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:


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

  • $\begingroup$ @user6786: Instead of modifying an existing answer by another user, please comment on the answer instead. $\endgroup$ Mar 20 '11 at 14:20
  • $\begingroup$ so i guess all my additions were lost. $\endgroup$
    – BBSysDyn
    Mar 20 '11 at 14:30
  • $\begingroup$ Yes, you're right that you need to use the quotient rule and your calculations are correct. $\endgroup$ Mar 24 '11 at 13:33
  • $\begingroup$ got it thanks @Jesse $\endgroup$
    – BBSysDyn
    Mar 24 '11 at 15:40

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 $(x_0,y_0)$.

  • $\begingroup$ @Jesse: The link doesn't work at the moment, and anyway, user6786 also suggested "other possible derivations". $\endgroup$ Mar 20 '11 at 11:25
  • $\begingroup$ both great answers @Christian, @Jesse thank you. $\endgroup$
    – BBSysDyn
    Mar 20 '11 at 14:49

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.