Problem Statement
A water trough of length $5\text{ m}$ has a cross-section in the shape of an inverted isosceles triangle with base width $2\text{ m}$ and vertical height $1.5\text{ m}$. Water is pumped into the trough at a constant rate of $0.2\text{ m}^3/\text{min}$. Find the rate at which the water level is rising when the depth of water is $0.8\text{ m}$.
Verified Solution & Marking Scheme
Express water surface width w in terms of water depth h
By similar triangles in the cross-section:
$\frac{w}{h} = \frac{2}{1.5} = \frac{4}{3} \implies w = \frac{4}{3}h$
Express volume of water V as a function of depth h
Cross-sectional area of water $A = \frac{1}{2} w h = \frac{1}{2}\left(\frac{4}{3}h\right)h = \frac{2}{3}h^2$.
Total volume in the $5\text{ m}$ trough:
$V = A \times 5 = 5 \left( \frac{2}{3}h^2 \right) = \frac{10}{3}h^2$
Differentiate with respect to time t
$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{10}{3}h^2\right) = \frac{20}{3}h \frac{dh}{dt}$
Given $\frac{dV}{dt} = 0.2\text{ m}^3/\text{min}$ and $h = 0.8\text{ m}$:
$0.2 = \frac{20}{3}(0.8) \frac{dh}{dt} = \frac{16}{3} \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{0.2 \times 3}{16} = \frac{0.6}{16} = \frac{3}{80} = 0.0375\text{ m/min}$