# Curvature of an ellipse

• Oct 15th 2012, 04:07 PM
MajMinor
Curvature of an ellipse
I'm asked to calculate the curvature of an ellipse:

(x^2/a^2)+(y^2/b^2)=1

The professor is setting x(t) = acos(t) and y(t) = bsin(t)

but I don't know why.

I think I can calculate the curvature from there.

Any help is appreciated.
• Oct 15th 2012, 04:36 PM
TheEmptySet
Re: Curvature of an ellipse
Quote:

Originally Posted by MajMinor
I'm asked to calculate the curvature of an ellipse:

(x^2/a^2)+(y^2/b^2)=1

The professor is setting x(t) = acos(t) and y(t) = bsin(t)

but I don't know why.

I think I can calculate the curvature from there.

Any help is appreciated.

You can use this defintion to to calculate the curvature or a parametrically defined curve.

$\kappa = \frac{|x'y''-y'x''|}{((x')^2+(y')^2)^{\frac{3}{2}}}$

This gives
$\kappa = \frac{|(-a\sin(t))(-b\sin(t))-(b\cos(t))(-a\cos(t))|}{((-a\sin(t))^2+(b\cos(t))^2)^{\frac{3}{2}}}=\frac{2ab }{(a^2\sin(t)+b^2\cos(t))^{\frac{3}{2}}}$
• Oct 15th 2012, 04:53 PM
MajMinor
Re: Curvature of an ellipse
Why is x(t)= acos(t) and y(t)=bsin(t), though?
• Oct 15th 2012, 04:58 PM
TheEmptySet
Re: Curvature of an ellipse
Quote:

Originally Posted by MajMinor
Why is x(t)= acos(t) and y(t)=bsin(t), though?

That is the parametric represention of an ellipse.

Note that

$\frac{x}{a}=\cos(t) \quad \frac{y}{b}=\sin(t)$

Now if square both equations and add them you get

$\left( \frac{x}{a} \right)^2=\cos^2(t) \quad \left( \frac{y}{b}\right)^2=\sin^2(t)$

$\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos^2(t) +\sin^2(t)=1$
• Oct 15th 2012, 05:08 PM
MajMinor
Re: Curvature of an ellipse
Got it. Thanks a lot!
• Sep 24th 2013, 11:02 AM
keithl
Re: Curvature of an ellipse
The empty set made a couple of small mistakes reducing the parameteric equation. I'm a newbie here, so I may get the markup wrong, but:

http://latex.codecogs.com/gif.latex?...{3 \over 2 } }

which, because http://latex.codecogs.com/gif.latex?...( t ) \equiv 1, simplifies to:

http://latex.codecogs.com/gif.latex?... \over 2 } } }

Test: if a = b = r for a circle,

http://latex.codecogs.com/gif.latex?... { 1 \over r }

For a circle, the curvature becomes 1/r as expected.

Thank you "Mr. Set" for showing me how to solve this, and an opportunity to help.