Question

x2+y24=1Find the length of arc of the ellipse${\mathit{x}}^{\mathbf{2}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{4}}\mathbf{=}\mathbf{1}$between$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{)}$and$\mathbf{(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{)}$. (Note that here$\mathit{b}\mathbf{>}\mathit{a}$; see Example 4.).

Short answer

The length of arc of ellipse is $2E(\frac{\mathrm{\pi }}{2},\frac{\sqrt{3}}{2})-2E(\varphi ,\frac{\sqrt{3}}{2})\approx 0.29161$.

Step-by-step solution

Given Information

The equation of ellipse is $x^2+\frac{y^2}{4}=1$.

Definition of circumference of ellipse.

The circumference of ellipse defined as $\mathbf{4}\mathit{a}\underset{\mathbf{0}}{\overset{\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{\int }}}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}\mathit{d}\mathit{\theta }$.

Find the value of Integral.

The equation of ellipse is $x^2+\frac{y^2}{4}=1$.

The circumference of ellipse defined as$C=4a\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{1-\frac{a^2-b^2}{a^2}{\sin }^2\theta }d\theta$.

Substitute the values given below in the equation mentioned above.

role="math" localid="1664304134132" $$\begin{gathered} a^2=1 \\ {b}^2=4 \end{gathered}$$

$$\begin{gathered} C=4a\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{1-\frac{a^2-b^2}{a^2}{\sin }^2\theta }d\theta \\ {C}_1=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{1-\frac{-1+4}{1}{\sin }^2\theta }d\theta -2\underset{0}{\overset{\varphi }{\int }}\sqrt{1-\frac{-1+4}{1}{\sin }^2\theta }d\theta \\ C=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{1-\frac{3}{4}{\sin }^2\theta }d\theta -2\underset{0}{\overset{\varphi }{\int }}\sqrt{1-\frac{3}{4}{\sin }^2\theta }d\theta \end{gathered}$$

The equation becomes as follows and the limits are mentioned in the formula.

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

Equate the above equation with the value of C₁, the value of C₁ becomes follows.

$$\begin{gathered} C_1=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{1-\frac{3}{4}{\sin }^2\theta }d\theta -2\underset{0}{\overset{\varphi }{\int }}\sqrt{1-\frac{3}{4}{\sin }^2\theta }d\theta \\ {C}_1=2E(\frac{\pi }{2},\frac{\sqrt{3}}{2})-2E(\varphi ,\frac{\sqrt{3}}{2}) \\ {C}_1\approx 0.29161 \end{gathered}$$

Hence, the solution is mentioned below.

The length of arc of ellipse is $2E(\frac{\pi }{2},\frac{\sqrt{3}}{2})-2E(\varphi ,\frac{\sqrt{3}}{2})\approx 0.29161$.