Radius of curvature for a plane curve 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?

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.
