1. ## arc length

I have been working on this problem for a bit and obviously missing something simple. I am finding arc length

Given the equation $\displaystyle y=\frac{3{x^{2/3}}}{2}+4$. I am coming up with $\displaystyle L=\int_{1}^{27}\sqrt{1+x^{1/9}}~dx$. Can I get some help?

2. Originally Posted by Possible actuary
I have been working on this problem for a bit and obviously missing something simple. I am finding arc length

Given the equation $\displaystyle y=\frac{3{x^{2/3}}}{2}+4$. I am coming up with $\displaystyle L=\int_{1}^{27}\sqrt{1+x^{1/9}}~dx$. Can I get some help?
Arc length is:

$\displaystyle L = \int_a^b \sqrt{1+[f'(x)]^2}\ dx$

here:

$\displaystyle f(x)=\frac{3{x^{2/3}}}{2}+4$

so:

$\displaystyle f'(x)=x^{-1/3}$

so:

$\displaystyle L = \int_a^b \sqrt{1+x^{-2/3}}\ dx$

RonL

3. ## Re:

Originally Posted by Possible actuary
I have been working on this problem for a bit and obviously missing something simple. I am finding arc length

Given the equation $\displaystyle y=\frac{3{x^{2/3}}}{2}+4$. I am coming up with $\displaystyle L=\int_{1}^{27}\sqrt{1+x^{1/9}}~dx$. Can I get some help?
RE:

4. Originally Posted by qbkr21
RE:
Yes, but the exact value can be found using the u substitution.

5. Originally Posted by curvature
Yes, but the exact value can be found using the u substitution.
Aside from having my (y`)^2 wrong, this is what I was having trouble with. Thanks! I have a graphing calculator, TI 85, but do not know if it can compute arc length. I am sure it can just have not messed with it to find out.