Estimate the instantaneous rate of change in the equation 18999(0.93)^t when a = 5

Would it start like this?

(a+5)-(5)/h

18999(0.93)^(5+h) - 18999(0.93)^5 / h <<<not sure about this

13217^h - 13217 / h <<<<<not sure about this

- Feb 26th 2011, 08:01 PMDevi09Finding instantaneous rate of exponetial equation using (a+h)-(a)/h method
Estimate the instantaneous rate of change in the equation 18999(0.93)^t when a = 5

Would it start like this?

(a+5)-(5)/h

18999(0.93)^(5+h) - 18999(0.93)^5 / h <<<not sure about this

13217^h - 13217 / h <<<<<not sure about this - Feb 26th 2011, 09:58 PMragnar
I'm not totally sure. What is a here? Should that be t? If so, then what you want is for very small .

- Feb 26th 2011, 10:20 PMDevi09
Yeah the original equation is 18999(0.93)^t

So you don't have to simplify the the equation anymore after your 2nd one? Just plug in the very small h's right? - Feb 26th 2011, 10:46 PMragnar
Sure, the original function is that. But what's a? You say a = 5, but what does that mean? Do you mean to say that t = 5, i.e. you're investigating the instantaneous rate of change near where the function takes the value 5? If so, then the above is correct and there isn't a simplification of the last expression I wrote. So yeah, just plug in intsy-tintsy .

- Feb 27th 2011, 04:18 AMHallsofIvy
You mean "when t= 5".

Quote:

Would it start like this?

(a+5)-(5)/h

Quote:

18999(0.93)^(5+h) - 18999(0.93)^5 / h <<<not sure about this

13217^h - 13217 / h <<<<<not sure about this

Here, A= 18999 and B= .93 so [tex]AB^5= (18999)(0.6956883693)= 13217.3833283307 so that

.

But you are not finished. The "instantaneous rate of change" is the limit as h goes to 0. Assuming that limit exists, you can get a good approximation by using a value of h very close to 0, as ragnar suggests. It can, in fact, be shown that the limit does exist and is exactly equal to ln(.93).