## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\frac{1}{2} \cos t-\frac{1}{4} \cos 2 t \\
y=\frac{1}{2} \sin t-\frac{1}{4} \sin 2 t
\end{array}\right. \\
& \frac{\pi}{2} \leq t \leq \frac{2 \pi}{3}
\end{aligned}
$$
## 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:

$$
\begin{aligned}
& x=\frac{1}{2} \cos t-\frac{1}{4} \cos 2 t ; x^{\prime}(t)=\frac{1}{4}(-2 \sin t+2 \sin 2 t)=\frac{1}{2}(\sin 2 t-\sin t) \\
& y=\frac{1}{2} \sin t-\frac{1}{4} \sin 2 t ; y^{\prime}(t)=\frac{1}{4}(2 \cos t-2 \cos 2 t)=\frac{1}{2}(\cos t-\cos 2 t)
\end{aligned}
$$

Then, using the above formula, we get:

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

- Last edited: 13:21, 21 May 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals $18-9$

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