Curvature of planar implicit curves

Question

I am trying to understand how the curvature equation

$$\kappa = -\frac{f_{xx} f_y^2-2f_{xy} f_x f_y + f_x^2 f_{yy}}{(f_x^2+f_y^2)^{3/2}}$$

for implicit curves is derived. These curves arise from equalities such as $f(x,y)=0$. I found this on the net:

http://www.cad.zju.edu.cn/home/zhx/GM/001/00-rep_dg.pdf

I can follow almost everything here until pg 49, then the author jumps to the final equation and I have no idea how he's done it.

Can anyone help, or point to other possible derivations? I understand the parametric form of curvature equation which is $\kappa = | \frac{d\vec{T}}{ds} |$ where $\vec{T}$ is unit tangent, if any parallels need to be made to that subject, just in case.

And one more question: How do I expand the term below?

$$\frac{\partial}{\partial x} \bigg( \frac{f_y}{\sqrt{f_x^2 + f_y^2}} \bigg)$$

Do I have to use the Quotient Rule?

$$\frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$

and in that case, I guess I would need to derive $\frac{\partial}{\partial x}(\sqrt{f_x^2+f_y^2})$. Would this be $\frac{1}{2}\frac{2f_x f_{xx} + 2f_y f_{yx}}{\sqrt{f_x^2+f_y^2}}$

Thanks again

Answer 1

Apply the formula $$\frac{d}{ds} = \frac{1}{|\nabla f|}\left(f_y \frac{\partial}{\partial x} - f_x \frac{\partial}{\partial y} \right)$$ to the very right hand side of $$\kappa = \left| \frac{dT}{ds} \right| = \left|\frac{d}{ds} \left(\frac{dx}{ds}, \frac{dy}{ds} \right)\right| = \left| \frac{d}{ds} \frac{(f_y, -f_x)^T}{\sqrt{f_x^2 + f_y^2}} \right| = \left| \frac{d}{ds}\left( \frac{f_y}{\sqrt{f_x^2 + f_y^2}}, \frac{-f_x}{\sqrt{f_x^2 + f_y^2}} \right)^T \right|$$ So: $$\frac{d}{ds}\left( \frac{f_y}{\sqrt{f_x^2 + f_y^2}} \right) = \frac{1}{|\nabla f|} \left[f_y \frac{\partial}{\partial x}\left(\frac{f_y}{\sqrt{f_x^2 + f_y^2}}\right) - f_x \frac{\partial}{\partial y}\left(\frac{f_y}{\sqrt{f_x^2 + f_y^2}}\right) \right]$$ and $$\frac{d}{ds}\left( \frac{-f_x}{\sqrt{f_x^2 + f_y^2}} \right) = \frac{1}{|\nabla f|} \left[f_y \frac{\partial}{\partial x}\left(\frac{-f_x}{\sqrt{f_x^2 + f_y^2}}\right) - f_x \frac{\partial}{\partial y}\left(\frac{-f_x}{\sqrt{f_x^2 + f_y^2}}\right) \right],$$ and I hope you don't mind if I leave the rest of the details to you.

Update: Yes, you're right that you need to use the quotient rule, and your calculations above are correct.

[18th.Dec.2019] There was a clerical error in the LaTeX code.

Answer 2

Let $(x_0,y_0)$ be a point of the curve $\gamma$ defined by $f(x,y)=0$, and let $s\mapsto(x(s),y(s))$ with $(x(0),y(0))=(x_0,y_0)$ be the parametric representation of $\gamma$ by arc length. Note that the sense of direction of $\gamma$ is not determined a priori, whence its curvature $\kappa$ is only determined up to sign.

From $f\bigl(x(s),y(s)\bigr)\equiv 0$ we get $f_x\dot x+ f_y\dot y\equiv 0$, and as $\dot x^2 +\dot y^2\equiv 1$ we see that (up to sign) $$\dot x={f_y\over\sigma},\quad \dot y=-{f_x\over\sigma}\qquad \left(\sigma:=\sqrt{f_x^2 + f_y^2}>0\right).\qquad(*)$$

To compute the curvature $\kappa$ we have to look at the polar angle of the tangent vector $(\dot x,\dot y)$, i.e., at $$\theta:=\arg(\dot x,\dot y)=\arg(f_y, -f_x).$$ The chain rule gives $$\kappa=\dot\theta={d\over ds}\arg(f_y,-f_x)=\nabla\arg(f_y,-f_x)\bullet\left({d\over ds}(f_y),{d\over ds} (-f_x)\right),$$ and using the formula $\nabla\arg(u,v)=\left({-v\over u^2+v^2}, {u\over u^2+v^2}\right)$ we obtain $$\kappa=\left({f_x\over\sigma^2},{f_y\over\sigma^2}\right)\bullet(f_{yx}\dot x+f_{yy}\dot y,\ -f_{xx}\dot x-f_{xy}\dot y)={-f_y^2 f_{xx}+2f_xf_yf_{xy}-f_x^2f_{yy}\over\sigma^3} $$ where we have used (*) and all partial derivatives of $f$ are to be evaluated at $(x_0,y_0)$.