# Arc Length Problem, check my answer?

• Jun 24th 2012, 07:16 PM
andvaka
[SOLVED] Arc Length Problem, check my answer?
Mind checking my solution anyone? I worry that I may have left out a step because the given coordinates include both x and y values.
• Jun 24th 2012, 07:19 PM
Prove It
Re: Arc Length Problem, check my answer?
• Jun 24th 2012, 07:27 PM
andvaka
Re: Arc Length Problem, check my answer?
My bad! Both are in the attachment.
• Jun 24th 2012, 07:37 PM
Prove It
Re: Arc Length Problem, check my answer?
Your derivative is incorrect. If \displaystyle \displaystyle \begin{align*} y = \frac{x^4 + 3}{6x} \end{align*}, this simplifies to \displaystyle \displaystyle \begin{align*} y = \frac{1}{6}x^3 + \frac{1}{2}x^{-1} \end{align*}, not \displaystyle \displaystyle \begin{align*} \frac{1}{6x^3} + \frac{1}{2}x^{-1} \end{align*}...
• Jun 24th 2012, 07:42 PM
andvaka
Re: Arc Length Problem, check my answer?
That aside, did I use the relationship properly when I just plugged in the x values from the coordinates provided for the definite integral?
• Jun 24th 2012, 07:53 PM
Prove It
Re: Arc Length Problem, check my answer?
Your logic seems correct. Now you need to fix the mistake, solve the integral and evaluate the price.