Calculating Length of a Cable Between Poles

[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

 

[HallsofIvy]

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

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.

 

[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

 

[HallsofIvy]

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.