009B Sample Midterm 1, Problem 3 Detailed Solution

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.

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

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