# Thread: Closest point on a curve

1. ## Closest point on a curve

Hi, I'm trying to solve the following problem:

Find the point on the parabola $y=x^2$ that is closest to the point (3, 0).

2. Hello Stroodle
Originally Posted by Stroodle
Hi, I'm trying to solve the following problem:

Find the point on the parabola $y=x^2$ that is closest to the point (3, 0).

We still seem to be having problems with LaTex at present, so I'll write it without. The distance, D, of the point (x, y) from (3, 0) is given by
D^2 = (x-3)^2 + y^2
Now replace y by x^2, and expand to get:
D^2 = x^4 + x^2 - 6x + 9
Differentiate and put the result equal to zero:
4x^3 + 2x - 6 = 0
which has a solution x = 1.

The second derivative is positive at this point, giving a minimum value of D^2 (and hence D) at the point (1, 1).

Hello StroodleWe still seem to be having problems with LaTex at present, so I'll write it without. The distance, D, of the point (x, y) from (3, 0) is given by
D^2 = (x-3)^2 + y^2
Now replace y by x^2, and expand to get:
D^2 = x^4 + x^2 - 6x + 9
Differentiate and put the result equal to zero:
4x^3 + 2x - 6 = 0
which has a solution x = 1.

The second derivative is positive at this point, giving a minimum value of D^2 (and hence D) at the point (1, 1).

I figured I'd take the time to write this out in latex to assit with clarity, and because I'm bored. I found a site to generate latex images, then I paste the URL here to get the latex equations. Its a pain but it works. So, as the Grandad pointed out above, from the pythagorean theorem and intuition (or the distance formula), the distance from the points given is:

$D&space;=&space;\sqrt{(x-3)^2&space;+&space;(y-0)^2}\;&space;\rightarrow&space;\;D^2&space;=&space;(x-3)^2+(y-0)^2\;&space;D^2&space;=&space;(x-3)^2+y^2$

Since by the definition of our equation $f(x)&space;=&space;x^2$ and the fact that the point can be represented as $(x,&space;f(x))$ we can replace the y with the x^2 as in the following:

$D^2&space;=&space;(x-3)^2&space;+(x^2)^2&space;=&space;(x-3)^2+x^4&space;=&space;x^2-3x-3x+9+x^4&space;=&space;x^4+x^2-6x+9$

Then we differentiate and set the derivitive equal to zero to find posibile soultions (i.e. the critical points). This is done here:

First note that:

$D=\sqrt{x^4+x^2-6x+9}$

And the derivitive is:

$\frac{dD}{dx}=\frac{1}{2}\;\frac{1}{\sqrt{x^4+x^2-6x+9}}(4x^3+2x-6)=\frac{4x^3+2x-6}{2\sqrt{x^4+x^2-6x+9}}$

I think this derivitive is diffirent than what the previous poster put before me, but this is because he took a different approach, the answer will be the same either way. Anyway, the zero of this function is 1. Thus, the distance is shortest at x=1, and this distance is the following:

$D&space;=&space;\sqrt{1^4+1^2-6(1)+9}&space;=&space;\sqrt{2-6+9}&space;=&space;\sqrt{3+2}&space;=&space;\sqrt{5}&space;=&space;2.2360679774998.....$

And thats the answer, I do believe.

4. Originally Posted by mfetch22
I figured I'd take the time to write this out in latex to assit with clarity, and because I'm bored. I found a site to generate latex images, then I paste the URL here to get the latex equations. Its a pain but it works.
FYI, LaTeX is back! Demo: $\displaystyle{\left(\beta mc^2 + \sum_{k = 1}^3 \alpha_k p_k \, c\right) \psi (\mathbf{x},t) = i \hbar \frac{\partial\psi(\mathbf{x},t) }{\partial t}}$

5. YESSSSSSS! Ahmen to latex.