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Given the plane curve $(x(t),y(t)) = (2 \cos t, \sin t)$, the task is to find the radius of curvature at $(0,1)$. (The given point corresponds to time $t=\pi/2$.)

The radius $R$ is given by $1/\kappa$ where $\kappa$ is the curvature, $$ \kappa = \frac{|v \times a |}{ |v|^3}, $$ with $v$ and $a$ being the velocity and acceleration vectors, respectively. Taking the first and second derivative, one can obtain $v=(-2\sin t, \cos t)$ and $a=(-2\cos t, -\sin t)$. At the given point, $v = (-2,0)$ and $a=( 0,-1)$, so that $v\times a = (0,0,2)$ and $$ \kappa = \frac{2}{8} = \frac{1}{4} \Rightarrow R = 4.$$

However, my solution manual says that $R = 2$ with no technical explanation. Can you help me figure out what's wrong?

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  • $\begingroup$ $v\times a$ should have only two components. $\endgroup$ – user9464 Jan 17 at 21:00
  • $\begingroup$ Maybe I should've clearly mentioned it, but $v$ and $a$ are 3-dimentional, their third coordinate being $0$. $\endgroup$ – User32563 Jan 17 at 21:03
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Let's do it in general: $\alpha(t) = (2\cos t, \sin t)$ implies $\alpha'(t) = (-2\sin t, \cos t)$ as well as $\alpha''(t) = -\alpha(t)$, and thus $$R(t) = \frac{1}{|\kappa(t)|} = \frac{\|\alpha'(t)\|^3}{\det(\alpha'(t),\alpha''(t))} = \frac{(4\sin^2t+\cos^2t)^{3/2}}{2\cos^2t +2\sin^2 t} = \frac{1}{2}(4\sin^2t+\cos^2t)^{3/2}.$$For $t = \pi/2$, we have $\cos \pi/2 = 0$ and $\sin \pi/2 = 1$, thus $$R(\pi/2) = \frac{4^{3/2}}{2} = 4.$$I think you have the right solution and the textbook has a mistake.

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    $\begingroup$ Thank you Ivo! That's what I've been second-guessing for days! $\endgroup$ – User32563 Jan 17 at 21:00
  • $\begingroup$ That always makes it harder. $\endgroup$ – marty cohen Jan 17 at 21:22
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Your solution is correct.

The given curve is an ellipse $(A\cos t, B\sin t)$.

At the extremities of axes, $\vec{v}$ and $\vec{a}$ are perpendicular. Then using $$a=\frac{v^2}{R}$$

known to high school students, it can be easily calculated $$R(t=\pm \frac{\pi}{2})=\frac{|(-A,0)|^2}{|(0,B)|}=\frac{A^2}{B}$$ $$R(t=0, \pi)=\frac{|(0,B)|^2}{|(-A,0)|}=\frac{B^2}{A}$$

A pleasing and symmetrical result, I first came across in a Physics book.

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  • $\begingroup$ Nice! Thanks. I learned the hard way that my textbook has errors! $\endgroup$ – User32563 Jan 18 at 6:20

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