# Thread: help derive exponential function from limited information

1. ## 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 ?

2. Originally Posted by cubew00t
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 $a\approx 0.1352585526$

3. 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?

4. Hello cubew00t

Welcome to Math Help Forum!
Originally Posted by cubew00t
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.

5. 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.

6. Hello cubew00t
Originally Posted by cubew00t
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.

7. 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!

8. Hello cubew00t
Originally Posted by cubew00t
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
$y = \frac{kx}{100+k-x}$