Question

Please help me find the arc length of the following curve over the given interval. y=x^(2/3) from x=1 to x=8

Answer 1

\[y=x^{\frac{ 2 }{ 3 }}\]
\[\frac{ dy }{ dx }=\frac{ 2 }{ 3 }x ^{-\frac{ 1 }{ 3 }}\]
\[L=\int\limits_{1}^{8}\sqrt{1+\frac{ 2 }{ 3 }x ^{-\frac{ 1 }{ 3 }}}dx\]

Answer 2

I might be wrong, I just derived it again, but isn't the form:

\[\Large L=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx \] So if I didn't confuse anything, you need to match your formula again

Answer 3

oh yes you're right i forgot to square the derivative

Answer 4

Ok this is where I am now...
\[\int\limits_{1}^{8}\frac{ \sqrt{4+9x ^{2/3}} }{ 3x ^{1/3} }dx\]
\[\frac{ du }{ dx }=\frac{ 6 }{ x ^{1/3} }\]

Answer 5

well, maybe like this
\[\Large \frac{1}{3} \int \frac{1}{x^{1/3}} \sqrt{9x^{2/3}+4}dx \] So your substitution should work, the term under the radical is linear.

\[ \large \frac{du}{dx}= \frac{6}{x^{1/3}} \] and now back substitution.

Answer 6

\[\Large \frac{1}{3} \int \frac{du}{6dx} \sqrt{u}dx =\frac{1}{18} \int u^{1/2}du \]