# Involute of a circle

1. Aug 23, 2007

### John O' Meara

Show that for the involute of a circle ($$x=a(\cos\theta + \theta\sin\theta) \\$$, $$y=a(\sin\theta - \theta\cos\theta)$$ for 0 <=$$\theta$$<=$$\pi \\$$) radius a, the radius of curvature, is $$\sqrt{2as}$$, where s is the arc length. The radius of curvature is $$\rho = \frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}\\ [\tex]; My attempt: [tex]\frac{dx}{d\theta} = a \theta\cos\theta \\$$ and $$\frac{dy}{d\theta} = a\theta\sin\theta \\$$,
that implies that $$\frac{dy}{dx} = \tan\theta \\$$, which implies that $$\frac{d^2y}{dx^2} = \sec^{2}\theta$$, therefore $$\rho = \sec\theta\\$$. 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

2. Aug 23, 2007

### dextercioby

Check again the second derivative of y wrt x. Be careful about using the chain rule. Remember that

$$\frac{d^2 y}{dx^2}=\frac{d}{d\Huge{\mathbf{x}}}\frac{dy}{dx}$$

Last edited: Aug 23, 2007
3. Aug 24, 2007

### John O' Meara

I checked $$\frac{d^2y}{dx^2}$$ and got the following $$\frac{d^2y}{dx^2} = \frac{d\tan\theta}{d\theta} \cdot \frac{d\theta}{dx} = \sec^{2}\theta \cdot \frac{d\theta}{dx} = \frac{\sec^{2}\theta}{\frac{dx}{d\theta}} \\$$. This gives $$\frac{d^2y}{dx^2} = \frac{\sec^{2}\theta}{a\theta \cos\theta} \\$$, which yields $$\rho =a\theta$$. Which is not the correct answer either, but the integral of $$a\theta$$ w.r.t, $$\theta$$ gives s. Thanks for the reply.

4. Aug 28, 2007

### John O' Meara

Is anyone interest as to why I can't get the correct answer. I get $$\rho = a\theta \mbox{ not the } \sqrt{2sa}$$? Thanks.

Last edited: Aug 29, 2007
5. Aug 28, 2007

### dextercioby

What is "s" equal to ? In terms of $\rho,\theta$, of course ?