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 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
 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

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garner123
 one year ago
Best ResponseYou've already chosen the best response.0Is this right? / if so how do I compute it?: \[\frac{ 10.186(0.997)^{20+h}10.186(0.997)^{20} }{ h }\]

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.1Did you learn the easier method of finding derivatives (the power rule, chain rule, etc.)

garner123
 one year ago
Best ResponseYou've already chosen the best response.0I am learning those now, but this question got away from me.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.1Ok well here you have an exponential form \(a^t\) , which can be reexpressed as \(e^{t*ln (a)}\) and then you can use the rules of differentiation for exponential functions of base e

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dx} e^x = e^x\]\[\frac{d}{dx}e^{ax} = ae^{ax}\]

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.1here "a" is your "ln(a)" and in your problem, a = 0.997

garner123
 one year ago
Best ResponseYou've already chosen the best response.0ok so: \[10.186e ^{t*\ln(0.997)}\]

garner123
 one year ago
Best ResponseYou've already chosen the best response.0Where do I go from there? I got: \[P = 10.186(0.997)^{0.003t}\]

garner123
 one year ago
Best ResponseYou've already chosen the best response.0sorry I meant e instead of 0.997

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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: \(10.186*ln(0.997)*0.997^t\)

garner123
 one year ago
Best ResponseYou've already chosen the best response.0Then plug in 20 for t and boom... Thanks a bunch. I am going to have to review the derivative rule I think...

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.1Yup that's all there is left :) no problem
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