MATH 100, HOMEWORK 4 SOLUTIONS
Section 3.7, #16: (a) The average rate of change of V when r changes
V (r2 ) − V (r1 )
, hence the answers are
from r1 to r2 is
r2 − r1
(i)
4
π83 − 34 π53
V (8) − V (5)
= 3
≈ 540.35µm,
8−5
8−5
4
π63 − 43 π53
V (6) − V (5)
3
(ii)
=
≈ 381.18µm,
6−5
6−5
4
π5.13 − 34 π53
V (5.1) − V (5)
(iii)
= 3
≈ 320.48µm.
5.1 − 5
5.1 − 5
(b) The instantaneous rate of change is
dV
4
= π · 3r2 = 4πr2 .
dr
3
At r = 5, this is equal to 4π · 52 = 100π = 314.16.
(c) The answer in (b) is identical to the formula for the area of the sphere of
radius r. To explain this, imagine that a thin layer of thickness ∆r has been
added to the surface of a ball of radius r. The volume ∆V of that layer is,
roughly, equal to the surface area times its thickness, i.e. 4πr2 · ∆r. Hence
∆V
≈ 4πr2 .
∆r
Section 3.8, #10: (a) Let m(t) denote the amount of tritium-3 at time
t (measured in years). Then m(t) = m0 ekt , where C = m(0) and k is a
constant that we need to find. We know that m(1) = m0 ek = 0.945m0 ,
hence
ek = 0.945, k = ln(.945) ≈ −0.05657.
(a) According to the formula we derived in class, the halflife is
− ln 2
≈ 12.2528 years.
k
(b) We want to find t such that m(t) = .2y(0), i.e. m0 ekt = .2m0 ,
ekt = .2, kt = ln .2, t =
1
ln .2
≈ 28.4502 years.
k
Section 3.8, #20: If the initial investment is A0 , then its value after t years
is A(t) = A0 e.06t .
(a) We want to find t such that A(t) = A0 e.06t = 2A0 :
e.06t = 2, .06t = ln 2, t =
ln 2
≈ 11.5525 years.
.06
(b) After 1 year, the investment is valued at A(1) = A0 e0.06 ≈ 1.0618A0
dollars. With annual compounding, the interest rate should be set at 6.18%
to achieve the same rate of growth.
Section 3.9, #14: We set up a coordinate system so that at time 0 ship
A is at (−150, 0) and ship B is at the origin. Then at time t ship A has
coordinates (x(t), 0) and ship B has coordinates (0, y(t)) (draw a picture!).
We are given the information that x(0) = −150, y(0) = 0, x0 (t) = 35,
y 0 (t) = 25. If r(t) is the distance between the ships at time t, then we have
the relation r2 = x2 + y 2 . Differentiate this:
2rr0 = 2xx0 + 2yy 0 , rr0 = xx0 + yy 0 .
At t = √4, we have x(4) √
= −150 + 4 · 35 = −10, y(4) = 4 · 25 = 100, and
2
2
r(4) = 10 + 100 = 10 101. Hence
√
10 101r0 (4) = −10 · 35 + 100 · 25 = 2150, r0 (4) =
2150
√
≈ 21.3933.
10 101
Section 3.9, #32: We are given that P V 1.4 = C. Differentiate this with
respect to t:
P 0 V 1.4 + P · 1.4V 0.4 V 0 = 0, V 0 =
−P 0 V 1.4
−P 0 V
=
.
1.4P V 0.4
1.4P
At the time when V = 400, P = 80, P 0 = −10:
V0 =
10 · 400
≈ 35.7143cm3 /min.
1.4 · 80
2