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Math 105: Review for Exam II - Solutions
- Find dy/dx
dy/dx dy/dx for each of the following.
(a) y = x
2
x
2
2 x
- ln 2 + ln (2x) + arctan 2
dy
dx
= 2x + (ln 2)
x
2 x
2 x
· 2 Note that e
2
, ln 2, and arctan 2 are constants.
(b) y =
x · arctan (5x)
dy
dx
x
− 1 / 2
arctan(5x) +
x ·
1 + (5x)
2
arctan(5x)
2 x
1 / 2
x
1 + 25x
2
(c) y = ln(tan(
cos(x
2
)
dy
dx
tan(
cos(x
2
)
· sec
2
cos(x
2
)
) · ln 2(
cos(x
2
)
) · (− sin(x
2
)) · 2 x
(d) y = sin
5
x + e
π
ln 4 + arcsin 6x
Note that sin
5
w = (sin w)
5
dy
dx
= 5 sin
4
x + e
π
ln 4 + arcsin 6x
·cos
x + e
π
ln 4 + arcsin 6x
(1)(ln 4 + arcsin 6x) − (x + e
π
1
1 −(6x)
2
(ln 4 + arcsin 6x)
2
- Consider the curve defined by x
3
3
xy
x
3
3
xy x
3
3
xy (known as the Folium of Descartes).
(a) Find dy/dx
dy/dx dy/dx. Use implicit differentiation.
3 x
2
2
dy
dx
y +
x
dy
dx
3 y
2
dy
dx
x
dy
dx
y − 3 x
2
dy
dx
3 y
2
x
y − 3 x
2
dy
dx
9
2
y − 3 x
2
3 y
2
9
2
x
(b) Find the equation of the tangent line at the point (1,2).
We want y = mx + b.
m =
9
2
2
2
9
2
, so y =
x + b.
Now plug in x = 1 and y = 2 to find b.
· 1 + b ⇒
= b
Therefore, we have y =
x +
- Suppose the domain of f(x)
f(x) f(x) is all reals and that f
f f has an inverse function f
− 1
(x)
f
− 1
(x) f
− 1
(x). Further,
suppose that f(2) = 5
f(2) = 5 f(2) = 5 and f
′
(2) = e
f
′
(2) = e f
′
(2) = e. Finally, let h(x) = 1/f(x)
h(x) = 1/f(x) h(x) = 1/f(x).
(a) What point must be on the graph of f
− 1
f (x)
− 1
f (x)
− 1
(x)?
Since f(2) = 5, we also know that f
− 1
(5) = 2; therefore, the point (5,2) is on the graph of f
− 1
(x).
(b) What point must be on the graph of hhh(((xxx)))?
Since f(2) = 5, we also know that h(2) = 1/5; therefore, the point (2,1/5) is on the graph of h(x).
The moral of the story in parts (a) and (b) is that inverses and reciprocals are not the same.
(c) Give an example of a point that cannot be on the graph of f(x)
f(x) f(x). Do not choose a
point with xxx-value of 2.
Since f
− 1
(x) is a function, we know that for each of its inputs (such as 5) there is only one output
(in this case, 2). This means that no point of the form (5, k) can be on the graph of f
− 1
(x)
(except for k = 2). Therefore, no point of the form (k, 5) can be on the graph of f(x) (except
for k = 2). [Note: this is the same as saying that f must be a one-to-one function; on f, no two
x-values correspond to the same y-value.]
(d) What is the value of the derivative of h(x)
h(x) h(x) at x = 2
x = 2 x = 2?
By the Quotient Rule (or Chain Rule if you prefer), h
′
(x) =
0 · f(x) − 1 · f
′
(x)
[f(x)]
2
f
′
(x)
[f(x)]
2
So, h
′
f
′
[f(2)]
2
e
2
e
- Suppose that y = f(t)
y = f(t) y = f(t) is a solution to the differential equation y
′
π
arcsin t + y
2
y
′
π
arcsin t + y
2
y
′
π
arcsin t + y
2
and that
f
f
f
. Find the equation of the tangent line to fff at
We want y = mx + b.
m =
π
arcsin
2
π
π
, so y =
x + b.
Now plug in x =
and y =
to find b.
= b
Therefore, we have y =
x +
- Find the following.
(a) an antiderivative of y =
1 − 9 x
2
3
3
y =
1 − 9 x
2
3
3
y =
1 − 9 x
2
3
3
5 arcsin 3x
x
4
sin 2x
3
x + C
(b) tan(arccostan(arccostan(arccos xxx))) (rewritten as an algebraic expression - no trigonometric functions)
Let θ = arccos x. That is, θ is the angle whose cosine is x.
x
y
θ
x
2
2
2
⇒ y =
1 − x
2
tan(arccos x) = tan θ =
opposite
adjacent
y
x
1 − x
2
x
- Circle always, sometimes, or never to make each statement below correct.
(a) If f
′
f
′
f
′
(1) = 0 then f
f f always /sometimes/never has a critical point at x = 1
x = 1 x = 1.
A critical point is where f
′
is 0 or undefined.
(b) If f
′
f(2) = 0
′
f(2) = 0
′
(2) = 0 then fff always/ sometimes /never has a local maximum or local minimum
at xxx = 2= 2= 2.
f might instead have a terrace point at x = 2; we need f
′
to change sign at x = 2 in order to
guarantee a local extremum there.
(c) If xxx = 3= 3= 3 is a critical point of fff , then f
′
f(3)
′
f(3)
′
(3) is always/ sometimes /never 0.
It may also be that f
′
(3) is undefined.
(d) If f
′′
f
′′
f
′′
(4) = 0, then f
f f always/ sometimes /never has an inflection point at x = 4
x = 4 x = 4.
We need f
′′
to change sign at x = 4 to guarantee an inflection point there.
For example, if f(x) = (x−4)
4
then f
′′
(4) = 0 but f has a local minimum rather than an inflection
point at x = 4.
However, if f(x) = (x − 4)
3
then f
′′
(4) = 0 and f does have an inflection point at x = 4.
Also see what happens at x = 0 in problem 7(d).
(e) If fff has a global maximum at xxx = 5= 5= 5, then f
′
f(5)
′
f(5)
′
(5) is always/ sometimes /never 0.
f
′
(5) might also be undefined, or x = 5 might be an endpoint of the domain.
(f) If f
′
f(6) = 0
′
f(6) = 0
′
(6) = 0 and f
′′
f(6) = − 2
′′
f (6) = − 2
′′
(6) = − 2 , then fff always /sometimes/never has a local maximum at
x = 6
x = 6 x = 6.
If f is concave down with a horizontal slope at x = 6, then f must have a local maximum there.
(g) If f
′
f(7) = 0
′
f(7) = 0
′
(7) = 0 and f
′′
f (7) = 0
′′
f(7) = 0
′′
(7) = 0, then fff always/ sometimes /never has a local extremum at
xxx = 7= 7= 7.
This means that the second derivative test is inconclusive, so you need to use a different test.
For example, if f(x) = (x − 7)
4
then f
′
(7) = 0 and f
′′
(7) = 0 and f has a local minimum at
x = 7.
However, if f(x) = (x − 7)
3
then f
′
(7) = 0 and f
′′
(7) = 0 and f has an inflection point but not a
local extremum at x = 7.
- The rate of change of a population P (t)
P (t) P (t) of eels is proportional to the size of the population.
When the population is 40000, it is growing at a rate of 400 eels per year. At time t = 0
t = 0 t = 0,
the population is 10000.
(a) Write a differential equation whose solution is P (t)
P (t) P (t).
Rate of change (P
′
) is (=) proportional to (k) size of population (P ) means P
′
= kP.
What’s the value of k? When P = 40000, we know P
′
= 400. That is, 400 = k · 40000, so k = .01.
Thus, we have P
′
=. 01 P.
(b) Solve your differential equation.
The general solution is P (t) = Ae
. 01 t
What’s the value of A? When t = 0, we know P = 10000. That is, 10000 = Ae
0
= A, so
A = 10000.
Thus, we have P (t) = 10000e
. 01 t
(c) When will the population reach 60000?
60000 = 10000e
. 01 t
6 = e
. 01 t
ln 6 = ln e
. 01 t
Take ln of each side.
ln 6 =. 01 t Recall that ln e
z
= z.
t = 100 ln 6 ≈ 179 .176 years
- You are designing an 18 ft
3
3 3
box that will have a square bottom and no top. The material
for the bottom costs 40 cents per square foot and the material for the sides costs 30 cents
per square foot. What dimensions give the least total cost? Be sure to show how you
know you have found the minimum.
x
x
y
Goal: minimize cost
Objective function: cost = C = 40 · x
2
We need to get this down to a function of just one variable, so we use the
constraint equation: volume = 18 = x
2
y.
Solving for y, we have y =
x
2
Substituting this back into the objective function gives C = 40 · x
2
x
2
= 40x
2
x
Now that we have C as a function of just one variable, we find its minimum.
C
′
(x) = 80x −
x
2
0 = 80x −
x
2
x
2
= 80x
= x
3
3 = x
Since C
′
is negative for 0 < x < 3 and positive for 3 < x, we know that the minimum occurs at x = 3.
And y =
x
2
2
= 2, so the dimensions are 3 by 3 by 2.
