## Problem Statement

Calculate the lengths of the arcs of the curves given by the equations in polar coordinates.

$$
\rho=6 e^{12 \varphi / 5},-\frac{\pi}{2} \leq \varphi \leq \frac{\pi}{2}
$$
## Solution

The length of the arc of a curve given by an equation in polar coordinates is determined by the formula

$$
L=\int_{\phi_{1}}^{\phi_{2}} \sqrt{\rho^{2}+\left(\frac{d \rho}{d \phi}\right)^{2}} d \phi
$$

Let's find $\frac{d \rho}{d \phi}$:

$$
\frac{d \rho}{d \phi}=6 \cdot \frac{12}{5} e^{12 \phi / 5}=\frac{72}{5} e^{12 \phi / 5}
$$

We obtain:

$$
\begin{aligned}
L & =\int_{-\pi / 2}^{\pi / 2} \sqrt{\left(6 e^{12 \phi / 5}\right)^{2}+\left(\frac{72}{5} e^{12 \phi / 5}\right)^{2}} d \phi= \\
& =\int_{-\pi / 2}^{\pi / 2} \sqrt{\left(36+\frac{5184}{25}\right) e^{24 \phi / 5}} d \phi= \\
& =\frac{78}{5} \int_{-\pi / 2}^{\pi / 2} e^{12 \phi / 5} d \phi=\left.\frac{78}{5} \cdot \frac{5}{12} e^{12 \phi / 5}\right|_{-\pi / 2} ^{\pi / 2}=\frac{13}{2} \cdot\left(e^{6 \pi / 5}-e^{-6 \pi / 5}\right)=13 \cdot \operatorname{sh} \frac{6 \pi}{5}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_19-5$ » Categories: Kuznetsov's Problem Book Integrals Problem 19| Integrals

- Last edited: 11:55, 27 May 2010.
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## Problem Kuznetsov Integrals 19-6

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