Show that for the involute of a circle ([tex] x=a(\cos\theta + \theta\sin\theta) \\ [/tex], [tex] y=a(\sin\theta - \theta\cos\theta) [/tex] for 0 <=[tex]\theta [/tex]<=[tex]\pi \\ [/tex]) radius a, the radius of curvature, is [tex] \sqrt{2as} [/tex], where s is the arc length. The radius of curvature is [tex]\rho = \frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}\\ [\tex];(adsbygoogle = window.adsbygoogle || []).push({});

My attempt: [tex]\frac{dx}{d\theta} = a \theta\cos\theta \\[/tex] and [tex]\frac{dy}{d\theta} = a\theta\sin\theta \\[/tex],

that implies that [tex] \frac{dy}{dx} = \tan\theta \\[/tex], which implies that [tex]\frac{d^2y}{dx^2} = \sec^{2}\theta[/tex], therefore [tex] \rho = \sec\theta\\[/tex]. Which is not the answer they suggest. Please help, thanks.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Involute of a circle

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