My Math Forum In 2009, the population of Hungary was approximated

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 May 9th, 2017, 01:10 PM #1 Member   Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0 In 2009, the population of Hungary was approximated In 2009, the population of Hungary was approximated by p=9.906(0.997)^t where P is in millions and t is in years since 2009. Assume the trend continues. A) what does the model predict for the population of Hungary in the year 2020? B) how fast (in people/year) does this model predict Hungary's population will be decreasing in 2020? Can anybody explain how to do A and B please? Answers do not need to be given. I just need to know how to solve. Last edited by skipjack; May 10th, 2017 at 09:48 AM.
 May 9th, 2017, 01:21 PM #2 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 A should be easy enough, it's just algebra with some interpretation. t is given in years since 2009 and they ask what is the expected population in 2020. So if t=0 is 2009, what is t when it's 2020? As for B, you are really wanting the derivative of the function and putting t in that represents 2020. I've seen your other questions on the forum and I get you are struggling to understand the derivative. Try to think about it in this context for this problem as t represents years, when you take the derivative you are saying what is the change in population per year. It's per year since t represents each year from 2009. Derivatives represent the change in a function with respect to an input. In this case, your input is years. Don't misinterpret what I'm saying here, p'(20) would not be change in population in twenty years, but the change in population per year at t=20 or 2029. Hopefully this helps, work it out a bit, and if you're still struggling put up your work and I or someone else will help you through it. Last edited by skipjack; May 10th, 2017 at 09:52 AM.
 May 10th, 2017, 03:50 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,484 Thanks: 628 $\displaystyle 0.997^t$ can be written as $\displaystyle e^{\ln(0.997^t)}= e^{t \ln(0.997)}$ so its derivative is $\displaystyle \ln(0.997) e^{t \ln(0.997)}= \ln(0.997) (0.997^t)$. Last edited by skipjack; May 10th, 2017 at 09:50 AM.

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