Q. 41 - Calculus

Question

Find the exact value of the arc length of each function f(x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

$f\left(x\right)={(4-x^{2/3})}^{3/2}$, $\left[a,b\right]=\left[0,2\right]$. 

Step-by-Step Solution

Answer

The arc length is $3\left(2^{2/3}\right)$.

Step 1. Given information.

Consider the function is $f\left(x\right)={(4-x^{2/3})}^{3/2}$, $\left[a,b\right]=\left[0,2\right]$.

Step 2. Use arc length formula.

The formula for a function to find the arc length from $x=a$ to $x=b$ is given by ${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Step 3. Find the arc length.

$\begin{array}{rcl}\mathrm{Arc} \mathrm{length}& =& {\int }_0^2\sqrt{1+{\left(\frac{d}{dx}{(4-x^{2/3})}^{3/2}\right)}^2}dx\\ & =& {\int }_0^2\sqrt{1+{\left(\left(\frac{3}{2}\right){(4-x^{2/3})}^{1/2}\left(0-\left(\frac{2}{3}\right)x^{-1/3}\right)\right)}^2}dx\\ & =& {\int }_0^2\sqrt{1+{\left(-\left(\frac{3}{2}\right){(4-x^{2/3})}^{1/2}\left(\frac{2}{3}\right)x^{-1/3}\right)}^2}dx\\ & =& {\int }_0^2\sqrt{1+{\left(-{(4-x^{2/3})}^{1/2}{x}^{-1/3}\right)}^2}dx\\ & =& {\int }_0^2\sqrt{1+(4-x^{2/3})x^{-2/3}}dx\\ & =& {\int }_0^2\sqrt{1+4x^{-2/3}-1}dx\\ & =& {\int }_0^{2}2x^{-1/3}dx\\ & =& 2{\left[\frac{x^{2/3}}{2/3}\right]}_0^2\\ & =& 2\left({\left[\frac{2^{2/3}}{2/3}\right]}_0^2-{\left[\frac{0^{2/3}}{2/3}\right]}_0^2\right)\\ & =& 3\left(2^{2/3}\right)\end{array}$