## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=e^{t}(\cos t+\sin t) \\
y=e^{t}(\cos t-\sin t)
\end{array}\right. \\
& 0 \leq t \leq \frac{3 \pi}{2}
\end{aligned}
$$
## 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 $t$:

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

Then, using the formula above, we have:

$$
\begin{aligned}
L & =\int_{0}^{3 \pi / 2} \sqrt{4 e^{2 t} \cos ^{2} t+4 e^{2 t} \sin ^{2} t} d t= \\
& =\int_{0}^{3 \pi / 2} 2 e^{t} \sqrt{\cos ^{2} t+\sin ^{2} t} d t= \\
& =\int_{0}^{3 \pi / 2} 2 e^{t} \sqrt{1} d t=2 \int_{0}^{3 \pi / 2} e^{t} d t=\left.2 e^{t}\right|_{0} ^{3 \pi / 2}=2\left(e^{3 \pi / 2}-e^{0}\right)=2\left(e^{3 \pi / 2}-1\right)
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_18-24»

Categories: Kuznetsov Problem Book Integrals Problem 18 | Integrals

- Last modified: 10:30, 21 May 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals $18-25$

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