Q.A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is $\tan^{-1}(0.5)$. Water is poured into it at a constant rate of $5$ cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is $4$ m.

The key idea is Related Rates: we connect $\frac{dV}{dt}$ (given) to $\frac{dh}{dt}$ (required) using the geometry of the cone.

Step 1: Relate radius and height.

Semi-vertical angle $\alpha$ satisfies $\tan \alpha = 0.5 = \frac{r}{h}$, so $r = \frac{h}{2}$.

Step 2: Express volume in terms of $h$ alone.

Volume of a cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$.

Step 3: Differentiate with respect to time $t$.

$\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt}$.

Step 4: Substitute known values.

Given $\frac{dV}{dt} = 5$ m³/h and $h = 4$ m:

$5 = \frac{\pi (4)^2}{4} \frac{dh}{dt} = 4\pi \frac{dh}{dt}$.

Thus $\frac{dh}{dt} = \frac{5}{4\pi}$ m/h.

This is a classic related rates problem. The core idea is simple: we know how fast the volume is changing ($dV/dt = 5$ m³/h), and we want to find how fast the depth is changing ($dh/dt$) at a specific moment. The bridge between these two rates is the geometric relationship between volume and depth for a cone.

The trick is that as water fills the cone, both the depth $h$ and the radius $r$ of the water's surface change together. But they aren't independent — the cone's fixed shape ties them together through the semi-vertical angle.

1. Set up the geometry. The cone has a semi-vertical angle $\alpha$ where $\tan \alpha = 0.5$. From the figure, $\tan \alpha = r/h$, so:

This is the crucial relation — at any depth $h$, the radius of the water surface is exactly half of $h$.

2. Write the volume in terms of $h$ only. The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. Substitute $r = h/2$:

3. Differentiate with respect to time. Both $V$ and $h$ are functions of time $t$. Differentiate both sides:

4. Plug in the known values. We are given $\frac{dV}{dt} = 5$ m³/h (constant), and we want $\frac{dh}{dt}$ when $h = 4$ m:

5. Solve for the rate.

Method: Related Rates for a Filling/Draining Container

The general five-step procedure for connecting a known rate to an unknown rate when two changing quantities are tied by a fixed geometric relationship.

Steps

Step 1: Identify the two rates and the fixed relationship

Here the given rate is $\frac{dV}{dt}$ and the required rate is $\frac{dh}{dt}$. The container's shape (a cone with a fixed semi-vertical angle) links the radius $r$ and height $h$ of the water surface at every instant, since $\tan(\text{semi-vertical angle}) = \frac{r}{h}$.

Step 2: Eliminate the extra variable using the geometry, before differentiating

A volume formula for a cone naturally involves two variables, $r$ and $h$. Use the fixed-angle relation to express $r$ in terms of $h$ (or vice versa) and substitute, so the volume becomes a function of a single variable:

Step 3: Differentiate both sides with respect to time

Apply the chain rule, since both $V$ and $h$ are functions of $t$:

Step 4: Substitute the given numerical values and solve

Plug in the known $\frac{dV}{dt}$ and the specific depth $h$ at the instant asked about — only after differentiating, never before — then solve algebraically for the unknown rate.

Common Mistakes

Mistake 1: Differentiating $V=\frac{1}{3}\pi r^2 h$ while treating $r$ as constant

Why it's wrong: as water fills the cone, the radius of the water's surface changes together with the depth — it is not a fixed number, so it cannot be dropped from the differentiation. Differentiating with $r$ held constant would produce an equation missing the crucial link between $\frac{dh}{dt}$ and the cone's geometry. Correct approach: use the semi-vertical angle to write $r$ in terms of $h$ (here $r=h/2$) and substitute into the volume formula before differentiating, so the volume is a function of $h$ alone.

Mistake 2: Substituting the numerical depth before differentiating

Why it's wrong: plugging in $h=4$ into the volume formula first turns $h$ into a constant, so the derivative with respect to time becomes $\frac{dV}{dt}=0$ — losing the very relationship the problem needs. Correct approach: differentiate the general relation between $V$ and $h$ symbolically first, and only substitute the specific numbers ($h=4$, $\frac{dV}{dt}=5$) into the resulting rate equation afterward.