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

• A. ${\int }_0^L\sqrt{1+64\frac{h^2{x}^2}{L^4}dx}$
• B. $2{\int }_0^{L/2}\sqrt{1+64\frac{h^3{x}^2}{L^4}dx}$
• C. ${\int }_0^{L/2}\sqrt{1+64\frac{h^2{x}^2}{L^4}dx}$
• D. $2{\int }_0^{L/2}\sqrt{1+64\frac{h^2{x}^2}{L^4}dx}$

Answer 1

The correct option is D $2{\int }_0^{L/2}\sqrt{1+64\frac{h^2{x}^2}{L^4}dx}$



$A,B\Rightarrow$ Supports
$\frac{dy}{dx}=\left(\frac{8h}{L^2}\right)x$
Total length of the cable
$={\int }_{-L/2}^{L/2}\sqrt{1+\left({\frac{dy}{dx}}^2\right)dx}$
$=2{\int }_0^{L/2}\sqrt{1+\frac{64h^2}{L^4}}{x}^{2}dx$