The time rate change of an aligator population P in aswamp is proportional to th

Question

The time rate change of an aligator population P in aswamp is proportional to the square of P. The swamp contained adozen aligators in 1988, two dozen in 1998. When will their be fourdozen alligators in the swamp? What happens thereafter? If possible use differential equ. to solve The time rate change of an aligator population P in aswamp is proportional to the square of P. The swamp contained adozen aligators in 1988, two dozen in 1998. When will their be fourdozen alligators in the swamp? What happens thereafter? If possible use differential equ. to solve If possible use differential equ. to solve

Explanation / Answer

QuestionDetails: The time rate change of an aligator population P in aswamp is proportional to the square of P. The swamp
contained a dozen aligators in 1988, two dozen in 1998. Whenwill their be four dozen alligators in the swamp?
What happens thereafter? If possible use differential equ. to solve
LET P BE THE POPULATION AT ANY TIME T FROM A ZERO DATE TO BE FOUNDOUT
WE ARE GIVEN THAT The time rate change of an aligator populationP in a swamp is proportional
to the square of P...SO
DP/DT = KP2..........................DP/P2 =KDT
WHERE K IS A CONSTANT
INTEGRATING
-1/P = KT
KPT = -1+C .........................1
WHERE C IS A CONSTANT.
LET US TAKE YEAR1988 AS REPRESENTING T=T1......P=12 AT T=T1
K*12*T1=-1+C......C=1+12KT1
KPT=-1+1+12KT1=12KT1
PT=12T1.....................................2
YEAR 1998 WILL BE T=T1+10.....P=24
24(T1+10)=12T1
12T1=-240.......T1=-20
PT=-240.............................3
SO WHEN P=48 WE SHALL HAVE
48T=-240
T=-5......
NOW WE FOUND YEAR 1988 =T1=-20 TH. YEAR
HENCE - 5TH. YEAR WILL BE 1988+15 = 2003 YEAR
SO IN YEAR 2003 THE POPULATION WILL BE 4 DOZEN.
WE FIND THAT AS TIME APPROACHES YEAR 2008 ...
CORRESPONDING TO T APPROACHING ZERO,
THE POPULATION HAS TO RAISE TO INFINITE LEVELS TO SATISFYEQN.3
If possible use differential equ. to solve
LET P BE THE POPULATION AT ANY TIME T FROM A ZERO DATE TO BE FOUNDOUT
WE ARE GIVEN THAT The time rate change of an aligator populationP in a swamp is proportional
to the square of P...SO
DP/DT = KP2..........................DP/P2 =KDT
WHERE K IS A CONSTANT
INTEGRATING
-1/P = KT
KPT = -1+C .........................1
WHERE C IS A CONSTANT.
LET US TAKE YEAR1988 AS REPRESENTING T=T1......P=12 AT T=T1
K*12*T1=-1+C......C=1+12KT1
KPT=-1+1+12KT1=12KT1
PT=12T1.....................................2
YEAR 1998 WILL BE T=T1+10.....P=24
24(T1+10)=12T1
12T1=-240.......T1=-20
PT=-240.............................3
SO WHEN P=48 WE SHALL HAVE
48T=-240
T=-5......
NOW WE FOUND YEAR 1988 =T1=-20 TH. YEAR
HENCE - 5TH. YEAR WILL BE 1988+15 = 2003 YEAR
SO IN YEAR 2003 THE POPULATION WILL BE 4 DOZEN.
WE FIND THAT AS TIME APPROACHES YEAR 2008 ...
CORRESPONDING TO T APPROACHING ZERO,
THE POPULATION HAS TO RAISE TO INFINITE LEVELS TO SATISFYEQN.3

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