# Arc length

• Oct 8th 2008, 12:56 PM
amiv4
Arc length
A cable hangs between two poles of equal height and http://webwork.math.uwyo.edu/webwork...471d66c821.png feet apart.
At a point on the ground directly under the cable and
http://webwork.math.uwyo.edu/webwork...dd0b8b8e91.png feet from the point on the ground halfway between the poles
the height of the cable in feet is
The cable weighs 17.3 pounds per linear foot.
Find the weight of the cable.
• Oct 8th 2008, 04:48 PM
shawsend
Arc length is:

$L=\int_0^a \sqrt{1+(f'(x))^2} dx$

right? And $f(x)=10+0.4x^{3/2}$

So total cable length is $2\int_0^{18}\sqrt{1+(.4(1.5 x^{1/2})^2}dx$
• Oct 8th 2008, 06:18 PM
amiv4
k i just wasnt sure about the limits. why is it 0 to 18
• Oct 9th 2008, 02:24 AM
shawsend
Quote:

Originally Posted by amiv4
k i just wasnt sure about the limits. why is it 0 to 18

Because f(x) measures the distance to the cable starting from the lowest point (call that the origin) and the two poles are 36 feet away.