Math 2250 Written HW #8 Solutions
1. Shown below is the graph of the function f (x) = x1/x . At what point does this curve have a
horizontal tangent line? Be sure to find both the x and the y coordinates of the point.
1.2
0.8
0.4
0
10
20
30
40
50
Answer: The goal is to find where the tangent line is horizontal, meaning where the slope
of the tangent line is zero, which means we’re looking for x so that f 0 (x) = 0.
Since the function f (x) is an exponential with a variable base and a variable power, to compute
f 0 (x) it is a very good idea to use logarithmic differentiation, since taking the logarithm turns
powers into products. So, following the logarithmic differentiation playbook, first we take the
natural log of both sides and simplify using the rules of logarithms:
ln(f (x)) = ln x1/x
1
ln(f (x)) ln(x).
x
Now we differentiate both sides with respect to x. If we let g(x) = ln(f (x)), then we can
write
g(x) = (h ◦ f )(x),
where h(u) = ln(u). Then h0 (u) = u1 , so the Chain Rule implies that
d
1
f 0 (x)
(ln(f (x))) = h0 (x) = g 0 (f (x)) · f 0 (x) =
· f 0 (x) =
.
dx
f (x)
f (x)
Therefore, applying the Product Rule on the right hand side, we have that
d
d 1
(ln(f (x))) =
ln(x)
dx
dx x
f 0 (x)
d 1
1 d
=
ln(x) +
(ln(x))
f (x)
dx x
x dx
f 0 (x)
−1
1 1
= 2 · ln(x) + ·
f (x)
x
x x
0
f (x)
1
= 2 (1 − ln(x)) .
f (x)
x
1
Multiplying both sides by f (x) and substituting in f (x) = x1/x yields
f (x)
(1 − ln(x))
x2
x1/x
f 0 (x) = 2 (1 − ln(x)) .
x
f 0 (x) =
Hence, the graph of f (x) will have a horizontal tangent line when f 0 (x) = 0, meaning when
0=
x1/x
(1 − ln(x)) .
x2
1/x
This can only happen if either xx2 = 0 or if 1 − ln(x) = 0. The first isn’t possible, but the
second will happen if ln(x) = 1, meaning that x = e. Therefore, the curve has a horizontal
tangent line when x = e, so the point on the curve is
(e, f (e)) = e, e1/e .
Since e ≈ 2.718 and e1/e ≈ 1.445, this seems to fit the picture.
2. After the end of the Cold War, the Rocky Flats nuclear weapons facility in Colorado was
decommissioned and, after years of cleanup, converted into a wildlife refuge. As wildlife reentered the area, burrowing species like prairie dogs were studied with particular interest by
wildlife ecologists since they were likely to redistribute contaminants left in the soil and even
to uncover buried waste.
(The above is all true...what follows is made up.) Ecologists estimated that there were 600
prairie dogs already on the grounds when their study started. After several years of monitoring
the population carefully, they estimated that the prairie dog population t days after the start
of the study could be estimated by the function
t − 1000
p(t) = 5000 arctan
+ 7000.
300
The graph of p is shown below and follows the classic pattern of population growth for species
entering a new environment: slow growth until some critical threshhold is reached, then rapid
growth until the “carrying capacity” of the environment is reached, upon which the growth
tails off.
How fast was the prairie dog population growing after 1000 days (about 33 months)? After
2200 days (about 6 years)?
12 600
9600
6600
3600
600
500
1000
2
1500
2000
2500
Answer: Since we’re trying to figure out how fast the prairie dog population is growing after
1000 days and after 2200 days, we need to compute p0 (1000) and p0 (2200). To do so, let’s first
compute p0 (t) for any time t. We’ll need to use the Chain Rule, so re-write p(t) = (q ◦ r)(t),
where
t − 1000
t
10
=
−
300
300
3
q(u) = 5000 arctan(u) + 7000.
r(t) =
Then
r0 (t) =
1
300
q 0 (u) = 5000 ·
1
5000
=
,
2
1+u
1 + u2
so the Chain Rule says that
p0 (t) = q 0 (r(t))r0 (t)
5000
p0 (t) =
· r0 (t)
1 + (r(t))2
5000
1
p0 (t) =
·
t−1000 2 300
1+
300
5000
p0 (t) =
h
t−1000 2
300
300 1 +
5000
p0 (t) =
300 +
(t−1000)2
300
i
.
Therefore, in particular,
5000
p0 (1000) =
300 +
(1000−1000)2
300
=
5000
50
= ,
300 − 0
3
and we conclude that the prairie dog population is growing at a rate of
dogs per day after 1000 days.
50
3
= 16.666 . . . prairie
Similarly,
5000
p0 (2200) =
300 +
(2200−1000)2
300
=
5000
300 +
12002
300
=
5000
50
=
5100
51
and we conclude that the prairie dog population is growing at a rate of 50
51 = 0.980 . . . prairie
dogs per day after 2200 days. Clearly, this is past the point of rapid growth and the population
is pretty stable.
3. You’re flying a kite at a constant height of 300 ft and notice that the kite is being carried
horizontally away from you at 25 ft/sec. When the string is 500 ft long, how fast do you need
to be letting out the string to keep up?
Answer: First of all, let’s draw a picture of the situation. Letting y(t) be the height of the
kite above the ground and letting x(t) be its horizontal distance from you, here’s the picture:
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sH
tL
yHtL
xHtL
In the above picture I’ve let s(t) be the length of the string.
Now, what do we know? First of all, by the Pythagorean Theorem we know that
x(t)2 + y(t)2 = s(t)2 .
Second, we know that the kite is always 300 feet in the air, so y(t) = 300 for all t, and we can
simplify the above as
x(t)2 + 3002 = s(t)2
Also, we know that the wind is blowing at a constant speed of 25 ft/sec, which means that
x0 (t) = 25
for all times t. Finally, we know that at some particular time t0 that we’re interested in, the
length of the string is 500 feet. In symbols, this means that
s(t0 ) = 500.
In particular, notice that this means that
x(t0 )2 + 3002 = 5002
and we can solve for x(t0 ):
x(t0 ) =
p
p
5002 − 3002 = 160, 000 = 400.
That exhausts what we know. What are we trying to figure out? Well, we need to know
how fast to let the string out, meaning how fast the length of the string is changing, which of
course is the derivative of s(t). Since we’re interested in this quantity at time t0 , we actually
want to figure out
s0 (t0 ).
To get there, of course, we can differentiate both sides of the equation
x(t)2 + 3002 = s(t)2 .
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This gives
d
d
x(t)2 + 3002 =
s(t)2
dt
dt
0
2x(t)x (t) + 0 = 2s(t)s0 (t)
using the Chain Rule. Therefore,
s0 (t) =
2x(t)x0 (t)
x(t)x0 (t)
=
2s(t)
s(t)
for any time t. In particular, at time t0 we have that
s0 (t0 ) =
x(t0 )x0 (t0 )
(400)(25)
10, 000
=
=
= 20.
s(t0 )
500
500
So at this moment in time you’d better be letting out the string at 20 ft/sec (which is pretty
fast!).
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