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how can I compute the curvative of an ellipse given by $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

do i need to take $x=a \cos(t)$ and $y=b \sin(t)$?

please show me a way how to solve this? thank you for helping:)

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2 Answers 2

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We have $\alpha'(t) = \langle -a \sin t, b \cos t \rangle$ and $\alpha''(t) = \langle -a\cos t , - b \sin t\rangle$, thus $|\alpha'(t)| = \sqrt{a^2\sin^2t + b^2\cos^2t}$. We have that $T(t) = \frac{\alpha'(t)}{|\alpha'(t)|},$ which has length $1$ and is tangent to $\alpha (t).$ $$T(t) = \langle \frac{-a \sin t}{\sqrt{a^2\sin^2t + b^2\cos^2t}}, \frac{b\cos t}{\sqrt{a^2\sin^2t + b^2\cos^2t}}\rangle$$ which leads to $$\kappa = \frac{|T'(t)|}{|\alpha'(t)| }= \frac{ab}{(\sqrt{a^2\sin^2t + b^2\cos^2t})^3}.$$

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    $\begingroup$ Since $\alpha$ is not parameterized by arc length, you cannot naively calculate the curvature that way. In fact, $κ = \frac{x’y’’- x’’y’}{((x’)^2 + (y’)^2)^{3/2}}$ $\endgroup$ Commented Mar 13, 2018 at 17:13
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    $\begingroup$ @Jsevillamol Check your formula it will match this answer if you plug in the values and simplify $\endgroup$
    – Makogan
    Commented May 28, 2020 at 7:18
  • $\begingroup$ @Jsevillamol Your comment is wrong. If the parameter is arc length $s$, then $\kappa(s) = |T'(s)|$. For a general parameter $t$ and parametrization $\alpha$, then the chain rule gives $\kappa(t) = |T'(t)|/s'(t) = |T'(t)|/|\alpha'(t)|$, as in their answer. $\endgroup$ Commented Jun 2, 2023 at 12:48
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You don't need the unit tangent to get the curvature or parameterization by arc length. It is much simpler to use the following formula: $$ \kappa = \frac{||v \times v'|| }{||v||^{3}},$$ where $$ v = (-a\sin(t),b \cos(t)) \qquad \text{and} \qquad v' = (-a\cos(t),-b\sin(t)) $$ and hence $$ ||v|| = \sqrt{(a^{2}\sin^{2}(t) + b^{2}\cos^{2}(t))} $$ and \begin{align*} v \times v' & = (-a\sin(t), b \cos(t),0) \times (-a \cos(t), -b \sin(t),0) \\ & = (0, 0, a b \sin^{2}(t)+ab\cos^{2}(t)) =(0,0,a b), \end{align*} so $$ ||v \times v'|| = ab.$$

Therefore

$$ \kappa = \frac{||v \times v'|| }{||v||^{3}} = \frac{ab}{\left(\sqrt{a^{2}\sin^{2}(t) + b^{2}\cos^{2}(t)}\right)^{3}} $$

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