Chapter 6.4, Problem 10E

To determine

Find the length of the curve $y=\sqrt{x-x^2}+{\sin }^{-1}\left(\sqrt{x}\right)$

Answer to Problem 10E

The length of the curve is $2$

Explanation of Solution

Given information: Given curve $y=\sqrt{x-x^2}+{\sin }^{-1}\left(\sqrt{x}\right)$

Calculation:

The domain of the function is $\left[0,1\right]$ .

(The expression under the root must be positive. It represents a parabola,

Opening down, with the graph above the axis being between the roots $0,1$

The other part of $y,{\sin }^{-1}\left(\sqrt{x}\right)$ is defined there as well.)

The curve has endpoints $\left[0,0\right]$ and $\left[1,\frac{\pi }{2}\right]$ ,

Its length is given by

  $L={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{{\left(\frac{dy}{dx}\right)}^2+1}}dx$

  $\begin{array}{l}y=\sqrt{x-x^2}+{\sin }^{-1}\left(\sqrt{x}\right)\\ \frac{dy}{dx}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{x}\sqrt{1-x}}\\ =\frac{2-2x}{2\sqrt{x}\sqrt{1-x}}\\ =\frac{2\sqrt{{\left(1-x\right)}^2}}{2\sqrt{x}\sqrt{1-x}}\\ =\sqrt{\frac{1-x}{x}}\\ 1+{\left(\frac{dy}{dx}\right)}^2=1+\frac{1-x}{x}=\frac{1}{x}\\ L={\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{\frac{1}{x}}}dx\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{-1/2}}dx\\ ={\left[2\sqrt{x}\right]}_0^1\\ =2\end{array}$

Thus the length of the curve is $2$