MHB A curvature problem (differentiation)

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In the discussion on curvature in the Euclidean plane, the relationship between the curvature $$\kappa$$ of a differentiable function $$y=f(x)$$ and the angle $$\phi$$ formed by the tangent line at a point is established. The curvature is defined as the rate of change of $$\phi$$ with respect to arc length $$s$$, leading to the formula $$\kappa = \frac{d\phi}{ds}$$. The participants derive that $$\kappa$$ can be expressed as $$\kappa = \frac{ \left[ 1+ \left( \frac{dy}{dx} \right)^2 \right]^{3/2} }{ \frac{d^2y}{dx^2} }$$. The discussion also clarifies the differentiation process, linking the second derivative of the function to the angle's derivative and the arc length. Ultimately, the derived formula for curvature is confirmed as $$\kappa= \frac{f''(x)}{\left [1+f'(x)^2 \right]^{3/2}}$$.
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In the Euclidean plane, assume a differentiable function $$y=f(x)$$ exists. At any given point, say $$(x_0,y_0)$$, the line tangential to $$y=f(x)$$ at this point intersects the x-axis at an angle $$\phi$$.

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

$$\kappa = \frac{d\phi}{ds} $$Problem:Prove that

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

$$\kappa = \frac{\left[ 1+\left( f'(x) \right)^2 \right]^{3/2}}{f''(x)}$$
 
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Here's an optional, visual aid, for a generic curve. Just in case... (Bandit)

 

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Isn't

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

The solution follows.

First, $$\frac{d \varphi}{ds}= \frac{\frac{ d \varphi}{dx}}{ \frac{ds}{dx}}$$.

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

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

Now, remembering the definition of arc length,

[math]\frac{ds}{dx}= \sqrt{1+f'(x)^2}[/math],

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

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