## garner123 one year ago P=10.186(0.997)^t Calculate how fast population is decreasing at t=20 How do I set this up? P is population in millions and t is time in years after 2000. I need to find the rate at which population is decreasing at exactly t=20

1. garner123

Is this right? / if so how do I compute it?: $\frac{ 10.186(0.997)^{20+h}-10.186(0.997)^{20} }{ h }$

2. kirbykirby

Did you learn the easier method of finding derivatives (the power rule, chain rule, etc.)

3. garner123

I am learning those now, but this question got away from me.

4. kirbykirby

Ok well here you have an exponential form $$a^t$$ , which can be re-expressed as $$e^{t*ln (a)}$$ and then you can use the rules of differentiation for exponential functions of base e

5. kirbykirby

$\frac{d}{dx} e^x = e^x$$\frac{d}{dx}e^{ax} = ae^{ax}$

6. kirbykirby

here "a" is your "ln(a)" and in your problem, a = 0.997

7. garner123

ok so: $10.186e ^{t*\ln(0.997)}$

8. kirbykirby

yes exactly :)

9. garner123

Where do I go from there? I got: $P = 10.186(0.997)^{-0.003t}$

10. garner123

sorry I meant e instead of 0.997

11. kirbykirby

Ok so by the derivative rule for exponentials there... you will get: $10.186 [\ln(0.997)*e^{t*\ln(0.997)}]$I just kept it in "log" form rather than evaluating it to -0.003 just yet) And now to make the answer a bit cleaner, you know $$e^{t*ln(0.997)}=0.997^t$$ so: $$10.186*ln(0.997)*0.997^t$$

12. garner123

Then plug in 20 for t and boom... Thanks a bunch. I am going to have to review the derivative rule I think...

13. kirbykirby

Yup that's all there is left :) no problem