Solution for my equation using matlab is different from what i get manually.

The equation above is the one I'm working on. I wrote the equation in Matlab as follows;

s = sqrt ( (x-10)^2 + (sqrt(9-x^2)-2)^2 ) + sqrt( (10-x)^2 + (-2-sqrt(9-x^2))^2 )

Then I differentiate s in term of x, and equate ds/dx = 0. When I solve for x, I expected the value to be 3, as that's the solution I got manually. However Matlab return the value x = 10.

Did I make any mistake in writing the equation for s?

Answer 1

[Walter Roberson]

No, definitely not 3. 3 leads to 0/0 for diff(s(x),x), requiring much finer analysis to show that ds at 3 is -1036*sqrt(53)*(1/2809) . The 10 value is correct.

You can plot the derivative to see that it is not 0 at 3.

[Haziq Shibli]

Thanks for your reply. Really appreciate it, and I get what you mean. However, I'm still stucked at the problem, and I think it might be clearer to you if I explain the problem from the beginning.

I am looking for the point R on the circle, that results in the shortest total distance between S and R, and R and F. I used the equation above to determine the value x of point R.

As we can see, point x should be equal to 3 as we expected, not 10. Can you point out any flaws in my workings?

Thank you in advance.

[Walter Roberson]

Calculus does not care if some of your intermediate values go complex. 10 is where the minimum distance occurs. It just happens that the minimum distance occurs with complex y.

To restrict to real values then you need to consider ds over the range of values for which sqrt(9-x^2) is real, namely x = -3 to +3. Check to see which edge of the boundary gets you closer to zero for ds, and it becomes apparent it is x = 3.

[Image Analyst]

You forgot to include your formula for the derivative of s, and the MATLAB code you used to solve for setting that to 0.

[Walter Roberson]

The formula they used for s is correct. It is sum of cartesian distances between point (x,y) and points S and F, followed by the substitution x^2 + y^2 = 9 implies y = sqrt(9 - x^2)

It is just that using calculus to minimize by looking for derivative being zero does not take into account constraints that y must be real-valued. The point with x = 10 that is found really does have the minimum total distance... it just has imaginary y.