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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
y = x + ln 2 + ln (2x) + arctan 2
2
x
2
2 x
y = x + ln 2 + ln (2x) + arctan 2
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 =
y = x · arctan (5x)
y = x · arctan (5x)
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
)
y = ln(tan(
cos(x
2
)
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 =
x + e
π
cos 4 + sin
5
(6x)
y =
x + e
π
cos 4 + sin
5
(6x)
y =
x + e
π
cos 4 + sin
5
(6x)
Note that e
π
and cos 4 are constants.
dy
dx
(1)(cos 4 + sin
5
(6x)) − (x + e
π
)(5 sin
4
(6x) · cos(6x) · 6)
(cos 4 + sin
5
(6x))
2
Recall that sin
5
(6x) = (sin(6x))
5
- 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) Verify that the point (1,2) is on the curve above.
We must check to see if the values x = 1 and y = 2 satisfy the equation above.
x
3
3
?
xy
3
3
?
?
Thus, the point (1,2) is on the curve.
(c) 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 +
- Evaluate the following limits.
Throughout this solution, the symbol F will stand for whatever notation your instructor prefers for
using L’Hopital’s Rule on the indeterminate form 0/0; this may be
“0/0”
= or
L
′
H
= or
H
= or = “0/0” or
“has the form ‘
’ and so, by L’Hopital’s Rule, is equal to” or something else. The symbol ♥ will serve
the same purpose for the indeterminate forms ∞/∞ and −∞/∞.
(a) lim
x→ 1
x
3
7 − 7 x
lim
x→ 1
x
3
7 − 7 x
lim
x→ 1
x
3
7 − 7 x
F lim
x→ 1
3 x
2
(b) lim
x→ 0
1 − cos 2x
x
lim
x→ 0
1 − cos 2x
x
lim
x→ 0
1 − cos 2x
x
= 0 Can’t use (and don’t need) L’Hopital’s Rule!
(c) lim
x→ 0
x
2
ln x
lim
x→ 0
x
2
ln x lim
x→ 0
x
2
ln x
This is of the indeterminate form 0 · (−∞) so we rewrite the function as a fraction in order to use
L’Hopital’s Rule.
lim
x→ 0
x
2
ln x = lim
x→ 0
ln x
1
x
2
♥ lim
x→ 0
1
x
− 2
x
3
= lim
x→ 0
x
x
3
= lim
x→ 0
x
2
(d) lim
x→ 0
1 − cos 4x
5 x
2
lim
x→ 0
1 − cos 4x
5 x
2
lim
x→ 0
1 − cos 4x
5 x
2
F lim
x→ 0
4 sin 4x
10 x
F lim
x→ 0
16 cos 4x
(e) lim
x→∞
x
2
x
lim
x→∞
x
2
x
lim
x→∞
x
2
x
♥ lim
x→∞
2 x
ln 2 · 2
x
♥ lim
x→∞
ln 2 · ln 2 · 2
x
- 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(arccos x)
tan(arccos x) tan(arccos x) (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
- How would your answers to the previous question change if the domain of f
f f were [− 10 , 10]
[− 10 , 10]
[− 10 , 10]?
There would be a global maximum at (10, 10
4
e
10
). (And the graph would be restricted to − 10 ≤ x ≤
- Use Newton’s Method with an initial guess of x 0
x
0
x 0
= − 1 to find the next three approxima-
tions to a solution of x
3
x
3
3
- x + 1 = 0. Then test your final approximation to see if it appears
to be close to a root.
Recall that x
n+
= x
n
f(x n
f
′
(x
n
= x
n
x
3
n
3 x
2
n
Using a calculator with initial guess x 0
= −1, we get the following.
x
0
x 1
x
2
x
3
Check: (− 0 .68233)
3
- (− 0 .68233) + 1 ≈ 0 .0000526, so our approximation seems to be a good one.
- The rate of change of a population PPP (((ttt))) 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 ttt = 0= 0= 0,
the population is 10000.
(a) Write a differential equation whose solution is PPP (((ttt))).
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 planning to build a box-shaped aquarium with no top and with two square ends.
Your budget is $288. If the glass for the sides costs $12 per square foot and the opaque
material for the bottom costs $3 per square foot, what dimensions will maximize the
volume? Be sure to show how you know you have found the maximum.
x
x
y
Goal : Maximize volume
Objective function: volume = V = x · x · y = x
2
y
We need to get this down to a function of just one variable, so we use the constraint equation:
total cost =(cost of base) + (cost of two square ends) + (cost of two other sides)
288 = 3 xy + 12 · 2 x
2
288 = 27xy + 24x
2
288 − 24 x
2
= 27xy
288 − 24 x
2
27 x
= y
Substituting this back into the objective function gives
V = x
2
y = x
2
288 − 24 x
2
27 x
= x ·
288 − 24 x
2
(288x − 24 x
3
Now that we have V as a function of just one variable, we find its maximum.
V
′
(x) =
(288 − 72 x
2
(288 − 72 x
2
0 = (288 − 72 x
2
72 x
2
x
2
x = 2 We discard x = −2 because lengths must be nonnegative.
Since V
′
is positive for x < 2 and negative for 2 < x, we know that the maximum occurs at x = 2.
And y =
288 − 24 x
2
27 x
2
, so the dimensions are 2 by 2 by
