Question

A parabolic cable is held between two supports at the same level. The horizontal span between the supports is L. The sag at the mid span is h. The equation of the parabola is \(y=4h\frac{x^2}{L^2}\), where x is the horizontal coordinate and y is the vertical coordinate with the origin at the centre of the cable. The expression for the total length of the cable is:
1. \(\mathop \displaystyle \int \nolimits_0^{\rm{L}} \sqrt {\begin{array}{*{20}{c}} {1 + 64\dfrac{{{{\rm{h}}^2}{{\rm{x}}^2}}}{{{{\rm{L}}^4}}}}\\ \; \end{array}} {\rm{\;dx}}\)
2. \(2\mathop \displaystyle \int \nolimits_0^{{\rm{L}}/2} \sqrt {\begin{array}{*{20}{c}} {1 + 64\dfrac{{{{\rm{h}}^3}{{\rm{x}}^2}}}{{{{\rm{L}}^4}}}}\\ \; \end{array}} {\rm{\;dx}}\)
3. \(\mathop \displaystyle \int \nolimits_0^{{\rm{L}}/2} \sqrt {\begin{array}{*{20}{c}} {1 + 64\dfrac{{{{\rm{h}}^2}{{\rm{x}}^2}}}{{{{\rm{L}}^4}}}}\\ \; \end{array}} {\rm{\;dx}}\)
4. \(2\mathop \displaystyle \int \nolimits_0^{{\rm{L}}/2} \sqrt {\begin{array}{*{20}{c}} {1 + 64\dfrac{{{{\rm{h}}^2}{{\rm{x}}^2}}}{{{{\rm{L}}^4}}}}\\ \; \end{array}} {\rm{\;dx}}\)
5.

Answer 1

Correct Answer - Option 4 : \(2\mathop \displaystyle \int \nolimits_0^{{\rm{L}}/2} \sqrt {\begin{array}{*{20}{c}} {1 + 64\dfrac{{{{\rm{h}}^2}{{\rm{x}}^2}}}{{{{\rm{L}}^4}}}}\\ \; \end{array}} {\rm{\;dx}}\)

Concept: 

Length of the curve f(x) between x = a and x = b is given by

Length of curve f(x) \(= \mathop \displaystyle \int \nolimits_{\rm{a}}^{\rm{b}} \sqrt {1 + {{\left( {\dfrac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^2}} {\rm{dx}} \)

\(y=4h\dfrac{{{x^2}}}{{{L^2}}}\)

y - Vertical coordinate with the origin at the centre of the cable (Rise of cable)

h – Sag at mid span, x – Horizontal coordinate, L - Horizontal span between the supports

Calculation:

Given:

We know

\(y=4h\dfrac{{{x^2}}}{{{L^2}}}\)

\(\dfrac{{dy}}{{dx}} = 8h \times \dfrac{x}{{{L^2}}}\)

At supports x =0, y = 0

At mid span x = L/2, y =h

The cable is symmetrical about centre, we consider the left portion of cable

Length of the cable \(= 2 \times \mathop \displaystyle \int \nolimits_0^{\frac{{\rm{L}}}{2}} \sqrt {1 + {{\left( {\dfrac{{8 \times {\rm{h}} \times {\rm{x}}}}{{{{\rm{L}}^2}}}} \right)}^2}} {\rm{dx}} \)

⇒ Length of the cable \(= 2\mathop \displaystyle \int \nolimits_0^{{\rm{L}}/2} \sqrt {\begin{array}{*{20}{c}} {1 + 64\dfrac{{{{\rm{h}}^2}{{\rm{x}}^2}}}{{{{\rm{L}}^4}}}}\\ \; \end{array}} {\rm{\;dx}} \)