Cable length

1. Oct 26, 2006

blumfeld0

A cable hangs between two poles of equal height and 37 feet apart.
At a point on the ground directly under the cable and
x feet from the point on the ground halfway between the poles
the height of the cable in feet is
h(x)=10 +(0.4)( x^{1.5})
The cable weighs 15.1 pounds per linear foot.
Find the weight of the cable.

so i find h'(x) i square it and add one to find the length of the curve
i.e
(1+ h'(x)^2 ) ^(1/2)

my question is is that right? and what are the limits of integration??

blumfeld0

2. Oct 26, 2006

HallsofIvy

Staff Emeritus
Well, more correctly "i find h'(x) i square it and add one" , take the square root and integrate "to find the length of the curve"

Since your variable, x, is the "from the point on the ground halfway between the poles", x= 0 there. At one pole x= -37/2 and at the other x= 37/2. Integrate with respect to x from -37/2 to 37/2.

3. Oct 26, 2006

blumfeld0

ok so i integrate (1 + h'(x)^2 )^(1/2) with repect to x from -37/2 to 37/2
i use mathematica and i get 39.25+ 24.9 i

where "i" is imaginary
why do i get an imaginary answer??

also what does the 15.1 pounds per linear foot have to do with it?

thanks

4. Oct 26, 2006

HallsofIvy

Staff Emeritus
Ah! Misinterpretation on my part. Since x is a distance it is always positive. The height "h(x)=10 +(0.4)( x^{1.5})" isn't even defined for x negative. Integrate from 0 to 37/2 to find the length of 1/2 and then double.