Q. 36 - 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. 

$f\left(x\right)=2{(1-x)}^{3/2}+3$, $\left[a,b\right]=\left[-2,0\right]$

Step-by-Step Solution

Answer

The arc length is $\begin{array}{rcl}& & \frac{2}{27}{\left(28\right)}^{3/2}-\frac{2}{27}{\left(10\right)}^{3/2}\end{array}$.

Step 1. Given information.

Consider the given function $f\left(x\right)=2{(1-x)}^{3/2}+3$, $\left[a,b\right]=\left[-2,0\right]$.

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 ${\int }_a^b\sqrt{1+(f^{\text{'}}{\left(x\right))}^2}dx$.

Step 3. Find the arc length.

$\begin{array}{rcl}\mathrm{Arc} \mathrm{length}& =& {\int }_{-2}^0\sqrt{1+{\left(\frac{\mathrm{d}}{\mathrm{dx}}(2{(1-\mathrm{x})}^{3/2}+3)\right)}^2}\mathrm{dx}\\ & =& {\int }_{-2}^0\sqrt{1+{\left(\left(2{\left(\frac{3}{2}\right)(1-\mathrm{x})}^{1/2}\right(-1)+0)\right)}^2}\mathrm{dx}\\ & =& {\int }_{-2}^0\sqrt{1+{\left(-3{(1-\mathrm{x})}^{1/2})\right)}^2}\mathrm{dx}\\ & =& {\int }_{-2}^0\sqrt{1+9(1-\mathrm{x})}\mathrm{dx}\\ & =& {\int }_{-2}^0\sqrt{10-9\mathrm{x}}\mathrm{dx}\end{array}$

Step 4. Simply obtained integral by simple substitution.

Substitute $10-9x=u$, $dx=-\frac{1}{9}du$ into $\begin{array}{rcl}& & {\int }_{-2}^0\sqrt{10-9\mathrm{x}}\mathrm{dx}\end{array}$.

$\begin{array}{rcl}{\int }_{28}^{10}-\frac{\sqrt{u}}{9}du& =& \left(-\frac{1}{9}\right){\int }_{28}^{10}\sqrt{u}du\\ & =& -\left(-\frac{1}{9}\right){\int }_{10}^{28}\sqrt{u}du\\ & =& \left(\frac{1}{9}\right){\int }_{10}^{28}\left(\frac{2}{3}{u}^{3/2}\right)\\ & =& \left(\frac{1}{9}\right){\left[\frac{2}{3}{u}^{3/2}\right]}_{10}^{28}\\ & =& \frac{2}{27}{\left(28\right)}^{3/2}-\frac{2}{27}{\left(10\right)}^{3/2}\end{array}$