anonymous
  • 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
Calculus1
jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • kirbykirby
Did you learn the easier method of finding derivatives (the power rule, chain rule, etc.)
anonymous
  • anonymous
I am learning those now, but this question got away from me.

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kirbykirby
  • 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
  • kirbykirby
\[\frac{d}{dx} e^x = e^x\]\[\frac{d}{dx}e^{ax} = ae^{ax}\]
kirbykirby
  • kirbykirby
here "a" is your "ln(a)" and in your problem, a = 0.997
anonymous
  • anonymous
ok so: \[10.186e ^{t*\ln(0.997)}\]
kirbykirby
  • kirbykirby
yes exactly :)
anonymous
  • anonymous
Where do I go from there? I got: \[P = 10.186(0.997)^{-0.003t}\]
anonymous
  • anonymous
sorry I meant e instead of 0.997
kirbykirby
  • 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
  • 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
  • kirbykirby
Yup that's all there is left :) no problem

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