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.

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- Oct 15th 2012, 04:07 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, 04:36 PMTheEmptySetRe: Curvature of an ellipse

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

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

This gives

$\displaystyle \kappa = \frac{|(-a\sin(t))(-b\sin(t))-(b\cos(t))(-a\cos(t))|}{((-a\sin(t))^2+(b\cos(t))^2)^{\frac{3}{2}}}=\frac{2ab }{(a^2\sin(t)+b^2\cos(t))^{\frac{3}{2}}}$ - Oct 15th 2012, 04:53 PMMajMinorRe: Curvature of an ellipse
Why is x(t)= acos(t) and y(t)=bsin(t), though?

- Oct 15th 2012, 04:58 PMTheEmptySetRe: Curvature of an ellipse
That is the parametric represention of an ellipse.

Note that

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

Now if square both equations and add them you get

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

$\displaystyle \ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos^2(t) +\sin^2(t)=1$ - Oct 15th 2012, 05:08 PMMajMinorRe: Curvature of an ellipse
Got it. Thanks a lot!

- Sep 24th 2013, 11:02 AMkeithlRe: 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.