## Problem Statement

Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.

$$
y=-\arccos x+\sqrt{1-x^{2}}+1, \quad 0 \leq x \leq \frac{9}{16}
$$
## Solution

The length of the arc of a curve given by the equation $y=f(x) ; a \leq x \leq b$ is determined by the formula

$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$

Let's find the derivative of the given function:

$$
f^{\prime}(x)=\left(-\arccos x+\sqrt{1-x^{2}}+1\right)^{\prime}=-\frac{-1}{\sqrt{1-x^{2}}}-\frac{2 x}{2 \sqrt{1-x^{2}}}=\frac{1-x}{\sqrt{1-x^{2}}}=\sqrt{\frac{(1-x)^{2}}{(1-x)(1+x)}}=\sqrt{\frac{1-x}{1+x}}
$$

Then, using the above formula, we get:

$$
\begin{aligned}
L & =\int_{0}^{9 / 16} \sqrt{1+\left(\sqrt{\frac{1-x}{1+x}}\right)^{2}} d x= \\
& =\int_{0}^{9 / 16} \sqrt{1+\frac{1-x}{1+x}} d x= \\
& =\int_{0}^{9 / 16} \sqrt{\frac{1+x+1-x}{1+x}} d x= \\
& =\int_{0}^{9 / 16} \frac{\sqrt{2} d x}{\sqrt{1+x}}= \\
& =\left.2 \sqrt{2} \sqrt{1+x}\right|_{0} ^{9 / 16}=2 \sqrt{2}\left(\frac{5}{4}-1\right)=\frac{1}{\sqrt{2}}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\� $\% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\% D0 \% 98 \% D0 \% BD \% D1 \% 82 \% D0 \% B5 \% D0 \% B3 \% D1 \% 80 \% D0 \% B0 \% D0 \% BB}$ $\%$ D1\%8B_17-20»

Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals | Problems for Checking

- Last modified: 09:43, 26 May 2010.
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## Problem Kuznetsov Integrals 17-21

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