Math 181 - Practice Questions for Final solutions
Problem 1 The Asian Ladybird has been introduced in the US as biological pest control. Unfortunately, it does so well here that it out-competes
native Ladybird species. Assume that a population of Ladybirds changes
according to the law
F (P ) = 0.03P (400 − P )
a) Find all equilibrium values for P (here and later, round all answers to one
digit after the decimal point).
b) Determine for each equilibrium found in part a) whether it is stable or
unstable. You can either use a graphical method, or use a theorem from
class.
c) Under consideration is an effort to get rid of these beetles again. How
many units of Ladybirds would have to be harvested each year at least in
order to drive the population to extinction?
a) Solve F (P ) = P which gives 11P − 0.03P 2 = 0, so equilibria are P = 0
and P = 11/0.03 ≈ 366.7 units.
b) Using the theorem: F 0 (P ) = 12 − 0.06P at P = 0 is 12, so P = 0
is unstable. Since F 0 (366.7) = 12 − 22 = −10 < −1, P = 366.7 is also
unstable. Using the graphical method you will see that P ’dances around’
the equilibrium 366.7 but does not approach it.
c) This is a maximal sustainable harvest question! So we maximize
H = F (P ) − P = 11P − 0.03P 2 ,
which leads to solving H 0 (P ) = 11 − 0.06P = 0. The answer P = 183.3 is a
breeding population for the maximal sustainable harvest, which is
H(183.3) ≈ 1008
units! the reason for this oddly large number is that even during the course
of one year, the beetles multiply so quickly that 183.3 units become 1191.7
units. Harvesting more than this answer of 1008.3, for every year, will drive
the beetle population to extinction.
Problem 2 The Northern Right Whale is one of the most endangered
marine mammals. Suppose that its population in the North Atlantic currently numbers 500 animals. According to the International Whaling Consortium (IWC), those populations of the Northern Right Whale where data
are available have a yearly growth rate of 7 percent.
a) Find the difference equation for this whale population P (t).
b) How many years will it take until the population reaches 5000 animals?
For the rest of this question, assume that every year, 20 additional Right
Whales are harvested or accidentally killed.
c) Find the equilibrium value for this situation.
d) Find the general solution. In how many years will the population reach
5000 under these conditions? What if in fact there were currently not 500,
but just 250 Northern Right Whales?
(see section 6.4, 6.5)
a) P (t + 1) − P (t) = 0.07P (t).
b) First cascade the equation from a), so
P (t) = 1.07t P (0) = 1.07t · 500.
Then solve P (t) = 5000 for t, leading to a time of (in years)
t=
ln 10
≈ 34
ln 1.07
(logarithms to any base work here).
c) E = 20/0.07 ≈ 286 whales.
d) We know the general solution has the form
P (t) = C · 1.07t + E
(1)
Substituting t = 0 and P (0) = 500, we get C = 214. To solve P (t) = 5000
under this law, we subtract 286 from both sides, then divide by 214 to get
1.07t =
5000 − 286
.
214
We take logarithms on both sides, divide by ln 1.07 and finally get
t=
ln[(5000 − 286)/214]
≈ 46
ln 1.07
(in years) for the time it will take for the population to reach 5000 whales.
Finally, if P (0) = 250 then the shape of the general solution would be the
same as (1) but we would get a negative value for C, namely C = 250−286 =
−36. Therefore
lim P (t) = −∞
t→∞
which does not make sense - but it can only mean the population will die
out. The scarce data available paints an even bleaker picture: for such tiny
populations scattered over incredibly vast areas, it becomes nearly impossible
for the whales to find a mate and reproduce, hence the birth rate crashes.
That is why some authors believe the Northern Right Whale is bound for
extinction, regardless of conservation efforts.
Problem 3 Consider the function
h(x) = sin(3x) +
3x
2
a) Find h0 (x).
b) Find all critical numbers of h(x) in the interval 0 ≤ x ≤ 4, and tell which
of them is a relative maximum and which a relative minimum
(see 5.7).
c) Unrelated problem: solve
exp(4t2 − 2t − 1) = 1.
exactly and to 3 digits after the decimal point (see 5.6).
a) h0 (x) = 3 cos(3x) + 3/2.
b) We need to solve h0 = 0, or
cos 3x = −
1
2
We know cos u = −1/2 for u = ±2π/3 + 2kπ, for any integer k. Substituting
u = 3x, we get
±2π + 6kπ
x=
9
Systematic trial and error tells us that those critical numbers which are in
the interval (0, 4) are
2π 4π 8π 10π
x=
, , ,
.
9 9 9 9
The sign pattern of h0 in the interval (0, 4) is
x
sign of h0 +
2π
9
|
4π
9
−
|
8π
9
+
|
10π
9
−
|
+
You find this either by substituting values for x in every one of these intervals, or at least in one interval and noting that the sign flips at every
boundary, evident from looking at a plot of h0 . So h has a relative maximum
at 2π/9, 8π/9 and a relative minimum at 4π/9, 10π/9.
c) Take logarithms on both sides to get
4t2 − 2t − 1 = ln(1) = 0
√
√
which has solutions t = (2 ± 4 + 16)/8 = (1 ± 5)/4.
5
4
3
2
1
0
0.0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
4.0
x
Problem 4 A drug is present in a patient’s blood, in a concentration of
25 mg per liter. Every hour, the kidneys reduce the amount of the drug by
12 percent. Assume that the patient has 18 liters of blood.
(see 6.2)
a) What will the concentration be after 4 hours?
b) When will the concentration drop to 1 mg per liter?
Let A(t) the amount of the drug present at time t. This decreases according
to
A(t + 1) − A(t) = −0.12A(t).
a) First, cascade this equation, so
A(t) = (1 − 0.12)t A(0) = 0.88t A(0)
The concentration of 25 mg per liter means
A(0) = 25
mg
· 18L = 450mg.
L
After four hours, we have the amount
A(4) = 0.884 · 450.
This gives the concentration
A(4)
= 0.884 · 25 = 14.99
18
in mg/L
b) The concentration will be 1 mg/L at the time when
A(t)
= 0.88t · 25 = 1
18
which gives after dividing by 25 and taking logarithms on both sides
t=
ln 1/25
≈ 25.18
ln 0.88
(in hours).
Problem 5 Consider the function
(see 5.6)
g(x) = 5 ln(x2 + 1) − 3x
a) Find all intervals where this function is increasing.
b) Find all intervals where this function is decreasing.
c) Find all relative maxima and minima.
d) Find the equation of the tangent line to the graph of g at x = 2.
We need the derivative g 0 for all parts,
g 0 (x) =
10x
10x − 3(x2 + 1)
−
3
=
x2 + 1
x2 + 1
a), b) First find all critical numbers. These are exactly the solutions of
−3x2 + 10x − 3 = 0
√
which are x = (−10 ± 100 − 36)/(−6), so x = 1/3 or x = 3. The sign
pattern of g 0 is − + − (the numerator is a parabola opening towards the
bottom). So g is increasing for 1/3 ≤ x ≤ 3 and decreasing outside of this
interval.
c) x = 1/3 is a relative minimum, x = 3 a relative maximum (no additional
work needed beyond parts a), b)).
d) At x = 2, g 0 (x) = 1, so the tangent line has an equation
y = 1 · (x − 2) + 5 ln 5 − 6 = x + 5 ln 5 − 8.
4.0
3.6
3.2
2.8
2.4
2.0
1.6
1.2
0.8
0.4
0.0
ï0.4
0.0
0.4
0.8
1.2
1.6
2.0
x
2.4
2.8
3.2
3.6
4.0