Question

Is this right? :/

Drawn inside:)

Answer 1

hahhaa @jdoe0001 :P

OS promptly crashed once I posted :/

but I figured that problem out!! do you get this one though? i'm not sure how to solve...

With yearly inflation of 3%, prices are given by

P(t)=P0(1.03)^t

where P0 is the price in dollars when t=0 and t is time in years.

Suppose P0=1. How fast (in cents/year) are prices rising when t=10 ?

**round to two decimal places!**

do you get this one?

Answer 2

they want you to evaluate
\[ \frac{d P}{dt} \]at t=10

Answer 3

ohh.. how do I do that?

Answer 4

with P0=1
\[ P= (1.03)^t \]
when you see the variable is an exponent, think "e"
\[ 1.03= e^{\ln(1.03)} \]
\[ P= e^{\ln(1.03)t} \]

now use
\[ \frac{d e^{ax}}{dx}= a e^{ax} \]

Answer 5

okay

so would it be like this?

(1.03)e^ln(1.03)10

?

Answer 6

no wait. the constant is ln(1.03) not just 1.03

Answer 7

ahh okay... so is this correct?

e^ln(1.03) = 1.03

ln(1.03)(1.03)10 = 0.0304(10)=0.304455 ?

did i enter that in correctly? if so, what happens now?

Answer 8

ln(1.03)* 1.03^10

1.03^10 = 1.3439..
ln(1.03)= 0.0295588...

0.0295588 * 1.3439 = 0.0397...

Answer 9

ohh okay oops i entered it in my calculator wrong!

so what happens next? :)

Answer 10

just to re-cap
\[ P= e^{\ln(1.03)t} \\ \frac{dP}{dt} = \ln(1.03) \cdot (1.03)^t

\]
evaluate at t=10 to find
\[ \frac{dP}{dt} = \ln(1.03) \cdot 1.3439= 0.0397 \]
P is changing are a rate of $0.04 per unit time or 4 cents/year

Answer 11

ohh okay i see... :)

awesome!!! Thank you!! this makes sense now! woo!! :)