Math 2260 Written HW #6 Solutions
1. A pot of soup is left to simmer for a long time over steady heat; as a result, the soup has
equilibrated and its temperature isn’t changing. You want to check the temperature of the
soup, so you pull out a digital thermometer. Before putting the thermometer into the soup
it reads 70◦ F, and after 1 second in the soup it reads 90◦ F. If the temperature of the soup is
actually 150◦ F, how long will it take until the thermometer reading passes 145◦ F?
Answer: Let H(t) denote the temperature of the thermometer after t seconds. Then we
know, by Newton’s Law of Cooling, that
H(t) = Hs + H0 ekt
where Hs = 150 is the temperature of the soup and H0 and k are constants. Moreover, we
know that H(0) = 70 and that H(1) = 90.
Therefore, at time t = 0 we have that
H(t) = 150 + H0 ek(0)
70 = 150 + H0 ,
so we see that H0 = 70 − 150 = −80. Hence, the temperature of the thermometer is given by
H(t) = 150 − 80ekt .
Now, plugging in t = 1 yields
H(1) = 150 − 80ek(1)
90 = 150 − 80ek ,
so we see that
−60 = −80ek ,
and so
ek =
3
4
and hence
k = ln(3/4).
Therefore,
H(t) = 150 − 80eln(3/4)t .
Then the temperature of the thermometer will pass 145◦ at the time t0 determined by
H(t0 ) = 150 − 80eln(3/4)t0
145 = 150 − 80eln(3/4)t0 .
Solving for t0 we see that
−5 = −80eln(3/4)t0 ,
1
meaning that
1
= eln(3/4)t0 .
16
Therefore,
ln(1/16) = ln(3/4)t0
and so
t0 =
ln(1/16)
≈ 9.63,
ln(3/4)
so the thermometer reading will pass 145◦ in just under 10 seconds.
2. A painting attributed to Vermeer (who lived from 1632 to 1675) suddenly appears, seemingly
out of nowhere, on lot list for an upcoming Sotheby’s auction. Skeptical art historians from
the National Gallery in London take a small sample of the paint from the painting and
determine that it contains 99.1% of its original 14 C. Is the painting authentic, or is it a
forgery? (Remember that the half-life of 14 C is 5730 years.)
Answer: Let C(t) denote the percentage of 14 C in the painting after t years. Then we know
that
C(t) = C0 ekt
for some constants C0 and k. Moreover, at time t = 0 the painting still has 100% of its
so we know that C0 = 100. Therefore,
C(t) = 100ekt .
Since the half-life of
14 C
is 5730 years, we know that C(5730) = 50, so we have
C(5730) = 100ek(5730)
50 = 100e5730k ,
so
1
= e5730k
2
and we can solve for k:
k=
Hence, the percentage of
14 C
ln(1/2)
.
5730
after t years is given by
C(t) = 100e
ln(1/2)
t
5730
.
If the current time is t0 years after the painting was painted, then we know that
C(t0 ) = 100e
99.1 = 100e
ln(1/2)
t
5730 0
ln(1/2)
t
5730 0
Therefore,
0.991 = e
2
ln(1/2)
t
5730 0
.
14 C,
and so
ln(0.991) =
ln(1/2)
t0 ,
5730
meaning that
t0 =
5730 ln(0.991)
≈ 74.7.
ln(1/2)
Hence, the painting is less than 75 years old. Since Vermeer died 338 years ago, we can pretty
confidently say that this painting is a fake.
3. Assuming n is a positive integer, evaluate the indefinite integral
Z
xn ln(x) dx.
Answer: I will integrate by parts, using
u = ln(x)
du =
1
x
dv = xn
v=
xn+1
.
n+1
Then the above integral is equal to
Z
xn+1
1
1
1
ln(x) −
xn dx =
xn+1 ln(x) −
xn+1 + C
n+1
n+1
n+1
(n + 1)2
xn+1
1
=
ln(x) −
+ C.
n+1
n+1
3