If the instantaneous rate of change of a population is #50t^2 - 100t^(3/2)# (measured in individuals per year) and the initial population is 25000 then what is the population after t years?

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Answer 1 · 2015-03-26T23:56:12

#P(t)= 50/3t^3-40t^(5/2)+25000#

To answer this question, you need to find a particular antiderivative.

Let #P(t)# be the population #t# years after the initial population was counted. (So #P(0)=25000#)

We are told that the population is changing at a rate -- stop! that means we're going to be told something about the derivative (the rate of change).

The population is changing at a rate of #50t^2-100t^(3/2)#.

This tells us that
#P'(t)=50t^2-100t^(3/2)#.

Our task is to find #P(t)#

If #f'(x)=x^n#, the #f(x)=x^(n+1)/(n+1)#

So #P(t)=50t^3/3-100*2/5 t^(5/2)+C# . . (See below if needed)

#P(t)=50/3t^3-40 t^(5/2)+C#

We also know that #P(0)=25000#, so we can substitute to find #C#

#25000=50/3(0)^3-40 (0)^(5/2)+C#, so #C=25000#
#P(t)= 50/3t^3-40t^(5/2)+25000#

I think writing all of the following clutters up the work:
To find the antiderivative of #t^(3/2)#:

#t^(3/2+1)/(3/2+1)=t^(5/2)/(5/2)=2/5 t^(5/2)#

(dividing is multiplying by the reciprocal.)