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

  1. #1
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    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
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  2. #2
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    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.
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  3. #3
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    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?
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    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.
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  5. #5
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    Quote Originally Posted by Devi09 View Post
    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 \frac{f(x+h)- f(x)}{h}. 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 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.

    Here, A= 18999 and B= .93 so [tex]AB^5= (18999)(0.6956883693)= 13217.3833283307 so that
    \frac{13217.3833283307(.93)^h- 13217.3833283307}h= 13217.3833283307\frac{(.93)^h- 1}{h}.

    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).
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