009B Sample Midterm 1, Problem 3 Detailed Solution

A population grows at a rate

$P'(t)=500e^{-t}$

where  $P(t)$  is the population after  $t$  months.

(a)   Find a formula for the population size after  $t$  months, given that the population is  $2000$  at  $t=0.$

(b)   Use your answer to part (a) to find the size of the population after one month.

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Background Information:
Recall:
$P(t)=\int P'(t)~dt$

Solution:

(a)

Step 1:
To find  $P(t)$  we need to integrate  $P'(t)$.
We get
${\begin{array}{rcl}\displaystyle {P(t)}&=&\displaystyle {\int P'(t)~dt}\\&&\\&=&\displaystyle {\int 500e^{-t}~dt.}\end{array}}$
Now, we use  $u$-substitution.
Let  $u=-t.$  Then,  $du=-dt.$
So, we have
${\begin{array}{rcl}\displaystyle {P(t)}&=&\displaystyle {\int -500e^{u}~du}\\&&\\&=&\displaystyle {-500e^{u}+C}\\&&\\&=&\displaystyle {-500e^{-t}+C.}\end{array}}$

Step 2:
Now, we need to find the constant  $C.$
From the information provided in the problem, we know  $P(0)=2000.$
So, we have
${\begin{array}{rcl}\displaystyle {2000}&=&\displaystyle {P(0)}\\&&\\&=&\displaystyle {-500e^{0}+C}\\&&\\&=&\displaystyle {-500+C.}\end{array}}$
Solving for  $C,$  we get  $C=2500.$
Hence, we have
$P(t)=-500e^{-t}+2500.$

(b)
We need to calculate  $P(1).$
Using our answer from part (a), we have
${\begin{array}{rcl}\displaystyle {P(1)}&=&\displaystyle {-500e^{-1}+2500}\\&&\\&=&\displaystyle {{\frac {-500}{e}}+2500}\\&&\\&=&\displaystyle {{\frac {-500+2500e}{e}}.}\end{array}}$

Final Answer:
(a)     $P(t)=-500e^{-t}+2500$
(b)     ${\frac {-500+2500e}{e}}$

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