# The population of a town a town after t weeks is given by p(t)=1200(2^-t). a) what is the initial population of the town? b) how many people are there after 1 week? c) what is the rate of change of people after 1 week?

Apr 22, 2018

a) $1200$ b) $600$ c) -600ln2approx-415.99

#### Explanation:

a) The initial population of the town is simply $p \left(t\right)$ evaluated at the smallest possible value of $t$, that is, $t = 0 :$

$p \left(0\right) = 1200 \left({2}^{0}\right) = 1200$

The initial population is $1200$ people.

b) The amount of people in the town after $1$ week is $p \left(t\right)$ evaluated at $t = 1 :$

$p \left(1\right) = 1200 \left({2}^{-} 1\right) = \frac{1200}{2} = 600$

There are $600$ people after a week.

c) The rate of change of people after $1$ week will be the first derivative of $p \left(t\right)$ evaluated at $t = 1$, IE, $p ' \left(1\right) .$

To differentiate, let's rewrite a little, recalling that $x = {e}^{\ln} x$:

$p \left(t\right) = 1200 \left({e}^{\ln} \left({2}^{-} t\right)\right)$

$\ln \left({a}^{b}\right) = b \ln a ,$ so $\ln \left({2}^{-} t\right) = - t \ln 2$:

$p \left(t\right) = 1200 \left({e}^{- t \ln 2}\right)$

Differentiate, using the Chain Rule:

$p ' \left(t\right) = 1200 \left({e}^{- t \ln 2}\right) \cdot \frac{d}{\mathrm{dt}} \left(- t \ln 2\right)$

$\frac{d}{\mathrm{dt}} \left(- t \ln 2\right) = - \ln 2 ,$ so:

$p ' \left(t\right) = - 1200 \left(\ln 2\right) \left({e}^{- t \ln 2}\right)$

Recalling that ${e}^{- t \ln 2} = {e}^{\ln {2}^{-} t} = {2}^{-} t$,

$p ' \left(t\right) = - 1200 \left(\ln 2\right) \left({2}^{-} t\right)$

$p ' \left(1\right) = - 1200 \left(\ln 2\right) \left({2}^{-} 1\right) = - 600 \ln 2$

The rate of change after a week is $- 600 \ln 2 \approx - 415.99$.