Hi all,

I'm doing my final year maths project, and it has appeared to be a bit beyond my limits... It would be great if you helped me somehow!

So, I have 3 points from a graph and a general equation. With this data, I have to find a, b, and c of this equation.

The equation is:

While the points I'm given are: (1,0) and (4.5, 0.5) and (7.15,0.7)

As far as I know, I have to substitute x and y with given coordinates of points and solve a three unknowns system of equations. But to do so, I have to (probably) transform the equations to a form of 'a+b+c = x'.

Can you please tell me if I'm correct and whether it can be done?

Best wishes!

What does $*$ mean? Is that multiplication? If so, then you wind up with a logical contradiction.

This is because $e^x > 0$ for all $x\in \mathbb{R}$. So, if $y=0$, then $c=0$. If $c=0$, then $y=0$ for all $x$. Are you missing a decimal?

Edit: I will assume that the first point you list is incorrect. Instead, you have the following three points:
$(1,y_0), (4.5,0.5),(7.15,0.7)$.

This gives the following equations:

Equation 1: $\dfrac{y_0}{c} = e^{-a(1-b)^2}$
Equation 2: $\dfrac{0.5}{c} = e^{-a(4.5-b)^2}$
Equation 3: $\dfrac{0.7}{c} = e^{-a(7.15-b)^2}$

Take ln of each side:

Equation 1a: $\ln y_0 - \ln c = -a(1-b)^2$
Equation 2a: $\ln 0.5 - \ln c = -a(4.5-b)^2$
Equation 3a: $\ln 0.7 - \ln c = -a(7.15-b)^2$

Thanks for such a quick answer!

Yes, * is multiplication. Also, you're absolutely right - I've rewritten a wrong point. All the coordinates are: (4.5, 0.5) and (6.33, 0.9) and (7.15,0.7).

Originally Posted by wtw
Thanks for such a quick answer!

Yes, * is multiplication. Also, you're absolutely right - I've rewritten a wrong point. All the coordinates are: (4.5, 0.5) and (6.33, 0.9) and (7.15,0.7).
Then, you wind up with the equations:

Eq 1: $0.5 = c\cdot e^{-a(4.5-b)^2}$
Eq 2: $0.9 = c\cdot e^{-a(6.33-b)^2}$
Eq 3: $0.7 = c\cdot e^{-a(7.15-b)^2}$

We can rewrite these as:

Eq 1a: $\ln 0.5 - \ln c = a(4.5-b)^2$
Eq 2a: $\ln 0.9 - \ln c = a(6.33-b)^2$
Eq 3a: $\ln 0.7 - \ln c = a(7.15-b)^2$

No, you don't want "a+ b+ c= x". For one thing a, b, and c are constants and x is a variable. Do exactly what you said, putting the numbers you are given into the general equation you get
First point: x= 4.5, y= 0.5 so this becomes $0.5= ce^{-a(4.5- b)^2}$
Second point: x= 6.33, y= 0.9 this becomes $0.9= ce^{-a(6.33- b)^2}$
Third point: x= 7.16, y= 0.7 so this becomes $0.7= ce^{-a(7.16- b)^2}$

Three equations to solve for three values, a, b, and c. Because of the exponentials I would start by taking the logarithm of both sides of each equation:
$log(0.5) log(c)- a(4.5- b)^2$
$log(0.9)= log(c)- a(6.33-b)^2$
$log(0.7)= log(c)- a(7.16- b)^2$

Subtracting the first equation from the second eliminates c:
$log(0.9)- log(0.5)= a((4.5- b)^2- (6.33- b)^2= a(20.25- 9b+ b^2- 40.0689+ 12.66b- b^2= a(3.66b-19.8189)$

Subtracting the third equation from the second also eliminates c:
$log(0.9)- log(0.7)= a((7.16- b)^2- (6.33- b)^2)= a(51.2656- 14.32b+ b^2- 40.0689+ 12.66b- b^2)= a(11.1967- 1.66b)$

Dividing the first of those two equations by the second eliminates a leaving an equation for b:
$\frac{log(0.9)- log(0.5)}{log(0.9)- log(0.7)}= \frac{3.66b- 19.8189}{11.1967- 1.66b}$

That should be easy to solve.