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AP Calculus 2 – Differential Equations and Exponential Growth.

a. On June 1, 2000, a state park in Colorado estimated that it had 500 deer on its land. Two years later, it

determines that there was 550 deer on the land. Assuming that the number of deer was changing

exponentially (i.e. its rate of change was proportional to the current amount), determine the equation that

predicts its growth. Complete the chart below through 2005 to the nearest deer.

dP

kt

= kP " P = Ce

dt

kt

P = 500e

t = 2, P = 550

2k

550 = 500e

#

55

&

ln

%

(

50

$

'

k =

2

b. On June 1, 2005, the Park service determines that the deer population is growing too quickly. So they decide

to remove (cull) 20 deer from the herd every year. Assuming that the growth rate of deer is the same as in (a),

!

determine the differential equation that is generated by this situation and solve it. Complete the chart below

through 2010 to the nearest deer.

dP

= kP " 20

dt

dP

= dt u = kP " 20, du = kdP

kP " 20

t = 5, P = 635

kdP

= kdt

635k = 20 + C

kP " 20

(

)

ln kP " 20 = k t " 5

+ C

C = 635k " 20 # 10.261

( )

k t"5

( )

20 + 10.261e

k t"5

kP " 20 = Ce

P =

k

k t"5

( )

kP = 20 + Ce

( )

k t"5

20 + Ce

P =

k

!

c. On June 1, 2010, the Park service determines that the deer population is still rampant so that they decide to

remove 10% of the deer population yearly. The growth rate is still the same as in (a). Determine the

!

differential equation that is generated by this situation and solve it. Complete the chart below through 2015

to the nearest deer.

dP

= ".1

dt

dP = ".1dt

ln P = ".1t + C

P = Ce

t = 0, P = 693

".1t

P = 693e

".1t

Stu Schwartz

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