Arc Length Problem, check my answer?

**andvaka** · June 24th 2012, 07:16 PM

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.

**Prove It** · June 24th 2012, 07:19 PM

Where is your solution? And your question, for that matter?

**andvaka** · June 24th 2012, 07:27 PM

My bad! Both are in the attachment.

**Prove It** · June 24th 2012, 07:37 PM

Your derivative is incorrect. If $\displaystyle \begin{align*} y = \frac{x^4 + 3}{6x} \end{align*}$ , this simplifies to $\displaystyle \begin{align*} y = \frac{1}{6}x^3 + \frac{1}{2}x^{-1} \end{align*}$ , not $\displaystyle \begin{align*} \frac{1}{6x^3} + \frac{1}{2}x^{-1} \end{align*}$ ...

**andvaka** · June 24th 2012, 07:42 PM

That aside, did I use the relationship properly when I just plugged in the x values from the coordinates provided for the definite integral?

**Prove It** · June 24th 2012, 07:53 PM

Your logic seems correct. Now you need to fix the mistake, solve the integral and evaluate the price.

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