Question

a water trough is 10 m long and a cross - section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. if the trough is being filled with water at a rate of.2 m³/min, how fast is the water level rising when the water is 30 cm deep?

Explanation:

Step1: Find the formula for the volume of water

The cross - section is an isosceles trapezoid. The area of a trapezoid is \(A=\frac{(a + b)h}{2}\), where \(a\) and \(b\) are the lengths of the parallel sides and \(h\) is the height. For the water in the trough, let the depth of water be \(h\) (in meters). The width of the water surface at height \(h\) (by similar triangles) is \(w=0.3+\frac{(0.8 - 0.3)}{0.5}h=0.3 + h\). The length of the trough \(L = 10m\). So the volume of water \(V=\frac{(0.3+(0.3 + h))h}{2}\times10=5h(0.6 + h)=3h+5h^{2}\) (since \(h\) is in meters, \(0.3m = 30cm\), \(0.5m=50cm\), \(0.8m = 80cm\)).

Step2: Differentiate the volume with respect to time

Differentiate \(V\) with respect to \(t\) using the power rule \(\frac{dV}{dt}=\frac{d}{dt}(3h + 5h^{2})\). By the chain rule \(\frac{dV}{dt}=(3 + 10h)\frac{dh}{dt}\). We know that \(\frac{dV}{dt}=0.2m^{3}/min\).

Step3: Substitute \(h = 0.3m\) into the differentiated equation

When \(h = 0.3m\), we substitute into \(\frac{dV}{dt}=(3 + 10h)\frac{dh}{dt}\). So \(0.2=(3+10\times0.3)\frac{dh}{dt}\). Since \(3 + 10\times0.3=3 + 3=6\), then \(\frac{dh}{dt}=\frac{0.2}{6}=\frac{1}{30}m/min\).

Answer:

\(\frac{1}{30}m/min\)