Calculation of curvature of a curve

Question

For a curve having an equation of $y=2x^2$, can somebody tell me how to find curvature using $$k(s)=|\frac{dT}{ds}|=|\frac{dT}{dx}/\frac{ds}{dx}|$$ Where $T$ is unit tangent and $s$ is arc length

Answer 1

The unit tangent vector is defined as,

$$\vec T=\frac{\vec r'}{|\vec r'|}$$

Where $\vec r'$ is the derivative of the position vector we are looking at.

Here $\vec r=\langle x,2x^2 \rangle$. So, $\vec r'=\langle 1,4x \rangle$. It has magnitude $\sqrt{1+16x^2}$. Then you can continue to find $\frac{d \vec T}{dx}$.

As for $\frac{ds}{dx}$, note that (up to some constant depending on how you define $s$):

$$s(u)=\int_{0}^{u} \sqrt{1+(\frac{dy}{dx})^2} dx$$

This leads to (by the fundamental theorem of calculus),

$$\frac{ds}{dx}=s'=\sqrt{1+(y')^2}$$

In our case using the chain rule, product rule, and power rule we get,

$$\vec T'=\langle -\frac{16x}{(1+16x^2)^{3/2}},\frac{4}{(1+16x^2)^{3/2}} \rangle$$

Whose magnitude is,

$$\frac{\sqrt{256x^2+16}}{(1+16x^2)^{3/2}}$$

$$=4\frac{\sqrt{16x^2+1}}{(16x^2+1)\sqrt{16x^2+1}}$$

$$=\frac{4}{16x^2+1}$$

Because $s'=\sqrt{1+(y')^2}=\sqrt{16x^2+1}$ we get,

$$k=\frac{4}{(16x^2+1)^{3/2}}$$

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The same exact method but with a more general form for $\vec r$,

$$\vec r=\langle x,y(x) \rangle$$

Gives,

$$\vec T'=\langle \frac{-y'y''}{(1+y'^2)^{3/2}}, \frac{y''}{(1+y'^2)^{3/2}} \rangle$$

$$=\frac{y''}{(1+y'^2)^{3/2}} \langle -y',1 \rangle$$

It's surprising to me that after a bit of manipulation the second coordinate has a nicer looking form then the first. Anyways, all this is leading to,

$$k=\frac{|y''|}{(1+y'^2)^{3/2}}$$

Answer 2

HINT:

For the graph of a function $f$, the curvature at the point $(x, f(x))$ is $$\frac{ f''(x)}{(1 + (f'(x))^2)^{\frac{3}{2}} }$$

Answer 3

This is nicely explained in the complex plane. First of all, it's well known that the tangent angle is related to the curvature in the natural coordinates as

$$\int \kappa~ds=\theta\\ \text{or}\\ \kappa=\frac{d\theta}{ds} $$

where $\kappa$ is the curvature, $\theta$ is the tangent angle, and $s$ is the arc length. The arc length is given by

$$s=\int |\dot z|~du$$

Moreover, the tangent angle is given by

$$\theta=\arg\dot z=\tan^{-1}\frac{\mathfrak{Im} \dot z}{\mathfrak{Re} \dot z}$$

If we now convert the to Cartesian coordinate (i.e., $u=x, z=x+iy$), we can show that

$$\kappa=\frac{d\theta}{ds}=\frac{y''}{(1+y'^2)^{3/2}}$$