## Problem Statement

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

$$
\rho=3 e^{3 \varphi / 4}, 0 \leq \varphi \leq \frac{\pi}{3}
$$
## 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 $\frac{d \rho}{d \phi}$:

$$
\frac{d \rho}{d \phi}=3 \cdot \frac{3}{4} e^{3 \phi / 4}=\frac{9}{4} e^{3 \phi / 4}
$$

We get:

$$
\begin{aligned}
L & =\int_{0}^{\pi / 3} \sqrt{\left(3 e^{3 \phi / 4}\right)^{2}+\left(\frac{9}{4} e^{3 \phi / 4}\right)^{2}} d \phi= \\
& =\int_{0}^{\pi / 3} \sqrt{\left(9+\frac{81}{16}\right) e^{3 \phi / 2}} d \phi= \\
& =\frac{15}{4} \int_{0}^{\pi / 3} e^{3 \phi / 4} d \phi=\left.\frac{15}{4} \cdot \frac{4}{3} e^{3 \phi / 4}\right|_{0} ^{\pi / 3}=5 \cdot\left(e^{\pi / 4}-1\right)
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_19-6$ »

Categories: Kuznetsov Problem Book Integrals Problem 19 | Integrals

- Last modified: 11:56, 27 May 2010.
- Content is available under CC-BY-SA 3.0.

Kuznetsov Integrals Problem 19-6 — PlusPi

Created by GeeTeatoo

The

![](https://cdn.mathpix.com/cropped/2024_05_22_36ed20bbfe1105944322g-12.jpg?height=48&width=46&top_left_y=29&top_left_x=2033)$\qquad$

The

佂

- 
- 
- 

$\qquad$ $\qquad$ $\qquad$ $\qquad$ $\qquad$ $\qquad$ $\qquad$ $\qquad$ $\qquad$ $\qquad$ $\qquad$ $\qquad$

## Kuznetsov Integrals Problem 19-7

## Material from PlusPi