Practice 2: related rates in an unfamiliar context

Answer:

(a) The cross-section of the water is a triangle similar to the cross-section of the trough, because the sides are straight and it fills from the point upwards.

At full depth 1 m the width is 1.2 m, so by similar triangles the width $w$ at depth $h$ satisfies

$\frac{w}{h} = \frac{1.2}{1} \quad \Rightarrow \quad w = 1.2h$

The triangular cross-sectional area is

$A = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2}(1.2h)(h) = 0.6h^2$

The trough is a prism of length 3 m, so

$V = 3 \times 0.6h^2 = 1.8h^2 \ \text{m}^3$

as required.

(b) We know $\dfrac{dV}{dt} = 0.05$ and want $\dfrac{dh}{dt}$. These are linked through $V$, so use the chain rule:

$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$

Differentiating $V = 1.8h^2$:

$\frac{dV}{dh} = 3.6h$

Substituting:

$0.05 = 3.6h \times \frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = \frac{0.05}{3.6h}$

At $h = 0.4$:

$\frac{dh}{dt} = \frac{0.05}{3.6 \times 0.4} = \frac{0.05}{1.44} \approx 0.0347$

The depth is rising at about 0.035 m per minute (2 significant figures), or roughly 3.5 cm per minute.

(c) From $\dfrac{dh}{dt} = \dfrac{0.05}{3.6h}$, the rate is inversely proportional to $h$.

So as the trough fills and $h$ increases, $\dfrac{dh}{dt}$ decreases — the water level rises more and more slowly.

Physically this is because the trough is wider higher up, so each extra centimetre of depth requires more water than the one before. As $h \to 0$ the formula predicts an arbitrarily fast rise, which is the model breaking down at the very bottom where the trough comes to a point.