A curvature problem (differentiation)

[DreamWeaver]

In the Euclidean plane, assume a differentiable function \(\displaystyle y=f(x)\) exists. At any given point, say \(\displaystyle (x_0,y_0)\), the line tangential to \(\displaystyle y=f(x)\) at this point intersects the x-axis at an angle \(\displaystyle \phi\).

The curvature of this curve, \(\displaystyle \kappa\), is the rate of change of \(\displaystyle \phi\) with respect to arc length, \(\displaystyle s\):

\(\displaystyle \kappa = \frac{d\phi}{ds} \)Problem:Prove that

\(\displaystyle \kappa = \frac{ \left[ 1+ \left( \frac{dy}{dx} \right)^2 \right]^{3/2} }{ \frac{d^2y}{dx^2} }\)Or equivalently

\(\displaystyle \kappa = \frac{\left[ 1+\left( f'(x) \right)^2 \right]^{3/2}}{f''(x)}\)

 

[DreamWeaver]

Here's an optional, visual aid, for a generic curve. Just in case... (Bandit)

Spoiler

View attachment 2614

 

[jacobi1]

Isn't

\(\displaystyle \kappa = \frac{f''(x)}{ \left [ 1+ (f'(x))^2 \right ]^{3/2}}\)?

The solution follows.

First, \(\displaystyle \frac{d \varphi}{ds}= \frac{\frac{ d \varphi}{dx}}{ \frac{ds}{dx}}\).

If the tangent line to f(x) at \(\displaystyle x=x_0\) intersects the x-axis at an angle \(\displaystyle \varphi\), then \(\displaystyle f'(x_0) = \tan \varphi\). Differentiating both sides with respect to \(\displaystyle x_0\) and assuming \(\displaystyle \varphi\) is a function of \(\displaystyle x_0\) gives \(\displaystyle f''(x_0)=\varphi' \sec^2 \varphi \). Solving and remembering that \(\displaystyle 1+\tan^2 (x)=\sec^2 (x)\), we have

\(\displaystyle \frac{f''(x_0)}{1+f'(x_0)^2}= \varphi'=\frac{d \varphi}{dx}\).

Now, remembering the definition of arc length,

\(\displaystyle \frac{ds}{dx}= \sqrt{1+f'(x)^2}\),

substituting into the first equation, and dropping subscripts gives our result,

\(\displaystyle \kappa= \frac{d \varphi}{ds}= \frac{f''(x)}{\left [1+f'(x)^2 \right]^{3/2}}\).