1. ## Population Growth

It is projected that, t, years from now, the population of a certain country will become $\displaystyle P(t) = 50e^{0.002t}$ million.

a) At what rate will the population be changing with respect to time 10 years from now?

...I thought taking the first derivative of the equation would give me the rate of change... and it did... but it's not the answer the question is looking for.

$\displaystyle P'(t) = e^{0.02t}$

...and I solved for 10 years and compared the population to the present...

$\displaystyle P(10) = 61$ million
$\displaystyle P(0) = 50$ million

$\displaystyle P(10) - P(0) = 11$ million

rate of change $\displaystyle = \frac{P(10) - P(0)}{10 - 0}$

rate of change $\displaystyle = 1.1$

...and that's also not the right answer...

So... now I don't know what to do.

2. Originally Posted by Macleef
It is projected that, t, years from now, the population of a certain country will become $\displaystyle P(t) = 50e^{0.002t}$ million.

a) At what rate will the population be changing with respect to time 10 years from now?

...I thought taking the first derivative of the equation would give me the rate of change... and it did... but it's not the answer the question is looking for.

$\displaystyle P'(t) = e^{0.02t}$

...and I solved for 10 years and compared the population to the present...

$\displaystyle P(10) = 61$ million
$\displaystyle P(0) = 50$ million

$\displaystyle P(10) - P(0) = 11$ million

rate of change $\displaystyle = \frac{P(10) - P(0)}{10 - 0}$

rate of change $\displaystyle = 1.1$

...and that's also not the right answer...

So... now I don't know what to do.
$\displaystyle P(t) = 50e^{0.002t}$

$\displaystyle P'(t) = 0.002 \times 50 e^{0.002t} = \frac{e^{0.002t}}{10}$

$\displaystyle P'(10) = \frac{e^{0.002(10)}}{10} \approx 0.1$

3. textbook answer is population is increasing at the rate of 1.22 million people/year... and don't know how to get this