Question

A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of \({\bf{12}}\;{\bf{f}}{{\bf{t}}^{\bf{3}}}/{\bf{min}}\), how fast is the water level raising when the water is 30 cm deep?

Short answer

The rate of change of height of water in the trough is \(\frac{4}{5}\;{\rm{ft/min}}\).

Step-by-step solution

Find the expression for the rate of change of volume of trough

The figure below represents the schematic of the trough.

Using the property of similar triangles.

\(\begin{aligned}\frac{3}{1} &= \frac{b}{h}\\b &= 3h\end{aligned}\)

The volume of the trough is given by the expression.

\(V = \frac{1}{2}\left( {bh} \right)l\)

Therefore,

\(\begin{aligned}V &= \frac{1}{2}\left( {3h} \right)h\left( {10} \right)\\ &= 15{h^2}\end{aligned}\)

Differentiate the equation of volume with respect to t.

\(\begin{aligned}\frac{{{\rm{d}}V}}{{{\rm{d}}t}} &= \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {15{h^2}} \right)\\ &= 15\left( {2h} \right)\left( {\frac{{{\rm{d}}h}}{{{\rm{d}}t}}} \right)\end{aligned}\)

Find the rate of change of height of water in the trough

Substitute 2 for \(\frac{{{\rm{d}}V}}{{{\rm{d}}t}}\) and \(\frac{1}{2}\) for h in the equation \(\frac{{{\rm{d}}V}}{{{\rm{d}}t}} = 15\left( {2h} \right)\left( {\frac{{{\rm{d}}h}}{{{\rm{d}}t}}} \right)\).

\(\begin{aligned}\frac{{{\rm{d}}V}}{{{\rm{d}}t}} &= 15\left( {2h} \right)\left( {\frac{{{\rm{d}}h}}{{{\rm{d}}t}}} \right)\\12 &= 30\left( {\frac{1}{2}} \right)\left( {\frac{{{\rm{d}}h}}{{{\rm{d}}t}}} \right)\\\frac{{{\rm{d}}h}}{{{\rm{d}}t}} &= \frac{4}{5}\;{\rm{ft/min}}\end{aligned}\)

Thus, the rate of change of height of water in the trough is \(\frac{4}{5}\;{\rm{ft/min}}\).