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Consider the plane curve $\gamma$ in polar coordinates: $$ r=r_0+e^{\lambda\theta}, \quad \theta_1 \le \theta \le \theta_2, $$ where $r_0,\lambda,\theta_1>0$. Is it possible to compute explicitly the length of $\gamma$?

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Use $\int_a^b\|\gamma'(t)\|dt$, where $\gamma(t)=(x(t),y(t))$, with the usual change of coordinates. – Sigur Jul 17 '12 at 21:43
@Sigur I know that formula, it doesn't answer my question! – Mercy King Jul 17 '12 at 21:46
Why not? What did you think? – Sigur Jul 17 '12 at 21:49
@Sigur The question seems clear enough! I'm not saying that I need the formula for the length a curve. – Mercy King Jul 17 '12 at 21:52
@Mercy Actually, your question is "Is it possible to compute explicitly the length of $\gamma$?" The answer is "yes" and some posters have given some clues as how to go about the calculation. – Code-Guru Jul 17 '12 at 23:29

There is a general way of doing this, but I assume you are not familiar with it so let's look at your particular example. In Cartesian coordinates, your curve looks like

$$ \gamma (\theta) = \begin{pmatrix} r\cos\theta\\ r\sin\theta \end{pmatrix} = \begin{pmatrix} \left(r_0 +e^{\lambda \theta}\right)r\cos\theta\\ \left(r_0 +e^{\lambda \theta}\right)r\sin\theta \end{pmatrix}$$ Therefore, the tangent is $$ \frac{d\gamma}{d\theta}= \left( \begin{array}{c} e^{\theta \lambda } \lambda \cos\theta-\sin\theta \left(e^{\theta \lambda }+r_0\right) \\ e^{\theta \lambda } \lambda \sin\theta+\cos\theta \left(e^{\theta \lambda }+r_0\right) \end{array} \right)\ ,$$ and its norm is calculated easily using trigonometrical identities to be $$\left|\frac{d\gamma}{d\theta}\right|=\sqrt{e^{2 \theta \lambda } \left(1+\lambda ^2\right)+2 e^{\theta \lambda } r_0+r_0^2}$$

The length is given by integrating $\left|\frac{d\gamma}{d\theta}\right|$ from $\theta_1$ to $\theta_2$.

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What do you get when you integrate? that's precisely what I'm asking for! – Mercy King Jul 17 '12 at 22:04
@Mercy I plugged it in Mathematica. It gives a nasty result which is much too long and cumersome to copy here, but it is a "closed form" as you hoped. – yohBS Jul 17 '12 at 22:19

This can be very slightly less excruciating. Every polar curve $r = f(\theta)$ can be written as the parametric equations $x(\theta) = f(\theta) \cos(\theta)$, $y(\theta) = f(\theta)\sin(\theta)$. You can do a chain rule calculation with a little trigonometric magic to crunch it down to $$ds = \sqrt{x'(\theta)^2 + y'(\theta)^2}\,d\theta = \sqrt{r^2 + r'(\theta)^2}\,d\theta.$$

You have $r(\theta) = r_0 + e^{\lambda\theta}.$ Then $r'(\theta) = \lambda e^{\lambda \theta},$ so $$ds = \sqrt{(r_0 + e^{\lambda\theta})^2 + \lambda^2 e^{2\lambda\theta}}\,d\theta = \sqrt{(1 + \lambda^2)e^{2\lambda\theta} + 2r_0 e^{\lambda\theta} + r_0^2}\,d\theta$$ This agreees with the form obtained by yohBS above.

You then need to embark down this path $$\int ds = \int \sqrt{(1 + \lambda^2)e^{2\lambda\theta} + 2r_0 e^{\lambda\theta} + r_0^2}\,d\theta = \int {\sqrt{(1 + \lambda^2)e^{2\lambda\theta} + 2r_0 e^{\lambda\theta} + r_0^2}\over e^{-\lambda x}} e^{\lambda\theta}d\theta$$ Now put $z = e^{\lambda \theta}$ and you get left with $$\int ds = {\sqrt{(1 + \lambda^2)z^2 + 2r_0 z + r^2_0}\over z}\, dz $$ Now try completing the square and doing a trigonometric substitution.

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