Ap Calculus 2 - Differential Equations And Exponential Growth Page 2
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3AP 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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