Question

Find the arc length of one arch of $\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$.

Short answer

The length of one arc is $\sqrt{2}E\left(\pi ,\frac{1}{\sqrt{2}}\right)\approx 3.82019$.

Step-by-step solution

Given Information

The function is $y=\sin x$.

Definition of the differential element of displacement.

The differential element of displacement is defined as $\mathbf{\int }\mathit{d}\mathit{s}\mathbf{=}\mathbf{\int }\sqrt{{\mathbf{dx}}^{\mathbf{2}}\mathbf{+}{\mathbf{dx}}^{\mathbf{2}}}$.

Find the value of Integral

The formula of differential element of displacement is $\int ds=\int \sqrt{d{x}^2+d{y}^2}$.

$$\begin{gathered} \int ds=\int \sqrt{d{x}^2\left(1+\frac{d{y}^2}{d{x}^2}\right)} \\ \int ds=\int \sqrt{\left(1+\frac{d{y}^2}{d{x}^2}\right)}dx \end{gathered}$$

Substitute the value given below in the equation mentioned above.

$$\begin{gathered} y=\sin x \\ \frac{dy}{dx}=\cos x \end{gathered}$$

The equation becomes as follows:

$$\begin{gathered} \int ds=\int \sqrt{\left(1+{\cos }^{2}x\right)}dx \\ \int ds=\int \sqrt{\left(1+(1-{\sin }^{2}x)\right)}dx \\ \int ds=\int \sqrt{\left(2-{\sin }^{2}x\right)}dx \\ \int ds=\sqrt{2}\int \sqrt{\left(1-\frac{{\sin }^{2}x}{2}\right)}dx \end{gathered}$$

The formula for the elliptical function is $E\left(\frac{\pi }{2},K\right)=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{1-k^2{\sin }^2\theta }d\theta$.

The equation becomes as follows:

$$\begin{gathered} \int ds=\sqrt{2}\int \sqrt{\left(1-\frac{{\sin }^{2}x}{2}\right)}dx \\ \int ds=\sqrt{2}E\left(\pi ,\frac{1}{\sqrt{2}}\right) \\ \int ds\approx 3.82019 \end{gathered}$$

The length of one arc is $\sqrt{2}E\left(\pi ,\frac{1}{\sqrt{2}}\right)\approx 3.82019$.