MATH 100, Section 110 (CSP)
Week 6: Marked Homework Solutions
2010 Oct 21
1. [3] After t seconds, the radius of the circular ripple is
r = 50t cm,
and the area within the circle is
A(t) = πr 2 = π(50t)2 = 2500πt2
cm2 .
The rate at which the area is increasing is
A0 (t) = 5000πt cm2 /s,
so (a) A0 (1) = 5000π cm2 /s, (b) A0 (3) = 15000π cm2 /s, (c) A0 (5) = 25000π cm2 /s.
2. [6] (a) When t = 0, the population is N = 4200 and the birth rate is
B = b N = (0.025)(4200) = 105 individuals·yr−1 .
(b) When t = 0, the death rate is
D = d N = (0.016)(4200) = 67.2 individuals·yr−1 .
(c) When t = 0, the (net) growth rate is
dN
= B − D = 105 − 67.2 = 37.8 individuals·yr−1 ,
dt
or
dN
= r N = (b − d)N = (0.025 − 0.016)(4200) = (0.009)(4200) = 37.8 individuals·yr −1 .
dt
(d) By Theorem 2 (p. 234), the differential equation
dN
= r N,
dt
with initial condition
N (0) = 4200
has the solution
N (t) = 4200ert ,
where
r = b − d = 0.025 − 0.016 = 0.009 yr−1 .
1
To find the time it takes for the population to quadruple, set N (t) = 4 · 4200 and solve for
t: ert = 4, rt = ln 4,
ln 4
ln 4
t=
=
( ≈ 154 ) yr.
r
0.009
(Notice that we don’t really need the specific information N (0) = 4200 to answer part (d).
The initial population could be N (0) = N0 with N0 unspecified.)
3. [5] Let m(t) denote the amount of uranium-238 at time t years, and assume that the
amount decays at a rate proportional to the amount present. Then the differential equation
for m(t) is
dm
= k m,
dt
which by Theorem 2 (p. 234) has the solution
m(t) = m(0)ekt ,
where m(0) is the amount of uranium-238 at time t = 0. To find k we use the half-life: when
t = 4.5×109 years, m(4.5×109 ) = 21 m(0). On the other hand, the solution to the differential
9
9
equation gives m(4.5 × 109 ) = m(0)ek(4.5×10 ) . Therefore e4.5×10 k = ln(1/2) = − ln 2,
k=−
ln 2
4.5 × 109
yr−1 .
9
(a) When t = 109 years, m(109 ) = m(0)e10 k = m(0)e−(ln 2)/4.5 and the percentage remaining
is
m(109 )/m(0) × 100% = e−(ln 2)/4.5 × 100% (≈ 85.7%).
(b) We solve for t when m(t) = 0.75m(0) = 43 m(0) (one quarter of the initial amount decayed
means three quarters of the initial amount is present at this time). ekt = 34 gives
t = (ln(3/4))/k = 4.5 × 109 (ln 4 − ln 3)/ ln 2 (≈ 1.87 × 109 ) years.
4. [10] Newton’s Law of Cooling (or warming, in this case) gives the differential equation
dT
= k(T − 200)
dt
and the initial condition is
T (0) = 20,
where T (t) is the temperature of the turkey in degrees Celcius, at time t in minutes. To
solve this, make the change of variables
y(t) = T (t) − 200,
then (as shown in the lecture and textbook) y(t) must solve the differential equation
dy
= k y,
dt
2
with initial condition
y(0) = 20 − 200 = −180.
Then by Theorem 2 (p. 234) the solution of the initial-value problem for y(t) is
y(t) = −180ekt .
We are also given that y(30) = 30 − 200 = −170, and we are asked to determine t such
that y(t) = 80 − 200 = −120.
To determine t, we need k, which we get by setting t = 30,
−180ek(30) = −170,
and solving:
e30k =
then
k=
170
,
180
30k = ln
17
,
18
1
17
1
18
ln
= − ln .
30 18
30 17
We set y(t) = −120 and solve for t:
−180ekt = −120,
and
t=
ekt =
ln(2/3)
1 2
ln = 30
k 3
ln(17/18)
2
120
= ,
180
3
(≈ 213)
min.
5. [7] We solve this differential equation in the same way as we solve the differential equation
for Newton’s Law of Cooling. Using the hint, write the differential equation as
dy
b
=a y+
,
dt
a
and make the change of variables
b
u(t) = y(t) + .
a
Differentiate with respect to t and use the differential equation for y:
du
dy
b
=
= a y+
= au
dt
dt
a
so u(t) must solve the differential equation
du
= a u,
dt
with initial condition
u(0) = y(0) +
3
b
b
=c+ .
a
a
Then by Theorem 2 (p. 234) the solution of the initial value problem for u(t) is
b
u(t) = c +
eat ,
a
then y(t) = u(t) − (b/a), so
y(t) =
Then the solution to
b
c+
a
dy
= −3y + 10,
dt
b
eat − .
a
y(0) = −99,
is (a = −3, b = 10, c = −99)
y(t) =
10
−99 −
3
4
e−3t +
10
.
3