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Math Help - Average rate and instantaneous rate?

  1. #1
    Member nautica17's Avatar
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    Average rate and instantaneous rate?

    Hi everyone! I happen to be stumped on this question below.

    A function y=f(x) and values of x0 and x1 are given.

    y= x^3 , x0= 1 , x1= 2

    a. Find the average rate of change of y with respect to x.

    b. Find the instantaneous rate of change of y with respect to x at x0.

    For a. I got 7 as my answer. I'm curious what everyone else gets. For b. I am stumped and know that you have to find a limit probably. Showing steps would be appreciated as I'd like to see how this is done.

    (NOTE: We are not allowed to use derivatives yet, so it all has to be limits and stuff.)
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  2. #2
    Member eXist's Avatar
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    For instantaneous rate of change:

    \lim_{h\rightarrow0} \frac{f(x + h) - f(x)}{h}

    But now your setting f(x) = x^3

    So you have something like this:

    \lim_{h\rightarrow0} f(x) = \frac{(x + h)^3 - x^3}{h}

    Hope this helps. By the way, this is the definition of a derivative.
    Last edited by eXist; August 30th 2009 at 12:34 PM.
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  3. #3
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    You probably know that to find the average rate you simple find the change in y and divide it by the change in x:

    \mbox{average rate}=\frac{f(x_1) - f(x_0)}{x_1 - x_0}

    Replacing the values we get:

    \mbox{average rate}=\frac{2^3 - 1^3}{2 - 1}=7

    To get the instantaneous rate you do need to find a limit. It is intuitive that we should take the limit of x1 going to x0, to find the instantaneous rate at x0.

    \mbox{instantaneous rate}=\begin{array}{c}lim\\x_1\rightarrow x_0\end{array} \frac{f(x_1) - f(x_0)}{x_1 - x_0}

    That is:

    \mbox{instantaneous rate}=\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1}

    Can you find this limit?
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  4. #4
    Member nautica17's Avatar
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    Okay so I got two answers. I don't know which is correct. If I go by what eXist says, then I get 343, but if I go by what pedrosorio says then I get 0. I sort of get the concept but I don't know what to use now. None of the answers make sense to me.

    Can anyone nudge me a bit more in the right direction, please?
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  5. #5
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    You are doing something wrong when you find the limits. Our approaches are equivalent.

    The 'h' in eXist's approach is equal to my 'x1 - x0', eXist has made a mistake and put lim x->0, when he meant lim h->0.
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  6. #6
    Member nautica17's Avatar
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    Okay I got 3 as my answer now. Is this correct or did I mess up again?
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  7. #7
    Member eXist's Avatar
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    Quote Originally Posted by pedrosorio View Post
    You are doing something wrong when you find the limits. Our approaches are equivalent.

    The 'h' in eXist's approach is equal to my 'x1 - x0', eXist has made a mistake and put lim x->0, when he meant lim h->0.
    Thanks for the correction.
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  8. #8
    Member eXist's Avatar
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    You should be getting 3 as your answer (Sorry, 0 was a typo, I had 0 up here earlier, so your answer is correct).

    If you actually take the derivative (which I know you can't use right now) you get:

    \frac{d}{dx}(x^3) = 3x^2\,dx = 3(1)^2 = 3

    Now, to do it without derivatives, you need to factor the top of this:

    \begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1}

    Now to factor two cubes (this is for any case):

    a^3 - b^3 = (a - b)(a^2 + ab + b^2)

    So what you have is:


    \begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1} = <br />
\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{(x_1 - 1)(x_1^2 + x_1 + 1)}{x_1 - 1} =<br />
\begin{array}{c}lim\\x_1\rightarrow 1\end{array} (x_1^2 + x_1 + 1) = 3<br />

    Sorry, I had a few typos . But your answer of 3 is correct. Hopefully all my work is correct now. Sorry again for any confusion.
    Last edited by eXist; August 30th 2009 at 12:48 PM.
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  9. #9
    Member nautica17's Avatar
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    Cool, thank-you guys! You're the best! Finally, I'm learning something in calculus. Great explanations too.
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  10. #10
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    Quote Originally Posted by eXist View Post
    You should be getting 3 as your answer (Sorry, 0 was a typo, I had 0 up here earlier, so your answer is correct).

    If you actually take the derivative (which I know you can't use right now) you get:

    \frac{d}{dx}(x^3) = 3x^2\,dx = 3(1)^2 = 3

    Now, to do it without derivatives, you need to factor the top of this:

    \begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1}

    Now to factor two cubes (this is for any case):

    a^3 - b^3 = (a - b)(a^2 + ab + b^2)

    So what you have is:


    \begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{x_1^3 - 1^3}{x_1 - 1} = <br />
\begin{array}{c}lim\\x_1\rightarrow 1\end{array} \frac{(x_1 - 1)(x_1^2 + x_1 + 1)}{x_1 - 1} =<br />
\begin{array}{c}lim\\x_1\rightarrow 1\end{array} (x_1^2 + x_1 + 1) = 3<br />

    Sorry, I had a few typos . But your answer of 3 is correct. Hopefully all my work is correct now. Sorry again for any confusion.
    Should use the right notation of \begin{array}{c}lim\\x\rightarrow 1\end{array} \frac{x^3 - 1^3}{x - 1} (for all other limits as well, since x_1 is a constant).

    :P
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  11. #11
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    Oops
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