Vector equation
$r(t)=2\cos(t)\mathbf i + 3\sin(t)\mathbf j\ \ (0 \le t \le 2\pi)$
represents ellipse.
I need to find curvature of this ellipse on endpoints of x and y axis that are given with $(2,0),(-2,0)$ and $(0,3),(0,-3)$ respectively.
Work so far:
- I need to evaluate curvature $k$ at $t=0$ and $t=\pi/2$
- $k = {{x'y''-y'x''}\over{(x'^2 + y'^2)^{3/2}}}$
- x and y for plugging to above equation are $x(t)=2\cos(t)$ and $y=3\sin(t)$
- After differentiating we have $k = {{6}\over{(5\cos(t)+4)^{3/2}}}$
- After evaluating for $t=0$ and $t=\pi/2$ we have $2\over9$ and $3\over4$ respectively