# 009B Sample Midterm 1, Problem 3 Detailed Solution

A population grows at a rate

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

where  ${\displaystyle P(t)}$  is the population after  ${\displaystyle t}$  months.

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

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

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

Solution:

(a)

Step 1:
To find  ${\displaystyle P(t)}$  we need to integrate  ${\displaystyle P'(t)}$.
We get
${\displaystyle {\begin{array}{rcl}\displaystyle {P(t)}&=&\displaystyle {\int P'(t)~dt}\\&&\\&=&\displaystyle {\int 500e^{-t}~dt.}\end{array}}}$
Now, we use  ${\displaystyle u}$-substitution.
Let  ${\displaystyle u=-t.}$  Then,  ${\displaystyle du=-dt.}$
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {P(t)}&=&\displaystyle {\int -500e^{u}~du}\\&&\\&=&\displaystyle {-500e^{u}+C}\\&&\\&=&\displaystyle {-500e^{-t}+C.}\end{array}}}$
Step 2:
Now, we need to find the constant  ${\displaystyle C.}$
From the information provided in the problem, we know  ${\displaystyle P(0)=2000.}$
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {2000}&=&\displaystyle {P(0)}\\&&\\&=&\displaystyle {-500e^{0}+C}\\&&\\&=&\displaystyle {-500+C.}\end{array}}}$
Solving for  ${\displaystyle C,}$  we get  ${\displaystyle C=2500.}$
Hence, we have
${\displaystyle P(t)=-500e^{-t}+2500.}$

(b)
We need to calculate  ${\displaystyle P(1).}$
Using our answer from part (a), we have

${\displaystyle {\begin{array}{rcl}\displaystyle {P(1)}&=&\displaystyle {-500e^{-1}+2500}\\&&\\&=&\displaystyle {{\frac {-500}{e}}+2500}\\&&\\&=&\displaystyle {{\frac {-500+2500e}{e}}.}\end{array}}}$

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