Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
garner123
Group Title
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
garner123 Group Title
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

This Question is Closed

garner123 Group TitleBest 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

kirbykirby Group TitleBest 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

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

kirbykirby Group TitleBest 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

kirbykirby Group TitleBest 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

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

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

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
yes exactly :)
 one year ago

garner123 Group TitleBest 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

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

kirbykirby Group TitleBest 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

garner123 Group TitleBest 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

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
Yup that's all there is left :) no problem
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.