- #1
B3NR4Y
Gold Member
- 170
- 1
I don't understand why finding the arc length is always difficult. I understand the formula and know pretty much all the integration methods, but whenever I try to find the arc length of a function like 8x2 = 27y3 from 1 to 8 it's unusually difficult.
I would start by solving for x
\begin{equation}
\begin{split}
&\frac{2}{3} x^{\frac{2}{3}} = y\\
&\frac{4}{9x^{\frac{1}{3}}} = y'\\
\end{split}
\end{equation}
Now I add this into the integration formula \int_a^b \sqrt{1+(y')^{2}} \, dx
\begin{equation}<br /> \begin{split}<br /> L^{8}_{1} &= \int_1^8 \, \sqrt{1+ (\frac{4}{9x^{\frac{1}{3}}})^{2}} \, dx \\<br /> &= \int_1^8 \, \sqrt{1+ (\frac{16}{81x^{\frac{2}{3}}})} \, dx \\ <br /> &= \int_1^8 \sqrt{\frac{81x^{\frac{2}{3}}+16}{81x^{\frac{2}{3}}}} \, dx \\<br /> &= \int_1^8 \frac{\sqrt{81x^{\frac{2}{3}}+16}}{9x^{\frac{1}{3}}} \, dx<br /> \end{split}<br /> \end{equation}
From here I would do a u-substitution of u=81x^{\frac{2}{3}}+16 and then take the derivative to find \frac{1}{54} du = \frac{1}{x^{\frac{1}{3}}} dx
My new limits would be 97 and 340
\begin{equation}
\begin{split}
L^{8}_{1} &= \frac{1}{9*54} \int_{97}^{340} \sqrt{u} \, du \\
&= \frac{1}{486} \Bigg(\frac{3}{2} u^{\frac{3}{2}} |^{97}_{340} \Bigg) \\
&=7.28... \\
\end{split}
\end{equation}
I got the question right, by going slowly and typing it all out, derp. Anyway, is there a special trick for these types of problems? Or is it to just go slow and remember rules.
I would start by solving for x
\begin{equation}
\begin{split}
&\frac{2}{3} x^{\frac{2}{3}} = y\\
&\frac{4}{9x^{\frac{1}{3}}} = y'\\
\end{split}
\end{equation}
Now I add this into the integration formula \int_a^b \sqrt{1+(y')^{2}} \, dx
\begin{equation}<br /> \begin{split}<br /> L^{8}_{1} &= \int_1^8 \, \sqrt{1+ (\frac{4}{9x^{\frac{1}{3}}})^{2}} \, dx \\<br /> &= \int_1^8 \, \sqrt{1+ (\frac{16}{81x^{\frac{2}{3}}})} \, dx \\ <br /> &= \int_1^8 \sqrt{\frac{81x^{\frac{2}{3}}+16}{81x^{\frac{2}{3}}}} \, dx \\<br /> &= \int_1^8 \frac{\sqrt{81x^{\frac{2}{3}}+16}}{9x^{\frac{1}{3}}} \, dx<br /> \end{split}<br /> \end{equation}
From here I would do a u-substitution of u=81x^{\frac{2}{3}}+16 and then take the derivative to find \frac{1}{54} du = \frac{1}{x^{\frac{1}{3}}} dx
My new limits would be 97 and 340
\begin{equation}
\begin{split}
L^{8}_{1} &= \frac{1}{9*54} \int_{97}^{340} \sqrt{u} \, du \\
&= \frac{1}{486} \Bigg(\frac{3}{2} u^{\frac{3}{2}} |^{97}_{340} \Bigg) \\
&=7.28... \\
\end{split}
\end{equation}
I got the question right, by going slowly and typing it all out, derp. Anyway, is there a special trick for these types of problems? Or is it to just go slow and remember rules.
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