91Ó°ÊÓ

The derivative of a function of a real variable in mathematics describes the sensitivity of the function value (output value) to changes in its argument (input value).

Calculus uses derivatives as a fundamental tool. When a derivative of a single-variable function exists at a given input value, it is the slope of the tangent line to the function's graph at that point.

Let the volume of water in the tank at any moment be\(V(t) = {A_w}h\).

Here,\({A_w}\)(in feet) is the constant area of the upper surface of the water. So,

\(\begin{array}{c}\frac{{dV(t)}}{{dt}} = {A_w}\frac{{dh}}{{dt}}\\\frac{1}{{{A_w}}}\frac{{dV(t)}}{{dt}} = \frac{{dh}}{{dt}}\end{array}\)

Using by Torricelli’s law, then the equation becomes,

\(\frac{{dV(t)}}{{dt}} = - {A_h}\sqrt {2gh} \)

Obtain the equation from the above two equations.

\(\frac{{dh}}{{dt}} = - \frac{{{A_h}}}{{{A_w}}}\sqrt {2gh} \)

Let the surface area of the hole be,

\(\begin{array}{c}{A_h} = \pi {r^2}\\ = \pi {\left( {\frac{{0.5}}{{12}}} \right)^2}\\ = \frac{\pi }{{576}}f{t^2}\end{array}\)

Let the area of the cross section of the right-circular cylinder be,

\(\begin{array}{c}{A_w} = \pi {R^2}\\ = \pi {(2)^2}\\ = 4\pi f{t^2}\end{array}\)

Substitute all the known values in the differential equation for the height.

\(\begin{array}{c}\frac{{dh}}{{dt}} = - \frac{{\frac{\pi }{{576}}}}{{4\pi }}\sqrt {64h} \\ = - \frac{8}{{2304}}\sqrt h \\\frac{{dh}}{{dt}} = - \frac{{\sqrt h }}{{288}}\end{array}\)