Arc length

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October 8th 2008, 11:56 AM
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

http://webwork.math.uwyo.edu/webwork...db3c42f1c1.png
The cable weighs 17.3 pounds per linear foot.
Find the weight of the cable.
October 8th 2008, 03:48 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 $2\int_0^{18}\sqrt{1+(.4(1.5 x^{1/2})^2}dx$
October 8th 2008, 05:18 PM
amiv4

k i just wasnt sure about the limits. why is it 0 to 18
October 9th 2008, 01:24 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.

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