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Find the curvature of the ellipse $c(t)=(a\cos(t),b\sin(t))$ for $t \in [0,2\pi)$ and $a \neq b$

Now I understand that $k=\frac{||c' \times c''||}{||c'||^3}$

$c'=(-a\sin(t), b\cos(t))$

$c''=(-a\cos(t), -b\sin(t))$

$c' \times c''=ab\sin^2(t)+ab\cos^2(t)=ab$ so $||c' \times c''||=ab$

$||c'||^3=(\sqrt{a^2\sin^2(t)+b^2\cos^2(t)})^3=(a^2\sin^2(t)+b^2\cos^2(t))^{\frac{3}{2}}$

Therefore $k=\frac{ab}{(a^2\sin^2(t)+b^2\cos^2(t))^{\frac{3}{2}}}$

IS this now complete?

  • 3
    $|c'|^3$ should read $$\left( a^2\sin^2 t+b^2\cos^2 t \right)^{\frac{3}{2}}$$ – 2017-01-11
  • 0
    Why you have eliminated the square rot in $||c'||^3$? That's your mistake. – 2017-01-11

0 Answers 0