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.

Printable View

- Oct 15th 2012, 05:07 PMMajMinorCurvature 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, 05:36 PMTheEmptySetRe: Curvature of an ellipse
- Oct 15th 2012, 05:53 PMMajMinorRe: Curvature of an ellipse
Why is x(t)= acos(t) and y(t)=bsin(t), though?

- Oct 15th 2012, 05:58 PMTheEmptySetRe: Curvature of an ellipse
- Oct 15th 2012, 06:08 PMMajMinorRe: Curvature of an ellipse
Got it. Thanks a lot!

- Sep 24th 2013, 12:02 PMkeithlRe: 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.