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.
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.
$\displaystyle \kappa = \frac{|x'y''-y'x''|}{((x')^2+(y')^2)^{\frac{3}{2}}}$
This gives
$\displaystyle \kappa = \frac{|(-a\sin(t))(-b\sin(t))-(b\cos(t))(-a\cos(t))|}{((-a\sin(t))^2+(b\cos(t))^2)^{\frac{3}{2}}}=\frac{2ab }{(a^2\sin(t)+b^2\cos(t))^{\frac{3}{2}}}$
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
$\displaystyle \left( \frac{x}{a} \right)^2=\cos^2(t) \quad \left( \frac{y}{b}\right)^2=\sin^2(t)$
$\displaystyle \ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos^2(t) +\sin^2(t)=1$
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:
which, because , simplifies to:
Test: if a = b = r for a circle,
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.