Final Exam Review Solution Sheet
Tom Sharland∗
May 13, 2013
1. Compute the following limits.
(a) limx→1
ex −e
sin(x−1) .
We first note that this is an indeterminate form of type 00 . This means we may apply
L’Hospital’s rule, and so, taking f (x) = ex − x and g(x) = sin(x − 1), we get f ′ (x) = ex
and g ′ (x) = cos(x − 1). This means
ex − e
f (x)
f ′ (x)
ex
e
= lim
= lim ′
= lim
= = e.
x→1 sin(x − 1)
x→1 g(x)
x→1 g (x)
x→1 cos(x − 1)
1
lim
1
(b) limx→0 (x3 + 2x + 1) x
1
This is an indeterminate form of type 1∞ . So let y = (x3 + 2x + 1) x , so that ln(y) =
3
ln(x3 +2x+1)
1
3
. Now limx→0 ln(x +2x+1)
is an indeterminate form of
x ln(x + 2x + 1) =
x
x
type 00 , so we can apply L’Hospital’s rule to it.
3x2 +2
3
3
x +2x+1
3x2 + 2
2
ln(x + 2x + 1)
= lim
= lim 3
= =2
lim ln(y) = lim
x→0
x→0 x + 2x + 1
x→0
x→0
x
1
1
Since limx→0 ln(y) = 2, it follows that limx→0 y = limx→0 eln(y) = e2 .
√
(c) limx→0+ ( x) ln(x)
This is an indeterminate form of type 0 · ∞. We need to convert it into a type
∞
∞ form before we can apply L’Hospital’s rule. So we write
lim
x→0+
√
ln(x)
x ln(x) = lim+ x→0
√1
x
which is now an indeterminate form of type ∞
∞ and so
√
1
√
ln(x)
−2x x
x
= lim
lim+ = lim+ = lim+ −2 x = 0.
+
−1
x
x→0
x→0
x→0
x→0
√1
3
x
2x 2
√
Hence we have shown limx→0+ x ln(x) = 0.
∗ with
thanks to Kevin Link for corrections and comments
1
0
0
or type
(d) limx→∞
x3
2x .
∞
. So we wheel out L’Hospital’s rule (a number
This is an indeterminate form of type ∞
d x
of times)! But first we recall that dx 2 = 2x ln(2) and take a deep breath. . .
3x2
6
6x
x3
=
lim
= lim x
=0
= lim x
x→∞ 2x ln(2)
x→∞ 2 (ln(2))3
x→∞ 2 (ln(2))2
x→∞ 2x
lim
2. Use derivative rules to compute f ′ (x) for each of the following functions.
(a) f (x) = e2x cos(x) + sin−1 (x).
The first term requires use of the product rule (and chain rule):
d
d 2x d
(cos(x)) + cos(x)
e
+
sin−1 (x)
dx
dx
dx
1
2x
2x
= e (− sin(x)) + cos(x) 2e
+√
1 − x2
1
= e2x (2 cos(x) − sin(x)) + √
.
1 − x2
f ′ (x) = e2x
(b) f (x) =
(product rule)
(chain rule)
x2 +3x4
x5 +2x .
We need to use the quotient rule.
′
f (x) =
=
=
=
=
d
d
(x5 + 2x) dx
x2 + 3x4 − (x2 + 3x4 ) dx
x5 + 2x
(x5 + 2x)2
(x5 + 2x)(2x + 12x3 ) − (x2 + 3x4 )(5x4 + 2)
(x5 + 2x)2
6
2
(2x + 4x + 12x8 + 24x4 ) − (5x6 + 15x8 + 2x2 + 6x4 )
x2 (x4 + 2)2
8
6
4
−3x − 3x + 18x + 2x2
x2 (x4 + 2)2
6
−3x − 3x4 + 18x2 + 2
.
(x4 + 2)2
(c) f (x) = xln(x) .
First we need to rewrite the equation as
2
f (x) = xln(x) = e(ln(x)) .
2
Now we need to carefully apply the chain rule
2 d
d (ln(x))2 e
= e(ln(x))
(ln(x))2
dx
dx
d
(ln(x))2
=e
(2 ln(x))
(ln(x))
dx
2
1
= e(ln(x)) (2 ln(x))
x
2 2 ln(x)
= e(ln(x))
x
2
ln(x)
= xln(x)
x
= (2 ln(x))xln(x)−1
2
(d) f (x) = etan(x ) .
This requires repeated application of the chain rule.
2
d
tan x2 etan(x )
dx
2
d
tan x2
= etan(x )
dx
d
tan(x2 )
x2
=e
sec2 x2
dx
tan(x2 )
2
2
=e
sec x 2x
f ′ (x) =
(e) f (x) = tan−1 (ex ).
Fortunately, this requires just one application of the chain rule!
1
d x
(e )
2
1 + (ex ) dx
ex
=
1 + e2x
f ′ (x) =
3. Set f (x) = 2x7 + 3x3 + x.
(a) Explain why f (x) has an inverse f −1 (x).
Let’s consider the derivative f ′ (x) = 14x6 + 9x2 + 1. We notice that f ′ (x) > 0 for
all values of x, and so f is increasing on (−∞, ∞). This means that it will pass the
horizontal line test and so must have an inverse.
(b) Compute (f −1 )′ (6). (Hint: f (1) = 6)
This is quite tricky, so first we need to think about the relationship between f (x) and
f −1 (x). Recall that, by definition of the inverse, we have
f f −1 (x) = x.
3
We differentiate both sides of this equation (the left hand side by using the chain rule),
to get
′
f ′ f −1 (x) f −1 (x) = 1;
which when rearranged gives
(f −1 )′ (x) =
1
f′
(f −1 (x))
.
We want to find (f −1 )′ (6), so we need to substitute x for 6 in the above equation. But
what is f −1 (6)? Fortunately, we are told that f (1) = 6, and so it follows that f −1 (6) = 1.
Furthermore, f ′ (1) = 14 + 9 + 1 = 24. So now we get
(f −1 )′ (6) =
1
1
1
= ′
=
.
f ′ (f −1 (6))
f (1)
24
4. The equation x3 y 5 + xy 3 = −10 defines y as an implicit function of x, i.e. y = f (x). Find an
equation for the tangent line of the graph of f (x) at the point (2, −1).
We need to differentiate the equation
x3 y 5 + xy 3 = −10
implicitly with respect to x. Recall that this means we consider y as a function of x, and so
if g is a function of y, then the chain rule tells us that
d
dg dy
dy
g(y) =
= g ′ (y) .
dx
dy dx
dx
In other words, we take the derivaitve of g with respect to y and then multiply it by
dy
ease of notation, I will henceforth write y ′ instead of dx
. So we differentiate:
dy
dx .
For
d
d
(−10)
x3 y 5 + xy 3 =
dx
dx
which gives us (through use of the product rule and the chain rule as mentioned above, with
g(y) = y 5 in the first term and g(y) = y 3 in the second term)
3x2 y 5 + 5y 4 y ′ x3 + y 3 + 3y 2 y ′ x = 0.
We now collect all the terms containing y ′ on one side, and move all the other terms to the
other side. This leaves us with
y ′ 5y 4 x3 + 3y 2 x = −3x2 y 5 − y 3
and so
3x2 y 5 + y 3
.
5y 4 x3 + 3y 2 x
The question wants us to find the tangent line at (2, −1), so we put x = 2 and y = −1 into
the above equation, to get
−12 − 1
13
dy 3(2)2 (−1)5 + (−1)3
=−
=
.
=−
dx x=2,y=−1
5(−1)4 (2)3 + 3(−1)2 (2)
40 + 6
46
y′ = −
4
So the tangent line is the line through (2, −1) with slope
y+1=
13
46 ,
so it has equation
13
(x − 2)
46
which in slope intercept form is
y=
72
13
36
13
x−
=
x− .
46
46
46
23
5. Related Rates Problems.
(a) A plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly
over a radar station. Find the rate at which the distance from the plane to the station
is increasing when it is 2 mi away from the station.
First, we draw a picture of the situation.
x
1
dx
dt
z
= 500
We use the following notation: x is the horizontal distance (in miles) travelled by the
plane and z is the distance (in miles) between the plane and the radar station, whilst t
will be used to denote time (in hours). Looking at the diagram, we see that x and y are
related by Pythagoras’ theorem; namely
x2 + 1 = z 2 .
The information given to us also tells us that
equation with respect to time gives
2x
and so
dx
dt
dx
dz
= 2z
dt
dt
dz
x dx
=
.
dt
z dt
5
= 500. Differentiating the above
We are asked to find the rate that z is changing when z = 2 (since z is the distance
2
2
between the plane and the radar
√ station). From the equation z = x + 1, we see that
when z = 2, we must have x = 3. So we can now calculate
√
√
3
dz =
· 500 = 250 3
√
dt z=2,x= 3
2
√
So the plane is moving away from the radar station at a rate of 250 3 ≈ 433.013 mi/h.
(b) A ladder 10 ft long rests against a vertical wall. If the lader slides away from the wall
at a rate of 1 ft/s, how fast is the angle between the ladder and ground changing when
the bottom of the ladder is 6 ft from the wall?
Again, we start with a picture.
10
y
θ
x
dx
dt
=1
We introduce the following notation: x is the distance from the wall to the ladder (in
feet), y is the height of the ladder on the wall1 (in feet), θ is the angle (in radians)
between the ladder and the ground and t is time (in seconds). We then see that
cos(θ) =
x
10
and so, differentiating both sides of this equation with respect to t (using the chain rule),
we see that
1
dθ
=
,
− sin(θ)
dt
10
from which it follows that
1
dθ
=−
.
dt
10 sin(θ)
We know that dx
dt = 1, but we need to find the value of sin(θ) when x = 6. But
6
when x = 6, we have cos(θ) = 10
= 53 and so since cos2 (θ) + sin2 (θ) = 1, we see that
1 It
turns out we don’t need y in this problem, but it doesn’t hurt to add the symbol just in case!
6
sin(θ) = 4/5 (this follows from the fact that {3, 4, 5} is a Pythagorean triple). So now
we get
1
dθ 1
=− .
=−
4
dt x=6
8
10 5
So the angle between the ladder and the ground is decreasing a rate of
second.
1
8
radians per
(c) A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3ft across
at the top and have a height of 1 ft. If the trough is being filled with water at a rate of
12 ft3 /min, how fast is the water level rising when the water is 6 inches deep?
Here, we start with a picture of the cross-section of the trough, remembering in the
meanwhile that it is 10 ft long.
3
w
1
h
dV
dt
= 12
Now we introduce the notation. Let h be the height of the water (in feet), w the width
of the water in the trough (in feet), V be the volume of water in the trough (in ft3 /min)
and t be the time (in minutes). We need to relate V and h – so how do we do that? We
notice that the water is in the shape of an “inverted prism”, with length 10 ft and cross
section which is an isosceles triangle with height h ft and w ft across. The formula for
the volume of a prism is its cross-sectional area times its length; in other words, we can
write
wh
V = 10
= 5wh.
(1)
2
This formula gives the volume V in terms of the variables w and h; we want a formula
for the volume just in terms of the variable h – in other words, we need to find a formula
relating h and w. Here we have to use the notion of similar triangles. By similar triangles,
we see that
3
w
=
⇔ w = 3h.
h
1
We now put this formula for w into equation (1), which gives us
V = 5 (3h) h = 15h2 .
7
We now differentiate this final equation with respect to t, and get
dh
dh
dV
= 15(2h)
= 30h ,
dt
dt
dt
which rearranges to
dh
1 dV
=
.
dt
30h dt
We need to find out dh
dt when the height of the water is 6 inches. . . which (to those of
you who are not so used to the imperial system of measurements) is 0.5 ft. So
1
12
4
dh =
12 =
= .
dt h=0.5
30(0.5)
15
5
So the water is rising at a rate of
6. Define
4
5
= 0.8ft/min.


−x
f (x) = sin(x)


(x − 3π)2 − π 2
if x < 0
if 0 ≤ x ≤ 2π
if x > 2π
(a) Find all critical numbers for f (x).
Recall that a critical number is a number c where f ′ (c) = 0 or f ′ (c) does not exist. For
x 6= 0, 2π, the derivative is given by


if x < 0
−1
′
f (x) = cos(x)
if 0 < x < 2π


2(x − 3π) if x > 2π.
We see that
• f (x) = −x has no critical numbers for x < 0, since f ′ (x) = −1.
• f (x) = sin(x) has critical numbers when f ′ (x) = cos(x) = 0, which are the points
x = π2 and x = 3π
2 in the interval (0, 2π).
2
• f (x) = (x − 3π) − π 2 has critical numbers when f ′ (x) = 2(x − 3π) = 0, which is
when x = 3π.
We now need to check at the points where the formulas change.
• At 0: The derivative from the left is −1 and the derivative from the right is cos(0) =
1, so the derivative does not exist, hence 0 is a critical number of f .
• At 2π: The derivative from the left is cos(2π) = 1 and the derivative from the right
is 2(2π − 3π) = −2π, so the derivative at 2π does not exist and so 2π is a critical
number of f .
In summary, the critical numbers of f are 0,
π 3π
2, 2 ,
2π and 3π.
(b) At each critical number c determine whether f (c) is a local minimum value, a local
maximum value or neither.
We handle each critical number in turn.
8
• c = 0: The derivative goes from negative to positive as x passes through 0, so f (0)
is a local minimum by the first derivative test.
• c = π2 : The second derivative at this point is − sin π2 = −1 and so f π2 is a local
maximum by the second derivative test.
3π
= 1 and so f 3π
is a local
• c = 3π
2 : The second derivative at this point is − sin 2
2
minimum by the second derivative test.
• c = 2π: The derivative goes from positive to negative as x passes through 2π, and
so f (2π) is a local maximum by the first derivative test.
• c = 3π: The second derivative at this point is f ′′ (3π) = 2 > 0, and so f (3π) is a
local minimum by the second derivative test.
(c) Draw a picture of the graph of f (x), indicating the coordinates of all points of intersection
with the x-axis and y-axis, and also indicating the points on the graph corresponding to
the critical numbers of f (x).
30
20
10
J , 1N
Π
2
2Π
Π
-10
0
-5
5
J
3Π
2
10
4Π
15
20
, -1N
I3 Π, -Π 2 M
-10
7. Set
4
f (x) = ex
+4x3 +4x2
.
Find the maximum value and the minimum value which f (x) takes on the closed interval
[−3, 1].
After a little bit of thinking, we see that f is a continuous function (it is the composition of
an exponential and a polynomial, which are continuous) on a closed interval, and so we can
9
use the closed interval method to find the absolute minimum and maximum for f . So first we
need to use the chain rule to compute f ′ (x), in order to locate the critical numbers of f .
4
3
2
d
x4 + 4x3 + 4x2 ex +4x +4x
dx
4
3
2
= 4x3 + 12x2 + 8x ex +4x +4x
f ′ (x) =
4
= 4x(x + 2)(x + 1)ex
4
3
+4x3 +4x2
.
2
Since ex +4x +4x > 0 for all values of x, the only points where f ′ (x) = 0 occur when x = 0,
x + 1 = 0 and x + 2 = 0, which is when x = 0, x = −1 and x = −2. Quick calculations yield
f (−2) = e(−2)
4
f (−1) = e(−1)
4
4
f (0) = e0
+4(−2)3 +4(−2)2
3
+4(−1) +4(−1)
+4(0)3 +4(0)2
2
= e16−32+16 = e0 = 1
= e1−4+4 = e1 = e
= e0+0+0 = e0 = 1
We now need to find the values taken by f at the endpoints of the interval [−3, 1], which are
the values f (−3) and f (−1). Again, quick calculations yield
f (−3) = e(−3)
f (1) = e(1)
4
4
+4(−3)3 +4(−3)2
+4(1)3 +4(1)2
= e81−108+36 = e9
= e1+4+4 = e9
Hence, by the closed interval method, for the function f (x) on [−3, 1], the absolute maximum
is f (−3) = f (1) = e9 and the absolute minimum is f (−2) = f (0) = 1. As a check, the graph
of the function is exhibited below.
40
30
20
10
-3
-2
-1
10
1
8. Optimisation Problems.
(a) If 1200 cm2 of material is available to make a box with a square base and an open top,
find the largest possible volume of the box.
We will label the side length of the box by ℓ and the height of the box by h. We also
denote the area by A and the volume by V . Below is a diagram of the box before it is
constructed, as well as the formulae for the surface area of the box and its volume.. The
ℓ
ℓ
h
Area = ℓ2 + 4ℓh = 1200
Volume = ℓ2 h
surface area of the box is We want to maximise V , but the equation for V contains both
the variables h and ℓ. So we need to find a relationship between h and ℓ. Fortunately,
the formula for the area gives such a relationship, so we find
1200 = ℓ2 + 4ℓh
⇔
1200 − ℓ2 = 4ℓh
⇔
1200 − ℓ2
= h.
4ℓ
We now substitute this into the equation for V to get
ℓ(1200 − ℓ2 )
ℓ3
1200 − ℓ2
2
2
=
= 300ℓ − .
V =ℓ h=ℓ
4ℓ
4
4
So we have an equation for V = V (ℓ) in terms of ℓ, where the domain of ℓ is [0, ∞). We
11
now differentiate this formula with respect to ℓ to get
3ℓ2
dV
= 300 −
dℓ
4
dV
and so we see that dV
dℓ = 0 when ℓ = 20. Furthermore, since dℓ > 0 for ℓ < 20 and
dV
dℓ < 0 for ℓ > 20, it follows from the first derivative test for absolute extreme values
that V (20) is an absolute maximum. Putting the value of ℓ = 20 back into the formula
for the h, we see that
1200 − 202
800
h=
=
= 10.
4(20)
80
Therefore, the value of V when ℓ = 20 and h = 10 is
V = 202 × 10 = 4000cm3 .
(b) Find the points on the ellipse 4x2 + y 2 = 4 that are farthest away from the point (1, 0).
y
(x, y)
y
d
x
1−x
(1, 0)
For a point (x, y), let d denote its distance from the point (1, 0). Looking at the picture,
Pythagoras’ theorem tells us that
d2 = (1 − x)2 + y 2 .
12
We want to find the points (x, y) on the ellipse which maximise the distance to (1, 0).
However, it will be more convenient (in terms of the formulae used) to find the points
which maximise D = d2 .2 At the moment, we have D in terms of both x and y, so first
we need to find a relationship between x and y. We are told that the point (x, y) lies on
the ellipse 4x2 + y 2 = 4, and so y 2 = 4(1 − x2 ). So now we can write
D = (1 − x)2 + y 2 = (1 − x)2 + 4(1 − x2 ) = (1 − 2x + x2 ) + (4 − 4x2 ) = 5 − 2x − 3x2 .
So now D = D(x) is a function of x, with x in [−1, 1]. So now we find the absolute
maximum value of D.
dD
1
= −2 − 6x = 0 when x = − .
dx
3
1
The value D − 3 is the absolute maximum of D, since D′′ (x) = −6 so D is concave
downward on [−1, 1]. Putting x = − 13 into the equation for the ellipse, we get
√
2 !
1
32
4 2
1
2
=4 1−
=
⇒y=±
.
y =4 1− −
3
9
9
3
√ So the points which maximise the distance to (1, 0) are (x, y) = −y 13 , ± 4 3 2 .
(c) A cylindrical can without a top is made to contain V cm3 of liquid. Find the dimensions
that will minimise the cost of the metal to make the can.
r
h
Volume = V = πr2 h
Area = A = πr2 + 2πrh
2 you should convince yourself that any point which maximises d will also maximise d2 , so this will not affect our
solutions
13
This problem seems more difficult, as there are no actual numbers involved3 . We use the
notation as shown in the diagram: h is the height of the can, r is the radius of the can,
A is the surface area and V is the volume. We want to minimise A, for a fixed value of
V . Currently, the equation for A contains both h and r, so we need to remove one of
them from this equation by using the equation for V .
V = πr 2 h
=⇒
h=
V
.
πr 2
We now substitute this formula for h into the equation for A:
V
2V
A = πr 2 + 2πrh = πr 2 + 2πr
= πr 2 +
.
πr 2
r
So now we have A = A(r) as a function of the radius r, with r in (0, ∞). So we
differentiate
r
2V
V
dA
3 V
3
= 2πr − 2 = 0 when r =
⇒r=
.
dr
r
π
π
We now need to find the corresponding value of h, so we insert this value of r into the
equation for h to get
r
V
3 V
h = q 2 =⇒ h =
= r.
π
π 3 Vπ
So in order to minimise the manufacture of the can, we need to make the height equal
to the radius (we did such an example in class with V = 500).
9. Find all the antiderivatives for each of the following functions f (x).
(a) f (x) = cos(x).
Since
d
dx
sin(x) = cos(x), the antiderivatives of f (x) = cos(x) are F (x) = sin(x) + C.
(b) f (x) = 3x2 − 8x + 1.
d
(x3 − 4x2 + x) = 3x2 − 8x + 1, the antiderivatives of f (x) = 3x2 − 8x + 1 are
Since dx
F (x) = x3 − 4x2 + x + C.
(c) f (x) = x2 .
2
d
1
Recall that dx
|x| = ln(x). Hence the antiderivatives of f (x) = x is ln(x) + C on any
interval not containing 0. Thus the most general antiderivatives are given by the formula
(
ln(−x) + C1 if x < 0
F (x) =
ln(x) + C2
if x > 0.
(d) f (x) = xn , if n 6= −1.
3 Actually,
I prefer it that way, as it makes calculations a lot easier!
14
By the power law, we see that a general antiderivative of f (x) = xn (for n 6= −1) is given
by
xn+1
F (x) =
+ C.
n+1
If n ≥ 0, this is valid on the whole of (−∞, ∞),
interval not containing 0, so if n is negative then
( n+1
x
+ C1
F (x) = xn+1
n+1
n+1 + C2
whereas if n < 0, this is valid on any
a general antiderivative is
if x < 0
if x > 0.
(e) f (x) = ax , for a > 0.
Recall that
d x
dx a
= ax (ln(a)) and so the antiderivatives of f (x) = ax are
F (x) =
ax
+ C.
ln(a)
10. Acceleration Problems.
(a) A particle moves with the acceleration function
a(t) = 5 + 4t − 2t2 .
It’s initial velocity is v(0) = 3 m/s and its initial displacement is s(0) = 10 m. Find its
position after t seconds.
Recall that the derivaitve of the velocity v(t) is a(t), and the derivative of the position
function s(t) is the velocity function v(t). So first we calculate the velocity function.
The antiderivative of a(t) is
2
v( t) = 5t + 2t2 − t3 + C.
3
We know that v(0) = 3 and so 3 = v(0) = C, hence the equation for v(t) is
2
v(t) = 3 + 5t + 2t2 − t3 .
3
Now we calculate the position function by finding the antiderivative of v(t). This gives
5
2
1
s(t) = 3t + t2 + t3 − t4 + D,
2
3
6
and since s(0) = D, and we are told that s(0) = 10, so D = 10. Thus the position of the
particle after t seconds is
2
1
5
s(t) = 10 + 3t + t2 + t3 − t4 .
2
3
6
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(b) Show that for motion in a straight line with constant acceleration a, initial velocity v0 ,
and initial displacement s0 , the displacement after time t is
s=
1 2
at + v0 t + s0 .
2
Hurrah! Another question without numbers! This is not as bad as it seems. First, we
know that the acceleration function is a(t) = t and so, by computing its antiderivative,
we find that
v(t) = at + C.
We are told that v(0) = v0 , and so C = v0 , thus
v(t) = at + v0 .
Now we find the antiderivative of this to find the formula for s(t). This gives
s(t) =
1 2
at + v0 t + D,
2
and since the initial displacement is s(0) = s0 , we see that D = s0 . This gives the
formula
1
s = s(t) = at2 + v0 t + s0
2
as required.
16