Chapter 12
Vector-Valued Functions
12.1
Vector Functions
1. Since the square root function is only defined for nonnegative values, we must have t2− 9 ≥ 0.
So the domain is (−∞, −3)S[3, ∞).
2. Since the natural logarithm is only defined for positive values, we must have 1 − t2 > 0. So
the domain is (−1, 1).
3. Since the inverse sine function is only defined for values between -1 and 1, the domain is [−1, 1].
4. The vector function is defined for all real numbers. 5. r(t) = sin πti + cos πtj − cos2πtk
6. r(t) = cos2πti + 2 sin2πtj + t2k
7. r(t) = e−ti + e2tj + e3tk
8. r(t) = −16t2i + 50tj + 10k
9. x = t2, y = sin t, z = cos t
10. r(t) = t sin t(i + k) = t sin ti + 0j + t sin tk so x = t sin t, y = 0, z = t sin t
11. x = ln t, y = 1 + t, z = t3
12. x = 5 sin t sin 3t, y = 5 cos 3t, z = 5 cos t sin 3t 48
13. z y x 4 14. z y x 15. z y x 16. z y x 17. y x 2 2 18. y x 19. z y x 20. z y x 21. z y x
Note: the scale is distorted in this graph. For t = 0, the graph starts at (1, 0, 1). The upper loop shown intersects the xz-plane at about (286751, 0, 286751).
22. z y x 10 10 10 23. z y x 24. z y x 25. r(t) = h4, 0i + h0 − 4, 3 − 0it = (4 − 4t)i + 3tj, 0 ≤ t ≤ 1 x y 26. r(t) = h0, 0, 0i + h1 − 0, 1 − 0, 1 − 0it = ti + tj + tk, 0 ≤ t ≤ 1
27. x = t, y = t, z = t2+ t2= 2t2; r(t) = ti + tj + 2t2k z y x 28. x = t, y = 2t, z = ±√t2+ 4t2+ 1 = ±√5t2− 1; r(t) = ti + 2tj ±√5t2− 1k z y x
29. x = 3 cos t, z = 9 − 9 cos2t = 0 sin2t; y = 3 sin t; r(t) = 3 cos ti + 3 sin tj + 9 sin2tk
z
y x
z y x 31. x = t, y = t, z = 1 − 2t; r(t) = ti + tj + (1 − 2t)k z y x 32. x = 11, y = t, z = 3 + 2t; r(t) = i + tj + (3 + 2t)k z y x
33. (b); Notice that the y and z values consistently increase while the x values oscillate rapidly between -1 and 1. The only vector fucntion that describes this behavior is (b).
34. (c); The trace of the graph on the xy−plane would look like a circle, while the z value oscillates between 0 and 1. The only vector function that describes this behavior is (c).
35. (d); Notice that the z value is contant. The only vector function that satisfies this constraint is (d).
36. (a); Notice that the x values consistently increase while the trace of the graph on the yz-plane would look like a circle. The only vector function that describes this behavior is (a).
37. Letting x = at cos t, y = bt sin t, and z = ct, we have z2 c2 = c2t2 c2 = t 2= t2cos2t + t2sin2t =a 2t2cos2t a2 + b2t2sin2 b2 =x 2 a2 + y2 b2 38. z y x
39. Letting x = aektcos t, y = bektsin t, and z = cekt, we have
z2 c2 = c2ekt c2 = e 2kt= e2ktcos2t + e2ktsin2t = a 2e2ktcos2t a2 + b2e2ktsin2t b2 x2 a2 + y2 b2 40. z y x
41. x2+ y2+ z2= a2sin2kt cos2t + a2sin2kt sin2t + a2cos2kt = a2sin2kt + a2cos2kt
42. k = 1 k = 2 k = 3 z y x z y x z y x k = 4 k = 10 k = 20 z y x z y x z y x 43. (a) z y x (b) r1(t) = ti + tj + (4 − t2)k r2(t) = ti − tj + (4 − t2)k (c) z y x
45. 46. k = 0.1 k = 0.2 k = 0.3 z y x z y x z y x 47. k = 2 k = 4 z y x z y x 48. k = 101 k = 1 z y x z y x
12.2
Calculus of Vector Functions
1. lim
t→2[t
2. r(t) =sin 2t
t i + (t − 2)
5k + ln t
1/tk. Using L’Hˆopital’s Rule,
lim t→0+r(t) = 2 cos 2t 1 i + (t − 2) 5j + 1/t −1/t2k = 2i − 32j
3. Using L’Hˆopital’s Rule, we have
lim t→1 t2− 1 t − 1, 5t − 1 t + 1 , 2et−1− 2 t − 1 = lim t→1= 2t 1, 5t − 1 t + 1 , 2et−1 1 i = h2, 2, 2 4. Since lim t→∞tan −1t = π 2, we have lim t→∞ e2t 2e2t+ t, e−1 2e−t+ 5, tan −1t = lim t→∞ 1 2 + te−2t, 1 2 + 5et, tan −1t = 1 2, 0, π 2
The last equality follows from using L’Hˆopital’s Rule to get
lim t→∞te −2t= lim t→∞ t e2t = limt→∞ 1 2e2t = 0 5. lim t→α[−4r1(t) + 3r2(t)] = −4(i − 2j + k) + 3(2i + 5j + 7k) = 2i + 23j + 17k 6. lim t→αr1(t) · r2(t) = (i − 2j + k) = (i − 2j + k) · (2i + 5j + 7k) = −1
7. Notice that the k component ln(t − 1) is not defined at t = 1. Therefore, r(t) is not continuous at t = 1.
8. Notice that sin πt, tan πt, and cos πt are each continuous at t = 1 since the sine, cosine, and tangent function are continuous on their domains. Therefore, since each of the component functions are continuous at t = 1, we know that r(t) is continuous at t = 1.
9. r0(t) = 3i + 8tj + (10t − 1)k so r0(1) = 3i + 8j + 9k = h3, 8, 9i while r(1.1) − r(1) 0.1 = h3(1.1) − 1, 4(1.1)2, 5(1.1)2− (1.1)i − h3(1) − 1, 4(1)2, 5(1)2− (1)i 0.1 = h2.3, 4.84, 4.95i − h2, 4, 4i 0.1 = h0.3, 0.84, 0.95i 0.1 = h3, 8.4, 9.5i 10. r0(t) = −5 (1 + 5t)2i + (6t + 1)j − 3(1 − t) 2k so r0(0) = −5 1 i + j + 3k = h−5, 1, −3i
while r(0.05) − r(0) 0.05 = 1 1 + 5(0.05), 3(0.05) 2+ (0.05), (1 + 0.05)3 − 1 1 + 5(0), 3(0) 2+ (0), (1 − 0)3 0.05 =h0.8, 0.0575, 0.857375i − h1, 0, 1i 0.05 =h−0.2, 0.0575, −0.142625i 0.05 = h−4, 1.15, −2.8525i 11. r0(t) = 1 ti − 1 t2j; r 00(t) = −1 t2i + 2 t3j
12. r0(t) = h−t sin t, 1 − sin ti; r00(t) = h−t cos t − sin t, − cos ti 13. r0(t) = h2te2t+ e2t, 3t2, 8t − 1i; r00(t) = h4te2t+ 4e2t, 6t, 8i
14. r0(t) = 2ti + 3t2j + 1 1 + t2k; r 00(t) = 2i + 6tj − 2t (1 + t2)2k 15. r0(t) = −2 sin ti + 6 cos tj r0(π/6) = −i + 3√3j x y 16. r0(t) = 3t2i + 2tj r0(−1) = 3i − 2j y x 17. r0(t) = j − 8t (1 + t2)2k r0(−1) = j − 2k z y x 18. r0(t) = −3 sin ti + 3 cos tj + 2k r0(π/4) = −3 √ 2 2 i + 3√2 2 j + 2k z x y
19. r(t) = ti +1 2j + 1 3t 3k; r(2) = 2i + 2j + 8 3k; r 0(t) = i + tj + t2k; r0(2) = i + 2j + 4k
Using the point (2, 2, 8/3) and the direction vector r0(2), we have x = 2 + t, y = 2 + 2t, z =
8/3 + 4t. 20. r(t) = (t3−t)i+ 6t t + 1j+(2t+1) 2k; r(1) = 3j+9k; r0(t) = (3t2−1)i+ 6 (t + 1)2j+(8t+4)k; r0(1) = 2i + 3
2j + 12k. Using the point (0, 3, 9) and the direction vector r
0(1), we have x =
2t, y = 3 +32, z = 9 + 12t.
21. r0(t) = het+ tet, 2t + 2, 3t2− 1i so r0(0) = h1, 2, −1i and |r0(0)| =p12+ 22+ (−1)2=√6
The unit tangent vector at t = 0 is given by r
0(0) |r0(0)| = h1, 2, −1i √ 6 = 1 √ 6, 2 √ 6, −1 √ 6
To find the parametric equations of the tangent line at t = 0, we first compute r(0) = h0, 0, 0i. The tangent line is then given in vector form as p(t) = h0, 0, 0i + t
1 √ 6, 2 √ 6, −1 √ 6 = 1 √ 6t, 2 √ 6t, −1 √ 6t or in parametric form as x = √1 6t, y = 2 √ 6t, z = −1 √ 6t.
22. r0(t) = h3 cos 3t, 2 sec22t, 1i so r0(π) = h−3, 2, 1i and |r0(π)| =p(−3)2+ (2)2+ (1)2 =√14.
The unit tangent vector at t = π is given by r
0(π) |r0(π)| = h−3, 2, 1i √ 14 = −3 √ 14, 2 √ 14, 1 √ 14
To find the parametric equations of the tangent line at t = π, we first compute r0(π) = h1, 0, πi. The tangent line is then given in vector form as
p(t) = h1, 0, πi + t −3√ 14, 2 √ 14, 1 √ 14 = 1 − −3√ 14t, 2 √ 14t, π + 1 √ 14t or in parametric form as x = 1 −√−3 14t, y = 2 √ 14t, z = π + 1 √ 14t 23. r(π/3) = * 1 2, √ 3 2 , π 3 + r0(t) = h− sin t, cos t, 1i r0(π/3) = * − √ 3 2 , 1 2, 1 +
so the tangent line is given by p(t) = * 1 2, √ 3 2 , π 3 + + t * − √ 3 2 , 1 2, 1 + = * 1 2 − √ 3 2 t, √ 3 2 + 1 2t, π 3 + t + 24. r(0) = h6, 1, 1i r0(t) = h−3e−t/2, 2e2t, 3e3ti
r0(0) = h−3, 2, 3i So the tangent line is given by r(t) = h6, 1, 1i + th−3, 2, 2i = h6 − 3t, 1 + 2t, 1 + 3ti 25. d dt[r(t) × r 0(t)] = r(t) × r00(t) + r0(t) × r0(t) = r(t) × r00(t) 26. d dt[r(t) · (tr(t))] = r(t) · d dt(tr(t))+ = r(t) · (tr 0(t) + r(t)) + r0(t) · (tr(t)) = r(t) · (tr0(t)) + r(t) · r(t) + r0(t) · (tr(t)) = 2t(r(t) · r0(t)) + r(t) · r(t) 27. d dt[r(t) · (r 0(t) × r00(t))] = r(t) · d dt(r 0(t) × r00(t)) + r0(t) · (r0(t) × r00(t)) = r(t) · (r0(t) × r000(t) + r00(t) × r00(t)) + r0(t) · (r0(t) × r00(t)) = r(t) · (r0(t) × r000(t)) 28. d dt[r1(t) × (r2(t) × r3(t))] = r1(t) × d dt(r2(t) × r3(t)) + r 0(t) × (r 2(t) × r3(t)) = r1(t) × (r2(t) × r03(t) + r02(t) × r3(t) + r01(t) × (r2(t) × r3(t)) = r1(t) × (r2(t) × r03(t)) + r1(t) × (r02(t) × r3(t)) + r1(t) × (r2(t) × r3(t)) 29. d dt[r1(2t) + r2( 1 t)] = 2r 0(2t) − 1 t2r 0 2( 1 t) 30. d dt[t 3r(t2)] = t3(2t)r0(t2) + 3t2r(t2) = 2t4r0(t2) + 3t2r(t2) 31. Z 2 −1 r(t)dt = Z 2 −1 tdt i + Z 2 −1 3t2dt j + Z 2 −1 4t3dt k = 1 2t 2 2 −1 i + t3 2 −1j + t 4 2 −1k = 3 2i + 9j + 15k 32. Z 4 0 r(t)dt = Z 4 0 √ 2t + 1dt i + Z 4 0 −√tdt j + Z 4 0 sin πtdt k = 1 3(2t + 1) 3/2 4 0 i − 2 3t 3/2 4 0 j − 1 πcos πt 4 0 k = 26 3 i − 16 3 j 33. Z r(t)dt = Z tetdt i + Z −e−2tdt j + Z tet2dt k = [tet− et+ c 1]i + 1 2e −2t+ c 2 j + 1 2e t2+ d 3 k = et(t − 1)i + 1 2e −2tj +1 2e t2k + c, where c = c1i + c2j + c3k. 34. Z r(t)dt = Z 1 1 + t2dt i + Z t 1 + t2dt j + Z t2 1 + t2dt k = [tan−1t + c1]i + 1 2ln(1 + t 2) + c 2 j + Z 1 − 1 1 + t2 k = [tan−1t + c1]i + 1 2ln(1 + t 2) + c 2 j + [t − tan−1t + c3]k = tan−1ti + 1 2ln(1 + t 2)j + (t − tan−1t)k + c,
where c = c1i + c2j + c3k.
35. r(t) =R r0(t)dt =R 6dt i + R 6tdt j + R 3t2dt k = [6t + c
1]i + [3t2+ c2]j + [t3+ c3]k
Since r(0) = i + 2j + k = c1i + c2j + c3k, c1− 1, c2= −2, and c3= 1. Thus,
r(t) = (6t + 1)i + (3t2− 2)j + (t3+ 1)k
36. r(t) =R r0(t)dt =R t sin t2dt i + R − cos 2tdt j = − 1 2cos t
2+ c
1 i + −12sin 2t + c2 j
Since r(0) = 32= (−12+ c1)i + c2j, c1= 2, and c2= 0. Thus,
r(t) = −1 2cos t 2+ 2 i −1 2sin 2tj. 37. r0(t) = R r00(t)dt =R 12tdt i + R −3t−1/2dt j + R 2dt k = [6t2+ c 1]i + [−6t1/2+ c2]j + [2t + c3]k
Since r0(1) = j = (6 + c1)i + (−6 + c2)j + (2 + c3)k, c1= −6, c2= 7, and c3= −2. Thus,
r0(t) = (6t2− 6)i + (−6t1/2+ 7)j + (2t − 2)k. r(t) = Z r0(t)dt = Z (6t2− 6)dt i + Z (−6t1/2+ 7)dt j + Z (2t − 2)dt k = [2t3− 6t + c4]i + [−4t3/2+ 7t + c5]j + [t2− 2t + c6]k. Since r(1) = 2i − k = (−4 + c4)i + (3 + c5)j + (−1 + c6)k, c4= 6, c5= −3, and c6= 0. Thus, r(t) = (2t3− 6t + 6)i + (−4t3/2+ 7t − 3)j + (t2− 2t)k. 38. r0(t) = Z r00(t)dt = Z sec2tdt i + Z cos tdt j + Z − sin tdt k = [tan t + c1]i + [sin t + c2]j + [cos t + c3]k
Since r0(0) = i + j + k = c1i + c2j + c3k, c1= 1, c2= 1, and c3= 0. Thus,
r0(t) = (tan t + 1)i + (sin t + 1)j + cos tk.
r(t) = Z r0(t)dt = Z (tan t + 1)dt i + Z (sin t + 1)dt j + Z cos tdt k = [ln | sec t| + c4]i + [− cos t + t + c5]j + [sin t + c6]k
.
Since r(0) = −j + 5k = (−1 + c5)j + (c6)k, c4= 0, c5= 0, and c6= 5. Thus,
r(t) = (ln | sec t| + t)i + (− cos t + t)j + (sin t + 5)k.
39. r0(t) = −a sin ti + a cos tj + ck; |r0(t)| =p(−a sin t)2+ (a cos t)2+ c2=√a2+ c2
s =R2π 0 √ a2+ c2dt = √a2+ c2t 2π 0 = 2π √ a2+ c2
40. r0(t) = i + (cos t − t sin t)j + (sin t + t cos t)k
|r0(t)| =p12+ (cos t − t sin t)2+ (sin t + t cos t)2=√2 + t2
s =R0π√2 + t2dt = t 2 √ 2 + t2+ ln |t +√2 + t2| π 0 = π 2 √ 2 + π2+ ln(π +√2 + π2) − ln√2
41. r0(t) = (−2etsin 2t + etcos 2t)i + (2etcos 2t + etsin 2t)j + etk |r0(t)| =p 5e2tcos22t + 5e2tsin22t + e2t =√6e2t =√6et s =R3π 0 √ 6etdt =√6et 3π 0 =√6(e3π− 1) 42. r0(t) = 3i + 2√3tj + 2t2k; |r0(t)| =q32+ (2√3t)2+ (2t2)2=√9 + 12t2+ 4t4= 3 + 2t2 s =R01(3 + 2t2)dt = 3t +23t3 1 0= 3 + 2 3 = 11 3
43. From r0(t) = h9 cos t, −9 sin ti, we find |r0(t)| = 9. Therefore, s =Rt
09du = 9t so that t =
s 9. By substituting for t in r(t), we obtain r(s) =D9 sins
9, 9 cos s 9 E
. Note that r0(s) =Dsins 9, cos s 9 E so that r0(s) = r sin2 s 9 + cos s 9 = 1.
44. From r(t) = h−5 sin t, 12, 5 cos ti, we find |r0(t)| =√169 = 13. Therefore, s =Rt
013du = 13t
so that t = s
13. By substituting for t in r(t), we obtain r(s) = h5 cos
s 13, 12 13s, 5 13cos s 13i.
Note that r0(t) =−135 sin13s,1213,135 cos13s so that |r0(s)| =r 25 169sin 25 13+ 144 169+ 25 169cos 2 s 13 = 1
45. From r0(t) = h2, −3, 4i, we find |r0(t)| = √29. Therefore, s = Rt
0
√
29du = √20t so that
t = √s
29. By substituting for t in r(t), we obtain r(s) =
1 +√2 29s, 5 − 3 √ 29s, 2 + 4 √ 29s . Note that r0(s) =D√2 29, − 3 √ 29, 4 √ 29 E so that r0(s) =r 4 29+ 9 29+ 16 29 = 1.
46. From r0(t) = hetcos t − etsin t, etsin t + etcos t, 0i we find
|r0(t)| =pe2tcost−2e2tcos t sin t + e2tsin2t + e2tsin2t + 2e2tsin t cos t + e2tcos2t =√2e2t = et√2.
Therefore, s =Rt
0e
u√2du =√2(et− 1) so that t = ln√s 2+ 1 . By substituting for t in r(t), we obtain r(s) =D√s 2+ 1 cos(ln s √ 2+ 1 ,√s 2+ 1 sinln√s 2+ 1 , 1ENote that r0(s) =D√1 2cos ln√s 2+ 1 −√1 2sin ln√s 2+ 1 ,√1 2sin ln√s 2 + 1 +√1 2cos ln√s 2+ 1 , 0E so that |r0(s)| = v u u u t 1 2cos 2ln√s 2+ 1 − cosln√s 2+ 1 sinln√s 2+ 1 +12sin2ln√s 2+ 1 +12sin2ln√s 2+ 1 + sinln√s 2+ 1 cosln√s 2+ 1 +12cos2ln√s 2+ 1 = s cos2 ln s √ 2 + 1 + sin2 ln s √ 2 + 1 = 1
47. Since d dt(r · r) = d dt|r| 2= d dtc 2 = 0 and d dt(r · r) = r · r 0+ r0· r = 2r · r0, we have r · r0 = 0. Thus, r0 is perpendicular to r.
48. Let v = ai + bj and r(t) = x(t)i + y(t)j. Then
Z b a v · r(t)dt = Z b a [ax(t) + by(t)]dt = a Z b a x(t)dt + b Z b a y(t)dt = v · Z b a r(t)dt.
49. From r(t) = r0+ tv, we get r0(t) = v so that |r0(t)| = |v|. Therfore s =
Rt
0|r
0(t)|du =
Rt
0|v|du = |v|t which gives t =
s
|v|. Substituting for t in r(t), we have r
0(s) = r 0 + s |v|v = r0+ s v |v| . Note that r0(s) = v |v| so that |r 0(s)| = |v| |v|= 1. 50. (a) |h3, −4i| =p32+ (−4)2= 5 so r(s) = h1, 2i + s 5h3, −4i = h1, 2i + s 3 5, −4 5
(b) r(t) = h1, 1, 10i + th1, 2, −1i and |h1, 2, −1i| = √1 + 4 + 1 = √6 so r(s) = h1, 1, 10i +
s 1 √ 6, 2 √ 6, −1 √ 6
12.3
Motion on a Curve
y x a v 1. v(t) = 2ti + t3j; v(1) = 2i + j; |v(1)| =√4 + 1 =√5; a(t) = 2i + 3t2j; a(1) = 2i + 3jy
x
a
v
2. v(t) = 2ti − 2 t3j; v(1) = 2i − 2j; |v(1)| = √ 4 + 4 = 2√2; a(t) = 2i + 6 t4j; a(1) = 2i + 6jy
x a
v
3. v(t) = −2 sinh 2ti + 2 cosh 2tj; v(1) = 2j; |v(0)| = 2; a(t) = −4 cosh 2ti + +4 sinh 2tj; a(0) = −4i
y x a v 4. v(t) = −2 sin ti + cos tj; v(π/3) = −√3i +1 2j;
|v(π/3)| =p3 + 1/4 =√13/2; a(t) = −2 cos ti − sin tj;
a(π/3) = −i − √ 3 2 j y z x a v 5. v(t) = (2t − 2)i + k; v(2) = 2j + k; |v(2)| =√4 + 1 =√5; a(t) = 2j; a(2) = 2j y z x a v 6. v(t) = i + j; v(2) = i + j + 12k; |v(2)| =√1 + 1 + 144 =√146; a(t) = 6tk; a(2) = 12k
y z x a v 7. v(t) = i + 2tj + 3t2k; mathbf v(1) = i + 2j + 3k; |v(1)| =√1 + 1 + 9 =√14; a(t) = 2j + 6tk; a(1) = 2j + 6k y z x a v 8. v(t) = i + 3t2j + k; v(1) = i + 3j + k; |v(1)| =√1 + 9 + 1 =√11; a(t) = 6tj; a(1) = 6j
9. The particle passes through the xy-plane when z(t) = t2−5t = 0 or t = 0, 5 which gives us the
points (0, 0, 0) and (25, 115, 0). v(t) = 2ti + (3t2− 2)j + (2t − 5)k; v(0) = −2j − 5k, v(5) =
10i + 73j + 5k; a(t) = 2i + 6tj + 2k; a(0) = 2i + 30j + 2k
10. If a(t) = 0, then v(t) = c1and r(t) = c1t + c2. The graph of this equation is a straight line.
11. Initially we are given s0= 0 and v0= (480 cos 30◦)i + (480 cos 30◦)j = 240
√ 3i + 240j. Using a(t) = −32j we find v(t) = Z a(t)dt = −32tj + c 240√3i + 240j = v(0) = c v(t) = −32tj + 240√3i + 240j = 240√3i + (240 − 32t)j r(t) = Z v(t)dt = 240√3ti + (240t − 16t2)j + b 0 = r(0) = b.
(a) The shell’s trajectory is given by r(t) = 240√3ti + (240t − 16t2)j or x = 240√3t, y =
240 − 16t2.
(b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum altitude is y(15/2) = 900 ft.
(c) Solving y(t) = 240t − 16t2 = 16t(15 − t) = 0, we see that the shell is at ground level
(d) From (c), impact is when t = 15. The speed at impact is
|v(15)| = |240√3i + (240 − 32 · 15)j| =p2402· 3 + (−240)2= 480 ft/s.
12. Initially we are given s0 = 1600j and v0 = (480 cos 30◦)i + (480 sin 30◦)j = 240
√
3i + 240j. Using a(t) = −32j we find
v(t) = Z a(t)dt = −32tj + c 240√3i + 240j = v(0) = c v(t) = −32tj + 240√3i + 240j = 240√3i + (240 − 32t)j r(t) = Z v(t)dt = 240√3ti + (240t − 16t2)j + b 1600j = r(0) = b.
(a) The shell’s trajectory is given by r(t) = 240√3ti+(240t−16t2+1600)j or s = 240√3t, y =
240t − 16t2+ 1600.
(b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum altitude is y(15/2) = 2400 ft.
(c) Solving y(t) = −16t2+ 240t + 1600 = −16(t − 20)(t + 5) = 0, we see that the shell hits
the ground when t = 20. The range of the shell is x(20) = 4800√3 ≈ 8314 ft.
(d) From (c), impact is when t = 20. The speed at impact is
|v(20)| = |240√3i + (240 − 32 · 20)j| =p2402· 3 + (−400)2= 160√13 ≈ 577 ft/s.
13. We are given s0= 81j and v0= 4i. Using a(t) = −32j, we have
v(t) = Z a(t)dt = −32tj + c 4i = v(0) = c v(t) = 4i − 32tj r(t) = Z v(t)dt = 4ti − 16t2j + b 81j = r(0) = b r(t) = 4ti + (81 − 16t2)j.
Solving y(t) = 81 − 16t2= 0, we see that the car hits the water when t = 9/4. Then
14. Let θ be the angle of elevation. Then v(0) = 98 cos θi + 98 sin θj. Using a(t) = −9.8j, we have v(t) =
Z
a(t)dt = −9.8tj + c 98 cos θi + 98 sin θj = v(0) = c v(t) = 98 cos θi + (98 sin θ − 9.8t)j r(t) = 98t cos θi + (98t sin θ − 4.9t2)j + b.
Since r(0) = 0, b = 0 and r(t) = 98t cos θi + (98t sin θ − 4.9t2)j. Setting y(t) = 98t sin θ −
4.9t2 = t(98 sin θ − 4.9t) = 0, we see that the projectile hits the ground when t = 20 sin θ.
Thus, using x(t) = 98t cos θ, 490 = s(t) = 98(20 sin θ) cos θ or sin 2θ = 0.5. Then 2θ = 30◦ or
150◦. The angles of elevation are 15◦ and 75◦.
15. Let s be the initial speed. Then v(0) = s cos 45◦i + s sin 45◦j = s √ 2 2 i + s√2 2 j. Using a(t) = −32j, we have v(t) = Z a(t)dt = −32j + c s√2 2 i + s√2 2 j = v(0) = c v(t) = s √ 2 2 i + s√2 2 − 32t ! j r(t) = s √ 2 2 ti + s√2 2 t − 16t 2 ! j + b. Since r(0) = 0, b = 0 and r(t) = s √ 2 2 ti + s√2 2 t − 16t 2 ! j.
Setting y(t) = s√2t/2 − 16t2= t(2√2/2 − 16t) = 0 we see that the ball hits the ground when
t = √2s/32. Thus, using x(t) = s√2t/2 and the fact that 100 yd = 300 ft, 300 = x(t) =
s√2 2 ( √ 2s/32) = s 2 32 and s = √ 9600 ≈ 97.98 ft/s.
16. Let s be the initial speed and θ the initial angle. Then v() = s cos θi + s sin θj. Using a(t) = −32j, we have
v(t) = Z
a(t)dt = −32tj + c s cos θi + s sin θj = v(0) = c v(t) = s cos θi + (s sin θ − 32t)j r(t) = st cos θi + (st sin θ − 16t2)j + b.
Since r(0) = 0, b = 0 and r(t) = st cos θi + (st sin θ − 16t2)j. Setting y(t) = st sin θ −
16t2 = t(s sin θ − 16t) =, we see that the ball hits the ground when t = (s sin θ)/16. Using
x(t) = st cos θi, we see that the range of the ball is x s sin θ 16 = s 2sin θ cos θ 16 = s2sin 2θ 32 .
For θ = 30◦, the range is s2sin 60◦/32 =√3s2/64 and for θ = 60◦the range is s2sin 120◦/32 =
√
3s2/64. In general, when the angle is 90◦− θ then range is
[s2sin 2(90◦− θ)]/32 = s2[sin(180◦− 2θ)]/32 = s2(sin 2θ)/32.
Thus, for angles θ and 90◦− θ, the range is the same.
17. r0(t) = v(t) = −r0ω sin ωti + r0ω cos ωtj; v = |v(t)| =
q r2
0ω2sin
2ωt + r2
0ω2cos2ωt = r0ω
ω = v/r0; a(t) = r00(t) = −r0ω2cos ωti − r0ω2sin ωtj
a = |a(t)| = q r2 0ω4cos2ωt + r02ω4sin 2ωt = r 0ω2= r0(v/r0)2= v2/r0.
18. (a) v(t) = −b sin ti + b cos tj + ck; |v(t)| =pb2sin2
t + b2cos2t + c2=√b2+ c2 (b) s =Rt 0|v(t)|du = Rt 0 √ b2+ c2du = t√b2+ c2; ds dt = √ b2+ c2 (c) d 2s
dt2 = 0; a(t) = −b cos ti−b sin tj; |a(t)| =
p
b2cos2t + b2sin2t = |b|. Thus, d2s/dt26=
|a(t)|. y x x0 θ x0tanθ (x0,y0)
19. Let the initial speed of the projectile be s and let the target be at (x0, y0). Then vp(0) = s cos θi + s sin θj and vt(0) = 0. Using
a(t) = −32j, we have vp(t) =R a dt = −32tj + c
s cos θi + s sin θj = vp(0) = c
vp(t) = s cos θi + (s sin θ − 32t)j
rp(t) = st cos θi + (st sin θ − 16t2)j + b.
Since rp(0) = 0, b = 0 and rp(t) = st cos θi + (st sin θ − 16t2)j. Also, vt(t) = −32tj + c and since
vt(0) = 0, c = 0 and vt(t) = −32tj. Then rt(t) = −16t2tj + b. Since rt(0) = x0i + y0j, bx0i + y0j
and rt(t) = x0i + (y0− 16t2)j. Now, the horizontal component of rp(t) will be x0when t = x0/s cos θ
at which time the vertical component of rp(t) will be
(sx0/s cos θ) sin θ − 16(x0/s cos θ)2= x0tan θ − 16(x0/s cos θ)2= y−) − 16(x0/s cos θ)2.
Thus, rp(x0/s cos θ) = rt(x0/s cos θ) and the projectile will strike the target as it falls.
20. The initial angle is θ = 0, the initial height is 1024 ft, and the initial speed is s = 180(5280)/3600 =
264 ft/s. Then x(t) = 264t and y(t) = −16t2+ 1024. Solving y(t) = 0 we see that the pack
hits the ground at t = 8 seconds. The horizontal distance tranvelled is x(8) = 2112 feet. From
the figure in the text, tan α = 1024/2112 = 16/33 and α ≈ 0.45 radian or 25.87◦.
21. By Problem 17, a = v2/v
0 = 15302/(4000 · 5280) ≈ 0.1108. We are given mg = 192, so
φ
<mv2/r0, 32 m>
<mv2/r0, 0>
< 0, 32m>
22. By problem 17, the centripetal acceleration is v2/r0. Then
the horizontal force is mv2/r
0. The vertical force is 32m.
The resultant force is U = (mv2/r
0)i + 32mj. From the
figure, we see that tan φ = (mv2/r
0)/32m = v2/32r0. Using
r0= 60 and v = 44 we obtain tan φ = 442/32(60) ≈ 1.0083
and φ ≈ 45.24◦.
23. Solving x(t) = (v0cos θ)t for t and substituting into y(t) −12gt2+ (v0sin θ)t + s0we obtain
y = −1 2g x v0cos θ 2 + (v0sin θ) x v0cos θ + s)= − g 2v2 0cos2θ x2+ (tan θ)x + s0,
which is the equation of a parabola.
24. Since the projectile is launched from ground level, s0 = 0. To find the maximum height
we maximize y(t) = −12gt2+ (v
0sin θ)t. Solving y0(t) = −gt + v0sin θ = 0, we see that
t = (v0/g) sin θ is a critical point. Since y00(t) = −g ≤ 0,
H = y v0sin θ g = 1 2g v2 0sin 2θ g2 + v0sin θ v0sin θ g = v2 0sin 2θ 2g
is the maximum height. To find the range we solve y(t) = −12gt2+ (v0sin θ)t = t(v0sin θ −
1
2gt) = 0. The positive solution to this equation is t = (2v0sin θ)/g. The range is thus
x(t) = (v0cos θ) 2v0sin θ g = v2 0sin 2θ g .
25. Letting r(t) = x(t)i + y(t)j + z(t)k, the equation dr/dt = v is equivalent to dx/dt =
6t2x, dy/dt = −4ty2, dz/dt = 2t(z + 1). Separating the variables and integrating, we
obtain x/x = 6t2dt, dy/y2 = −4tdt, dz/(z + 1) = 2tdt, and ln x = 2t3+ c1, −1/y =
2t2+ c2, ln(z + 1) + t2+ c3. Thus, r(t) = k1e2t 3 i + 1 2t2+ k 2 j + (k3et 2 − 1)k. 26. We require the fact that dr/dt = v. Then
dL dt = d dt(r × p = r dp dt + dr dt × p = τ + v × p = τ + v × mv = τ + m(v × v) = τ + 0 = τ.
27. (a) Since F is directed along r we have F = cr for some constant c. Then
τ = r × F = r × (cr) = c(r × r) = 0. (b) If τ = 0 then dL/dt = 0 and L is constant.
28. (a) Since the cannon is pointing directly to the left, tha parmetric equations describing the path of the cannon ball are given by
x(t) = v0t, y(t) = −
1 2gt
2+ s 0
The cannon ball will touch the groun when y = 0, which occurs at t =r 2s0
g . At that time, x is given by x =r 2s0 g = −v0 r 2s0
g . Notice that this x value will be farther
to the left with increasing values of v0. Therefore, the cannon ball travels farther with
more gunpowder.
(b) As shown in part (a), the cannon ball will touch the groun when t =r 2s0
g . This value
of t is independent of v0. This occurs because v0 has no vertical component.
(c) If the cannon ball is dropped, we have v0 = 0. Therefore, the parametric equations
describing the cannon ball motion are given by
x(t) = 0, y(t) = −1
2gt
2+ s 0.
As before, y = 0 when t =r 2s0
g . Therefore the cannon ball touches the ground at the
same time regardless of whether it is fired or dropped.
12.4
Curvature and Acceleration
1. r0(t) = −t sin ti + t cos tj + 2tk; |r0(t)| =p t2sin2t + t2cos2t + 4t2=√5t; T = −sin t√ 5 i + cos t √ 5 j + 2 √ 5k
2. r0(t) = et(− sin t + cos t)i + et(cos t + sin t)i +√2etk,
|r0(t)| = [et(sin2
t−2 sin t cos t+cos2t)+e2t(cos2t+2 sin t cos t+sin2t)+2e2t]1/2=√4e2t = 2et;
T(t) =1
2(− sin t + cos t)i + 1
2(cos t + sin t)j + √
2
2 k
3. We assume a > 0. r0(t) = −a sin ti + a cos tj + ck; |r0(t)| = p
a2sin2t + a2cos2t + c2 = √ a2+ c2; T(t) −√a sin t a2+ c2i + a cos t √ a2+ c2j + c √ a2+ c2k; dT dt = − a cos t √ a2+ c2i − a sin t √ a2+ c2j, dT dt = s a2cos2t a2+ c2 + a2sin2t a2+ c2 = a √ a2+ c2; N = − cos ti − sin tj; B = T × N = i j k −√a sin t a2+ c2 a cos t √ a2+ c2 c √ a2+ c2 − cos t − sin t 0 = √c sin t a2+ c2i − c cos t √ a2+ c2+ a √ a2+ c2k; κ = |dT/dt| r0(t) = a/√a2+ c2 √ a2+ c2 = a a2+ c2
4. r0(t) = i + tj + t2k; |r0(t)| =√1 + t2+ t4, |r0; (1)| =√3; T(t) = (1 + t2+ t4)−1/2(i + tj + t2k), T(1)√1 3(i + j + k); dT dt = − 1 2(1 + t 2+ t4)−3/2(2t + 4t3)i + [(1 + t2+ t)−1/2− t 2(1 + t 2+ t)−3/2(2t + 4t3)]j [2t(1 + t2+ t4)−1/2t 2 2(1 + t 2+ t4)−3/2(2t + 4t3)]k; d dtT(1) = − 1 √ 3i + 1 √ 3k, d dtT(1) =r 1 3 + 1 3 = √ 2 √ 3; N(1) = − 1 √ 2(i − k)k, B(1) = i j k 1/√3 1/√3 1/√3 −1/√2 0 1/√2 = √1 6(i − 2j + k); κ = d dtT(1) = |r0(1)| = √ 2/√3 √ 3 = √ 2 3
5. From Example 1 in the text, a normal to the osculating plane is B(π/4) = 261(3i − 3j + 2√2k).
The point on the curve when t = π/4 is (√2,√2, 3π/4). An equation of the plane is 3(x −
√
2) − 3(y −√2) + 2√2(z − 3π/4(= 0, 3x − 3y + 2√2z = 3π/2, or 3√2x − 3√2y + 4z = 3π.
6. From Problem 4, a normal to the osculating plane is B(1) =√1
6(i − 2j + k). The point on the
curve when t = 1 is (1, 1/2, 1/3). An equaiton of the plane is (x−1)−2(y −1/2)+(z −1/3) = 0 or x − 2y + z = 1/3. 7. v(t) = j + 2tk, |v(t)| =√1 + 4t2; a(t) = 2k; v · a = 4t, v × a = 2i, |v × a| = 2; aT= 4t √ 1 + 4t2, aN = 2 √ 1 + 4t2 8. v(t) = −3 sin ti + 2 cos tj + k,
|v(t)| =p9 sin2t + 4 cos2t = 1 =p5 sin2
t + 4 sin2t + 4 cos2t + 1 =√5psin2
+1; a(t) = −3 cos ti − 2 sin tj; v · a = 9 sin t cos t − 4 sin t cos t = 5 sin t cos t,
v × a = 2 sin ti − 3 cos tj + 6k, |v × a| =q4 sin2+(cos2t + 36 =√5√cos2t + 8;
aT √ 5 sin t cos t p sin2t + 1 , aN= s cos2t + 8 sin2t + 1 9. v(t) = 2ti + 2tj + 4tk, |v(t)| = 2√6t, t > 0; a(t) = 2i + 2j + 4k; v · a = 24t, v × a = 0; aT= 24t 2√6t = 2 √ 6, aN= 0, t > 0 10. v(t) = 2ti − 3t2j = 4t3k, |v(t)| = t√4 + 9t2+ 16t4, t >); a(t) = 2i − 6tj + 12t2k; v · a = 4t + 18t3+ 48t5; v × a = −12t4i − 16t3j − 6t2k, |v × a| = 2t2√36t4+ 64t2+ 9; aT = 4 + 18t2+ 48t4 √ 4 + 9t2+ 16t4, aN = 2t√36t4+ 64t2+ 9 √ 4 + 9t2+ 16t4 t > 0 11. v(t) = 2i + 2tj, |v(t)| = 2√1 + t2; a(t) = 2j; v × a = 4k, |v × a| = 4; aT = 2t √ 1 + t2, aN = 2 √ 1 + t2
12. v(t) = 1 1 + t2i + t 1 + t2j, |v(t)| = √ 1 + t2 1 + t2 ; a(t) = − 2t (1 + t2)2i + 1 − t2 (1 + t2)2j; v · a = − 2t (1 + t2)3 + t − t3 (1 + t2)3; v × a = 1 (1 + t2)2k, |v × a| = 1 (1 + t2)2; aT = − t/(1 + t2)3 √ 1 + t2)/(1 + t2 = − t (1 + t2)3/2, aN = a/(1 + t2)2 √ 1 + t2/(1 + t2) = 1 (1 + t2)3/2
13. v(t) = −5 sin ti + 5 cos tj, |v(t)| = 5; a(t) = −5 cos ti − 5 sin tj; v · a = 0,
v × a = 25k, |v × a| = 25; aT = 0, aN = 5
14. v(t) = sinh ti + cosh tj, |v(t)| =psinh t2+ cosh2t0 a(t) = cosh ti + sinh tj
v · a = 2 sinh t cosh t; v × a = (sinh2t − cosh2t)k = −k, |v × a| = 1;
aT = 2 sinh t cosh t p sinh2+ cosh2 , aN = 1 p sinh2+ cosh2 15. v(t) = et(i + j + k), |v(t)| =√3e−t; a(t) = e−t(i + j + k); v · a = −3e−2t; v × a = 0, |v × a| = 0; aT = − √ 3e−t, aN = 0 16. v(t) = i + 2j + 4k, |v(t)| =√21; a(t) = 0; v · a = 0, v × a = 0, |v × a| = 0; aT = 0, aN = 0
17. v(t) = −a sin ti + b cos tj + ck, |v(t)| =pa2sin2t + b2cos2+c2; a(t) = −a cos ti − b sin tj;
v × a = bc sin ti − ac cos tj + abk, |v × a| =pb2c2sin2t + a2c2cos2t + a2b2
κ = |v × a|
|v|3 =
p
b2c2sin2t + a2c2cos2t + a2b2
(a2sin2t + b2cos2t + c2)3/2
18. (a) v(t) = −a sin ti + b cos tj, |v(t)| =pa2sin2t + b2cos2t; a(t) = −a cos ti − b sin tj;
v × a = abk; |v × a| = ab; κ = ab
(a2sin2t + b2cos2t)3/2
(b) When a = b, |v(t)| = a, |v × a| = a2, and κ = a2/a3= 1/a.
19. The equation of a line is v(t) = b + tc, when b and c are constant vectors.
v(t) = c, |v(t)| = |c|; a(t) = 0; v × a = 0; κ = |v × a|/|v|3= 0
20. v(t) = a(1 − cos t)i + a sin tj; v(π) = 2ai, |v(π)| = 2a; a(t) = a sin ti + a cos tj,
a(π) = −aj; |v × a| = i j k 2a 0 0 0 −a 0 = −2a2k; |v × a| = 2a2; κ = |v × a| |v|3 = 2a2 8a3 = 1 4a
21. v(t) = f0(t)i + g0(t)j, |v(t)| =p[f0(t)]2+ [g0(t)]2; a(t) = f00(t)i + g00(t)j;
v × a = [f0(t)g00(t) − g0(t)f00(t)]k, |v × a| = |f0(t)g00(t) − g0(t)f00(t)|;
κ = |v × a|
|v|3 =
|f0(t)g00(t) − g0(t)f00(t)|
([f ”(t)]2+ [g0(t)]2)3/2
22. For y = F (x), r = xi + F (x)j. We identify f (x) = x and g(x) = F (x) in Problem 21. Then f0(x) = 1, f00(x) = 0, g0(x) = F0(x), g00(x) = F00(x), and κ = |F00(x)|/(1 + [F0(x)]2)3/2.
23. F (x) = x2, F (0) = 0, F (1) = 1; F0(x) = 2x, F0(0) = 0, F0(1) = 2; F00(x) = 2, F00(0) = 2, F00(1) = 2; κ(0) = 2 (1 + 02)3/2 = 2; ρ(0) = 1 2; κ(1) = 2 (1 + 22)3/2 = 2 5√5 ≈ 0.18; ρ(1) =5 √ 5 2 ≈ 5.59; Since 2 > 2/5 √
5, the curve is ”sharper” at (0, 0).
24. F (x) = x3, F (−1) = −1, F (1/2) = 1/8; F0(x) = 3x2, F0(−1) = 3, F0(1/2) = 3/4; F00(x) = 6x, F00(−1) = −6, F00(1/2) = 3; κ(−1) = (1+3|−6|2)3/2 = 6 10√10 = 3 5√10 ≈ 0.19; ρ(−1) = 5 √ 10 3 ≈ 5.27; κ(12) = 3 [1 + (3/4)2]3/2 = 3 125/64 ≈ 1.54; ρ( 1 2) = 125 192 ≈ 0.65
Since 1.54 > 0.19, the curve is ”sharper” at (1/2, 1/8).
25. Letting F (x) = x2, we can use Problem 22 to get κ(x) = |F
00(x)|
|1 + (F0(x))2|3/2.
Now, F0(x)2x, F00(x) = 2, and (F0(x))2= 4x2so that κ = 2
(1 + 4x2)3/2.
As x → ±∞, the denominator grows without bound. Therefore, κ(x) → 0 as x → ±∞.
26. (a)
y
x
(b) κ0(t) = 2t(t 2+ 2) (t4+ t2+ 1)3/2√t4+ 4t2+ 1 − 3t(2t2+ 1)√t4+ 4t2+ 1 (t4+ t2+ 1)5/2 ;critical numbers occur at t = −.271469, t = 0, and t = .271469. (c) Maximum of 1.017182 occurs at t = −.271469 and t = .271469.
27. Since (c, F (c)) is an inflection point and F00 exists on an interval containg c, we must have
F00(c) = 0. Therefore, using the formula from Problem 22, we see that the curvature is zero.
28. We use the fact that T · N = 0 and T · T = N · N = 1. Then |a(t)|2= a · a = (a nN + atT) · (anN + atT) = aN2N · N + 2anatN · T + a2TT · T = a 2 N + a 2 T.
Chapter 12 in Review
A. True/False
1. True; |v(t)| =√2
2. True; the curvature of a circle of radius a is κ = a1. 3. True
4. False; consider r(t) = t2i. In this case, v(t) = 2ti and a(t) = 2i. Since v · a = 4t, the velocity
and acceleration vectors are not orthogonal for t 6= 0. 5. True
6. False; see Problem 20c in Section 14.2 7. True 8. True 9. False; consider r1(t) = r2(t) = i. 10. True, d dt|r(t)| 2= d dt(r · r) = r · dr dt + dr dt · r = 2r · dr dt.
B. Fill in the Blanks
1. y = 4 2. 0 3. r0(t) = h1, 2t, t2i so r0(1) = h1, 2, 1i 4. r00(t) = h0, 2, 2ti so r00(1) = h0, 2, 2i 5. r0(1) × r00(1) = i j j 1 2 1 0 2 2 = h2, −2, 2i so r0(1) × r00(1) =√12. Since r0(1)| =√6, we have κ(1) = r 0(1) × r00(1) |r0(1)|3 = √ 12 6√6 = √ 2 6 . 6. T(1) = r 0(1) |r0(1)| = h1, 2, 1i √ 6 = 1 √ 6, 2 √ 6, 1 √ 6 7. T(t) = r 0(t) |r0(t)| = h1, 2t, t2i √ 1 + 4t2+ t4 = 1 √ 1 + 4t2+ t4, 2t √ 1 + 4t2+ t4, t2 √ 1 + 4t2+ t4 So T0(t) = −2(t2+ 2) (t4+ 4t2+ 1)3/2, −2(t4− 1) (t4+ 4t2+ 1)3/2, 2t(2t2+ 1) (t4+ 4t2+ 1)3/2 . This gives T0(1) = −6 63/2, 0, 6 63/2 = −1√ 6, 0, 1 √ 6 and |T0(1)| =q1 6+ 1 6 = 1 √ 3. Therefore N(1) = T 0(1) |T0(1)| = D −1 √ 6, 0, 1 √ 6 E (√1 3) = h−1√ 2, 0, 1 √ 2i.
8. B(1) = T(1) × N(1) = i j k 1 √ 6 2 √ 6 1 √ 6 −1 √ 2 0 1 √ 2 = 1 √ 3, −1 √ 3, 1 √ 3
9. A normal to the normal plane is T(1) =D√1
6, 2 √ 6, 1 √ 6 E
so we can use n = h1, 2, 1i as a vector normal to the plane. Since r(1) = h1, 1,13i, the point (1, 1,1
3) lies on the normal plane at t = 1.
Thus an equation of the normal plane is (x − 1) + 2(y − 1) + (z −13) = 0 or x + 2y + z = 1)3
or 3x + 6y + 3z = 10
10. A normal to the osculating plane is B(1) =D√1
3, −1 √ 3, 1 √ 3 E . So we can use n = h1, −1, 1i as a
normal vector. Using the point (1, 1,13), an equation of the osculating plane is (z − 1) − (y −
1) + (z −13) = 0 or x − y + z = 13 or 3x − 3y + 3z = 1.
C. Exercises
1. r0(t) = cos ti + sin tj + k; s =Rπ 0 p cos2t + sin2+1dt =Rπ 0 √ 2dt =√2π 2. r0(t) = 5i + j + 7k; s(t) =Rt 0 √ 25 + 1 + 49du = 5√3t; s(3) = 15√3. Solving 5√3t = 80√3,we see that the distance traveled will be 80√3 when t = 16 or at the point (80, 17, 112).
3. r(3) = −27i + 8j + k; r0(t) = −6ti = √ 2
t + 1+ k; r
0(2) = −18i + j + k. The tangent line
is x = −27 − 18t, y = 8 + t, z = 1 + t. 4. z y x 5. z y x 6. d dt[r1(t) × r2(t)] = r1(t) × d dtr2(t) + d dtr1(t) × r2(t)
= (t2i + 2tj + t3k) × (−i + 2tj + 2tk) + (2ti + 2j + 2t2k) × [−ti + t2j + (t2+ 1)k] = (4t2− 2t4)i − 3t3j + (2t3+ 2t)k + (2t2+ 2 − 3t4)i − (5t3+ 2t)j + (2t3+ 2t)k = (2 + 6t2− 5t4)i − (8t3+ 2t)j + (4t3+ 4t)k d dt[r1(t) × r2(t)] = d dt[(2t 3+ 2t − t5)i − (2t4+ t2)j + (t4+ 2t2)k] = (2 + 6t2− 5t4)i − (8t3+ 2t)j + (4t3+ 4t)k
7. d dt[r1(t) · r2(t)] = r1(t) · d dtr2(t) + d dtr1(t) · r2(t)
= (cos ti − sin tj + 4t3k) · (2ti + sin tj + 2e2tk) (− sin ti − cos tj + 12t2k) · (t2i + sin tj + e2tk)
= (2t cos t − sin t cos t + 8t3e2t− t2sin t − sin t cos t + 12t2e2t = 2t cos t − t2sin t − 2 sin t cos t + 8t3e2t+ 12t2e2t
d
dt[r1(t) · r2(t)] = d dt[t
2cos t − sin2t + 4t3e2t] = −t2sin t + 2t cos t − 2 sin t cos t + 8t3e2t+ 12t2e2t
8. d dt[r1(t) · (r2(t) × r3(t))] = r1(t) · d dt[r2(t) × r3(t)] + r 0(t) · [r 2(t) × r3(t)] = r1(t) · [(r2(t) × r03(t)) + (r02(t) × r3(t))] + r01(t) · (r2(t) × r3(t)) = r1(t) · (r2(t) × r03(t)) + r1(t) · r20(t) × r3(t)) = r01(t) · (r2(t) × r3(t))
9. We are given F = ma = 2j; v(0) = i + j + k. and r(0) = i + j. Then v(t) = Z a(t)dt = Z 2 mjdt = 2 mtj + c i = j + k = v(0) = c v(t) = i + 2 mt + 1 j + k r(t) = ti + 1 mt 2+ t j + tk + b i + j = r(0) = b r(t) = (t + 1)i + 1 mt 2+ t + 1 j + tk
The parametric equations are x = t, y = 1
mt 2+ t + 1, z = t. 10. y x a v v(t) = i − 3t2j, v(1) = i − 3j; a(t) = −6tj, a(1) = −6j |v(1)| = |i − 3j| =√1 + 9 =√10
11. v(t) = 6i + j + 2tk; a(t) = 2k. To find when the particle passes through the plane, we solve
−6t + t + t2= −4 or t2− 5t + 4 = 0. This gives t = 1 and t = 4. v(1) = 6i + j + 2k, a(1) = 2k;
v(4) = 6i + j + 8k, a(4) = 2k 12. We are given r(0) = i + 2j + 3k. r(t) = Z v(t)dt = Z (−10ti + (3t2− 4t)j + k)dt = −5t2i + (t3− 2t2)j + tk + c i + 2j + 3k = r(0) = c r(t) = (1 − 5t2)i + (t3− 2t2+ 2)j + (t + 3)k r(t) = −19i + 2j + 5k
13. v(t) =R a(t)dt = R (√2 sin ti +√2 cos tj)dt = −√2 cos ti +√2 sin tj + c;
−i + j + k = v(π/4) = −i + j + c, c = k; v(t) = −√2 cos ti +√2 sin tj + k;
r(t) = −√2 sin ti−√2 cos tj+tk+b; i+2j+(π/4)k = r(π/4) = −i−j+(π/4)k+b, b = 2i+3j;
r(t) = (2 − 2√2 sin t)i + (3 −√2 cos t)j + tk; r(3π/4) = i + 4j + (3π/4)k
14. v(t) = ti + t2j − tk; |v| = t√t2+ 2, t > 0; a(t) = i + 2tj − k; v · a = t + 2t3+ t = 2t + 2t3; v × a = t2bi + t2k, |v × at2√2; aT = 2t + 2t3 t√t2+ 2 = 2 + 2t2 √ t2+ 2, aN = t2√2 t√t2+ 2 = √ 2t √ t2+ 2; κ = t 2√2 t3(t2+ 2)3/2 = √ 2 t(t2+ 2)3/2
15. r0(t) = sinh ti + cosh tj + k, r0(1) = sinh 1i + cosh 1j + k; |r0(t)| =p
sinh2t + cosh2t + 1 = √
2 cosh2t =√2 cosh t; |r0(1)| =√2 cosh 1;
T =√1 2tanh ti + 1 √ 2j + 1 √ 2sech tk, T(1) = 1 √ 2(tanh 1i + j + sech 1k); dT dt = 1 √ 2sech 2 ti −√1 2sech t tanh tk; d dtT(1) = 1 √ 2sech 2 1i −√1 2sech 1 tanh 1k, d dtT(1) == sech 1√ 2 p sech21 + tanh2+1 =√1
2sech 1; N(1) = sech 1i − tanh 1k;
B(1) = T(1) × N(1) = −√1 2tanh 1i + 1 √ 2(tanh 2 1 + sech21)j − √1 2sech 1k =√1 2(− tanh 1i + j − sech 1k) κ = d dtT(1) /|r0(1)| = (sech 1)/ √ 2 √ 2 cosh 1 = 1 2sech 21
16. The parametric equations describing the path of the ball are
x(t) = 66 cos(30◦)t = 33√3ty(t) = −16t2+ 66 sin(30◦)t + 148 = −16t2+ 33t + 148
The ball touches the ground when y(t) = 0 or −16t2+ 33t + 148 = 0. This occurs when
t ≈ 4.243. The ball therefore strikes the ground at x(4.243) = 242.52 ft.
The velocity of the ball at time t is v(t) = h33√3, −32t + 33i. The impact velocity is given
by v(4.243) = h33√3, −32(4.243) + 33i ≈ h57.158, −102.776i. The impact speed is then
