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At item costs $200$ dollars at $t=0$ and costs $P$ in year $t$. When the inflation is $r\%$ per year, the price is given by

$$\displaystyle P = 200e^{\Large \frac{r t}{100}}$$

If $r$ increases at a rate of $.2$ per year, when $r=4$ and $t=3$, at what rate is the price rising (in dollars per year)?

I took the derivative of the equation and I got

$$\displaystyle \frac {dP}{dt}= 200 e^{\Large \frac{rt}{100}} \times \frac{t}{100} \times \frac{dr}{dt}.$$

I assumed that $t$ was constant but I am not really sure? I kept getting a wrong answer of $1.2e^{3/25}$. Can someone tell me what I am doing wrong? Thank you!

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It really helps readability if you format your questions using MathJax. Also, please make sure I edited everything properly. Regards –  Amzoti Feb 28 '13 at 21:41

1 Answer 1

The problem says that $\frac{dr}{dt}=0.2$, so $r$ is not constant.

For differentiating $\frac{rt}{100}$ you will need to use the Product Rule. Thus $$\frac{dP}{dt}=200 e^{\frac{rt}{100}}\frac{1}{100}\left(r+\frac{dr}{dt}\right).$$

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oh! Thank you so much! –  user56852 Feb 28 '13 at 23:32
    
You are welcome. Everything should work out fine now. –  André Nicolas Feb 28 '13 at 23:45

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