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Math Help - How to solve this simple equation involving an exponential function?

  1. #1
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    How to solve this simple equation involving an exponential function?

    So given these two functions:

    f(x) = e^(-2x + 3)
    g(x) = √(ax)

    how can you find the value of a, given that the two curves intersect perpendicularly??
    the way i see it is that you have to many unknowns (i equated the derivatives).


    Help much appreciated!!! thanks!!!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yehia View Post
    So given these two functions:

    f(x) = e^(-2x + 3)
    g(x) = √(ax)

    how can you find the value of a, given that the two curves intersect perpendicularly??
    the way i see it is that you have to many unknowns (i equated the derivatives).


    Help much appreciated!!! thanks!!!
    The solution to the intersection point can only be found numerically so it is next to impossible to find an expression for a that way. However we can be tricky.

    First we are looking for a point x such that f(x) = g(x). That is we have a point x = b on the functions f(x) and g(x) such that
    e^{-2b + 3} = \sqrt{ab}

    Now, perpendicular slopes follow the rule that m' = -1/m. So we are looking for a at the intersection point such that
    g'(x) = -1/f'(x)

    Evaluate this at x = b and see if you can use the intersection condition to write this equation as a function of a only.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    The solution to the intersection point can only be found numerically so it is next to impossible to find an expression for a that way. However we can be tricky.

    First we are looking for a point x such that f(x) = g(x). That is we have a point x = b on the functions f(x) and g(x) such that
    e^{-2b + 3} = \sqrt{ab}

    Now, perpendicular slopes follow the rule that m' = -1/m. So we are looking for a at the intersection point such that
    g'(x) = -1/f'(x)

    Evaluate this at x = b and see if you can use the intersection condition to write this equation as a function of a only.

    -Dan
    Dan, I did EXACTLY that. i ended up with BOTH those equations, BUT both happened to be the same. can you please show me the way forward or at least do it yourself and see if you get an answer?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yehia View Post
    Dan, I did EXACTLY that. i ended up with BOTH those equations, BUT both happened to be the same. can you please show me the way forward or at least do it yourself and see if you get an answer?
    f(x) = e^{-2x + 3} \implies f'(x) = -2e^{-2x + 3}

    g(x) = \sqrt{ax} \implies g'(x) = \frac{\sqrt{a}}{2\sqrt{x}}

    At the intersection point we have f(b) = g(b), or
    e^{-2b + 3} = \sqrt{ab}

    Also we know that g'(b) = -1/f'(b), so
    \frac{\sqrt{a}}{2\sqrt{b}} = \frac{1}{2 e^{-2b + 3}}

    But what does the intersection point equation say that e^{-2b + 3} is equal to?

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    f(x) = e^{-2x + 3} \implies f'(x) = -2e^{-2x + 3}

    g(x) = \sqrt{ax} \implies g'(x) = \frac{\sqrt{a}}{2\sqrt{x}}

    At the intersection point we have f(b) = g(b), or
    e^{-2b + 3} = \sqrt{ab}

    Also we know that g'(b) = -1/f'(b), so
    \frac{\sqrt{a}}{2\sqrt{b}} = \frac{1}{2 e^{-2b + 3}}

    But what does the intersection point equation say that e^{-2b + 3} is equal to?

    -Dan
    Ok, i went on and now i end up with:

    b^5 = e^(12-8b)
    Last edited by topsquark; May 21st 2011 at 07:51 AM. Reason: Removed begging
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yehia View Post
    Ok, i went on and now i end up with:

    b^5 = e^(12-8b)
    Please show us how you got that equation. That will help us pinpoint the error.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    Please show us how you got that equation. That will help us pinpoint the error.

    -Dan
    no
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yehia View Post
    no
    Then you no longer require help. Thread closed.

    -Dan
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