Math 113 Exam #2 Solutions
1. The equation
(4 − x)y 2 = x3 ,
determines a curve called a cissoid, pictured below. What is the equation of the tangent line to the
cissoid at the point (2, 2)?
3
2
1
0
1
2
3
4
-1
-2
-3
Answer: Differentiating the above equation on both sides yields
−1 · y 2 + (4 − x) · 2yy 0 = 3x2
or, equivalently,
(4 − x) · 2yy 0 = 3x2 + y 2 .
Solving for y 0 yields
y0 =
3x2 + y 2
.
2y(4 − x)
Therefore, at the point (2, 2), the slope of the tangent line is given by plugging in (x, y) = (2, 2) in the
above equation:
3(2)2 + 22
16
Slope =
=
= 2.
2(2)(2)
8
Therefore, the slope of the tangent line is 2, and so, by the point-slope formula, the tangent line is
given by
y − 2 = 2(x − 2)
or, equivalently,
y = 2x − 2.
2. Consider the function
f (x) =
√
5
sin x.
At which values of x does the graph of f have a vertical tangent line?
Answer: A line is vertical when its slope is infinite (either +∞ or −∞). Since the slope of the
tangent line to the graph is given by the derivative, the tangent line will be vertical when the derivative
approaches ±∞. Now, we compute the derivative using the Chain Rule:
f 0 (x) =
1
cos x
(sin x)−4/5 · cos x =
.
5
5(sin x)4/5
1
Clearly, this approaches ±∞ when sin x = 0, so the graph of the function will have a vertical tangent
line whenever sin x = 0, which happens when
x = nπ for any integer n.
3. Estimate tan(0.05) using an appropriate linearization.
Answer: Since 0.05 is close to zero and we know that
tan(0) =
sin(0)
0
= = 0,
cos(0)
1
it makes sense to approximate tan(0.05) by plugging in 0.05 to the linearization of f (x) = tan(x) at
zero. By definition, this linearization is
L(x) = f (0) + f 0 (0)(x − 0).
As already indicated, f (0) = 0. Now, f 0 (x) = sec2 x =
f 0 (0) =
1
cos x ,
so
1
1
= 1 = 1.
(cos 0)2
1
Therefore, the linearization is
L(x) = 0 + 1(x − 0),
or L(x) = x. Therefore,
tan(0.05) ≈ L(0.05) = 0.05.
2
4. Suppose g(x) = xx . What is g 0 (x)?
Answer: Let’s use logarithmic differentiation. Taking the natural log of both sides:
2
ln g(x) = ln xx = x2 ln x.
Now, differentiating both sides yields
1 0
1
g (x) = 2x ln x + x2 ,
g(x)
x
so
g 0 (x)
= x(2 ln x + 1).
g(x)
Hence,
g 0 (x) = g(x)x(2 ln x + 1).
2
Remember that g(x) = xx ; substituting that in gives the final answer
2
g 0 (x) = xx
+1
(2 ln x + 1).
5. Hydrogen peroxide (H2 O2 ) spontaneously decomposes into water and oxygen gas at a rate proportional
to the quantity of hydrogen peroxide. Suppose you start with 100L of hydrogen peroxide in a tank, and
after 1 week there are 60L of hydrogen peroxide in the tank. How many liters of hydrogen peroxide
will there be in the tank after 2 weeks?
Answer: Let H(t) denote the amount of hydrogen peroxide (in liters) in the tank after t weeks. Then
we’re told that
H 0 (t) = kH(t),
2
so we know that
H(t) = Cekt
for some constants C and k.
To find C, evaluate the function H at t = 0:
100 = H(0) = Cek·0 = Ce0 = C,
so we have that C = 100 and we can say that H(t) = 100ekt for some constant k. To solve for k,
evaluate at t = 1:
60 = H(1) = 100ek·1 = 100ek .
Thus,
ek =
so
3
60
= ,
100
5
3
k = ln .
5
Therefore, the quantity of hydrogen peroxide in the tank after t weeks is given by
t
3 t
3
3
.
H(t) = 100eln 5 ·t = 100 eln 5 = 100
5
The quantity of hydrogen peroxide is given by
H(2) = 100
2
3
9
= 36,
= 100 ·
5
25
so there are 36 liters of hydrogen peroxide in the tank after 2 weeks.
6. A spherical balloon is inflated by an electric pump. To prevent strain on the material, you want to
inflate the balloon in such a way that the surface area is increasing at a constant rate of 20 square feet
per minute. At what rate (in cubic feet per minute) should air be pumped into the balloon when the
radius of the balloon is 2 feet?
Answer: Recall that, if r(t) gives the radius of the sphere at time t, then the volume of the balloon
is given by
4
3
V (t) = π (r(t))
3
and the surface area of the balloon is given by
2
A(t) = 4π (r(t)) .
The statement of the problem tells us that A0 (t) = 20 for all t, and that, at the time t0 we’re interested
in, r(t0 ) = 2.
We’re asked to determine the rate of change of the volume of the balloon at time t0 , which is just
V 0 (t0 ). Notice that
4
V 0 (t) = π · 3(r(t))2 r0 (t) = 4π(r(t))2 r0 (t).
3
Therefore,
V 0 (t0 ) = 4π(r(t0 ))2 r0 (t0 ).
Substituting in r(t0 ) = 2 gives that
V 0 (t0 ) = 4π(2)2 r0 (t0 ) = 16πr0 (t0 ),
3
so the problem boils down to determining r0 (t0 ).
Since we know A0 (t0 ) and since r0 (t0 ) will appear in the expression for A0 (t0 ), to do so we need to
differentiate A(t):
A0 (t) = 4π · 2r(t)r0 (t) = 8πr(t)r0 (t).
Therefore, at time t0 ,
20 = A0 (t0 ) = 8πr(t0 )r0 (t0 ) = 8π · 2r0 (t0 ) = 16πr0 (t0 ),
so we can solve for r0 (t0 ):
r0 (t0 ) =
5
20
=
.
16π
4π
Therefore, using the expression for V 0 (t0 ) given above, we have that
V 0 (t0 ) = 16πr0 (t0 ) = 16π
5
= 20,
4π
so we should be pumping air in at 20 cubic feet per minute when the radius of the balloon is 2 feet.
4