Fall 2014 – MATH 151, Sections 549-551
Quiz #5 Solutions
Problem 1 (4 pts). Let s(t) =
p
√
3
1 + t. Find s0 (4).
√
3
0
Solution. The correct
answer
is
(d).
To
see
this,
we
first
find
s
(t).
If
we
let
f
(t)
=
t=
√
1
1
t 3 and g(t) = 1 + t = 1 + t 2 , then we can write s(t) = f g(t) . By the chain rule,
s0 (t) = f 0 g(t) g 0 (t)
so we need to find f 0 (t) and g 0 (t). Using the power law, we have
1
1 2
1
f 0 (t) = t− 3 = 2 = √
.
3 2
3
3 t
3t 3
For g 0 (t), we first differentiate 1 with respect to t and add our answer to the derivative
1
1
of t 2 with respect to t. Since 1 is a constant, its derivative is just 0. For t 2 , we use the
power law again and find that
1 1
1
1
g 0 (t) = t− 2 = 1 = √ .
2
2 t
2t 2
Substituting these expressions for f 0 (t) and g 0 (t) into
!
1
q
s0 (t) =
1
3
3 (1 + t 2 )2
so that
s0 (4) =
1
√
3
3 9
!
1
√
2 4
!
=
This is not one of the choices (a), (b) or (c).
1
1
√
3
3 9
the chain rule, we see that
!
1
√
2 t
!
1
4
!
=
1
√ .
12 3 9
Math 151 Fall 2014
Quiz #5 Solutions
Problem 2 (4 pts). Let f (x) =
√
2
x2 − 7x. Which of the following statements is false?
Solution. The correct answer is (c). To see this, first note that √
g(x) = x2 − 7x is a
polynomial, which is continuous everywhere. Moreover, h(x) = x is continuous on
[0, ∞). Since f (x) = h(g(x)), is is continuous every where it is defined. Since 0 and 7
are in the domain of f (x), then f (x) is continuous as these points. Thus (a) and (d) are
1
true. Next, we take the derivative of f (x). Writing f (x) = (x2 − 7x) 2 and using the
power law and the chain rule, we get
− 1
1
2x − 7
f 0 (x) = x2 − 7x 2 2x − 7) =
1 .
2
2(x2 − 7x) 2
When x = 0 or x = 7, the denominator is 0, so f 0 (x) is not defined at 0 or 7 – that is,
f (x) is not differentiable as 0 or 7. Hence (b) is true and (c) is false.
Problem 3 (6 pts). Suppose that xf (x) = cot(xf (x)). If f (2) = 6, what is f 0 (2)? [Hint:
First find f 0 (x) by implicit differentiation.]
Solution. The correct answer is f 0 (2) = −3. To get this, first we take the derivative of
both sides of
xf (x) = cot(xf (x))
with respect to x. For the right hand side, we use the product rule and implicit differentiation to get
0
xf (x) = f (x) + xf 0 (x).
For the left hand side, the chain rule gives:
0
0
cot xf (x)
= − csc2 xf (x) xf (x) .
0
Sbstituting the expression we found for xf (x) into the above equation yields
0
cot xf (x)
= − csc2 xf (x) f (x) + xf 0 (x) .
Putting the derivatives for the right and left hand sides together, we see that
f (x) + xf 0 (x) = − csc2 xf (x) f (x) + xf 0 (x)
= − csc2 xf (x) f (x) − x csc2 xf (x) f 0 (x).
Solving for f 0 (x) gives:
f (x) − csc2 xf (x) − 1
f 0 (x) =
x 1 + csc2 xf (x)
2
−f (x) 1 + csc xf (x)
=
2
x 1 + csc xf (x)
=
−f (x)
.
x
Math 151 Fall 2014
Quiz #5 Solutions
3
Finally, substituting x = 2 into the last equation above, and using f (2) = 6, we see that
f 0 (2) = −f2(2) = −6
= −3.
2
Problem 4 (4 pts). A spherical balloon is inflating with helium at a rate of 400π cubic
feet per minute. How fast is the balloon’s radius increasing the instant the radius is 5
feet? (The volume of a sphere is given by V = 43 πr3 .)
Solution. The correct answer is 4 ft. / min. First, note that the radius of the balloon
changes with time, so r = r(t) (i.e., r is a function of time). We are given in the statement
of the problem that dV
= 400π. Thus we should differentiate the volume formula with
dt
respect to time and set the derivative equal to 400π. To do this, we need to use implicit
differentitation. We have:
4
dr
dV
= π(3r2 )
dt
3
dt
dV
which, since dt = 400π gives
dr
400π = 4πr2 .
dt
Solving for dr
gives
dt
dr
400π
=
.
dt
4πr2
To find the rate at which the radius is increasing the instant the radius is 5 feet, we plug
r = 5 into the above equation to get
400π
400π
dr
=
=
= 4.
2
dt
4π(5 )
100π
Bonus Problem (2 pts). For which values of x is the piriform y 2 = x3 (2 − x) (see the
graph below) not differentiable?
Math 151 Fall 2014
Quiz #5 Solutions
4
Solution. The correct answer is x = 0, 2. The derivative at a point on the curve is the
slope of the tangent line to the curve at that point. So we need to look for places on the
graph where the slope of the tangent line is undefined. Probably the easiest place to see
that this occurs is at x = 2, where we have a vertical tangent line. The tangent line is
also undefined at x = 1, since there is a “cusp” (i.e., a kink in the graph much like the
graph of y = |x|) where there are infinitely many possible tangent lines to the point at
x = 0.