Math 115 HW #8 Solutions
1. The function with the given graph is a solution of one of the following differential equations.
Decide which is the correct equation and justify your answer.
(a) y 0 = 1 + xy
(b) y 0 = −2xy
(c) y 0 = 1 − 2xy
Answer: If (b) were correct, then, when x = 0, we would have that y 0 = 0. However, the
slope of the tangent line is clearly positive when x = 0 in the picture, so (b) cannot be correct.
If (a) were correct, then y 0 would be positive for all positive values of x. However, the slope
of the tangent line is clearly negative for most positive x, so (a) cannot be correct.
Thus, (c) must be the equation which is solved by the function whose graph we see.
2. Find an equation of the curve that passes through the point (0, 1) and whose slope at (x, y)
is xy.
Answer: If the slope of the curve at the point (x, y) is xy, then we have that the curve is
the graph of a function which solves the differential equation
dy
= xy.
dx
This is a separable equation, so we can find solutions implicitly by separating variables and
integrating:
Z
Z
dy
= xdx.
y
Thus,
ln y =
x2
+ C.
2
Exponentiating both sides yields
y = eC ex
for A > 0 a constant.
1
2 /2
= Aex
2 /2
2
The functions y = Aex /2 give all the curves whose slopes at (x, y) are xy; however, only one
of these curves can pass through the point (0, 1), namely the one for which
1 = Ae0
2 /2
= A.
Therefore, the equation of the curve we’re looking for is
2 /2
y = ex
.
3. Find the orthogonal trajectories of the family of curves
y 2 = kx3 .
(An orthogonal trajectory of a family of curves is a curve that intersects each curve of the
family orthogonally, that is, at right angles)
Sketch several members of each family on a common graph (feel free to use Mathematica or
any other graphing utility to help create this sketch).
Answer: Since the orthogonal trajectories to this family are those curves whose tangent
lines are orthogonal to the tangent lines of the family, we should determine the slopes of the
tangent lines to the given family of curves. To do so, differentiate implicitly:
2y
Solving for
dy
dx ,
dy
= 3kx2 .
dx
we see that
Since y 2 = kx3 , we know that k =
dy
3kx2
=
.
dx
2y
y2
;
x3
plugging this into the above gives that
2
3 y 3 x2
dy
3y
= x
=
.
dx
2y
2x
Since the orthogonal trajectories should have perpendicular tangent lines, their slopes will
be negative reciprocals of the above. In other words, the orthogonal trajectories of the given
family are the solutions of the differential equation
dy
2x
=− .
dx
3y
This is a separable equation, so we separate variables and integrate:
Z
Z
3ydy = − 2xdx.
Thus,
3 2
y = −x2 + C,
2
so the orthogonal trajectories are the curves
3
x2 + y 2 = C.
2
2
3
2
1
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
Figure 1: Original family in red, orthogonal trajectories in blue
4. A glucose solution is administered intravenously into the bloodstream t a constant rate r. As
the glucose is added, it is converted into other substances and removed from the bloodstream
at a rate that is proportional to the concentration at that time. Thus a model for the
concentration C = C(t) of the glucose solution in the bloodstream is
dC
= r − kC
dt
where k is a positive constant.
(a) Suppose that the concentration at time t = 0 is C0 . Determine the concentration at any
other time t by solving the differential equation.
Answer: This is a separable equation, so separate and integrate:
Z
Z
dC
= dt.
r − kC
Thus,
1
− ln |r − kC| = t + D.
k
Multiplying both sides by −k and exponentiating gives
|r − kC| = e−Dk e−t = Ae−kt .
Allowing A to be either positive or negative lets us get rid of the absolute value signs;
solving for C gives
r − Ae−kt
r
A
C(t) =
= − e−kt .
k
k
k
3
If we plug in t = 0, then
C0 =
r
A
r
A
− e−k·0 = − ,
k
k
k
k
so A = r − kC0 .
Hence, the concentration at time t is given by the equation
C(t) =
or, equivalently,
C(t) =
r
r − kC0 −kt
−
e
k
k
r r
−
− C0 e−kt .
k
k
(b) Assuming that C0 < r/k, find limt→∞ C(t) and interpret your answer.
Answer: We take the limit
i r
hr r
−
− C0 e−kt =
lim C(t) = lim
t→∞
t→∞ k
k
k
since e−kt → 0 as t → ∞.
5. A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 6% alcohol is
pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate.
What is the percentage of alcohol after one hour?
Answer: Let a(t) be the volume of alcohol (in gallons) in the vat after t minutes. The vat
starts out with
0.04 × 500 = 20 gallons
of alcohol in it, so a(0) = 20. The rate of change of alcohol in the vat is
da
= (rate in) − (rate out).
dt
We know that the rate in is
0.06 × 5 = 0.3 gal/min.
On the other hand, the rate of alcohol going out of the vat is
a(t)
a(t)
×5=
gal/min.
500
100
Therefore,
da
a
30 − a
= 0.3 −
=
.
dt
100
100
This is a separable equation, so we can separate variables and integrate:
Z
Z
da
dt
=
.
30 − a
100
Thus,
− ln |30 − a| =
4
t
+ C.
100
If all the beer in the vat were 6% alcohol, then there would be 30 gallons of alcohol in the
vat. Hence, a(t) ≤ 30 for all t, so 30 − a ≥ 0, meaning that the absolute value signs above
are superfluous. Multiplying by −1 and exponentiating yields
30 − a = e−C e−t/100 = Ae−t/100 .
Therefore,
a(t) = 30 − Ae−t/100 .
Since a(0) = 20, we know that
20 = 30 − Ae−0/100 = 30 − A,
so A = 10.
Therefore, the volume of alcohol in the vat after t minutes is
a(t) = 30 − 10e−t/100 .
In particular, after 1 hour there will be
a(60) = 30 − 10e−60/100 ≈ 30 − 5.5 = 24.5 gallons of alcohol
in the tank. Hence, the percentage of alcohol in the tank is
24.5
× 100 = 4.9%
500
6. Biologists stocked a lake with 400 fish and estimated the carrying capacity (the maximal
population for the fish of that species in that lake) to be 10,000. The number of fish tripled
in the first year.
(a) Assuming that the size of the fish population satisfies the logistic equation
dP
P
= rP 1 −
,
dt
N
find an expression for the size of the population after t years.
Answer: As we saw in class, solutions of the logistic equation are of the form
P (t) =
P0 N
,
(N − P0 )e−rt + P0
where P0 is the initial population and N is the carrying capacity. Hence, in this case
P0 = 400 and N = 10, 000, so the fish population is given by
P (t) =
4, 000, 000
400 · 10, 000
=
.
−rt
(10, 000 − 400)e
+ 400
9600e−rt + 400
Since the fish population tripled in the first year, we know that P (1) = 3 · 400 = 1200,
so we have that
4, 000, 000
.
1200 =
9600e−r + 400
5
Therefore,
9600e−r + 400 =
so
4, 000, 000
40, 000
10, 000
=
=
,
1200
12
3
10,000
3
− 400
8800
11
=
= .
9600
3 · 9600
36
Taking the natural log of both sides gives that
11
−r = ln
.
36
e−r =
Hence, the population after t years is given by
P (t) =
4, 000, 000
11
9600eln( 36 )t + 400
P (t) =
4, 000, 000
.
t
9600 11
+
400
36
or
(b) How long will it take for the population to increase to 5000?
Answer: We want to solve for t0 such that P (t0 ) = 5000. Therefore,
5000 =
so
9600
11
36
4, 000, 000
,
11 t0
9600 36
+ 400
t0
+ 400 =
4, 000, 000
= 800.
5000
Hence,
11
36
t0
=
1
400
= .
9600
24
Taking the natural log of both sides,
11
1
t0 ln
= ln
,
36
24
so
ln
t0 =
ln
1
24 11
36
≈ 2.68
so the population will increase to 5000 in about 2.7 years.
7. (a) Show that if P satisfies the logistic equation, then
d2 P
P
2P
2
=r P 1−
1−
.
dt2
N
N
Answer: If P satisfies the logistic equation, then
dP
P
= rP 1 −
.
dt
N
6
Differentiating this equation gives,
Substituting
dP
dt
dP
P
1 dP
d2 P
=r
1−
+ rP −
dt2
dt
N
N dt
dP
P
P
=r
1−
−
dt
N
N
2P
dP
1−
.
=r
dt
N
P
into the right side of the above yields
= rP 1 − N
d2 P
= r2 P
dt2
2P
P
1−
,
1−
N
N
as desired.
(b) Deduce that a population grows fastest when it reaches half its carrying capacity.
Answer: The population will be growing fastest when dP
dt reaches its maximum. To
2
dP
find the maximum, we note that the critical points of dt occur when ddtP2 = 0; i.e. when
P
2P
0=r P 1−
1−
.
N
N
2
P
This can only occur when P = 0, 1 − N
= 0 or 1 − 2P
N = 0. The first two never happen
unless the initial population is either 0 or N (in which case the population never grows
at all). In the third case,
2P
= 1,
N
Meaning that
N
P = ,
2
so, indeed, the population is growing fastest when it reaches half its carrying capacity.
8. Consider a population P = P (t) with constant relative birth and death rates α and β,
respectively, and a constant emigration rate m, where α, β, and m are positive constants.
Assume that α > β. Then the rate of change of the population at time t is modeled by the
differential equation
dP
= rP − m
where r = α − β
dt
(a) Find the solution of this equation that satisfies the initial condition P (0) = P0 .
Answer: This is a separable equation, so separate variables and integrate:
Z
Z
dP
= dt.
rP − m
Thus,
1
ln |rP − m| = t + C,
r
7
so
rP − m = erC ert = Aert
where A is some constant. Therefore,
P (t) =
A rt m
e + .
r
r
When t = 0, P (0) = P0 , so
P0 =
A r·0 m
A m
e +
= + .
r
r
r
r
Hence,
A = rP0 − m,
so we see that
P (t) =
or, equivalently,
rP0 − m rt m
e +
r
r
m rt m
P (t) = P0 −
e + .
r
r
(b) What condition on m will lead to an exponential expansion of the population?
Answer: The above equation for P (t) yields exponential growth when
P0 −
m
> 0.
r
In other words, if
m < rP0 ,
then there will be an exponential expansion of the population.
(c) What condition on m will result in a constant population? A population decline?
Answer: The condition
m
P0 −
=0
r
will yield constant population, so the condition on m is that
m = rP0 .
Likewise, when m > rP0 , the population will decline.
(d) In 1847, the population of Ireland was about 8 million and the difference between the
relative birth and death rates was 1.6% of the population. Because of the potato famine
in the 1840s and 1850s, about 210,000 inhabitants per year emigrated from Ireland. Was
the population expanding or declining at that time?
Answer: In this model, P0 = 8, 000, 000, r = 0.016. Hence,
rP0 = 0.016 × 8, 000, 000 = 128, 000.
Since m = 210, 000, we see that m > rP0 , so the population was declining.
8