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
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$\int\limits_0^L \sqrt{1 + 64 \frac{h^2x^2}{L^4}}dx$
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$2\int\limits_0^{L/2} \sqrt{1 + 64 \frac{h^3x^2}{L^4}}dx$
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$\int\limits_0^{L/2}\sqrt{1 + 64 \frac{h^2x^2}{L^4}}dx$
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$2\int\limits_0^{L/2}\sqrt{1 + 64 \frac{h^2x^2}{L^4}}dx$
D
Correct answer
Explanation
The arc length of a curve y = f(x) from a to b is the integral of sqrt(1 + (dy/dx)^2) dx. Given y = 4h * x^2 / L^2, dy/dx = 8h * x / L^2. Thus, (dy/dx)^2 = 64 * h^2 * x^2 / L^4. The total length is the integral from -L/2 to L/2, which is 2 * integral from 0 to L/2.