Arclength

**ch2kb0x** · March 17th 2008, 10:00 PM

I am getting nowhere with this problem. I would appreciate it if one could show the work, including the answer. thank you.

y = 1/3x^(3/2) - x^(1/2) , [2,8]

Thanks again!

**earboth** · March 17th 2008, 10:36 PM

Originally Posted by ch2kb0x

I am getting nowhere with this problem. I would appreciate it if one could show the work, including the answer. thank you.

y = 1/3x^(3/2) - x^(1/2) , [2,8]

Thanks again!

According to your headline I assume that you want to calculate the length of the curve of the function from 2 to 8 (?).

You know that the length l of the arc can be calculated by:

$l = \int \sqrt{1+(y')^2}\ dx$

1. Determine the first derivative. I've got: $y'=\frac12 \cdot x^{\frac12}-\frac12 \cdot x^{-\frac12}$

2. Calculate (y')²

3. Calculate $1 + (y')^2= \frac14 x + \frac12 + \frac14 x^{-1} = \left( \frac12x^{\frac12} + \frac12 x^{-\frac12} \right)^2$

4. Now calculate the arc length:

$l = \int_2^8 \sqrt{\left( \frac12x^{\frac12} + \frac12 x^{-\frac12} \right)^2}\ dx$

5. I've got: $l = \frac{17}3 \cdot \sqrt{2} \approx 8.01...$

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