# Thread: Curvature of an ellipse

1. ## 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.

2. ## Re: Curvature of an ellipse

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
$\kappa = \frac{|(-a\sin(t))(-b\sin(t))-(b\cos(t))(-a\cos(t))|}{((-a\sin(t))^2+(b\cos(t))^2)^{\frac{3}{2}}}=\frac{2ab }{(a^2\sin(t)+b^2\cos(t))^{\frac{3}{2}}}$

3. ## Re: Curvature of an ellipse

Why is x(t)= acos(t) and y(t)=bsin(t), though?

4. ## Re: Curvature of an ellipse

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

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

$\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos^2(t) +\sin^2(t)=1$

5. ## Re: Curvature of an ellipse

Got it. Thanks a lot!

6. ## 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:

$\kappa = { { { | ( -a \sin ( t ) ) ( -b \sin ( t ) ) - ( b \cos ( t ) ) ( - a \cos ( t ) ) | } \over { ( ( - a \sin ( t ) ) {^2}+( b \cos ( t ) ) ^2) { ^ {3 \over 2 } } } } } ={{ a b ( \sin{^2} ( t ) +\cos{^2} ( t )) }\over { ( { a^2} \sin{^2} ( t )+{ b^2} \cos{^2}( t ) ) ^ {3 \over 2 } }$

which, because $\sin{^2} ( t ) +\cos{^2} ( t ) \equiv 1$, simplifies to:

$\kappa = { { a b }\over { ( { a^2} \sin{^2} ( t )+{ b^2} \cos{^2}( t ) ) ^ {3 \over 2 } } }$

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

$\kappa = { { r r }\over { ( { r^2} \sin{^2} ( t )+{ r^2} \cos{^2}( t ) ) ^ {3 \over 2 } } } = { { r ^ 2 } \over { ( r ^ 2 ) ^ { 3\over 2 } } } = { 1 \over r }$

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.

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