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Math Help - help derive exponential function from limited information

  1. #1
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    Wink help derive exponential function from limited information

    Hello, nice to meet you all, I've been working on a spreadsheet that requires me to create a function to determine a value from a exponential curve where the area of the curve is variable . The terminal points, 0,0 and 100,100 remain constant and the curve is symmetrical at the 45degree angle. How can I derive a function knowing only these things. Thanks!

    from asking other places this is probably not an exponential function, this image better describes what i am looking for:

    can this be expressed mathematically to solve for x and y ?
    Last edited by cubew00t; December 5th 2009 at 11:15 AM.
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  2. #2
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    Quote Originally Posted by cubew00t View Post
    Hello, nice to meet you all, I've been working on a spreadsheet that requires me to create a function to determine a value from a exponential curve where the area of the curve is variable . The terminal points, 0,0 and 100,100 remain constant and the curve is symmetrical at the 45degree angle. How can I derive a function knowing only these things. Thanks!
    from asking other places this is probably not an exponential function, this image better describes what i am looking for:
    can this be expressed mathematically to solve for x and y ?
    I tried to get an exponential function, but I don't know if it is symmetric to the line y = -x+100

    1. Area A = 15% * 10,000 = 1500

    2. The general form of the exponential function is:

    f(x) = a \cdot e^{k \cdot x} + c

    3. Since f(0) = 0~\implies~ c = -a

    4. f(100) = 100:

    100=a \cdot e^{k \cdot 100} - a~\implies~k=\frac1{100} \cdot \ln\left(\frac{100}{a}+1\right)

    5. A=\int_0^{100} f(x) dx=1500

    A=\int_0^{100} \left(a \cdot e^{\frac1{100} \cdot \ln\left(\frac{100}{a}+1\right) \cdot x} - a  \right) dx=1500

    \int_0^{100} \left(a \cdot e^{\frac1{100} \cdot \ln\left(\frac{100}{a}+1\right) \cdot x}\right) dx - \int_0^{100}(a  ) dx=1500

    a \dfrac1{\frac1{100} \cdot \ln\left(\frac{100}{a}+1\right)} \int_0^{100} \left(\frac1{100} \cdot \ln\left(\frac{100}{a}+1\right) \cdot e^{\frac1{100} \cdot \ln\left(\frac{100}{a}+1\right) \cdot x}\right) dx - \int_0^{100}(a  ) dx=1500

     \dfrac{100a}{ \ln\left(\frac{100}{a}+1\right)} \left[ e^{\frac1{100} \cdot \ln\left(\frac{100}{a}+1\right) \cdot x}  \right]_0^{100}  - [ax]_0^{100}=1500

     \dfrac{100a}{ \ln\left(\frac{100}{a}+1\right)} \left[ \frac{100}{a}+1 - 1  \right]  - [100a]=1500

    Solve for a. I used a calculator.

    I've got a\approx 0.1352585526
    Attached Thumbnails Attached Thumbnails help derive exponential function from limited information-expo_spiegel.png  
    Last edited by earboth; December 5th 2009 at 12:29 PM.
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  3. #3
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    wow thank you, that was amazing, so amazing that it shot right over my head

    i have come to believe that it is not an exponential function that describes the second image but perhaps a parabola at a 45 degree angle? Is this possible and what would be the function derived from knowing just the area?
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  4. #4
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    Hello cubew00t

    Welcome to Math Help Forum!
    Quote Originally Posted by cubew00t View Post
    Hello, nice to meet you all, I've been working on a spreadsheet that requires me to create a function to determine a value from a exponential curve where the area of the curve is variable . The terminal points, 0,0 and 100,100 remain constant and the curve is symmetrical at the 45degree angle. How can I derive a function knowing only these things. Thanks!

    can this be expressed mathematically to solve for x and y ?
    I have a solution for you that seems to satisfy all your requirements including symmetry about the line x+y =100. It is:
    y = \frac{-kx}{x-(100+k)},\; k > 0,\; 0\le x\le 100
    The area under the curve is:
    \int^{100}_0\frac{-kx}{x-(100+k)}dx = k(100+k)\ln\left(\frac{100+k}{k}\right)-100k
    Here are a few sample values.
    k=1; area = 3.7%

    k=10; area = 16.4%

    k= 50; area = 32.4%
    The curve is not, of course, exponential. It's part of a rectangular hyperbola, with asymptotes x = 100+k and y = -k. The curve contains the points (0,0) and (100,100) and is symmetrical about the line x+y=100. (You can check this by replacing x by 100-x in the function: the same values of y will be produced in the reverse order.)

    I attach a screen shot of the graphs for k = 1, 10, 50 and an Excel spreadsheet with the calculations that produced these graphs.

    Grandad
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  5. #5
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    Thank you this is exactly what i was looking for!

    Edit: if all the information I have is Area, how can I deduce a function that works in excel? For example, I input Area and it on-the-fly changes the values that affect the shape hyperbola.

    Thank you for your efforts again.
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  6. #6
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    Hello cubew00t
    Quote Originally Posted by cubew00t View Post
    Thank you this is exactly what i was looking for!

    Edit: if all the information I have is Area, how can I deduce a function that works in excel? For example, I input Area and it on-the-fly changes the values that affect the shape hyperbola.

    Thank you for your efforts again.
    I attach a revised version of the spreadsheet. You'll see that cell H1 contains a value of k that can be varied to give the function values in column F, producing the % area in cell G24. You can use Tools -> Goal Seek to find the value of k for a given percentage area. In the spreadsheet attached, I've set this to 15%, giving a value of k as 8.513. I've also modified the chart so that just this single (variable) set of data is displayed.

    Grandad
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    MrGrandad you are truly a godsend, and I will take this opportunity to impose upon you for the last time (lol).

    I see the spreadsheet you have graciously created however it's like eating fish without having learned how to fish. It is too complicated for me to reverse engineer to get what I specifically need. Can you modify it slightly to allow me to input Area, that will affect k and not the other way around?

    =(H1*(100+H1)*LN((100+H1)/H1)-100*H1)/100

    This is beyond me to reverse engineer. While I await your response, I will keep trying!
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  8. #8
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    Hello cubew00t
    Quote Originally Posted by cubew00t View Post
    MrGrandad you are truly a godsend, and I will take this opportunity to impose upon you for the last time (lol).

    I see the spreadsheet you have graciously created however it's like eating fish without having learned how to fish. It is too complicated for me to reverse engineer to get what I specifically need. Can you modify it slightly to allow me to input Area, that will affect k and not the other way around?

    =(H1*(100+H1)*LN((100+H1)/H1)-100*H1)/100

    This is beyond me to reverse engineer. While I await your response, I will keep trying!
    It's not possible to produce an explicit formula for an equation like this. (It's not merely difficult, it really isn't possible!) So you'll have to rely on a tool like Goal Seek in a spreadsheet to do the job for you. It seems from you last posting that you're not familiar with this superb spreadsheet tool. (It's worth finding out how to use it!)

    I don't know which version of Excel you have, and whether the attached spreadsheet will work (I uninstalled Excel 2007 from my machine because I hated it so much, and went back to my ancient Excel 2000 which I love!), but I have now added some VBA code to do the Goal Seek for you, if you have a compatible version. Just press <Ctrl>+g to run the code; enter the area required (no percentage sign, just a number between 1 and 49), and the value of k will appear in cell D1.

    If the code doesn't work, you'll have to find out how to use Goal Seek and do it manually. It's worth the effort - it's the only way to solve a problem like this.

    I have also simplified the spreadsheet, and the function to the equivalent
    y = \frac{kx}{100+k-x}
    Grandad
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  9. #9
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    Thumbs up

    Thank you so much for your efforts, had I realized that it was impossible to reverse the formula I wouldn't have spent the couple hours on it that I did. Hahah. You are much too patient and generous with your time sir.
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