# Curvature of implicit curve

This is a problem from "Advanced Calculus" by Buck:

If $\gamma$ is a curve which satisfies the equation $f(x,y)=0$, show that the curvature at a point on $\gamma$ is given by:

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

Sincerely, I consider that the textbook does not provide enough theory to tackle problems of this nature. Can you give me a suggestion or at least a good reference?

Coincidentally, I just solved this problem a month or two ago (in a different context), so I'll take my notes from there and reconstruct my solution.

Starting with:

$$f(x,y)=0$$

Take the derivative of both sides, with respect to x:

$$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}=0$$

(This is because $\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}$, for any $f(x,y)$.) Solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx}=-{\frac{\partial f}{\partial x}\over\frac{\partial f}{\partial y}}$$

Taking the derivative of both sides again, we get:

$$\frac{d^2y}{dx^2}=-{\frac{\partial f}{\partial y}\frac d{dx}\left(\frac{\partial f}{\partial x}\right)-\frac{\partial f}{\partial x}\frac d{dx}\left(\frac{\partial f}{\partial y}\right)\over\left(\frac{\partial f}{\partial y}\right)^2}$$

But, since $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are still functions of $x$ and $y$, we can use the above rule:

$$\frac{d^2y}{dx^2}=-{\frac{\partial f}{\partial y}\left(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial x\partial y}\frac{dy}{dx}\right)-\frac{\partial f}{\partial x}\left(\frac{\partial^2f}{\partial x\partial y}+\frac{\partial^2f}{\partial y^2}\frac{dy}{dx}\right)\over\left(\frac{\partial f}{\partial y}\right)^2}$$

And since we already solved for $\frac{dy}{dx}$, we can substitute that in, too:

$$\frac{d^2y}{dx^2}=-{\frac{\partial f}{\partial y}\left(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial x\partial y}\left(-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial y}\right)\right)-\frac{\partial f}{\partial x}\left(\frac{\partial^2f}{\partial x\partial y}+\frac{\partial^2f}{\partial y^2}\left(-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial y}\right)\right)\over\left(\frac{\partial f}{\partial y}\right)^2}$$

which simplifies to:

$$\frac{d^2y}{dx^2}=-{\frac{\partial^2f}{\partial x^2}\left(\frac{\partial f}{\partial y}\right)^2-2\frac{\partial^2f}{\partial x\partial y}\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}+\frac{\partial^2f}{\partial y^2}\left(\frac{\partial f}{\partial x}\right)^2\over\left(\frac{\partial f}{\partial y}\right)^3}$$

Now, we can plug them into the curvature formula (which I already derived here):

$$\kappa={\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{dy}{dt}\frac{d^2x}{dt^2}\over\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}^3}$$

If we let $t=x$ (essentially turning the parametric function input into an implicit or explicit function input), we get:

$$\kappa={\frac{dx}{dx}\frac{d^2y}{dx^2}-\frac{dy}{dx}\frac{d^2x}{dx^2}\over\sqrt{\left(\frac{dx}{dx}\right)^2+\left(\frac{dy}{dx}\right)^2}^3}$$

which simplifies to:

$$\kappa={\frac{d^2y}{dx^2}\over\sqrt{1+\left(\frac{dy}{dx}\right)^2}^3}$$

Then we substitute:

$$\kappa={-({\frac{\partial^2f}{\partial x^2}\left(\frac{\partial f}{\partial y}\right)^2-2\frac{\partial^2f}{\partial x\partial y}\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}+\frac{\partial^2f}{\partial y^2}\left(\frac{\partial f}{\partial x}\right)^2)/\left(\frac{\partial f}{\partial y}\right)^3}\over\sqrt{1+\left(-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial y}\right)^2}^3}$$

which simplifies to:

$$\kappa=-{\frac{\partial^2f}{\partial x^2}\left(\frac{\partial f}{\partial y}\right)^2-2\frac{\partial^2f}{\partial x\partial y}\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}+\frac{\partial^2f}{\partial y^2}\left(\frac{\partial f}{\partial x}\right)^2\over\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}^3}$$

as desired (within a plus or minus sign). [For whatever reason, I couldn't get those parentheses in the second-to-last expression to work properly.]

(If $\frac{\partial f}{\partial y}=0$, then swap $x$ and $y$. If both are zero, then $\kappa$ is undefined to begin with, without the use of L'Hôpital's rule.)

Originally, I actually did this another way; if you like, I can post that way as well.

• You seem to assume that $f_y \neq 0$. Commented Nov 10, 2015 at 22:03
• @FlybyNight Then swap $x$ and $y$; the final expression doesn't change if you do so. If both are zero, then the curvature is undefined to begin with. Commented Nov 10, 2015 at 22:39
• Comments are to ask for more information and to suggest improvements (to answers). Telling me to swap $x$ and $y$ doesn't really improve your answer. Commented Nov 10, 2015 at 22:41
• @FlybyNight Edited, but you really didn't need the anger. Commented Nov 10, 2015 at 22:57
• Take a step back and re-read the comments. I don't see any anger. From my point of view, you using the imperative conjugation "Then swap..." was quite dismissive and rude. I guess the lesson to learn is that you can't judge a persons mood, or intentions from two sentence. Commented Nov 10, 2015 at 23:08

We can assume that the curve is smooth in a neighbourhood of the point that we are considering. If it weren't, then curvature would not make sense.

Our curve is given by an equation $f(x,y)=0$.

Let us assume that a curve $\gamma(t) = (x(t),y(t))$ parametrises our curve. This means what $$f(x(t),y(t)) \equiv 0$$

The formula for the curvature of a plane curve is:

$$\kappa = \frac{\dot x \ddot y - \ddot x \dot y}{(\dot x^2 + \dot y^2)^{3/2}}$$

If we can express $\dot x$, $\dot y$, $\ddot x$ and $\ddot y$ in terms of $f_x$, $f_y$, $f_{xx}$, $f_{xy}$ and $f_{yy}$ then we're done.

Differentiating the identity $f(x(t),y(t)) \equiv 0$ with respect to $t$ gives $\dot x f_x + \dot y f_y \equiv 0$. It follows that, for some constant $\lambda > 0$, we have $\dot x \equiv -\lambda f_y$ and $\dot y \equiv \lambda f_x$.

Differentiating $\dot x \equiv -\lambda f_y$ with respect to $t$ gives $\ddot x \equiv -\lambda(\dot x f_{xy} + \dot y f_{yy})$, while differentiating $\dot y \equiv \lambda f_x$ with respect to $t$ gives $\ddot y \equiv \lambda(\dot x f_{xx} + \dot y f_{xy})$.

We can now put these into the expression for $\kappa$. Firstly:

\begin{eqnarray*} \dot x \ddot y - \ddot x \dot y &\equiv& -\lambda^2 f_y(\dot x f_{xx} + \dot y f_{xy})+\lambda^2(\dot x f_{xy} + \dot y f_{yy})f_x \\ \\ &\equiv& -\lambda^2 f_y(-\lambda f_y f_{xx} + \lambda f_x f_{xy})+\lambda^2(-\lambda f_y f_{xy} + \lambda f_x f_{yy})f_x \\ \\ &\equiv& \lambda^3 \left( f_y^2f_{xx}-2f_xf_yf_{xy}+f_x^2f_{yy}\right) \end{eqnarray*}

Similarly, with the denominator:

\begin{eqnarray*} \dot x^2 + \dot y^2 &\equiv& (-\lambda f_y)^2 + (\lambda f_x)^2 \\ \\ &\equiv& \lambda^2(f_x^2 + f_y^2) \\ \\ \left(\dot x^2 + \dot y^2\right)^{3/2} &\equiv& \lambda^3 \left(f_x^2 + f_y^2\right)^{3/2} \end{eqnarray*}

Putting the numerator over the nominator gives the result

$$\frac{\dot x \ddot y - \ddot x \dot y}{(\dot x^2 + \dot y)^{3/2}} \equiv \frac{f_y^2f_{xx}-2f_xf_yf_{xy}+f_x^2f_{yy}}{\left(f_x^2 + f_y^2\right)^{3/2}}$$

The modulus signs can be added. Some authors insist that $\kappa \ge 0$, while others allow $\kappa \in \mathbb R$.