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?

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.
