I was wondering if anyone can help me figure out a mistake I'm making in this problem: calculate the arc length of $y=x^{1/3}$ over $[3,4]$. I am familiar with the formula and once I square the derivative and put it back into the formula, I attempted u-substituion. But now I am stuck.
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- $\begingroup$ I don't think your function has a nice antiderivative. $\endgroup$Martin Argerami– Martin Argerami2018-05-01 23:46:07 +00:00Commented May 1, 2018 at 23:46
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2 Answers
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You have $f(x)=\sqrt[3]{x}$ so that $f'(x)= \dfrac{1}{3\sqrt[3]{x^2}}$. Then the length is $$\int_a^b \sqrt{1+ \big(f'(x)\big)^2} \; dx= \int_3^4 \sqrt{1+ \left(\dfrac{1}{3\sqrt[3]{x^2}} \right)^2} \;dx= \int_3^4 \sqrt{1+ \dfrac{1}{9 x^{4/3}}} \;dx $$
This last integral has no 'nice' antiderivative, see it here. So we need to numerically find this, you can use WolframAlpha again here to find $$\int_3^4 \sqrt{1+ \dfrac{1}{9 x^{4/3}}} \;dx \approx 1.010510879128807\ldots$$
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1 Arc length is $$\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$
The derivative is $$\frac{dy}{dx} = \frac{1}{3x^{\frac{2}{3}}}$$
So arc length is $$\int_{3}^{4}\sqrt{1+\left(\frac{1}{3x^{\frac{2}{3}}}\right)^2}dx \approx1.010510879$$
- 1$\begingroup$ Your work as displayed is correct but your final answer is not. $\endgroup$mathematics2x2life– mathematics2x2life2018-05-01 23:52:19 +00:00Commented May 1, 2018 at 23:52