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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
 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
 one year ago

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garner123Best ResponseYou've already chosen the best response.0
Is this right? / if so how do I compute it?: \[\frac{ 10.186(0.997)^{20+h}10.186(0.997)^{20} }{ h }\]
 one year ago

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

garner123Best ResponseYou've already chosen the best response.0
I am learning those now, but this question got away from me.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
Ok 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
 one year ago

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

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

garner123Best ResponseYou've already chosen the best response.0
ok so: \[10.186e ^{t*\ln(0.997)}\]
 one year ago

garner123Best ResponseYou've already chosen the best response.0
Where do I go from there? I got: \[P = 10.186(0.997)^{0.003t}\]
 one year ago

garner123Best ResponseYou've already chosen the best response.0
sorry I meant e instead of 0.997
 one year ago

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

garner123Best ResponseYou've already chosen the best response.0
Then plug in 20 for t and boom... Thanks a bunch. I am going to have to review the derivative rule I think...
 one year ago

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