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?

• $v\times a$ should have only two components. – user9464 Jan 17 at 21:00
• Maybe I should've clearly mentioned it, but $v$ and $a$ are 3-dimentional, their third coordinate being $0$. – User32563 Jan 17 at 21:03

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.

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

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.

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