Math 2320Differential Equations

Worksheet #4

1a) Model the growth of the population of 50,000 bacteria in

a petri dish if the growth rate is k.

000,50)0(, PkPdt

dP

1b) Suppose 3 days later, the population has grown to about 80,000 bacteria. Find the growth rate and estimate the bacteria population after 5 days.

000,50)0(, PkPdt

dP

kPdt

dP

dtkP

dP

is separable.

ckteP

cktP

||ln

kt

c

ck

etP

e

eP

000,50)(

000,50

000,50)0( )0(

3/

3/1

3

3

5

8000,50)(

5

8

000,50

000,80

000,50000,80)3(

t

k

k

k

tP

e

e

eP

After 5 days:

438,1095

8000,50)5(

3/5

P

After 5 days, there will be approximately 109,438 bacteria in the petri dish.

2) A fish hatchery raises trout in ponds. At the beginning of the year, the ponds contain approximately 100,000 trout. The growth rate (birthrate minus deathrate) is estimated to be about 5 per 100 per week. The hatchery wants to harvest at a constant rate of R fish per week, and increase the population by 150,000 by the end of the year. Find the appropriate harvest rate, R.

000,250)52( and 000,100)0( PPInitial Conditions:

RPdt

dP

100

5Initial ODE:

20/

20/

20/20/

20/20/

20/20/20/

20/20

1

20)(

20

100

5

..,100

5

t

t

tt

tt

ttt

tdt

ceRtP

ceR

dteRPe

eRPedt

d

eRPedt

dPe

eeFIRPdt

dP

Solve as a Linear (or Separable) ODE:

Continued on the next slide.

2) A fish hatchery raises trout in ponds. At the beginning of the year, the ponds contain approximately 100,000 trout. The growth rate (birthrate minus deathrate) is estimated to be about 5 per 100 per week. The hatchery wants to harvest at a constant rate of R fish per week, and increase the population by 150,000 by the end of the year. Find the appropriate harvest rate, R.

000,250)52( and 000,100)0( PPInitial Conditions:

RPdt

dP

100

5Initial ODE:

20/20)( tceRtP

Solve as a Linear (or Separable) ODE:

Solve for R and c:

20

000,100

20000,100

20000,100)0( 20/0

cR

Rc

ceRP

20/

20/

000,100)(

20

000,10020)(

t

t

cectP

cec

tP

035,12

)1(

000,150

)1(000,150

000,100000,250)52(

5/13

5/13

20/52

c

ce

ec

cecP

20/035,12035,12000,100)( tetP

Continued on the next slide.

2) A fish hatchery raises trout in ponds. At the beginning of the year, the ponds contain approximately 100,000 trout. The growth rate (birthrate minus deathrate) is estimated to be about 5 per 100 per week. The hatchery wants to harvest at a constant rate of R fish per week, and increase the population by 150,000 by the end of the year. Find the appropriate harvest rate, R.

000,250)52( and 000,100)0( PPInitial Conditions:

RPdt

dP

100

5Initial ODE:

20/20)( tceRtP

Solve as a Linear (or Separable) ODE:

Solve for R and c:

20

000,100 cR

035,12c

20/035,12035,12000,100)( tetP

20

000,100 cR

035,12c

439820

035,12000,100

R

Trout should be harvested at a rate of approximately 4398 per week.

3) A tank, having a capacity of 3000 gallons, initially contains 20 pounds of salt, dissolved in 1000 gallons of water. A solution containing 0.4 pounds of salt per gallon flows into the tank at a rate of 5 gallons per min and the well-stirred solution flows out of the tank at a rate of 2 gallons per minute. 3a)How much time will elapse before the tank is filled to capacity?Current capacity = Current volume + (flow in – flow out):

3000 gallons = 1000 gallons + (5 gal/min – 2 gal/min)•t minutes

3000 = 1000 + 3t

2000 = 3t

T = 2000/3 = 666 2/3 minutes.

It will take 666 2/3 minutes for the tank to fill to capacity.

3b) What is the salt concentration in the tank when it contains 2000 gallons of solution?

20)0( AInitial Condition:

t

A

dt

dA

t

A

dt

dA

31000

22

min

gal2

gal)31000(

lb

min

gal 5

gal

lb4.

Initial ODE:

Solve as a Linear (or Separable) ODE:

231000

2

A

tdt

dA

3/231000ln31000

2

)31000(: 32

teeIF tdtt

Continued on the next slide.

3b) What is the salt concentration in the tank when it contains 2000 gallons of solution?

20)0( AInitial Condition:

t

A

dt

dA

31000

22

Initial ODE:

Solve as a Linear (or Separable) ODE:

231000

2

A

tdt

dA

3/231000ln31000

2

)31000(: 32

teeIF tdtt

3/252

3/552

3/23/2

3/23/2

3/23/2

3/2

)31000()31000(

)31000(

)31000(2)31000

)31000(2)31000

)31000(231000

)31000(2)31000(

tctA

ct

dttAt

tAtdt

d

tAt

t

dt

dAt

Continued on the next slide.

3b) What is the salt concentration in the tank when it contains 2000 gallons of solution?

20)0( AInitial Condition:

t

A

dt

dA

31000

22

Initial ODE:

Solve the ODE: 3/252 )31000()31000( tctA

Continued on the next slide.

Apply the Initial Condition:

000,38100

380

10040020

03100003100020)0( 3/2

52

c

c

c

cA

3/252 )31000(000,38)31000()( tttA

3b) What is the salt concentration in the tank when it contains 2000 gallons of solution?

20)0( AInitial Condition:

t

A

dt

dA

31000

22

Initial ODE:

Solve the ODE: 3/252 )31000()31000( tctA

Continued on the next slide.

Apply the Initial Condition: 3/252 )31000(000,38)31000()( tttA

Find the time when the tank contains 2000 gallons of solution.

3/1000

10003

2000)31000(

t

t

t

minutes

3b) What is the salt concentration in the tank when it contains 2000 gallons of solution?

20)0( AInitial Condition:

t

A

dt

dA

31000

22

Initial ODE:

Solve the ODE: 3/252 )31000()31000( tctA

Continued on the next slide.

Apply the Initial Condition: 3/252 )31000(000,38)31000()( tttA

Find the time when the tank contains 2000 gallons of solution. T = 1000/3 minutes

When the tank contains 2000 gallons, there will be 560.62 pounds of salt in the tank.

Find the salt concentration in the tank when t = 1000/3 minutes

62.560

)2000(000,38800

)2000(000,38)2000(

))(31000(000,38))(31000()(

3/2

3/252

3/23

10003

100052

31000

A

3c) What is the salt concentration at the instant that the tank is filled to capacity?

When the tank is filled to capacity, there will be 1017.32 pounds of salt in the tank.

From 3a), the tank is filled to capacity when t = 2000/3 minutes

32.1017

)3000(000,381200

)3000(000,38)3000(

))(31000(000,38))(31000()(

3/2

3/252

3/23

20003

200052

32000

A

4) At the instant that a cake is removed from an oven, its temperature is 375F. The cake is placed in a room whose temperature is 75F. After 2 minutes, the cake cools to a temperature of 175F. What is the temperature of the cake after 10 minutes? 175)2(,375)0( TTInitial

Conditions:)75( Tk

dt

dTInitial ODE:

Solve the separable (or linear) ODE:

ckt

ckt

eT

eT

cktT

dtkT

dT

75

75

75ln75

Continued on the next slide.

4) At the instant that a cake is removed from an oven, its temperature is 375F. The cake is placed in a room whose temperature is 75F. After 2 minutes, the cake cools to a temperature of 175F. What is the temperature of the cake after 10 minutes? 175)2(,375)0( TTInitial

Conditions:)75( Tk

dt

dTInitial ODE:

Solve the separable (or linear) ODE:

cktetT 75)(

Apply the initial conditions:

kt

cck

ckt

etT

ee

eT

30075)(

300

75375)0()0(

2/

31

2/1

31

231

2

)2(

30075)(

300100

30075175)2(

t

k

k

k

k

tT

e

e

e

eT

Find the temperature of the cake after 10 minutes.

23.7630075)10( 2/10

31 T

The temperature of the cake after 10 minutes is approximately 76 degrees.