Chapter 7.4, Problem 18E

To determine

To calculate: Analytically, the curve's exact length can be determined by algebraically differentiating the integrated.

Answer to Problem 18E

The length of the curve from

  $\begin{array}{c}\\ L=\frac{7\sqrt{3}}{3}\end{array}$

Explanation of Solution

Given Information: The function $y\left(x\right)$ is inverse to the function $f\left(x\right)=\sqrt{3x^4-1}$ hence, a derivation of $y\left(x\right)$ at $x$ is equal to $f\left(x\right)$ .

The integral is divided by $-2$ as the bottom limit of integration.

Calculation:

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left({\displaystyle \underset{-2}{\overset{x}{\int }}\sqrt{3t^4-1}dt}\right)\\ =\frac{d}{dx}\left({\displaystyle \underset{-2}{\overset{0}{\int }}\sqrt{3t^4-1}dt+{\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{3t^4-1dt}}}\right)\\ =\frac{d}{dx}\left({\displaystyle \underset{-2}{\overset{0}{\int }}\sqrt{3t^4-1dt}+\frac{d}{dx}\left({\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{3t^4-1}}\right)}\right)\\ =\frac{d}{dx}\left({\displaystyle \underset{-2}{\overset{0}{\int }}\sqrt{3t^4-1}dt}\right)+\frac{d}{dx}\left({\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{3t^4-1}dt}\right)\\ =\sqrt{3x^4}-1\end{array}$

Find the length of the curve given the definition in order to $x=-2$ to $x=-1$

  $\begin{array}{c}L={\displaystyle \underset{-2}{\overset{-1}{\int }}{\sqrt{1+\left(\frac{dy}{dx}\right)}}^{2}dx}\\ ={\displaystyle \underset{-2}{\overset{-1}{\int }}\sqrt{1+{\left(\sqrt{3x^4-}1\right)}^{2}dx}}\\ ={\displaystyle \underset{-2}{\overset{-1}{\int }}\sqrt{1+3x^4-1}}dx\\ ={\displaystyle \underset{-2}{\overset{-1}{\int }}\sqrt{3x^4}}dx\\ =\sqrt{3}{\displaystyle \underset{-2}{\overset{-1}{\int }}{x}^{2}dx}\\ =\sqrt{3}.\frac{x^3}{3}\left|{}_{-2}^{-1}\right.\\ =\frac{\sqrt{3}}{3}\left[-1-\left(-8\right)\right]\\ L=\frac{7\sqrt{3}}{3}\end{array}$