# a population p(t) is changing at a rate of P(t)=10,000e^(-t/4) with t in years. If the population is initially 80,000 at t=0, how large is the population as t--> + infinity

The rate of change of a function is the first derivative of the function.

If p(t) is the population function, then `p'(t)=P(t)=10000e^(- t /4)`

To find the population function we integrate P(t) with respect to t:

`p(t)=int10000e^(-t/4)dt`

`=10000int e^(-t/4)dt` Let `u=-t/4,du=-1/4dt` then

`=-40000inte^u du`

`=-40000e^u+C_1`

`=-40000e^(-t/4)+C`

Now p(0)=80000 so `80000=-40000e^0+C==>C=120000`

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The rate of change of a function is the first derivative of the function.

If p(t) is the population function, then `p'(t)=P(t)=10000e^(- t /4)`

To find the population function we integrate P(t) with respect to t:

`p(t)=int10000e^(-t/4)dt`

`=10000int e^(-t/4)dt` Let `u=-t/4,du=-1/4dt` then

`=-40000inte^u du`

`=-40000e^u+C_1`

`=-40000e^(-t/4)+C`

Now p(0)=80000 so `80000=-40000e^0+C==>C=120000`

Then `p(t)=-40000e^(-t/4)+120000`

We are asked to find the limit of p(t) as t goes to infinity:

`lim_(t->oo)(-40000e^(-t/4)+120000)`

`=lim_(t->oo)-40000/e^(t/4)+120000`

`=120000`

The population approaches 120000 as t goes to infinity.

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