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 \[{\text{y}} = 4{\text{h}}\left( {\frac{{{{\text{x}}^2}}}{{{{\text{L}}^2}}}} \right)\] , 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. \[\int\limits_0^{\text{L}} {\sqrt {1 + 64\frac{{{{\text{h}}^2}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}\] B. \[2\int\limits_0^{\frac{{\text{L}}}{2}} {\sqrt {1 + 64\frac{{{{\text{h}}^3}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}\] C. \[\int\limits_0^{\frac{{\text{L}}}{2}} {\sqrt {1 + 64\frac{{{{\text{h}}^2}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}\] D. \[2\int\limits_0^{\frac{{\text{L}}}{2}} {\sqrt {1 + 64\frac{{{{\text{h}}^2}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}\]

” option2=”\[2\int\limits_0^{\frac{{\text{L}}}{2}} {\sqrt {1 + 64\frac{{{{\text{h}}^3}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}\]” option3=”\[\int\limits_0^{\frac{{\text{L}}}{2}} {\sqrt {1 + 64\frac{{{{\text{h}}^2}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}\]” option4=”\[2\int\limits_0^{\frac{{\text{L}}}{2}} {\sqrt {1 + 64\frac{{{{\text{h}}^2}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}\]” correct=”option3″]

The correct answer is $\boxed{\text{C. }\int\limits_0^{\frac{{\text{L}}}{2}} {\sqrt {1 + 64\frac{{{{\text{h}}^2}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}}$.

The total length of the cable is the sum of the lengths of the two legs of the parabola. The length of each leg is equal to the arc length of the parabola from the origin to the point where it intersects the x-axis. The arc length of a parabola is given by the following formula:

$$L = \int_0^a \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$

In this case, the derivative of the equation of the parabola is $dy/dx = 8h x/L^2$. Substituting this into the formula for the arc length, we get:

$$L = \int_0^a \sqrt{1 + \left(\frac{8h x}{L^2}\right)^2} dx = \int_0^a \sqrt{1 + 64\frac{{{{\text{h}}^2}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}$$

The integral is evaluated from $x=0$ to $x=L/2$, since the parabola intersects the x-axis at $x=L/2$. Therefore, the total length of the cable is:

$$L = 2 \int_0^{\frac{{\text{L}}}{2}} {\sqrt {1 + 64\frac{{{{\text{h}}^2}{{\text{x}}^2}}}{{{{\text{L}}^4}}}} } {\text{dx}}$$