## Problem Statement

Calculate the lengths of the arcs of the curves given by the parametric equations.

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=e^{t}(\cos t+\sin t) \\
y=e^{t}(\cos t-\sin t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$
## Solution

The length of the arc of a curve defined by parametric equations is determined by the formula

$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}} d t
$$

Let's find the derivatives with respect to $t$:

$$
\begin{aligned}
& x^{\prime}(t)=e^{t}(\cos t+\sin t)+(\cos t-\sin t) e^{t}=2 e^{t} \cos t \\
& y^{\prime}(t)=e^{t}(\cos t-\sin t)+(-\cos t-\sin t) e^{t}=-2 e^{t} \sin t
\end{aligned}
$$

Then, using the formula above, we have:

$$
\begin{aligned}
L & =\int_{0}^{\pi} \sqrt{4 e^{2 t} \cos ^{2} t+4 e^{2 t} \sin ^{2} t} d t= \\
& =\int_{0}^{\pi} 2 e^{t} \sqrt{\cos ^{2} t+\sin ^{2} t} d t= \\
& =\int_{0}^{\pi} 2 e^{t} \sqrt{1} d t=2 \int_{0}^{\pi} e^{t} d t=\left.2 e^{t}\right|_{0} ^{\pi}=2\left(e^{\pi}-e^{0}\right)=2\left(e^{\pi}-1\right)
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+18-6$ »

Categories: Kuznetsov's Problem Book Integrals Problem 18 | Integrals

- Last modified: 10:21, 21 May 2010.
- Content is available under CC-BY-SA 3.0.

Kuznetsov Integrals Problem 18-6-PlusPi

Kuznetsov Integrals Problem 18-6 - PlusPi
Created by GeeTeatoo

(

![](https://cdn.mathpix.com/cropped/2024_05_22_b045d7f8347528b5fcdfg-10.jpg?height=52&width=39&top_left_y=-6&top_left_x=2045)$\qquad$
.

Table

2. 

- 

.

- 
- 
- 
- 
- 

is

佂

- 


## Kuznetsov Integrals Problem 18-7

## Material from PlusPi