# Finding instantaneous rate of exponetial equation using (a+h)-(a)/h method

• Feb 26th 2011, 07:01 PM
Devi09
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

• Feb 26th 2011, 08:58 PM
ragnar
I'm not totally sure. What is a here? Should that be t? If so, then what you want is $\frac{f(5+h)-f(5)}{h} = \frac{18999(0.93)^{5+h} - 18999(0.93)^{5}}{h}$ for very small $h$.
• Feb 26th 2011, 09:20 PM
Devi09
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, 09:46 PM
ragnar
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 $h$.
• Feb 27th 2011, 03:18 AM
HallsofIvy
Quote:

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".

Quote:

Would it start like this?

(a+5)-(5)/h
No, the "difference quotient" is $\frac{f(x+h)- f(x)}{h}$. You forgot the "f"! Also, you have the "a" where you should have "h"

Quote:

No, the first line above is correct but the second line is not. You cannot write $A(B^{5+h)}= (AB^5)^h$- that's just not true. What you want is A(B^{5+ h})= A(B^5)*B^h.
$\frac{13217.3833283307(.93)^h- 13217.3833283307}h= 13217.3833283307\frac{(.93)^h- 1}{h}$.