13
CALCULUS OF
VECTOR-VALUED
FUNCTIONS
13.1
Vector-Valued Functions
(LT Section 14.1)
Preliminary Questions
1. Which one of the following doesnotparametrize 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 parameters=t3we get:
t3,−7t3, t3= s,−7s, s =s1,−7,1
This is a parametrization of the line through the origin, with the direction vectorv= −1,7,1.
(c) The parametrization8−4t3,2+5t2,9t3does not parametrize a line. In particular, the points(8,2,0)(att=0), (4,7,9)(att=1), and(−24,22,72)(att=2) are not collinear.
2. What is the projection ofr(t)=ti+t4j+etkonto thexz-plane?
solution The projection of the path onto thexz-plane is the curve traced byti+etk =t,0, et. This is the curve z=exin thexz-plane.
3. Which projection ofcost,cos 2t,sintis a circle? solution The parametric equations are
x=cost, y=cos 2t, z=sint
The projection onto thexz-plane iscost,0,sint. Sincex2+z2=cos2t+sin2t=1, the projection is a circle in the xz-plane. The projection onto thexy-plane is traced by the curvecost,cos 2t,0. Therefore,x=costandy=cos 2t. We expressyin terms ofx:
y=cos 2t=2 cos2t−1=2x2−1
The projection onto thexy-plane is a parabola. The projection onto theyz-plane is the curve0,cos 2t,sint. Hence y=cos 2tandz=sint. We findyas a function ofz:
y=cos 2t=1−2 sin2t=1−2z2 The projection onto theyz-plane is again a parabola.
4. What is the center of the circle with parametrization
r(t)=(−2+cost)i+2j+(3−sint)k? solution The parametric equations are
x= −2+cost, y=2, z=3−sint Therefore, the curve is contained in the planey=2, and the following holds:
(x+2)2+(z−3)2=cos2t+sin2t=1
We conclude that the curver(t)is the circle of radius 1 in the planey=2 centered at the point(−2,2,3).
S E C T I O N 13.1 Vector-Valued Functions (LT SECTION 14.1) 251
5. How do the pathsr1(t)= cost,sintandr2(t)= sint,costaround the unit circle differ?
solution The two paths describe the unit circle. However, astincreases from 0 to 2π, the point on the path sinti+costj moves in a clockwise direction, whereas the point on the path costi+sintjmoves in a counterclockwise direction.
6. Which three of the following vector-valued functions parametrize the same space curve? (a) (−2+cost)i+9j+(3−sint)k (b) (2+cost)i−9j+(−3−sint)k (c) (−2+cos 3t)i+9j+(3−sin 3t)k (d) (−2−cost)i+9j+(3+sint)k (e) (2+cost)i+9j+(3+sint)k
solution All the curves except for (b) lie in the vertical planey=9. We identify each one of the curves (a), (c), (d) and (e).
(a) The parametric equations are:
x= −2+cost, y=9, z=3−sint Hence,
(x+2)2+(z−3)2=(cost)2+(−sint)2=1 This is the circle of radius 1 in the planey=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 planey=9, centered at(−2,9,3).
(d) In this curve we have:
x= −2−cost, y=9, z=3+sint Hence,
(x+2)2+(z−3)2=(−cost)2+(sint)2=1 Again, the circle of radius 1 in the planey=9, centered at(−2,9,3).
(e) In this parametrization we have:
x=2+cost, y=9, z=3+sint Hence,
(x−2)2+(z−3)2=(cost)2+(sint)2=1 This is the circle of radius 1 in the planey=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 ofr(t)=eti+1
tj+(t+1)− 3k?
solution r(t)is defined fort=0 andt= −1, hence the domain ofr(t)is: D= {t∈R:t=0, t = −1}
What is the domain ofr(s)=esi+√sj+cossk? 3. Evaluater(2)andr(−1)forr(t)=sinπ2t, t2, (t2+1)−1. solution Sincer(t)=sinπ2t, t2, (t2+1)−1, then
r(2)=sinπ,4,5−1 = 0,4,1 5 and r(−1)= sin−π 2 ,1,2 −1 = −1,1,1 2
Does either ofP =(4,11,20)orQ=(−1,6,16)lie on the pathr(t)=1+t,2+t2, t4? 5. Find a vector parametrization of the line throughP =(3,−5,7)in the directionv= 3,0,1. solution We use the vector parametrization of the line to obtain:
r(t)=−→OP+tv= 3,−5,7 +t3,0,1 = 3+3t,−5,7+t or in the form:
r(t)=(3+3t)i−5j+(7+t)k, −∞< t <∞
Find a direction vector for the line with parametrizationr(t)=(4−t)i+(2+5t)j+ 12tk. 7. Match the space curves in Figure 8 with their projections onto thexy-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 thexy-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 thexy-plane is a vertical line, hence the corresponding projection is (ii). The projection of curve (B) onto thexy-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,cost,sint
(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, e0.08tcost, e0.08tsint (b) r(t)=cost,sint,sin 12t (c) r(t)=
t, t, 25t
1+t2
(d) r(t)=cos3t,sin3t,sin 2t (e) r(t)=t, t2,2t (f) r(t)=cost,sint,costsin 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)
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 thexy-plane?
(a) r1(t)=t, t2, et (b) r 2(t)= et, t2, t (c) r 3(t)= t, t2,cost 11. Match the space curves (A)–(C) in Figure 11 with their projections (i)–(iii) onto thexy-plane.
y
y
x
x
(A) (B) (C)
(i) (ii) (iii)
z y x z y x z z y x z y x z FIGURE 11
solution Observing the curves and the projections onto thexy-plane we conclude that: Projection (i) corresponds to curve (C); Projection (ii) corresponds to curve (A); Projection (iii) corresponds to curve (B).
Describe the projections of the circler(t)= sint,0,4+costonto the coordinate planes.
In Exercises 13–16, the functionr(t)traces a circle. Determine the radius, center, and plane containing the circle. 13. r(t)=(9 cost)i+(9 sint)j
solution Sincex(t)=9 cost,y(t)=9 sintwe have:
x2+y2=81 cos2t+81 sin2t=81(cos2t+sin2t)=81 This is the equation of a circle with radius 9 centered at the origin. The circle lies in thexy-plane.
r(t)=7i+(12 cost)j+(12 sint)k 15. r(t)= sint,0,4+cost
solution x(t)=sint,z(t)=4+cost, hence:
x2+(z−4)2=sin2t+cos2t=1
y =0 is the equation of thexz-plane. We conclude that the function traces the circle of radius 1, centered at the point (0,0,4), and contained in thexz-plane.
r(t)= 6+3 sint,9,4+3 cost 17. LetCbe the curver(t)= tcost, tsint, t. (a) Show thatClies on the conex2+y2=z2.
(b) Sketch the cone and make a rough sketch ofCon the cone. solution x=tcost,y=tsintandz=t, hence:
x2+y2=t2cos2t+t2sin2t=t2cos2t+sin2t =t2=z2.
x2+y2= z2is the equation of a circular cone, hence the curve lies on a circular cone. As the heightz= tincreases linearly with time, thexandycoordinates trace out points on the circles of increasing radius. We obtain the following curve:
x
y z
Use a computer algebra system to plot the projections onto the xy- andxz-planes of the curve r(t) = tcost, tsint, tin Exercise 17.
In Exercises 19 and 20, let
r(t)= sint,cost,sintcos 2t as shown in Figure 12. y x z y x z FIGURE 12
19. Find the points wherer(t)intersects thexy-plane.
solution The curve intersects thexy-plane at the points wherez=0. That is, sintcos 2t=0 and so either sint=0 or cos 2t=0. The solutions are, thus:
t=πkort= π 4 +
πk
2 , k=0,±1,±2, . . . The valuest=πkyield the points:(sinπk,cosπk,0)=0, (−1)k,0
. The valuest= π
4+ πk2 yield the points:
k=0: sinπ 4,cos π 4,0 = 1 √ 2, 1 √ 2,0 k=1: sin3π 4 ,cos 3π 4 ,0 = 1 √ 2,− 1 √ 2,0 k=2: sin5π 4 ,cos 5π 4 ,0 = −√1 2,− 1 √ 2,0 k=3: sin7π 4 ,cos 7π 4 ,0 = −√1 2, 1 √ 2,0
(Other values ofkdo not provide new points). We conclude that the curve intersects thexy-plane at the following points: (0,1,0),(0,−1,0), 1 √ 2, 1 √ 2,0 , 1 √ 2,− 1 √ 2,0 , −√1 2,− 1 √ 2,0 , −√1 2, 1 √ 2,0
Show that the projection ofr(t)onto thexz-plane is the curve
z=x−2x3 for −1≤x≤1 21. Parametrize the intersection of the surfaces
y2−z2=x−2, y2+z2=9 usingt=yas the parameter (two vector functions are needed as in Example 3).
solution We solve forzandxin terms ofy. From the equationy2+z2=9 we havez2=9−y2orz= ±9−y2. From the second equation we have:
x=y2−z2+2=y2−9−y2 +2=2y2−7
Takingt=yas a parameter, we havez= ±9−t2,x=2t2−7, yielding the following vector parametrization: r(t)=2t2−7, t,±9−t2, for −3≤t≤3.
Find a parametrization of the curve in Exercise 21 using trigonometric functions. 23. Viviani’s Curve Cis the intersection of the surfaces (Figure 13)
x2+y2=z2, y=z2
(a) Parametrize each of the two parts ofCcorresponding tox≥0 andx≤0, takingt=zas parameter. (b) Describe the projection ofConto thexy-plane.
(c) Show thatClies 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)].
S E C T I O N 13.1 Vector-Valued Functions (LT SECTION 14.1) 255
y
y=z2 Viviani's curve
(A) (B) Viviani’s curve viewed from the negative y-axis. x
x2+y2=z2 z
FIGURE 13 Viviani’s curve is the intersection of the surfacesx2+y2=z2andy=z2. solution
(a) We must solve foryandxin terms ofz(which is a parameter). We get: y=z2
x2=z2−y2 ⇒ x= ±z2−y2= ±z2−z4
Here, the±fromx = ±z2−z4represents the two parts of the parametrization:+forx ≥ 0, and− forx ≤ 0. Substituting the parameterz=twe get:
y=t2, x= ±t2−t4= ±t1−t2. We obtain the following parametrization:
r(t)=±t1−t2, t2, tfor −1≤t≤1 (1) (b) The projection of the curve onto thexy-plane is the curve on thexy-plane obtained by setting thez-coordinate of r(t)equal to zero. We obtain the following curve:
±t1−t2, t2,0, −1≤t≤1
We also note that sincex= ±t1−t2, thenx2=t2(1−t2), but alsoy=t2, so that gives us the equationx2=y(1−y) for the projection onto thexyplane. 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 thexyplane, with radius 1/2, centered at thexypoint(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 thatClies on this sphere, we show that the coordinates of the points onC(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=±t1−t22+(t2−1)2+t2=t2(1−t2)+(t2−1)2+t2 =(t2−1)(t2−1−t2)+t2=1
We conclude that the curveClies on the sphere of radius 1 with center(0,1,0).
Show that any point onx2+y2=z2can be written in the form(zcosθ, zsinθ, z)for someθ. Use this to find a parametrization ofViviani’s curve (Exercise 23) withθas parameter.
25. Use sine and cosine to parametrize the intersection of the cylindersx2+y2 = 1 and x2+z2 = 1 (use two vector-valued functions). Then describe the projections of this curve onto the three coordinate planes.
solution The circlex2+z2=1 in thexz-plane is parametrized byx=cost,z=sint, and the circlex2+y2=1 in thexy-plane is parametrized byx=coss,y=sins. Hence, the points on the cylinders can be written in the form:
x2+z2=1: cost, y,sint, 0≤t≤2π x2+y2=1: coss,sins, z, 0≤t≤2π
The points(x, y, z)on the intersection of the two cylinders must satisfy the following equations: cost=coss
y=sins z=sint
The first equation implies thats= ±t+2πk. Substituting in the second equation givesy=sin(±t+2πk)=sin(±t)= ±sint. Hence,x=cost,y= ±sint,z=sint. We obtain the following vector parametrization of the intersection:
r(t)= cost,±sint,sint
The projection of the curve on thexy-plane is traced bycost,±sint,0which is the unit circle in this plane. The projection of the curve on thexz-plane is traced bycost,0,sintwhich is the unit circle in thexz-plane. The projection of the curve on theyz-plane is traced by0,±sint,sintwhich is the two segmentsz=yandz= −yfor−1≤y≤1.
Use hyperbolic functions to parametrize the intersection of the surfacesx2−y2=4 andz=xy. 27. Use sine and cosine to parametrize the intersection of the surfacesx2+y2=1 andz=4x2(Figure 14).
y
x z
FIGURE 14 Intersection of the surfacesx2+y2=1 andz=4x2.
solution The points on the cylinderx2+y2=1 and on the parabolic cylinderz=4x2can be written in the form: x2+y2=1: cost,sint, z
z=4x2: x, y,4x2
The points(x, y, z)on the intersection curve must satisfy the following equations: x=cost
y=sint z=4x2 ⇒
x=cost, y=sint, z=4 cos2t
We obtain the vector parametrization:
r(t)=cost,sint,4 cos2t, 0≤t≤2π Using the CAS we obtain the following curve:
z –2 x y 1 –1 2 4 2 1 –1
S E C T I O N 13.1 Vector-Valued Functions (LT SECTION 14.1) 257
In Exercises 28–30, two pathsr1(t)andr2(t)intersect if there is a pointP lying on both curves. We saythatr1(t)and r2(t)collideifr1(t0)=r2(t0)at some timet0.
Which of the following statements are true? (a) Ifr1andr2intersect, then they collide. (b) Ifr1andr2collide, then they intersect.
(c) Intersection depends only on the underlying curves traced byr1andr2, but collision depends on the actual parametrizations.
29. Determine whetherr1andr2collide or intersect: r1(t)=
t2+3, t+1,6t−1 r2(t)=4t,2t−2, t2−7
solution To determine if the paths collide, we must examine whether the following equations have a solution: ⎧ ⎪ ⎨ ⎪ ⎩ t2+3=4t t+1=2t−2 6 t =t 2−7 We simplify to obtain: t2−4t+3=(t−3)(t−1)=0 t=3 t3−7t−6=0
The solution of the second equation ist =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 =s 2−7
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 ⇒ s
1=1, s2=3 Substitutings1=1 ands2=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, s 2 1−7=12−7= −6 ⇒ 6 t1 = s12−7 6 t2 = 6 3 =2, s 2 2−7=32−7=2 ⇒ 6 t2 =s 2 2−7
We conclude that the paths intersect at the endpoints of the vectorsr1(−1)andr1(3)(or equivalentlyr2(1)andr2(3)). That is, at the points(4,0,−6)and(12,4,2).
Determine whetherr1andr2collide or intersect: r1(t)=
t, t2, t3, r2(t)=
4t+6,4t2,7−t
In Exercises 31–40, find a parametrization of the curve. 31. The vertical line passing through the point(3,2,0)
solution The points of the vertical line passing through the point(3,2,0)can be written as(3,2, z). Usingz=tas parameter we get the following parametrization:
The line passing through(1,0,4)and(4,1,2)
33. The line through the origin whose projection on thexy-plane is a line of slope 3 and whose projection on theyz-plane is a line of slope 5 (i.e.,z/y=5)
solution We denote by(x, y, z)the points on the line. The projection of the line on thexy-plane is the line through the origin having slope 3, that is the liney=3xin thexy-plane. The projection of the line on theyz-plane is the line through the origin with slope 5, that is the linez=5y. Thus, the points on the desired line satisfy the following equalities:
y=3x
z=5y ⇒ y=3x, z=5·3x=15x
We conclude that the points on the line are all the points in the form(x,3x,15x). Usingx=tas parameter we obtain the following parametrization:
r(t)= t,3t,15t, −∞< t <∞. The horizontal circle of radius 1 with center(2,−1,4)
35. The circle of radius 2 with center(1,2,5)in a plane parallel to theyz-plane
solution The circle is parallel to theyz-plane and centered at(1,2,5), hence thex-coordinates of the points on the circle arex = 1. The projection of the circle on theyz-plane is a circle of radius 2 centered at(2,5). This circle is parametrized by:
y=2+2 cost, z=5+2 sint
We conclude that the points on the required circle can be written as(1,2+2 cost,5+2 sint). This gives the following parametrization: r(t)= 1,2+2 cost,5+2 sint, 0≤t≤2π. The ellipse x 2 2 +y 3 2
=1 in thexy-plane, translated to have center(9,−4,0)
37. The intersection of the planey= 1
2with the spherex2+y2+z2=1 solution Substitutingy= 1
2in the equation of the sphere gives: x2+ 1 2 2 +z2=1 ⇒ x2+z2= 3 4 This circle in the horizontal planey= 1
2has the parametrizationx= √
3
2 cost,z= √
3
2 sint. Therefore, the points on the intersection of the planey= 1
2and the spherex2+y2+z2=1, can be written in the form
√ 3 2 cost,12, √ 3 2 sint , yielding the following parametrization:
r(t)= √ 3 2 cost, 1 2, √ 3 2 sint , 0≤t≤2π.
The intersection of the surfaces
z=x2−y2 and z=x2+xy−1 39. The ellipse x 2 2 +z 3 2
=1 in thexz-plane, translated to have center(3,1,5)[Figure 15(A)]
(A) 3 1 (B) y x z z y x 3 1
FIGURE 15 The ellipses described in Exercises 39 and 40.
solution The translated ellipse is in the vertical planey= 1, hence they-coordinate of the points on this ellipse is y=1. Thexandzcoordinates satisfy the equation of the ellipse:
x−3 2 2 + z−5 3 2 =1. This ellipse is parametrized by the following equations:
S E C T I O N 13.1 Vector-Valued Functions (LT SECTION 14.1) 259
Therefore, the points on the translated ellipse can be written as(3+2 cost,1,5+3 sint). This gives the following parametrization: r(t)= 3+2 cost,1,5+3 sint, 0≤t≤2π. The ellipse y 2 2 +z 3 2
=1, translated to have center(3,1,5)[Figure 15(B)]
Further Insights and Challenges
41. Sketch the curve parametrized byr(t)= |t| +t,|t| −t. solution We have: |t| +t= 0 t≤0 2t t >0; |t| −t= 2t t≤0 0 t >0
Ast increases from−∞to 0, thex-coordinate is zero and they-coordinate is positive and decreasing to zero. Ast increases from 0 to+∞, they-coordinate is zero and thex-coordinate is positive and increasing to+∞. We obtain the following curve:
x
y
r(t)= |t| +t,|t| −t
Find the maximum height above thexy-plane of a point onr(t)=et,sint, t (4−t).
43. LetCbe the curve obtained by intersecting a cylinder of radiusrand a plane. Insert two spheres of radius rinto the cylinder above and below the plane, and letF1andF2be the points where the plane is tangent to the spheres [Figure 16(A)]. LetKbe the vertical distance between the equators of the two spheres. Rediscover Archimedes’s proof thatCis an ellipse by showing that every pointP onCsatisfies
P F1+P F2=K 2
Hint:If two lines through a pointP are tangent to a sphere and intersect the sphere atQ1andQ2as in Figure 16(B), then the segmentsP Q1andP Q2have equal length. Use this to show thatP F1=P R1andP F2=P R2.
(A) R2 (B) Q1 Q2 F2 F1 P P K R1 FIGURE 16
solution To show thatCis an ellipse, we show that every pointP onCsatisfies: F1P+F2P =K
We denote the points of intersection of the vertical line throughP with the equators of the two spheres byR1andR2(see figure).
R2 F2 F1 P K R1
We denote byO1andO2the centers of the spheres.
F1 O1
P r
SinceF1is the tangency point, the radiusO1F1is perpendicular to the plane of the curveC, and therefore it is orthogonal to the segmentP F1on this plane. Hence,O1F1P is a right triangle and by Pythagoras’ Theorem we have:
O1F12+P F12=O1P2 r2+P F 1 2 =O1P 2 ⇒ P F1= O1P 2 −r2 (1) R1 O1 P r
O1R1P is also a right triangle, hence by Pythagoras’ Theorem we have: O1R12+R1P2=O1P2
r2+R
1P2=O1P2 ⇒ P R1=
O1P2−r2 (2)
Combining (1) and (2) we get:
P F1=P R1 (3)
Similarly we have:
P F2=P R2 (4)
We now combine (3), (4) and the equalityP R1+P R2=Kto obtain: F1P+F2P =P R1+P R2=K
Thus, the sum of the distances of the pointsP onCto the two fixed pointsF1andF2is a constantK >0, henceCis an ellipse.
Assume that the cylinder in Figure 16 has equationx2+y2=r2and the plane has equationz=ax+by. Find a vector parametrizationr(t)of the curve of intersection using the trigonometric functions costand sint.
45. Now reprove the result of Exercise 43 using vector geometry.Assume that the cylinder has equationx2+y2= r2and the plane has equationz=ax+by.
(a) Show that the upper and lower spheres in Figure 16 have centers C1=
0,0, ra2+b2+1 C2=0,0,−ra2+b2+1
S E C T I O N 13.2 Calculus of Vector-Valued Functions (LT SECTION 14.2) 261
(b) Show that the points where the plane is tangent to the sphere are F1= r a2+b2+1 a, b, a2+b2 F2= −r a2+b2+1 a, b, a2+b2 Hint:Show thatC1F1andC2F2have lengthrand are orthogonal to the plane.
(c) Verify, with the aid of a computer algebra system, that Eq. (2) holds with K=2ra2+b2+1
To simplify the algebra, observe that sinceaandbare arbitrary, it suffices to verify Eq. (2) for the pointP =(r,0, ar). solution
(a) and (b)SinceF1is the tangency point of the sphere and the plane, the radius toF1is orthogonal to the plane. Therefore to show that the center of the sphere is atC1and the tangency point is the given point we must show that:
−−−→C1F1 =r (1)
−−−→
C1F1is orthogonal to the plane. (2)
We compute the vector−−−→C1F1: −−−→ C1F1= ra a2+b2+1, rb a2+b2+1, r(a2+b2) a2+b2+1−r a2+b2+1 = r a2+b2+1a, b,−1 Hence, −−−→C1F1 = r a2+b2+1 a, b,−1 = r a2+b2+1 a2+b2+(−1)2=r
We, thus, proved that (1) is satisfied. To show (2) we must show that−−−→C1F1is parallel to the normal vectora, b,−1to the planez=ax+by(i.e.,ax+by−z=0). The two vectors are parallel since by (1)−−−→C1F1is a constant multiple of a, b,−1. In a similar manner one can show (1) and (2) for the vector−−−→C2F2.
(c) This is an extremely challenging problem. As suggested in the book, we useP = (r,0, ar), and we also use the expressions forF1andF2as given above. This gives us:
P F1= 1+2a2+b2−2a1+a2+b2r2 P F2= 1+2a2+b2+2a1+a2+b2r2 Their sum is not very inspiring:
P F1+P F2=
1+2a2+b2−2a1+a2+b2r2+1+2a2+b2+2a1+a2+b2r2 Let us look, instead, at(P F1+P F2)2, and show that this is equal toK2. Since everything is positive, this will imply thatP F1+P F2=K, as desired.
(P F1+P F2)2=2r2+4a2r2+2b2r2+2
r4+2b2r4+b4r4
=2r2+4a2r2+2b2r2+2(1+b2)r2=4r2(1+a2+b2)=K2
13.2
Calculus of Vector-Valued Functions
(LT Section 14.2)
Preliminary Questions
1. State the three forms of the Product Rule for vector-valued functions.
solution The Product Rule for scalar multiplef (t)of a vector-valued functionr(t)states that: d
dtf (t)r(t)=f (t)r(t)+f(t)r(t) The Product Rule for dot products states that:
d
Finally, the Product Rule for cross product is d
dtr1(t)×r2(t)=r1(t)×r2(t)+r1(t)×r2(t).
In Questions 2–6, indicate whether the statement is true or false, and if it is false, provide a correct statement.
2. The derivative of a vector-valued function is defined as the limit of the difference quotient, just as in the scalar-valued case.
solution The statement is true. The derivative of a vector-valued functionr(t)is defined a limit of the difference quotient:
r(t)= lim t→0
r(t+h)−r(t) h in the same way as in the scalar-valued case.
3. There are two Chain Rules for vector-valued functions: one for the composite of two vector-valued functions and one for the composite of a vector-valued and a scalar-valued function.
solution This statement is false. A vector-valued functionr(t)is a function whose domain is a set of real numbers and whose range consists of position vectors. Therefore, ifr1(t)andr2(t)are vector-valued functions, the composition “(r1·r2)(t)= r1(r2(t))” has no meaning sincer2(t)is a vector and not a real number. However, for a scalar-valued functionf (t), the compositionr(f (t))has a meaning, and there is a Chain Rule for differentiability of this vector-valued function.
4. The terms “velocity vector” and “tangent vector” for a pathr(t)mean one and the same thing. solution This statement is true.
5. The derivative of a vector-valued function is the slope of the tangent line, just as in the scalar case.
solution The statement is false. The derivative of a vector-valued function is again a vector-valued function, hence it cannot be the slope of the tangent line (which is a scalar). However, the derivative,r(t0)is the direction vector of the tangent line to the curve traced byr(t), atr(t0).
6. The derivative of the cross product is the cross product of the derivatives. solution The statement is false, since usually,
d
dtr1(t)×r2(t)=r1(t)×r2(t) The correct statement is the Product Rule for Cross Products. That is,
d
dtr1(t)×r2(t)=r1(t)×r2(t)+r1(t)×r2(t)
7. State whether the following derivatives of vector-valued functionsr1(t)andr2(t)are scalars or vectors: (a) d dtr1(t) (b) d dt r1(t)·r2(t) (c) d dt r1(t)×r2(t) solution (a) vector, (b) scalar, (c) vector.
Exercises
In Exercises 1–6, evaluate the limit. 1. lim t→3 t2,4t,1 t
solution By the theorem on vector-valued limits we have: lim t→3 t2,4t,1 t = lim t→3t 2,lim t→34t,tlim→3 1 t = 9,12,1 3 . lim
t→πsin 2ti+costj+tan 4tk
3. lim t→0e
2ti+ln(t+1)j+4k
solution Computing the limit of each component, we obtain: lim t→0 e2ti+ln(t+1)j+4k=lim t→0e 2ti+lim t→0ln(t+1) j+ lim t→04 k=e0i+(ln 1)j+4k=i+4k
S E C T I O N 13.2 Calculus of Vector-Valued Functions (LT SECTION 14.2) 263 lim t→0 1 t+1, et−1 t ,4t 5. Evaluate lim h→0 r(t+h)−r(t) h forr(t)= t−1,sint,4.
solution This limit is the derivativeddtr. Using componentwise differentiation yields:
lim h→0 r(t+h)−r(t) h = dr dt = d dt t−1, d dt(sint) , d dt(4) = −1 t2,cost,0 . Evaluate lim t→0 r(t)
t forr(t)= sint,1−cost,−2t.
In Exercises 7–12, compute the derivative. 7. r(t)=t, t2, t3
solution Using componentwise differentiation we get: dr dt = d dt(t), d dt(t 2), d dt(t 3) =1,2t,3t2 r(t)=7−t,4√t,8 9. r(s)=e3s, e−s, s4
solution Using componentwise differentiation we get: dr ds = d ds(e 3s), d ds(e− s), d ds(s 4) =3e3s,−e−s,4s3 b(t)=e3t−4, e6−t, (t+1)−1 11. c(t)=t−1i−e2tk
solution Using componentwise differentiation we get:
c(t)=t−1 i−e2t k= −t−2i−2e2tk
a(θ)=(cos 3θ)i+(sin2θ)j+(tanθ)k 13. Calculater(t)andr(t)forr(t)=t, t2, t3.
solution We perform the differentiation componentwise to obtain: r(t)=(t), (t2), (t3)=1,2t,3t2 We now differentiate the derivative vector to find the second derivative:
r(t)= d dt
1,2t,3t2= 0,2,6t.
Sketch the curver(t)=1−t2, tfor−1≤t≤1. Compute the tangent vector att=1 and add it to the sketch. 15. Sketch the curver1(t)=
t, t2together with its tangent vector att=1. Then do the same forr 2(t)=
t3, t6. solution Note thatr1(t)= 1,2tand sor1(1)= 1,2. The graph ofr1(t)satisfiesy=x2. Likewise,r
2(t)=
3t2,6t5and sor
2(1)= 3,6. The graph ofr2(t)also satisfiesy=x2. Both graphs and tangent vectors are given here.
2 r2(t) 1
r1(t)
Sketch the cycloidr(t)=t−sint,1−costtogether with its tangent vectors att= π
3 and 3π
4. In Exercises 17–20, evaluate the derivative byusing the appropriate Product Rule, where
r1(t)= t2, t3, t, r 2(t)= e3t, e2t, et 17. d dt r1(t)·r2(t) solution d dt(r1(t)·r2(t))=r1(t)·r2(t)+r1(t)·r2(t) =t2, t3, t·3e3t,2e2t, et+2t,3t2,1·e3t, e2t, et =3t2e3t+2t3e2t+tet+2te3t+3t2e2t+et =(3t2+2t)e3t+(2t3+3t2)e2t+(t+1)et
d dt t4r1(t) 19. d dt r1(t)×r2(t) solution d dt(r1(t)×r2(t))=r1(t)×r2(t)+r1(t)×r2(t) =t2, t3, t×3e3t,2e2t, et+2t,3t2,1×e3t, e2t, et = i j k t2 t3 t 3e3t 2e2t et + i j k 2t 3t2 1 e3t e2t et =(t3et−2te2t)i+(3te3t−t2et)j+(2t2e2t−3t3e3t)k +(3t2et−e2t)i+(e3t−2tet)j+(2te2t−3t2e3t)k = [(t3+3t2)et−(2t+1)e2t]i+ [(3t+1)e3t−(t2+2t)et]j + [(2t2+2t)e2t−(3t3+3t2)e3t]k d dt r(t)·r1(t) t=2, assuming that r(2)= 2,1,0, r(2)= 1,4,3 In Exercises 21 and 22, let
r1(t)= t2,1,2t, r 2(t)= 1,2, et 21. Compute d dtr1(t)·r2(t) t=1in two ways: (a) Calculater1(t)·r2(t)and differentiate. (b) Use the Product Rule.
solution
(a) First we will calculater1(t)·r2(t):
r1(t)·r2(t)=
t2,1,2t·1,2, et =t2+2+2tet And then differentiating we get:
d dt(r1(t)·r2(t))= d dt(t 2+2+2tet)=2t+2tet+2et d dt(r1(t)·r2(t)) t=1=2+2e+2e=2+4e (b) First we differentiate: r1(t)= t2,1,2t, r 1(t)= 2t,0,2 r2(t)= 1,2, et, r2(t)=0,0, et Using the Product Rule we see:
d dt(r1(t)·r2(t))=r1(t)·r2(t)+r1(t)·r2(t) =t2,1,2t·0,0, et+ 2t,0,2 ·1,2, et =2tet+2t+2et d dt(r1(t)·r2(t)) t=1=2e+2+2e=2+4e Compute d dtr1(t)×r2(t) t=1in two ways: (a) Calculater1(t)×r2(t)and differentiate. (b) Use the Product Rule.
In Exercises 23–26, evaluate d
dtr(g(t))using the Chain Rule. 23. r(t)=t2,1−t, g(t)=et
solution We first differentiate the two functions: r(t)= d dt t2,1−t= 2t,−1 g(t)= d dt(e t)=et
S E C T I O N 13.2 Calculus of Vector-Valued Functions (LT SECTION 14.2) 265
Using the Chain Rule we get: d
dtr(g(t))=g(t)r(g(t))=e
t2et,−1=2e2t,−et
r(t)=t2, t3, g(t)=sint 25. r(t)=et, e2t,4, g(t)=4t+9
solution We first differentiate the two functions: r(t)= d dt et, e2t,4=et,2e2t,0 g(t)= d dt (4t+9)=4 Using the Chain Rule we get:
d
dtr(g(t))=g(t)r(g(t))=4
e4t+9,2e2(4t+9),0=4e4t+9,8e8t+18,0
r(t)= 4 sin 2t,6 cos 2t, g(t)=t2
27. Letr(t) = t2,1−t,4t. Calculate the derivative ofr(t)·a(t)att = 2, assuming thata(2) = 1,3,3 and a(2)= −1,4,1.
solution By the Product Rule for dot products we have d dtr(t)·a(t)=r(t)·a(t)+r(t)·a(t) Att=2 we have d dtr(t)·a(t) t=2 =r(2)·a(2)+r(2)·a(2) (1) We compute the derivativer(2):
r(t)= d dt
t2,1−t,4t= 2t,−1,4 ⇒ r(2)= 4,−1,4 (2) Also,r(2)=22,1−2,4·2= 4,−1,8. Substituting the vectors in the equation above, we obtain:
d
dtr(t)·a(t)
t=2= 4,−1,8 · −1,4,1 + 4,−1,4 · 1,3,3 =(−4−4+8)+(4−3+12)=13 The derivative ofr(t)·a(t)att=2 is 13.
Letv(s)=s2i+2sj+9s−2k. Evaluate d
dsv(g(s))ats=4, assuming thatg(4)=3 andg(4)= −9.
In Exercises 29–34, find a parametrization of the tangent line at the point indicated. 29. r(t)=t2, t4, t= −2
solution The tangent line has the following parametrization:
(t)=r(−2)+tr(−2) (1)
We compute the vectorsr(−2)andr(−2):
r(−2)=(−2)2, (−2)4= 4,16 r(t)= d dt t2, t4=2t,4t3 ⇒ r(−2)= −4,−32 Substituting in (1) gives: (t)= 4,16 +t−4,−32 = 4−4t,16−32t The parametrization for the tangent line is, thus,
x=4−4t, y=16−32t, −∞< t <∞.
To find a direct relation betweenyandx, we expresstin terms ofxand substitute iny=16−32t. This gives: x=4−4t⇒t= x−4
−4 . Hence,
y=16−32t=16−32·x−4
−4 =16+8(x−4)=8x−16. The equation of the tangent line isy=8x−16.
r(t)=cos 2t,sin 3t, t= π
4 31. r(t)=1−t2,5t,2t3, t=2
solution The tangent line is parametrized by:
(t)=r(2)+tr(2) (1)
We compute the vectors in the above parametrization:
r(2)=1−22,5·2,2·23= −3,10,16 r(t)= d
dt
1−t2,5t,2t3=−2t,5,6t2 ⇒ r(2)= −4,5,24 Substituting the vectors in (1) we obtain the following parametrization:
(t)= −3,10,16 +t−4,5,24 = −3−4t,10+5t,16+24t
r(t)=4t,5t,9t, t= −4 33. r(s)=4s−1i−8
3s−3k, s=2
solution The tangent line is parametrized by:
(s)=r(2)+sr(2) (1)
We compute the vectors in the above parametrization: r(2)=4(2)−1i−8 3(2) −3k=2i− 1 3k r(s)= d ds 4s−1i−8 3s −3k = −4s−2i+8s−4k ⇒ r(2)= −i+1 2k Substituting the vectors in (1) we obtain the following parametrization:
(t)= 2i−1 3k +s −i+1 2k =(2−s)i+ 1 2s− 1 3 k r(s)=(lns)i+s−1j+9sk, s=1 35. Use Example 4 to calculate d
dt(r×r), wherer(t)=
t, t2, et. solution In Example 4 it is proved that:
d
dtr×r=r×r (1)
We compute the derivativesr(t)andr(t):
r(t)= d dt t, t2, et=1,2t, et r(t)= d dt 1,2t, et=0,2, et Using (1) we get d dtr×r=r×r= t, t2, et×0,2, et= i j k t t2 et 0 2 et =t2et−2et i−0−tet j+2t−0 k =t2−2eti+tetj+2tk=t2−2t et, tet,2t
Letr(t)= 3 cost,5 sint,4 cost. Show thatr(t)is constant and conclude, using Example 7, thatr(t)and r(t)are orthogonal. Then computer(t)and verify directly thatr(t)is orthogonal tor(t).
37. Show that thederivative of the normis not equal to thenorm of the derivativeby verifying thatr(t)= r(t)for r(t)= t,1,1.
solution First let us computer(t)forr(t)= t,1,1: r(t)= d
dt(
t2+2)= t t2+2 Now, first let us compute the derivative,r(t):
r(t)= 1,0,0 and then computing the norm:
r(t) = 1,0,0 =√1=1 It is clear in this example, thatr(t)= r(t).
S E C T I O N 13.2 Calculus of Vector-Valued Functions (LT SECTION 14.2) 267
Show that d
dt(a×r)=a×rfor any constant vectora.
In Exercises 39–46, evaluate the integrals.
39.
3
−1
8t2−t,6t3+tdt
solution Vector-valued integration is defined via componentwise integration. Thus, we first compute the integral of each component. 3 −1 8t2−t dt= 8 3t 3−t2 2 3− 1 = 72− 9 2 − −8 3− 1 2 = 212 3 3 −1 6t3+t dt= 3 2t 4+t2 2 3−1= 243 2 + 9 2 − 3 2+ 1 2 =124 Therefore, 3 −1 8t2−t,6t3+tdt= 3 −1 8t2−t dt, 3 −1 6t3+t dt = 212 3 ,124 1 0 1 1+s2, s 1+s2 ds 41. 2 −2 u3i+u5j du
solution The vector-valued integration is defined via componentwise integration. Thus, we first compute the integral of each component. 2 −2 u3du= u4 4 2−2= 16 4 − 16 4 =0 2 −2 u5du= u6 6 2− 2 = 64 6 − 64 6 =0 Therefore, 2 −2 u3i+u5j du= 2 −2 u3du i+ 2 −2 u5du j=0i+0j 1 0 te−t2i+tln(t2+1)j dt 43. 1 0 2t,4t,−cos 3tdt
solution The vector valued integration is defined via componentwise integration. Therefore,
1 0 2t,4t,−cos 3tdt= 1 0 2t dt, 1 0 4t dt, 1 0 − cos 3t dt = t2 1 0 ,2t2 1 0 ,−sin 3t 3 1 0 = 1,2,−sin 3 3 1 1/2 1 u2, 1 u4, 1 u5 du 45. 4 1 t−1i+4√tj−8t3/2k dt
solution We perform the integration componentwise. Computing the integral of each component we get:
4 1 t−1dt =lnt 4 1 =ln 4−ln 1=ln 4 4 1 4√t dt=4·2 3t 3/24 1 = 8 3 43/2−1 = 56 3 4 1 −8 t3/2dt = −16 5t 5/24 1 = −16 5 45/2−1 = −496 5 Hence, 4 1 t−1i+4√tj−8t3/2kdt =(ln 4)i+56 3j− 496 5 k
In Exercises 47–54, find both the general solution of the differential equation and the solution with the given initial condition.
47. dr
dt = 1−2t,4t, r(0)= 3,1
solution We first find the general solution by integratingddtr:
r(t)= 1−2t,4tdt= (1−2t) dt, 4t dt =t−t2,2t2+c (1) Sincer(0)= 3,1, we have: r(0)=0−02,2·02+c= 3,1 ⇒c= 3,1 Substituting in (1) gives the solution:
r(t)=t−t2,2t2+ 3,1 =−t2+t+3,2t2+1
r(t)=i−j, r(0)=2i+3k 49. r(t)=t2i+5tj+k, r(1)=j+2k
solution We first find the general solution by integratingr(t):
r(t)= t2i+5tj+kdt= t2dt i+ 5t dt j+ 1dt k= 1 3t 3 i+ 5 2t 2 j+tk+c (1) The solution which satisfies the initial condition must satisfy:
r(1)= 1 3·1 3 i+ 5 2·1 2 j+1·k+c=j+2k That is, c= −1 3i− 3 2j+1k Substituting in (1) gives the following solution:
r(t)= 1 3t 3 i+ 5 2t 2 j+tk− 1 3i− 3 2j+k= 1 3t 3− 1 3 i+ 5t2 2 − 3 2 j+(t+1)k r(t)= sin 3t,sin 3t, t, rπ2 = 2,4,π 2 4 51. r(t)=16k, r(0)= 1,0,0, r(0)= 0,1,0
solution To find the general solution we first findr(t)by integratingr(t):
r(t)=
r(t) dt=
16kdt=(16t)k+c1 (1)
We now integrater(t)to find the general solutionr(t):
r(t)= r(t) dt= ((16t)k+c1) dt= 16(t) dt k+c1t+c2=(8t2)k+c1t+c2 (2) We substitute the initial conditions in (1) and (2). This gives:
r(0)=c1= 0,1,0 =j
r(0)=0k+c1·0+c2= 1,0,0 ⇒ c2= 1,0,0 =i Combining with (2) we obtain the following solution:
r(t)=(8t2)k+tj+i=i+tj+(8t2)k
r(t)=e2t−2, t2−1,1
,r(1)= 0,0,1,r(1)= 2,0,0 53. r(t)= 0,2,0,r(3)= 1,1,0,r(3)= 0,0,1
solution To find the general solution we first findr(t)by integratingr(t):
r(t)=
r(t) dt=
0,2,0dt = 0,2t,0 +c1 (1) We now integrater(t)to find the general solutionr(t):
r(t)= r(t) dt= (0,2t,0 +c1) dt= 0, t2,0+c 1t+c2 (2)