Chapter 7.4, Problem 25E

To determine

To calculate: The curve's length must be determined $y=x^{1/3}+x^{2/3}$ .

Answer to Problem 25E

The curve's length is $3.614$

Explanation of Solution

Given Information: Take into account this function in terms of $y$:

Calculation:

  $\begin{array}{c}0=x^{2/3}+x^{1/3}-y\\ \sqrt[3]{x}=\frac{-1\pm \sqrt{1+4y}}{2}\\ \sqrt[3]{x}=\frac{-1+\sqrt{1+4y}}{2}\left(\sqrt[3]{x>0}\right)\\ x=\frac{{\left(-1+\sqrt{1+4y}\right)}^3}{8}\\ y\in \left[0,\frac{{(2\sqrt[3]{2}+1)}^2-1}{4}\right]\approx \left[0,2.847\right]\end{array}$

Due to the lack of the derivative $\frac{dy}{dx}(0)$ , think of the curve in terms of $y.$

The calculation for determining length.

The following formula determines the length of the $\left(1\right):$

  $\begin{array}{c}\frac{dx}{dy}=\frac{3}{8}{\left(-1+\sqrt{1+4y}\right)}^2.\frac{4}{2\sqrt{1+4y}}\\ =\frac{3}{4}\frac{{\left(-1+\sqrt{1+4y}\right)}^2}{\sqrt{1+4y}}\end{array}$

When you square this derivative and put the result back into $\left(2\right),$ the result is challenging to integrate. Use the NINT in this case.

To calculate this integral, use the NINT formula.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2.847}{\int }}\sqrt{1+\frac{9}{16}}.\frac{{\left(-1+\sqrt{1+4y}\right)}^2}{1+4y}dy}\\ NINT\left(\sqrt{1+\frac{9}{16}.\frac{{\left(-1+\sqrt{1+4y}\right)}^4}{1+4y}},y,0,2.847\right)\approx 3.614\end{array}$

The curve's length is $3.614$