# Define curvature and curvature of a circle

Question:(a) Define the curvature function $\kappa$ of a plane curve.

The curvature of $\kappa$ of a plane curve is the amount of turning in the osculating plane. In other words it decribes the speed of rotation.

Is how I defined it okay? I feel like I need more or something.

Question:(b) Determine the curvature function of the circle $\alpha(t)=(r\cos t,r\sin t)$. To find our curvature we must find the derivative of $\alpha$;

$\alpha'(t)=(-r\sin t, r\cos t)$.

Now to find our constant we do the following:

$|\alpha'(t)|= \sqrt{r^2\sin^2 t+r^2\cos^2 t}= \sqrt{r^2(\cos^2 t+\sin^2 t)}= \sqrt{r^2}= r.$

Thus, we get the arc-length as,

$s= \int_{0}^{t}r dt= rt \bigg |^t_0= rt.$

So, we can write $t= \dfrac{s}{r}$. Using the value of $t$ we substitute it into $\beta$ such that it is a curve parametrized by arc length,

$\beta(s)=(r\cos\dfrac{s}{r}, r\sin\dfrac{s}{r}).$

Using this we find the derivative to be,

$\beta'(s)=(-\sin\dfrac{s}{r}, \cos\dfrac{s}{r}).$

Looking for our constant we get,

$|\beta'(s)|=\sqrt{(-\sin\dfrac{s}{r})^2+\cos^2\dfrac{s}{r}}= \sqrt{1}=1 ,\ \ \ \forall s.$

Finding our second derivative we get,

$\beta"(s)=(-\dfrac{1}{r}\cos\dfrac{s}{r}, -\dfrac{1}{r}\sin\dfrac{s}{r}).$

$|\beta"(s)|=\sqrt{(-\dfrac{1}{r}\cos\dfrac{s}{r})^2+ (-\dfrac{1}{r}\sin\dfrac{s}{r})^2}= \sqrt{\dfrac{1}{r^2}}= \dfrac{1}{r}$

Therefore, we see that the curvature $\kappa$ of the circle is a constant,$\dfrac{1}{r}$.

Is my work okay? I just want to make sure.