## Problem Statement

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

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

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

Then, using the formula above, we have:

$$
\begin{aligned}
L & =\int_{0}^{2 \pi} \sqrt{\left(t^{2} \cos t\right)^{2}+\left(t^{2} \sin t\right)^{2}} d t= \\
& =\int_{0}^{2 \pi} \sqrt{t^{4}\left(\cos ^{2} t+\sin ^{2} t\right)} d t=\int_{0}^{2 \pi} \sqrt{t^{4} \cdot 1} d t=\int_{0}^{2 \pi} t^{2} d t=\left.\frac{1}{3} \cdot t^{3}\right|_{0} ^{2 \pi}= \\
& =\frac{1}{3} \cdot\left((2 \pi)^{3}-0^{3}\right)=\frac{8 \pi^{3}}{3}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%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 $\% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_18-22$ »

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

Ukrainian Banner Network

- Last modified: 09:55, 21 May 2010.
- Content is available under CC-BY-SA 3.0.

Created by Geeteatoo

## Problem Kuznetsov Integrals 18-23

## Material from PlusPi