## Problem Statement

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

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

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

Then, using the above formula, we get:

$$
\begin{aligned}
& L=\int_{0}^{\pi / 3} \sqrt{(4(\sin 2 t-\sin t))^{2}+(4(\cos t-\cos 2 t))^{2}} d t= \\
& =\int_{0}^{\pi / 3} 4 \sqrt{\sin ^{2} 2 t-2 \sin 2 t \sin t+\sin ^{2} t+\cos ^{2} t-2 \cos 2 t \cos t+\cos ^{2} 2 t} d t= \\
& =\int_{0}^{\pi / 3} 4 \sqrt{1-2 \sin 2 t \sin t+1-2 \cos 2 t \cos t} d t= \\
& \pi / 3 \\
& =\int_{0}^{\pi / 3} 4 \sqrt{2} \sqrt{1-\sin 2 t \sin t-\cos 2 t \cos t} d t= \\
& =\int_{0}^{\pi / 3} 4 \sqrt{2} \sqrt{1-\sin t \cos t \sin t-\left(1-\sin ^{2} t\right) \cos t} d t= \\
& =\int_{0}^{\pi / 3} 4 \sqrt{2} \sqrt{1-\sin ^{2} t \cos t-\cos t+\sin ^{2} t \cos t} d t= \\
& =\int_{0}^{\pi / 3} 4 \sqrt{2} \sqrt{1-\cos t} d t=\int_{0}^{\pi / 3} 4 \sqrt{2} \sqrt{2} \sin \frac{t}{2} d t=\int_{0}^{\pi / 3} 8 \sin \frac{t}{2} d t= \\
& =\int_{0}^{\pi / 3} 8 \cdot 2 \sin \frac{t}{2} d\left(\frac{t}{2}\right)=\left.16\left(-\cos \frac{t}{2}\right)\right|_{0} ^{\pi / 3}=-16\left(\cos \frac{\pi}{6}-\cos 0\right)=-16\left(\frac{\sqrt{3}}{2}-1\right)=8(2-\sqrt{3})
\end{aligned}
$$

$\%$ D1\%8B_18-20»

Categories: Kuznetsov Problem Book Integrals Problem 18 | Integrals | Problems for Checking

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## Problem Kuznetsov Integrals 18-21

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