13 CALCULUS OF VECTOR-VALUED FUNCTIONS

[Pages:86]13 CALCULUS OF

VECTOR-VALUED FUNCTIONS

13.1 Vector-Valued Functions (LT Section 14.1)

Preliminary Questions

1. 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, 1 This 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) = ti + t4j + et k onto the xz-plane? solution The projection of the path onto the xz-plane is the curve traced by ti + et k = t, 0, et . This is the curve z = 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 t The 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 the xz-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 - 1 The projection onto the xy-plane is a parabola. The projection onto the yz-plane is the curve 0, cos 2t, sin t . Hence y = 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 t Therefore, the curve is contained in the plane y = 2, and the following holds:

(x + 2)2 + (z - 3)2 = cos2 t + sin2 t = 1 We 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 ti + cos tj moves in a clockwise direction, whereas the point on the path cos ti + 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)k

solution 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 t

Hence,

(x + 2)2 + (z - 3)2 = (cos t)2 + (- sin t)2 = 1

This 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 3t

Hence,

(x + 2)2 + (z - 3)2 = (cos 3t)2 + (- sin 3t)2 = 1

This 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 t

Hence,

(x + 2)2 + (z - 3)2 = (- cos t)2 + (sin t)2 = 1

Again, 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 t

Hence,

(x - 2)2 + (z - 3)2 = (cos t)2 + (sin t)2 = 1

This 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.

Exercises

1.

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

1 t

j

+

(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}

3. EvaWluhaatet irs(2th)eadnodmr(a-in1o)ffro(rsr)(t=) =es i s+in2stj,+t2c, o(ts2s+k?1)-1 .

solution

Since r(t) =

sin

2

t,

t2,

(t 2

+

1)-1

, then

r(2) = sin , 4, 5-1 = 0, 4, 1 5

and

r(-1) = sin - , 1, 2-1 = -1, 1, 1

2

2

May 16, 2011

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

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

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 + t

or in the form: r(t) = (3 + 3t)i - 5j + (7 + t)k, - < t <

7. Match the space curves in Figure 8 with their projections onto the xy-plane in Figure 9.

Find

a

direction

vector

for

the

line

with

parametrization

r(t )

=

(4

-

t )i

+

(2

+

5t )j

+

1 2

tk.

z

z

z

y

y

y

x

x

x

(A)

(B)

(C)

FIGURE 8

y

y

y

x

x

x

(i)

(ii)

(iii)

FIGURE 9

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

9. Match the vector-valued functions (a)?(f) with the space curves (i)?(vi) in Figure 10.

(a) (c) (e)

rrr(((ba(((cttt))))))Mrrr===123a(((tctttttt)))h,,+t===tt2h,1,e152ct1s,2,t+o,pe5csta0to,tc2.2st0et82,ctctu,corossvistne,ts2,sitienn0.Ft0i8gtusrien

8t

with

the

following(vbe)ctro(rt-)v=aluecdosfut,nscitniotn,ss:in 12t (d) r(t) = cos3 t, sin3 t, sin 2t (f) r(t) = cos t, sin t, cos t sin

12t

z y

x

z

y x

z y

x

(i) z

x

y

(ii)

(iii)

z

z

y

x

y

x

solution (a) (v) (d) (vi)

(iv)

(v)

FIGURE 10

(b) (i) (e) (iv)

(vi)

(c) (ii) (f) (iii)

May 16, 2011

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

11.

Match the Which

space of the

cfoulrlvoews i(nAg)c?u(Crv)eisnhFaivgeutrhee1s1awmiethprthoejeirctpioronjeocnttioonths e(ix)?y(-ipiil)anoen?to

the

xy-plane.

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

z

(b)

r2(t) =

et , t2, t

z

z (c) r3(t) = t, t2, cos t

y

y

y

x

(A) z

y

x (B) z

y

x (C) z

y

x

x

(i)

(ii)

FIGURE 11

x (iii)

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

In ExercDiseesscr1i3b?e1t6h,etphreofjuenctcitoionns orf(tt)hetrcaicrecslearc(itr)cl=e. Dsientetr,m0,in4e+thceorsatdiuosn,tocetnhteerc,oaonrddipnlaatneepclaonnetsa.ining the circle. 13. r(t) = (9 cos t)i + (9 sin t)j solution 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) = 81

This is the equation of a circle with radius 9 centered at the origin. The circle lies in the xy-plane.

15. r(t)r(=t) =sin7ti,+0,(412+ccoosst)tj + (12 sin t)k solution x(t) = sin t, z(t) = 4 + cos t, hence:

x2 + (z - 4)2 = sin2 t + cos2 t = 1

y = 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.

17. (a)

SLheotrwC(tbt)he=atthCe6cl+iuers3voesnirnt(htt,e) 9=c,o4nte+cxo32sc+to,styts2in=t ,zt2

. .

(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 increases linearly with time, the x and y coordinates trace out points on the circles of increasing radius. We obtain the following curve:

z

x

y

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)

In ExerciseUss1e9aancdom20p,ulteetr 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.

r(t) = sin t, cos t, sin t cos 2t

as shown in Figure 12.

z

x y

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 = 0 or cos 2t = 0. The solutions are, thus:

t = k or t = + k , k = 0, ?1, ?2, . . . 42

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

0, (-1)k, 0

. The values t

=

4

+

k 2

yield the points:

k = 0 : sin , cos , 0 = 1 , 1 , 0

44

22

k = 1 : sin 3 , cos 3 , 0 = 1 , - 1 , 0

4

4

22

k = 2 : sin 5 , cos 5 , 0 = - 1 , - 1 , 0

4

4

22

k = 3 : sin 7 , cos 7 , 0 = - 1 , 1 , 0

4

4

22

(Other values of k do not provide new points). We conclude that the curve intersects the xy-plane at the following points:

(0, 1, 0), (0, -1, 0), 1 , 1 , 0 , 1 , - 1 , 0 , - 1 , - 1 , 0 , - 1 , 1 , 0

22

2

2

2

2

22

21.

Parametrize the intersection Show that the projection

of of

rth(et )sounrftaoctehse

xz-plane

is

the

curve

y2 z-=z2x=-x2x-32, for

y2 -

+ 1

z2

x=91

using t = y as the parameter (two vector functions are needed as in Example 3).

solution We solve for z and x in terms of y. From the equation y2 + z2 = 9 we have z2 = 9 - y2 or z = ? 9 - y2. From the second equation we have:

x = y2 - z2 + 2 = y2 - 9 - y2 + 2 = 2y2 - 7

Taking t = y as a parameter, we have z = ? 9 - t2, x = 2t2 - 7, yielding the following vector parametrization:

r(t) = 2t2 - 7, t, ? 9 - t2 , for - 3 t 3.

23. Viviani's Curve C is the intersection of the surfaces (Figure 13) Find a parametrization of the curve in Exercise 21 using trigonometric functions. x2 + y2 = z2, y = z2

(a) Parametrize each of the two parts of C corresponding to x 0 and x 0, taking t = z as parameter. (b) Describe the projection of C onto the xy-plane. (c) Show that C lies on the sphere of radius 1 with center (0, 1, 0). This curve looks like a figure eight lying on a sphere [Figure 13(B)].

May 16, 2011

S E C T I O N 13.1

z x2 + y2 = z2

Vector-Valued Functions (LT SECTION 14.1) 255

x y

Viviani's curve

y = z2

(A)

(B) Viviani's curve viewed

from the negative y-axis.

FIGURE 13 Viviani's curve is the intersection of the surfaces x2 + y2 = z2 and y = z2.

solution (a) We must solve for y and x in terms of z (which is a parameter). We get:

y = z2

x2 = z2 - y2 x = ? z2 - y2 = ? z2 - z4

Here, the ? from x = ? z2 - z4 represents the two parts of the parametrization: + for x 0, and - for x 0. Substituting the parameter z = t we get:

y = t2, x = ? t2 - t4 = ?t 1 - t2.

We obtain the following parametrization:

r(t) = ?t 1 - t2, t2, t for - 1 t 1

(1)

(b) The projection of the curve onto the xy-plane is the curve on the xy-plane obtained by setting the z-coordinate of r(t) equal to zero. We obtain the following curve:

?t 1 - t2, t2, 0 , -1 t 1

We also note that since x = ?t 1 - t2, then x2 = t2(1 - t2), but also y = t2, so that gives us the equation x2 = y(1 - y) for the projection onto the xy plane. We rewrite this as follows.

x2 = y(1 - y) x2 + y2 - y = 0 x2 + y2 - y + 1/4 = 1/4 x2 + (y - 1/2)2 = (1/2)2

We can now identify this projection as a circle in the xy plane, with radius 1/2, centered at the xy point (0, 1/2). (c) The equation of the sphere of radius 1 with center (0, 1, 0) is:

x2 + (y - 1)2 + z2 = 1

(2)

To show that C lies on this sphere, we show that the coordinates of the points on C (given in (1)) satisfy the equation of the sphere. Substituting the coordinates from (1) into the left side of (2) gives:

x2 + (y - 1)2 + z2 = ?t 1 - t2 2 + (t2 - 1)2 + t2 = t2(1 - t2) + (t2 - 1)2 + t2

= (t2 - 1)(t2 - 1 - t2) + t2 = 1

We conclude that the curve C lies on the sphere of radius 1 with center (0, 1, 0).

v2e5c. topUras-rveaSamhsluioenewterdiztafhaunatnditocatncinoooysnfipsnV)oe.iiTvntoithaoenpnnia'xsdrae2cmsu+cerrvtiyrbei2ez(eE=thxtehezerp2criciosnajetenec2rbt3sieeo)cwnwtsirioitotnhtfetonhfaiisnsthcptehuaerrcvayfemoliroenmntdetero(.rzsthcxoes2th+r,ezeysc2inoo=r,dzi1n) afatonerdpsolxam2nee+s.

z2 = 1 (use two . Use this to find a

solution The circle x2 + z2 = 1 in the xz-plane is parametrized by x = cos t, z = sin t, and the circle x2 + y2 = 1

in the xy-plane is parametrized by x = cos s, y = sin s. Hence, the points on the cylinders can be written in the form:

x2 + z2 = 1: cos t, y, sin t , 0 t 2 x2 + y2 = 1: cos s, sin s, z , 0 t 2

May 16, 2011

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

The points (x, y, z) on the intersection of the two cylinders must satisfy the following equations:

cos t = cos s y = sin s z = sin t

The first equation implies that s = ?t + 2k. Substituting in the second equation gives y = sin (?t + 2 k) = sin (?t) = ? sin t. Hence, x = cos t, y = ? sin t, z = sin t. We obtain the following vector parametrization of the intersection:

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

The projection of the curve on the xy-plane is traced by cos t, ? sin t, 0 which is the unit circle in this plane. The projection of the curve on the xz-plane is traced by cos t, 0, sin t which is the unit circle in the xz-plane. The projection of the curve on the yz-plane is traced by 0, ? sin t, sin t which is the two segments z = y and z = -y for -1 y 1.

27.

Use sine and cosine to parametrize the intersection of the surfaces Use hyperbolic functions to parametrize the intersection of the

x2 + y2 surfaces

x=2

1-ayn2d

z= =4

4x2 and

z(F=iguxrye.

14).

z

y

x

FIGURE 14 Intersection of the surfaces x2 + y2 = 1 and z = 4x2.

solution The points on the cylinder x2 + y2 = 1 and on the parabolic cylinder z = 4x2 can be written in the form:

x2 + y2 = 1: cos t, sin t, z z = 4x2: x, y, 4x2

The points (x, y, z) on the intersection curve must satisfy the following equations:

x = cos t

y = sin t z = 4x2

x = cos t, y = sin t, z = 4 cos2 t

We obtain the vector parametrization: r(t) = cos t, sin t, 4 cos2 t , 0 t 2

Using the CAS we obtain the following curve:

z

4

2 ?1

y 2 1

?1 1

x

?2

r(t) = cos t, sin t, 4 cos2 t

May 16, 2011

S E C T I O N 13.1 Vector-Valued Functions (LT SECTION 14.1) 257

In Exercises 28?30, two paths r1(t) and r2(t) intersect if there is a point P lying on both curves. We say that r1(t) and r2(t) collide if r1(t0) = r2(t0) at some time t0.

29. DetWermhiicnheowf htheethfeorllro1wainndg rs2tacteomlliednetsoarrientterursee?ct: (a) If r1 and r2 intersect, then they collidre1.(t) = t2 + 3, t + 1, 6t-1

(b) If r1 and r2 collide, then they interserc2t.(t) = 4t, 2t - 2, t2 - 7 (c) Intersection depends only on the underlying curves traced by r1 and r2, but collision depends on the actual soluptariaomnetrTizoatdioetnesr.mine if the paths collide, we must examine whether the following equations have a solution:

t2 + 3 = 4t t + 1 = 2t - 2

6 t

=

t2

-

7

We simplify to obtain:

t2 - 4t + 3 = (t - 3)(t - 1) = 0

t =3 t3 - 7t - 6 = 0

The solution of the second equation is t = 3. This is also a solution of the first and the third equations. It follows that r1(3) = r2(3) so the curves collide. The curves also intersect at the point where they collide. We now check if there are other points of intersection by solving the following equation:

r1(t) = r2(s)

t2

+

3,

t

+

1,

6 t

= 4s, 2s - 2, s2 - 7

Equating coordinates we get:

t2 + 3 = 4s t + 1 = 2s - 2

6 t

=

s2

-

7

By the second equation, t = 2s - 3. Substituting into the first equation yields:

(2s - 3)2 + 3 = 4s

4s2 - 12s + 9 + 3 = 4s

s2 - 4s + 3 = 0 s1 = 1, s2 = 3

Substituting s1 = 1 and s2 = 3 into the second equation gives:

t1 + 1 = 2 ? 1 - 2 t1 = -1

t2 + 1 = 2 ? 3 - 2 t2 = 3

The solutions of the first two equations are:

t1 = -1, s1 = 1; t2 = 3, s2 = 3

We check if these solutions satisfy the third equation:

6 t1

=

6 -1

=

-6,

s12 - 7 = 12 - 7 = -6

6 t1

= s12 - 7

6 t2

=

6 3

=

2,

s22 - 7 = 32 - 7 = 2

6 t2

=

s22 - 7

We conclude that the paths intersect at the endpoints of the vectors r1(-1) and r1(3) (or equivalently r2(1) and r2(3)). That is, at the points (4, 0, -6) and (12, 4, 2).

In ExercDiseetser3m1i?n4e0w, fihnedthaerpra1raamndetrr2izacotilolindoefotrhiencteurrsveec.t:

31. The vertical line passing throughr1th(et)p=ointt,(t32,,2t,30,) solution The points of the vertical line passing through

r2(t) = the point

4t + 6, 4t2, 7 - t (3, 2, 0) can be written

as

(3, 2, z).

Using

z

=

t

as

parameter we get the following parametrization:

r(t) = 3, 2, t , - < t <

May 16, 2011

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