## Problem Statement

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

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

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

We obtain:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+18-27$ »

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

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## Problem Kuznetsov Integrals 18-28

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