Ap Calculus 2 - Differential Equations And Exponential Growth Page 3
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3d. Extra credit. On June 1, 2015, the Park service determines that the deer population is being lowered too
much. So it decides to add deer every year so that by the year 2020, the deer population will be what it was at
the start of its exploration in the year 2000 (500 deer). Assuming that the growth rate of deer stays the same
as in (a), determine the differential equation that is generated by this situation and solve it. Specifically, how
many deer will you need to add each year? Complete the chart through 2020 to the nearest deer.
dP
t = 15, P = 420
t = 20, P = 500
= kP + X
dt
5
C " X
Ce
" X
dP
420 =
500 =
= dt
k
k
kP + X
5
X = C " 420k
X = Ce
" 500k
kdP
= kdt
5
C " 420k = Ce
" 500k
kP + X
5
(
)
ln kP + X = k t "15
+ C
80k = Ce
" C
80k
k t"15
(
)
kP + X = Ce
C =
# 14.169
5
e
"1
k t"15
(
)
kP = Ce
" X
X = C = 420k # "5.846
(
)
k t"15
Ce
" X
k t"15
(
)
14.169e
+ 5.846
P =
$ P =
k
k
Add approximately 6 deer yearly
!
!
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010
524
550
577
605
635
646
657
668
680
693
2011
2012
2013
2014
2015
2016
2017
2018
2019
2020
627
567
513
465
420
435
450
466
482
500
Stu Schwartz
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