MATH 32A, BRIDGE 2013
M. Wang
Homework 7
Solution
1. Evaluate the limit (14.2 Exercises 1, 3)
a. limt→3 t2 , 4t, 1t
Ans: limt→3 t2 , 4t, 1t = 9, 12, 13
b. limt→0 e2t i + ln(t + 1)j + 4k
Ans: limt→0 e2t i + ln(t + 1)j + 4k = i + 4k
2. Compute the derivative. (14.2 Exercises 9, 11)
a. r(s) = he3s , e−s , s4 i
Ans: r0 (s) = h3e3s , −e−s , 4s3 i
b. c(t) = t−1 i − e2t k
Ans: c0 (t) = −t−2 i − 2e2t k
3. Evaluate the derivative by using the appropriate Product Rule, where. (14.2 Exercises 17, 18, 19)
r1 (t) = ht2 , t3 , ti,
r2 (t) = he3t , e2t , et i
r01 (t) = h2t, 3t2 , 1i,
a.
d
dt (r1 (t)
r02 (t) = h3e3t , 2e2t , et i
· r2 (t))
Ans:
d
(r1 (t) · r2 (t)) = r1 (t) · r02 (t) + r01 (t) · r2 (t)
dt
= ht2 , t3 , ti · h3e3t , 2e2t , et i + h2t, 3t2 , 1i · he3t , e2t , et i
= (3t2 e3t + 2t3 e2t + tet ) + (2te3t + 3t2 e2t + et )
= (3t2 + 2t)e3t + (2t3 + 3t2 )e2t + (t + 1)et
b.
d 4
dt (t r1 (t))
Ans:
d 4
(t r1 (t)) = t4 r01 (t) + 4t3 r1 (t)
dt
= t4 h2t, 3t2 , 1i + 4t3 ht2 , t3 , ti
= h2t5 , 3t6 , t4 i + h4t5 , 4t6 , 4t4 i
= h6t5 , 7t6 , 5t4 i
1
c.
d
dt (r1 (t)
× r2 (t))
Ans:
d
r1 (t) × r2 (t) = r1 (t) × r02 (t) + r01 (t) × r2 (t)
dt
= ht2 , t3 , ti × h3e3t , 2e2t , et i + h2t, 3t2 , 1i × he3t , e2t , et i
= (t3 et − 2te2t )i − (t2 et − 3te3t )j + (2t2 e2t − 3t3 e3t )k
+ (3t2 et − e2t )i − (2tet − e3t )j + (2te2t − 3t2 e3t )k
= t2 (t + 3)et − (2t + 1)e2t , −t(t + 2)et + (3t + 1)e3t , 2t(t + 1)e2t − 3(t3 + t2 )e3t
4. Evaluate
d
dt r(g(t))
using the Chain Rule (14.2 Exercises 23, 25)
a. r(t) = ht2 , 1 − ti, g(t) = et
d
r(g(t)) = g 0 (t)r0 (g(t)) = et h2g(t), −1i = h2e2t , −et i
Ans: dt
b. r(t) = het , e2t , 4i, g(t) = 4t + 9
d
Ans: dt
r(g(t)) = g 0 (t)r0 (g(t)) = 4heg(t) , 2eg(t) , 0i = h4e4t+9 , 8e4t+9 , 0i
5. Find a parametrization of the tangent line at the point indicated. (14.2 Exercises 29, 31)
a. r(t) = ht2 , t4 i, t = −2
Ans: r0 (t) = h2t, 4t3 i =⇒ L(t) = r(t0 ) + tr0 (t0 ) = h4, 16i + th−4, −32i
b. r(t) = h1 − t2 , 5t, 2t3 i, t = 2
Ans: r0 (t) = h−2t, 5, 6t2 i =⇒ L(t) = r(t0 ) + tr0 (t0 ) = h−3, 10, 16i + th−4, 5, 24i
6. Evaluate the integrals. (14.2 Exercises 41, 43)
a.
c.
R2
3
5
−2 (u i + u j)du
2
R2 3
Ans: −2 (u i + u5 j)du = ( 41 u4 + 61 u6 )−2 = 0
R1
0 h2t, 4t, − cos 3tidt
R1
Ans: 0 h2t, 4t, − cos 3tidt = ht2 , 2t2 , − 13 sin 3ti |10
2
= h1, 2, − 13 sin 3i