How to find the function #p(t)# which gives the total population of the deer after #t # years? See entire problem below.

A population of #50# deer are introduced into a state park. Biologist know that the rate of growth over time of the population of the deer is given by the function: #r(t)=500/(2+t^2)# deer/year

1 Answer
Gió
Jul 28, 2017

I got #p(t)=250*sqrt(2)arctan(t/sqrt(2))+50#
....but check my maths anyway!

Explanation:

Here you have the RATE of CHANGE that represents how your function #p(t)# changes with time or:
#r(t)=(dp(t))/dt#
but we know that:
#r(t)=500/(2+t^2)#

We can then write:

#(dp(t))/dt=500/(2+t^2)#

which is a Differential Equation that we can solve with some manipulations as follows:

1) separate:

#dp(t)=500/(2+t^2)color(red)(dt)#

2) integrate:

#intdp(t)=int500/(2+t^2)dt#

#p(t)=int500/(2+t^2)dt#

we can solve the right hand integral collecting #2# and manipulating to get

#int500/(2+t^2)dt=500/2int1/(1+t^2/2)dt=500/2int1/(1+(t/sqrt(2))^2)dt=500/2*sqrt(2)arctan(t/sqrt(2))+c=250*sqrt(2)arctan(t/sqrt(2))+c#

so we get:
#p(t)=250*sqrt(2)arctan(t/sqrt(2))+c#

But we still need to find #c#.
We use the initial condition where at #t=0# the population is #p(0)=50# and get:
#50=250*sqrt(2)arctan(0)+c#
rearranging:
#c=50#

so our complete function will be:

#color(blue)(p(t)=250sqrt(2)arctan(t/sqrt(2))+50)#

that graphically can be seen as:

enter image source here

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