Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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
share|cite|improve this question
Use formula 13 here. – J. M. Oct 22 '11 at 8:29
That ain't much work; you still remember how to differentiate a sine and a cosine, no? – J. M. Oct 22 '11 at 9:24
Are you sure about that formula you got for $k$? You seem to have lost a square in there... – J. M. Oct 22 '11 at 12:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.