I need to find the length of a spiral. The spiral start at a certain radius $R_1$ and ends at a smaller radius $R_2$. As the spiral spins inwards, the distance between each arm of the spiral decreases but not by a constant amount (as the radius decreases, the decrease in radius is smaller and smaller).
I tried using the archemedian spiral equation ($r = a + b\theta$) and making $b$ and $a$ function of $r$ (where $a$ is start radius and $2\pi b$ is the decrease in radius per revolution),and then using a standard spiral length integral formula but this seems to give the wrong answer. I think that the formula above doesn't apply to my situation, would anyone know a better way to model it?
Thanks!
Use the formula $\int ds$
Where $ds = \sqrt{(dx)^2+(dy)^2}$
We can write $$x = r\cos\theta = (a+b\theta)\cos\theta \Rightarrow dx = b\cos\theta -(a+b\theta)\sin\theta\hspace{1mm}d\theta$$
And $$y = r\sin\theta = (a+b\theta)\sin\theta \Rightarrow dx = b\sin\theta +(a+b\theta)\cos\theta\hspace{1mm}d\theta$$
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You will have $$ds = \sqrt{b^2+(a+b\theta)^2}\hspace{1mm} d\theta$$
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Integrate $$\int_{\theta_1}^{\theta_2} \sqrt{b^2+(a+b\theta)^2}\hspace{1mm} d\theta$$
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You need to find $\theta_1$ and $\theta_2$ as needed