Chapter 7.4, Problem 28E

To determine

To calculate: To determine the non-smooth curve's length $\sqrt{x}+\sqrt{y}=1$ .

Answer to Problem 28E

The length of the curve

  $1.62322$

Explanation of Solution

Given Information: Let's first translate $y$ into $x$ as

Calculation:

  $\begin{array}{l}\sqrt{x}+\sqrt{y}=1\\ y={\left(1-\sqrt{x}\right)}^2\end{array}$

Take note of the derivative $\frac{dy}{dx}\left(0\right)$ cannot be calculated using the smooth curves formula since it does not exist.

The given curve is symmetric with respect to the line $y=x$ , therefore this line divides the curve into two equal-length halves. Because the part below the line is a smooth curve, it should be evaluated first.

Finding the interval where the curve below the line is specified, furthermore

Since $y=x$ and $y={\left(1-\sqrt{x}\right)}^2$ is meet at the $x$ point, the curve below the line is given by

  $y={\left(1-\sqrt{x}\right)}^2$

Where $x\in \left[\frac{1}{4},1\right]$ .

The formula provides the length of this curve.

  ${{\displaystyle \underset{4}{\overset{1}{\int }}\sqrt{1+\left(\frac{d}{dx}\left({(1-\sqrt{x})}^2\right)\right)}}}^{2}dx$

The derivative is the next step.

The curve's derivative is

  $\begin{array}{c}\frac{d}{dx}\left({(1-\sqrt{x})}^2\right)=2(1-\sqrt{x}).\left(-\frac{1}{2\sqrt{x}}\right)\\ =\frac{1}{\sqrt{x}}-1\end{array}$

Get an integral to calculate this.

  ${{\displaystyle \underset{\frac{1}{4}}{\overset{1}{\int }}\sqrt{1+\left(\frac{1}{\sqrt{x}}-1\right)}}}^{2}dx={\displaystyle \underset{\frac{1}{4}}{\overset{1}{\int }}\sqrt{2+\frac{1}{x}}}-\frac{2}{\sqrt{x}}dx$

Use NINT, and then multiply the result by $22$ .

  $NINT\left(\sqrt{2+\frac{1}{x}}-\frac{2}{\sqrt{x}},x,\frac{1}{4},1\right)\approx 0.81161$

The length of the curve

  $\begin{array}{c}\sqrt{x}+\sqrt{y}=1\\ \mathrm{2.0.81161}=1.62322\end{array}$