Given $\sin \psi$ and $\cos \psi$ (equivalently $\frac{dy}{ds}$ and $\frac{dx}{ds}$) how to compute curvature $\kappa = \frac{d\psi}{ds}$?

Question

Take a curve $y = f(x)$ and let $\psi(x)$ be the angle between the curve and the horizontal at the point $x$.

Letting $ds$ be an arc-length element (so $ds^2 = dx^2 + dy^2$) we have that $\tan \psi = \frac{dy}{dx} = y'$, $\sin \psi = \frac{y'}{\sqrt{1+(y')^2}} = \frac{dy}{ds}$ and $\cos \psi = \frac{1}{\sqrt{1+(y')^2}} = \frac{dx}{ds}.$

How can we show from this that $\kappa := \frac{d\psi}{ds} = \frac{y''}{\left( 1 + (y')^2\right)^{3/2}}$?

This claim is made in lecture notes I'm reading.

I've thought about using the idea that $\frac{d\psi}{ds} = \frac{\partial\psi}{\partial x} \frac{dx}{ds} + \frac{\partial\psi}{\partial y} \frac{dy}{ds}$ but can't make this work because $\psi$ does not have explicit dependence on $x$.

Similarly, writing $\psi = \arcsin (\dots)$ leads to the need to calculate $\frac{d}{ds} F(y')$, for some function $F$ of the derivative $y'$, which I don't know how to do.

Is there a more straightforward approach?

Answer 1

You have $\tan\psi=y'$ and therefore $(1+\tan^2\psi)\psi'=y''$, or $$\psi'={y''\over1+\tan^2\psi}={y''\over 1+y'^2}\ .$$ Now by the chain rule $${d\psi\over ds}={d\psi\over dx}\>{dx\over ds}={\psi'\over s'}={\psi'\over\sqrt{1+y'^2}}\ ,$$ so that $$\kappa={d\psi\over ds}={y''\over(1+y'^2)^{3/2}}\ .$$

Answer 2

Let P and Q be two neighboring points on the curve AB. Let arc AP $=s~$ and arc AQ $=s+\delta s~$, so that arc PQ $=\delta s~$, A being a fixed point on the curve from which arc lengths are measured. Draw tangents at P and Q. Let $~\delta \psi~$ be the angle between these tangents.

The fraction $~\frac{\delta \psi}{\delta s}~$ is called the average curvature of the arc PQ. The limiting value of the average curvature when Q$~\to~$P is called the curvature of the curve at a point P.

Thus curvature at the point P $$\mathcal k=\lim_{\text{Q}\to\text{P}}\frac{\delta \psi}{\delta s}=\lim_{\delta s\to0}\frac{\delta \psi}{\delta s}=\frac{d\psi}{ds}$$and the radious of curvature at the point P$$=\frac{1}{\text{curvature}}=\frac{ds}{d\psi}=\rho=\frac{1}{\mathcal k}$$

The equation of the curve is $~y=f(x)~$(say).

Then $~\tan \psi=\frac{dy}{dx}~$.

Differentiating with respect to $~s~$, $$\sec^2\psi\frac{d\psi}{ds}=\frac{d}{ds}\left(\frac{dy}{dx}\right)$$ $$\implies \sec^2\psi\cdot \frac{1}{\rho}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\frac{dx}{ds}$$ $$\implies \sec^2\psi\cdot \frac{1}{\rho}=\frac{d^2y}{dx^2}\cdot \cos\psi$$ $$\implies \rho=\frac{\sec^3\psi}{\frac{d^2y}{dx^2}}$$ $$=\frac{(1+\tan^2\psi)^{3/2}}{\frac{d^2y}{dx^2}}$$ Hance$$\rho=\frac{\left\{1+\left(\frac{dy}{dx}\right)^2\right\}^{3/2}}{\frac{d^2y}{dx^2}}=\frac{(1+y_1^2)^{3/2}}{y_2}$$where $~y_1=\frac{dy}{dx}~$and$~y_2=\frac{d^2y}{dx^2}~$

Therefore $$\kappa = \frac{d\psi}{ds} = \frac{y''}{\left( 1 + (y')^2\right)^{3/2}}$$

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Note: $~\delta \psi~$ is called the angle of contingence of the arc PQ. It is also called the total curvature of the arc PQ.

• Also$$\cos\psi=\frac{dx}{ds}$$