Suggested languages for you:

Americas

Europe

Q. 37

Expert-verifiedFound in: Page 499

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

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.

$f\left(x\right)=\sqrt{9-{x}^{2}}$, $\left[a,b\right]=\left[-3,3\right]$

The arc length is $3\pi $.

Consider the given function $f\left(x\right)=\sqrt{9-{x}^{2}}$, $\left[a,b\right]=\left[-3,3\right]$.

The formula for a function to find the arc length from $x=a$ to $x=b$ is given by role="math" localid="1649165845561" ${\int}_{a}^{b}\sqrt{1+{\left({f}^{\text{'}}\left(x\right)\right)}^{2}}dx$.

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

94% of StudySmarter users get better grades.

Sign up for free