# Thread: Finding Maximum and Minimum Curvature

1. ## Finding Maximum and Minimum Curvature

Show that the curvature is greatest at the endpoints of the major axis, and is least at the endpoints of the minor axis, for the ellipse given by $x^2 +4y^2 = 4$

To solve this problem, do I need to compute $K = \frac{|y''|}{[1+(y')^2]^{3/2}}$; and then find $\frac{dK}{dx}$, to determine if the critical values coincide with the end points of the ellipse?

2. ## Re: Finding Maximum and Minimum Curvature

Yes, that looks like a good plan!

3. ## Re: Finding Maximum and Minimum Curvature

All right. Here is where I am stuck:

$K = \frac{\frac{|4y-4x|}{16y^2}}{[1+x^2/16y^2]^{3/2}}$

4. ## Re: Finding Maximum and Minimum Curvature

You should just solve for y as a function of x,

$y=\frac{1}{4}\sqrt{4-x^2}$ and then plug the derivatives y' and y'' into the equation for K. The definition of curvature should be a function of x right? Then differentiate it with respect to x.

5. ## Re: Finding Maximum and Minimum Curvature

Oh, so we take the principle root, because the absolute value bars and the squared term remove the negative root?

6. ## Re: Finding Maximum and Minimum Curvature

Okay, here is where I am stuck:

$K = \frac{\frac{16}{(4-x^2)^{3/2}}}{[1+x^2/16(4-x^2)^{-1}]^{3/2}}}$

7. ## Re: Finding Maximum and Minimum Curvature

Originally Posted by Bashyboy
Oh, so we take the principle root, because the absolute value bars and the squared term remove the negative root?
Well, there will be two solutions because it's an ellipse with two endpoints with maximum curvature. I should have a +/- in front of there.

8. ## Re: Finding Maximum and Minimum Curvature

Originally Posted by Bashyboy
Okay, here is where I am stuck:

$K = \frac{\frac{16}{(4-x^2)^{3/2}}}{[1+x^2/16(4-x^2)^{-1}]^{3/2}}}$
Well how did you get here? Are you trying to differentiate it now that you have K? It looks real messy and it is difficult for me to sanity check because I don't know what derivatives you plugged in. the first and second derivatives I get for the function $y=\frac{1}{4}\sqrt{4-x^2}$ are $y'=-\frac{x}{4\sqrt{4-x^2}}$ and then $y''=-\frac{1}{(4-x^2)^{\frac{3}{2}}}$

Ultimately you will have to plug those into the formula for K

9. ## Re: Finding Maximum and Minimum Curvature

Here's what it looks like when you plug them in:

$K = \frac{\frac{1}{(4-x^2)^{3/2}}}{[1+(\frac{-x}{4\sqrt{4-x^2}})^2]^{3/2}}}$

It can be toyed with, I won't go through all of it but I will try to help with the first bit of the algebra, although I wonder if there is an easier way to accomplish this?

$K = \frac{\frac{1}{(4-x^2)^{3/2}}}{[1+\frac{x^2}{16(4-x^2)}]^{3/2}}}$

$= \frac{[\frac{1}{(4-x^2)}]^{3/2}}{[\frac{16(4-x^2)+x^2}{16(4-x^2)}]^{3/2}}$

Now notice the denominator and numerator have the same power. So you can just multiply by the inverse of the denominator and cancel out a (4-x^2) factor:

$K=[\frac{16}{16(4-x^2)+x^2}]^{3/2}$

Looks more manageable now huh?

Check and make sure you agree with everything though, but this looks like a function you can readily differentiate and look at critical numbers.