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- anonymous

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

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- anonymous

- jamiebookeater

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- anonymous

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

- kirbykirby

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

- anonymous

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

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- 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

- kirbykirby

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

- kirbykirby

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

- anonymous

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

- kirbykirby

yes exactly :)

- anonymous

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

- anonymous

sorry I meant e instead of 0.997

- 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:
\(10.186*ln(0.997)*0.997^t\)

- anonymous

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

- kirbykirby

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

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