Thread: Finding 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

I'm not totally sure. What is a here? Should that be t? If so, then what you want is for very small .

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?

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 .

Originally Posted by Devi09

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

You mean "when t= 5".

Would it start like this?

(a+5)-(5)/h

No, the "difference quotient" is . You forgot the "f"! Also, you have the "a" where you should have "h"

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

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

No, the first line above is correct but the second line is not. You cannot write - that's just not true. What you want is A(B^{5+ h})= A(B^5)*B^h.

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).