Perimeter of a Broken Line Inscribed in a Logarithmic Spiral

Question

Exercise:

Find the limit of the perimeter of a broken line $M_0M_1...M_n$ inscribed in a logarithmic spiral $t = e^{-\phi}$ (as $n \to \infty$), if the vertices of this broken line have, respectively, the polar angles $\phi_0 = 0$, $\phi_1 = \frac{\pi}{2}$, $\cdots$, $\phi_n = \frac{n\pi}{2}$.

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Attempt:

I have no idea how to go about this. All I've been able to do is verify that there is a defined limit:

$\lim\limits_{n \to \infty}{e^{\frac{-\pi}{2}n}} = 0$, so the sum of smaller and smaller lengths will result in a defined number.

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Request:

Can I get a kickstart? Hints are welcome. (If I'm still lost I'll ask for the solution.)

Answer 1

I've worked this out and tested it. So let me get you started. We can express this logarithmic curve in the complex plane as

$$z=e^{(i-1)\theta}$$

So first we identify the discrete points on the line whose length you would like to determine, i.e,

$$z_n=e^{n(i-1)\pi/2}$$

Then the length we wish to determine is given simply by

$$\sum_{n=1}^\infty|z_n-z_{n-1}|$$

And you can take it from there. Just bear in mind that $|e^{i\theta}|=1$. I imagine this can generalized for any flair coefficient and $\Delta\theta$.