0, 4,

1

5

and

r(1) =sin

2

, 1, 21

=1, 1, 1

2

May 16, 2011

252 C H A P T E R 13 CALCULUS OF VECTOR-VALUED FUNCTIONS (LT CHAPTER 14)

Does either of P = (4, 11, 20) or Q = (1, 6, 16) lie on the path r(t) = 1 + t, 2 + t2, t4?5. Find a vector parametrization of the line through P = (3, 5, 7) in the direction v = 3, 0, 1.solution We use the vector parametrization of the line to obtain:

r(t) = OP + tv = 3, 5, 7 + t 3, 0, 1 = 3 + 3t, 5, 7 + tor in the form:

r(t) = (3 + 3t)i 5j + (7 + t)k, < t <

Find a direction vector for the line with parametrization r(t) = (4 t)i + (2 + 5t)j + 12 tk.7. Match the space curves in Figure 8 with their projections onto the xy-plane in Figure 9.

y

x

z

y

x

z

y

x

z

(A) (B) (C)

FIGURE 8

(i)

x

y

(ii)

x

y

(iii)

x

y

FIGURE 9

solution The projection of curve (C) onto the xy-plane is neither a segment nor a periodic wave. Hence, the correctprojection is (iii), rather than the two other graphs. The projection of curve (A) onto the xy-plane is a vertical line, hencethe corresponding projection is (ii). The projection of curve (B) onto the xy-plane is a periodic wave as illustrated in (i).

Match the space curves in Figure 8 with the following vector-valued functions:

(a) r1(t) = cos 2t, cos t, sin t(b) r2(t) = t, cos 2t, sin 2t(c) r3(t) = 1, t, t

9. Match the vector-valued functions (a)(f) with the space curves (i)(vi) in Figure 10.(a) r(t) = t + 15, e0.08t cos t, e0.08t sin t (b) r(t) = cos t, sin t, sin 12t (c) r(t) =

t, t,

25t

1 + t2

(d) r(t) = cos3 t, sin3 t, sin 2t (e) r(t) = t, t2, 2t (f) r(t) = cos t, sin t, cos t sin 12t

y

(i) (ii) (iii)

(iv) (v) (vi)

x

z

y

x

z

y

x

z

y

y

x

x

z

z

y

x

z

FIGURE 10

solution(a) (v) (b) (i) (c) (ii)(d) (vi) (e) (iv) (f) (iii)

May 16, 2011

S E C T I O N 13.1 Vector-Valued Functions (LT SECTION 14.1) 253

Which of the following curves have the same projection onto the xy-plane?

(a) r1(t) =t, t2, et

(b) r2(t) =

et , t2, t

(c) r3(t) =

t, t2, cos t

11. Match the space curves (A)(C) in Figure 11 with their projections (i)(iii) onto the xy-plane.

y

y

xx

(A) (B) (C)

(i) (iii)(ii)

z

y

x

z

y

x

z

z

y

x

z

y

x

z

FIGURE 11

solution Observing the curves and the projections onto the xy-plane we conclude that: Projection (i) corresponds tocurve (C); Projection (ii) corresponds to curve (A); Projection (iii) corresponds to curve (B).

Describe the projections of the circle r(t) = sin t, 0, 4 + cos t onto the coordinate planes.In Exercises 1316, the function r(t) traces a circle. Determine the radius, center, and plane containing the circle.13. r(t) = (9 cos t)i + (9 sin t)jsolution Since x(t) = 9 cos t , y(t) = 9 sin t we have:

x2 + y2 = 81 cos2 t + 81 sin2 t = 81(cos2 t + sin2 t) = 81This is the equation of a circle with radius 9 centered at the origin. The circle lies in the xy-plane.

r(t) = 7i + (12 cos t)j + (12 sin t)k15. r(t) = sin t, 0, 4 + cos tsolution x(t) = sin t , z(t) = 4 + cos t , hence:

x2 + (z 4)2 = sin2 t + cos2 t = 1y = 0 is the equation of the xz-plane. We conclude that the function traces the circle of radius 1, centered at the point(0, 0, 4), and contained in the xz-plane.

r(t) = 6 + 3 sin t, 9, 4 + 3 cos t17. Let C be the curve r(t) = t cos t, t sin t, t.(a) Show that C lies on the cone x2 + y2 = z2.(b) Sketch the cone and make a rough sketch of C on the cone.solution x = t cos t , y = t sin t and z = t , hence:

x2 + y2 = t2 cos2 t + t2 sin2 t = t2(cos2 t + sin2 t) = t2 = z2.x2 + y2 = z2 is the equation of a circular cone, hence the curve lies on a circular cone. As the height z = t increaseslinearly with time, the x and y coordinates trace out points on the circles of increasing radius. We obtain the followingcurve:

x

y

z

r(t) = t cos t, t sin t, t

May 16, 2011

254 C H A P T E R 13 CALCULUS OF VECTOR-VALUED FUNCTIONS (LT CHAPTER 14)

Use a computer algebra system to plot the projections onto the xy- and xz-planes of the curve r(t) =t cos t, t sin t, t in Exercise 17.

In Exercises 19 and 20, let

r(t) = sin t, cos t, sin t cos 2tas shown in Figure 12.

y

x

z

y

x

z

FIGURE 12

19. Find the points where r(t) intersects the xy-plane.

solution The curve intersects the xy-plane at the points where z = 0. That is, sin t cos 2t = 0 and so either sin t = 0or cos 2t = 0. The solutions are, thus:

t = k or t = 4

+ k2

, k = 0, 1, 2, . . .

The values t = k yield the points: (sin k, cos k, 0) =(

0, (1)k, 0)

. The values t = 4 + k2 yield the points:

k = 0 :(

sin

4, cos

4, 0

)=

(12,

12, 0

)

k = 1 :(

sin3

4, cos

3

4, 0

)=

(12, 1

2, 0

)

k = 2 :(

sin5

4, cos

5

4, 0

)=

( 1

2, 1

2, 0

)

k = 3 :(

sin7

4, cos

7

4, 0

)=

( 1

2,

12, 0

)

(Other values o