Mathematics 102 Review Questions
Problem 1: Multiple Choice Questions
1: Consider the function y = f (x) = 3e−2x − 5e−4x
(a) The function has a local maximum at x = (1/2) ln(10/3)
(b) The function has a local minimum at x = (1/2) ln(10/3)
(c) The function has a local maximum at x = (−1/2) ln(3/5)
(d) The function has a local minimum at x = (1/2) ln(3/5)
(e) The function has a local maximum at x = (−1/2) ln(3/20)
Solution to 1: For y = 3e−2x −5e−4x , the derivative is f ′ (x) = −6e−2x +20e−4x . The critical point is
at x = (1/2) ln(10/3). The second derivative is f ′′ (x) = 12e−2x − 80e−4x , which is negative. (At the
critical point, f ′′ = −18/5) so the correct answer is that there is a local maximum x = (1/2) ln(10/3).
Answer (a).
2: Let m1 be the slope of the function y = 3x at the point x = 0 and let m2 be the slope of the
function y = log3 x at x = 1 Then
(a) m1 = ln(3)m2 (b) m1 = m2 (c) m1 = −m2 (d) m1 = 1/m2 (e) m1 = m2 / ln(3)
Solution to 2: The functions are inverses and the points (0, 1) and (1, 0) are the images of one
another about the line y = x. Thus the slopes are reciprocals so that m1 = 1/m2 . Answer (d)
3: Consider the curve whose equation is x4 + y 4 + 3xy = 5. The slope of the tangent line, dy/dx,
at the point (1, 1) is
(a) 1 (b) -1 (c) 0 (d) -4/7 (e) 1/7
Solution to 3: Using implicit differentiation on x4 +y 4 +3xy = 5 leads to 4x3 +4y 3 y ′ +3(xy ′ +y) = 0
at (x, y) = (1, 1) this leads to 4 + 4y ′ + 3(y ′ + 1) = 0 so that y ′(4 + 3) = −(4 + 3) and y ′ = −1.
Answer (b)
4: Two kinds of bacteria are found in a sample of tainted food. It is found that the population size
of type 1, N1 and of type 2, N2 satisfy the equations
dN1
= −0.2N1 ,
dt
N1 (0) = 1000,
dN2
= 0.8N2 ,
dt
N2 (0) = 10.
Then the population sizes are equal N1 = N2 at the following time:
(a) t = ln (40) (b) t = ln (60) (c) t = ln (80) (d) t = ln (90) (e) t = ln (100)
Solution to 4:
The differential equations are of the form
dN1
= −k1 N1 ,
dt
dN2
= k2 N2 ,
dt
N1 (0) = N1,0
N2 (0) = N2,0
1
The solutions are
N1 (t) = N1,0 e−k1 t ,
N2 (t) = N2,0 ek2 t
The populations are equal when
N1,0 e−k1 t = N2,0 ek2 t
N1,0
= e(k2 +k1 )t
N2,0
tequal
N1,0
1
ln
=
(k2 + k1 )
N2,0
!
N1,0 = 1000, N2,0 = 10, k1 = 0.2, k2 = 0.8
so tequal = ln (100). Answer (e)
5: In a conical pile of sand the ratio of the height to the base radius is always r/h = 3. (Recall
that the volume of a cone with height h and radius r is V = (π/3)r 2 h.) If the volume is increasing
at rate 3 m3 /min, how fast (in m/min) is the height changing when h = 2m?
(a) 1/(12π) (b) (1/π)1/3 (c) 27/(4π) (d) 1/(4π) (e) 1/(36π)
Solution to 5: V = (π/3)r 2 h, and r = 3h so V = (π/3)(3h)2 h = 3πh3 . Thus V ′ (t) = 3π · 3h2 h′ (t).
Plugging in V ′ (t) = 3, h = 2 and rearranging leads to 3 = 3π · 3(22 )h′ (t) so h′ (t) = 1/(12π) m /min.
Answer (a)
6: Newton’s Law of cooling leads to a differential equation that predicts the temperature T (t) of an
object whose initial temperature is T0 in an environment whose temperature is E. The predicted
temperature is given by T (t) = E + (T0 − E)e−kt where t is time and k is a constant. Shown in
Fig 1 on the following page is some data points plotted as ln(T (t) − E) versus time in minutes. The
ambient temperature was E = 22◦ C. Also shown on the graph is the line that best fits those 11
points. According to this graph, the value of the constant k is approximately
(a) -1/27 (b) e1/27 (c) 1/27 (d) 4/27 (e) ln(1/27)
Solution to 5: The ambient temperature is E = 22◦ C. Since T (t) = E + (T0 − E)e−kt , we can
write (T (t) − E) = (T0 − E)e−kt . Thus ln(T (t) − E) = ln(T0 − E) + ln(e−kt ) = ln(T0 − E) − kt.
Thus the intercept is ln(T0 − E) and the slope is −k.
We see that −k ≈ rise/run = (4 − 3)/(0 − 27) so that k ≈ 1/27. Thus, the correct answer is
k ≈ 1/27. Answer (c).
2
ln ( T(t) - E )
4.1
4.0
3.5
Bestfitline
3.0
20
10
time in minutes
2.9
0.0
30.0
Figure 1: Figure for Multiple Choice problem 6
3
Long Answer Problems
Problem 2a: Fig. 2 shows a 1 km race track with circular ends. Find the values of x and y that
will maximize the area of the rectangle.
y
x
Figure 2: This shape is investigated in both problems 1 and 2.
Solution
The perimeter of the race track consists of the two straight segments plus the two semi-circular
ends. The radius of the circular ends is y/2, so that the total perimeter, which is constant (and
serves as a constraint for the problem) is
P = 2x + πy = 1km
The area of the rectangle is A = xy. We can eliminate one of the variables using x = (1/2)(P − πy)
so that
1
1
A(y) = y(P − πy) = (P y − πy 2)
2
2
The graph of A(y) is a parabola opening downwards, so that we know that a maximum (not a
minimum) will be found. The critical point is at y such that
1
A′ (y) = (P − 2πy) = 0,
2
y=
P
.
2π
The value of x can then be found from
1
P
1
1
x = (P − πy) = (P − ) = P.
2
2
2
4
Problem 2b: Now suppose that Fig 2 shows the shape of a leaf of some plant. If the plant grows
so that x increases at the rate 2 cm/year and y increases at the rate 1 cm/year, at what rate will
the leaf’s entire area be increasing?
Solution
The area of the shape shown in Fig 2 is the area of the rectangle plus two semi-circles. The
radius of the semi-circles is y/2, so that
Aleaf = xy + π
4
y2
.
4
The rate of change of the area per unit time is
y(t)2
d
dAleaf
x(t)y(t) + π
.
=
dt
dt
4
!
Using the chain rule and product rule leads to
dAleaf
2y(t) ′
= x′ (t)y(t) + x(t)y ′ (t) + π
y (t).
dt
4
SImplifying and using the information in the problem leads to
πy(t)
dAleaf
= 2y(t) + x(t) +
.
dt
2
(If we are given the values of x and y we can find the numerical value for this rate of change.)
Problem 3: Find the dimensions of the largest rectangle that can fit exactly into a circle whose
radius is 10 cm.
Solution
Problem 4: A cell of the bacterium E.coli has the shape of a cylinder with two hemispherical
r
h
Figure 3: Shape of the object described in Problem 4. Note: Useful volumes and surface areas: For
a hemisphere, V = (2/3)πr 3, S = 2πr 2. For a cylinder, V = πr 2 h and S = 2πrh (not including end
caps)
caps, as shown in Fig 3. Consider this shape, with h the height of the cylinder, and r the radius of
the cylinder and hemispheres. (a) Find the values of r and h that lead to the largest volume for a
fixed constant surface area, S= constant. (b) Describe or sketch the shape you found in (a). (c) A
typical E. coli cell has h = 1µm and r = 0.5µm. Based on your results in (a) and (b), would you
agree that E. coli has a shape that maximizes its volume for a fixed surface area? (Explain your
answer).
Solution The volume of the shape is the same as the volume of two hemispheres (i.e. one sphere)
and a cylinder, i.e.
4
V = πr 3 + πr 2 h.
3
5
The surface area, which is constant is
S = 4πr 2 + 2πrh.
This allows us to eliminate h from the volume; using
h=
we get
S − 4πr 2
,
2πr
S − 4πr 2
4
.
V = πr 3 + πr 2
3
2πr
!
4
r
2
r
V = πr 3 + S − 2πr 3 = − πr 3 + S .
3
2
3
2
′
To find critical points, set V (r) = 0.
S
V ′ (r) = −2πr 2 + = 0,
2
1
⇒r=
2
s
S
π
For this radius, we check what type of critical point it is by finding the second derivative,
V ′′ (r) = −4πr < 0.
Since the second derivative is negative, the critical point is a local maximum. The height is then
h=
S − 4πr 2
= 0.
2πr
This means that the shape that has largest volume for the given surface area is actually a sphere,
with no cylinder at all.
Problem 5: If the cell shown above in Fig 3 is growing so that the height increases twice as fast
as the radius. If the radius is growing at 1 µm per day at what rate will the volume of the cell
increase? (Leave your answer in terms of the height and radius of the cell.)
Solution
The volume of the cell is the sum of the cylindrical and spherical parts. The radius of both the
cylinder and the sphere is r so
4
V = πr 2 h + πr 2 .
3
The rate of increase of the volume is then
4
d
dV
πr(t)2 h(t) + πr(t)3 .
=
dt
dt
3
Using the product and chain rules we get
4
dV
= π[2r(t)r ′ (t)h(t) + r(t)2 h′ (t)] + 3πr(t)2 r ′ (t) .
dt
3
6
The radius increases at r ′ (t) = 1µm per day and the height at h′ (t) = 2µm per day (twice as fast).
Simplifying and using this information leads to
dV
= 2π[r(t)h(t) + r(t)2 ] + 4πr(t)2 .
dt
If we are given the values of r, h at a specific time, then we can further calculate the exact rate of
change of the volume (in µm3 .)
Problem 6:(a) It takes you 1 hrs (total) to travel to and from UBC every day to study Philosophy
101. The amount of new learning (in arbitrary units) that you can get by spending t hours at the
university is given approximately by
10t
LP (t) =
.
9+t
How long should you stay at UBC on a given day if you want to maximize your learning per time
spent? (Time spent includes travel time.) (b) If you take Math 10000 instead of Philosophy, your
learning at time t is
LM (t) = t2 .
How long should you stay at UBC to maximize your learning in that case?
Solution (a) You want to maximize the function
E(t) =
E ′ (t) = 10
E ′ (t) = 10
LP (t)
10t
=
1+t
(9 + t)(1 + t)
1 · [(9 + t)(1 + t)] − [(9 + t)(1 + t)]′ · t
[(9 + t)(1 + t)]2
9 − t2
1 · [(9 + t)(1 + t)] − [(9 + t) + (1 + t)] · t
=
10
[(9 + t)(1 + t)]2
(9 + t)2 (1 + t)2
critical points occur when t2 = 9, i.e. at t = ±3hs. We reject the negative. We still need to check
that this is a maximum for learning. It is easiest to use the first derivative test. The denominator
of E ′ (t) is always positive, so its sign depends only on the numerator. We see that E ′ (t) > 0 for
t < 3 and E ′ (t) < 0 for t > 3. Thus t = 3 is a local maximum, and we are done. In case you
consider using the second derivative test, you would find that
E ′′ (t) = 10
[9 − t2 ]′ · [(9 + t)(1 + t)]2 − [9 − t2 ] · ([(9 + t)(1 + t)]2 )′
(9 + t)4 (1 + t)4
Luckily, at the critical point t2 = 9, so the last bunch of terms are zero, so that
E ′′ (t) = 10
[9 − t2 ]′
−2t
[9 − t2 ]′ · [(9 + t)(1 + t)]2
=
10
=
10
(9 + t)4 (1 + t)4
(9 + t)2 (1 + t)2
(9 + t)2 (1 + t)2
so we get a negative second derivative, i.e. this confirms that the answer is a local maximum.
(b) In case (b) there is no local maximum, and you need to stay as long as possible (23 hrs) to
maximize your learning per unit time.
7
Note: This problem closely resembles the optimal foraging example discussed in Chapter 6 of
the lecture notes. It can also be solved geometrically in the same way.
Problem 7: The atoms of some radioactive material are known to have a probability k of decaying
per unit time. We will use y(t) to denote the amount of radioactivity remaining at time t. Suppose
that there is 100 gm of this radioactive substance initially, at time t = 0. Consider what happens
during a small time interval ∆t. How much radioactive material is left at time t1 = ∆t? At time
ts = 2∆t? Write down an equation that links y(tn+1) to y(tn ) where tn = n∆t. Convert this
equation to a differential equation. If the half life of this substance is 1 day, find out how much is
left after 5 days. What is the value of k in this case, and how much radioactivity is left at time t?
Solution: We are given that y(0) = 100gm. At any time t + ∆t we can write that
y(t + ∆t) = [Amt at time t] − [Amt decayed during ∆t] = y(t) − ky(t)∆t
Rearranging and dividing by ∆t leads to
y(t + ∆t) − y(t)
= −ky(t).
∆t
This can also be written as
y(tn+1 ) − y(tn )
= −ky(tn ).
∆t
For small ∆t the left hand side can be approximated by the derivative dy/dt leading to the differential equation
dy
= −ky, y(0) = y0 = 100gm.
dt
This is a differential equation whose solution is y(t) = y0 e−kt . We are told that the half-life is 1
day, so that k = ln(2)/1 = ln(2) = 0.693 per day. Thus
y(t) = 100e− ln(2)t .
After 5 days the amount remaining is
5
y(t) = 100e− ln(2)·5 = 100e− ln(2 ) =
100
100
100
=
=
= 3.135gm.
5
eln(2 )
25
32
This answer makes sense since in 5 days we have 5 half-lives, so the material has fallen to (1/2)5 of
its original value, as we have seen above.
Problem 8: Given a population of 6 billion people on Planet Earth, and using the approximate
growth rate of r = 0.0125 per year, how long ago was this population only 1 million? Assume that
the growth has been the same throughout history (which is not actually true).
Solution
If the growth rate is constant then the population satisfies
dN
= rN,
dt
N(t) = N0 ert
8
Using r = 0.0125 per year and N0 = 1 million and the present population N(t) = 6000 million, we
have that
6000 = 1 · e0.0125t , ln(6000) = 0.0125t
Thus
t=
ln(6000)
= 696ys ago.
0.0125
Problem 9: Find critical points for the function y = ex (1 − ln(x)) for 0.1 ≤ x ≤ 2 and classify
their types.
Solution y = ex (1 − ln(x)), so y ′ = ex (1 − ln(x) − (1/x)). Setting this equal zero leads to
1 − ln(x) − (1/x) = 0
We notice that x = 1 satisfies this equation and is a critical point since ln(1) = 0. To classify this,
note that
y ′′ = ex 1 − ln(x) −
1
1
1
+ ex − + 2
x
x x
=−
ex
(−x2 + x2 ln(x) + 2x − 1).
x2
Substituting x = 1 leads to y ′′ (1) = −e ln(1) = 0. This turns out to be an inflection point, not a
maximum or minimum.
x
0.4
0.8
1.2
1.6
2
3.6
-3
3.4
3.2
-4
3
-5
2.8
2.6
-6
2.4
0.4
0.8
1.2
1.6
2
x
Figure 4: The function y = ex (1 − ln(x)) and the function y = ln(x) − ex are shown for the interval
0.1 ≤ x ≤ 2. The first one has an inflection point and the second one has a local maximum on this
interval.
Problem 10: The function y = ln(x) − ex has a critical point in the interval 0.1 ≤ x ≤ 2. It is not
possible to solve for the value of x at that point, but it is possible to find out what kind of critical
point that is. Determine whether that point is a local maximum, minimum, or inflection point.
Solution Critical points satisfy y ′ = x1 − ex = 0. We cannot solve this explicitly, as this is a
transcendental equation. However, we can compute that
y ′′ = −
1 + ex x2
.
x2
9
It is evident that this is negative for x > 0, so the critical point is a local maximum.
Problem 11: (a) Consider the polynomial y = 4x5 − 15x4 . Find all local minima maxima, and
inflection points for this function.
(b) Find the global minimum and maximum for this function on the interval [-1,1].
Solution
(a) The function is y(x) = 4x5 − 15x4 and its derivatives are y ′ (x) = 20x4 − 60x3 and y ′′ (x) =
80x3 − 180x2 = 20x2 (4x − 9). The critical points are located at y ′(x) = 0, i.e. at 20x4 − 60x3 =
20x3 (x − 3) = 0. Solving this leads to x = 0, 3. The second derivative is zero at x = 0, 9/4 but only
the latter is a place where there is a sign change, and consequently an inflection point.
(b) Checking the critical point in [-1,1] and the points at the interval ends we find that y(0) =
0, y(1) = 4 − 15 = −11, y(−1) = −4 − 15 = −19. Hence, the global minimum is attained at x = −1
and the global maximum at x = 0.
Problem 12: Consider the polynomial y = −x5 − x4 + 3x3 . Use calculus to find all local minima
maxima, and inflection points for this function.
(b) Find the global minimum and maximum for the function in Problem (3) on the interval
[-1,1].
Solution
The derivatives are
y ′(x) = −5x4 − 4x3 + 9x2 ,
y ′′(x) = −20x3 − 12x2 + 18x
Critical points satisfy
0 = y ′ (x) = x2 (−5x2 − 4x + 9),
x = 0, 1, −9/5.
The second derivative can be written as
y ′′(x) = 2x(−10x2 − 6x + 9)
The second derivative is zero and changes sign at the inflection points
x = 0, −
3√
3
±
11.
10 10
To classify the critical points, we note that at x = 1, y ′′ (1) = 2(−10 − 6 + 9) < 0 implies a local
maximum. The critical point at x = −9/5 is a local minimum and x = 0 is an inflection point.
Problem 13: Find a polynomial of third degree that has a local maximum at x = 1, a zero and an
inflection point at x = 0, and goes through the point (1,2). Hint: assume p(x) = ax3 + bx2 + cx + d
and find the values of a, b, c, d.
Solution
We consider a polynomial of the form p(x) = ax3 + bx2 + cx + d. The polynomial has a zero at
x = 0 and hence p(0) = d = 0. The derivative are p′ (x) = 3ax2 + 2bx + c and p′′ (x) = 6ax + 2b.
10
Since there is an inflection point at x = 0, we have that p′′ (0) = 2b = 0 so b = 0. Having a local
maximum at x = 1 further means that p′ (1) = 3a + 0 + c = 0. The polynomial goes through (1,2)
so p(1) = 2 = a + c. We thus have two algebraic equations to solve for a and c:
3a + c = 0, a + c = 2,
⇒
a = −1, c = 3.
Thus the desired polynomial is
p(x) = −x3 + 3x.
Problem 14: Find a linear approximation to the function y = x2 at the point whose x coordinate
is x = 2. Use your result to approximate the value of (2.0001)2.
Solution We first find the equation of the tangent line to the function at the given point x = 2.
The derivative there is f ′ (2) = 4. Thus, the tangent line equation is
y = f (2) + f ′ (2)(x − 2) = 4 + 4(x − 2)
Evaluating at x = 2.0001 leads to the linear approximation (2.0001)2 ≈ 4 + 4(2.0001 − 2) = 4.0004.
Problem 15: The Lennard-Jones potential, V (x) is the potential energy associated with two
uncharged molecules a distance x apart, and is given by the formula
b
a
− 6
12
x
x
V (x) =
where a, b > 0. Molecules would tend to adjust their separation distance so as to minimize this
potential. Find any local maxima or minima of this potential. Find the distance between the
molecules, x, at which V (x) is minimized and use the second derivative test to verify that this is a
local minimum.
Solution
We find a critical point by setting V ′ (x) = 0. This leads to
V ′ (x) =
−12a 6b
+ 7 = 0,
x13
x
12a
6b
= 7
13
x
x
⇒
⇒
x6 = 2a/b
Thus x = (2a/b)1/6 . The second derivative is
12 · 13a 6 · 7b
1
V (x) =
− 8 = 8
14
x
x
x
′′
156a
− 42b .
x6
Using the fact that x6 = 2a/b and x > 0 we simplify to
1
V ′′ (x) = 8
x
!
1
156a
− 42b = 8
2a/b
x
!
156b
− 42b .
2
This is clearly always positive for b > 0 so the function V (x) is concave up and the critical point is
a local minimum.
11
Problem 16: Consider an object thrown upwards with initial velocity v0 > 0 and initial height
h0 > 0. Then the height of the object at time t is given by
1
y = f (t) = − gt2 + v0 t + h0 .
2
Find critical points of f (t) and use both the second and first derivative tests to establish that this
is a local maximum.
Solution The derivatives are
y ′(t) = −gt + v0 , y ′′(t) = −g.
The critical point is at t = g/v0 and since the second derivative is negative, this is a local maximum.
Problem 17: The figure (not drawn to scale) shows a tumor mass containing a necrotic (dead)
core (radius r2 ), surrounded by a layer of actively dividing tumor cells. The entire tumor can be
assumed to be spherical, and the core is also spherical1 .
(a) If the necrotic core increases at the rate 3 cm3 /year and the volume of the active cells increases
by 4 cm3 /year, at what rate is the outer radius of the tumor (r1 ) changing when r1 = 1 cm. (NOTE:
Show all your work, leave your answer as a fraction in terms of π; indicate units with your answer.)
necrotic
core
r
2
r
1
active cells
(b) At what rate (in cm2 /yr) does the outer surface area of the tumor increase when r1 = 1cm?
Solution to Problem 17:
(a) The volume of the active part of the tumor is
4
Vactive = πr13 − Vcore
3
Vactive , r1 , r2 , Vcore are all changing with time
dVactive
dr1 dVcore
= 4πr12
−
dt
dt
dt
dr1
−3
dt
dr1
4 + 3 = 4π
.
dt
4 = 4π(1)2 ·
1
Recall that the volume and surface area of a sphere are V = (4/3)πr3 , S = 4πr2
12
Solving for dr1 /dt leads to
7
dr1
=
cm/year.
dt
4π
(b) S = 4πr12 so
S ′ (t) = 4π(2r1 )
dr1
dt
Using the same values as above leads to
S ′ (t) = 8π ·
7
= 14.
4π
Problem 18: Shown below is a major artery, (radius R) and one of its branches (radius r). A
labeled schematic diagram is also shown (right). The length 0A is L, and the distance between 0 and
P is d, where 0P is perpendicular to 0A. The location of the branch point (B) is to be determined
so that the total resistance to blood flow in the path ABP is as small as possible. (R, r, d, L are
positive constants, and R > r.)
L
R
d
r
0
B
A
P
(a) Let the distance between 0 and B be x. What is the length of the segment BA and what is
the length of the segment BP?
(b) The resistance of any blood vessel is proportional to its length and inversely proportional to
its radius to the fourth power2 . Based on this fact, what is the resistance, T1 , of segment BA
and what is the resistance, T2 , of the segment BP?
(c) Find the value of the variable x for which the total resistance, T (x) = T1 + T2 is a minimum.
Solution to Problem 18:
√
x2 + d2 .
√
(b) The resistance of segment BA is (L−x)/R4 and the resistance of the segment BP is x2 + d2 /r 4
(a) The length of the segment BA is L − x and the length of the segment BP is
(c) The total resistance is
(L − x)
+
T (x) =
R4
2
√
x2 + d2
.
r4
“z is inversely proportional to y” means that z = k/y for some constant k
13
First we find critical points:
T ′ (x) =
Then T ′ (x) = 0 when
1
x
−1
√
+
.
R4
r 4 x2 + d2
x
r4
.
=√ 2
4
R
x + d2
x2
r 8
.
= 2
R
x + d2
For convenience, call α2 = (r/R)8 . Then α2 (x2 + d2 ) = x2 , or x2 (1 − α2 ) = α2 d2 and we have
that
4
αd
r
d
r4d
q
√
x = ±√
=
±
.
=
±
R
1 − α2
R8 − r 8
1 − r 8 /R8
We reject the negative value of x based on the geometry. We still have to check if this is a
minimum or maximum. We find that
i
h √
2 + d2 − x · √ x
x
1
·
1
2
2
x +d 
T ′′ (x) = 0 + 4 
.
r
x2 + d2
This can be written as

x2

1− 2 2
1√
1√
x2 + d2 − x2
T ′′ (x) = 4 x2 + d2  2 x +d2  = 4 x2 + d2
r
x +d
r
(x2 + d2 )2
!
1√
d2
= 4 x2 + d2
.
r
(x2 + d2 )2
From this last result we see that T ′′ (x) > 0 so that we have a minimum.
14
!