Math 115 HW #8 Solutions - Colorado State University

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¡ú¡Þ C(t) and interpret your answer.

Answer: We take the limit



i r

hr r

?

? C0 e?kt =

lim C(t) = lim

t¡ú¡Þ

t¡ú¡Þ k

k

k

since e?kt ¡ú 0 as t ¡ú ¡Þ.

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 ¡Á 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 ¡Á 5 = 0.3 gal/min.

On the other hand, the rate of alcohol going out of the vat is

a(t)

a(t)

¡Á5=

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) ¡Ü 30 for all t, so 30 ? a ¡Ý 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 ¡Ö 30 ? 5.5 = 24.5 gallons of alcohol

in the tank. Hence, the percentage of alcohol in the tank is

24.5

¡Á 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

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