Q.A cylindrical tank of radius $10$ m is being filled with wheat at the rate of $314$ cubic metre per hour. Then the depth of the wheat is increasing at the rate of (A) $1$ m/h (B) $0.1$ m/h (C) $1.1$ m/h (D) $0.5$ m/h

Concept: Related Rates — we relate the rate of change of volume to the rate of change of depth using the geometry of the cylinder.

The volume of a cylinder is $V = \pi r^2 h$. Here $r = 10$ m is constant, so $V = 100\pi h$.

Differentiate both sides with respect to time $t$:

We are given $\frac{dV}{dt} = 314$ m³/h. Using $\pi \approx 3.14$: …

This is a textbook related rates problem — one of the cleanest applications of implicit differentiation in calculus. The core idea: when two quantities are linked by a geometric formula (here, volume and height of a cylinder), their rates of change with respect to time are also linked. If you know how fast one is changing, you can find how fast the other is changing, provided the geometry doesn't change shape.

The tank is a cylinder with a fixed radius of $10$ m. That's crucial: the radius isn't changing, so the cross-sectional area is constant. That means the volume increases linearly with height — no complicated shape changes.

Let's walk through it step by step.

1. Write the relationship between volume and height. For a cylinder, $V = \pi r^2 h$. Here $r = 10$ m, so

This is the static formula. But we care about how $V$ and $h$ change over time.

2. Differentiate both sides with respect to time $t$. Since $r$ is constant, $\pi r^2$ is just a number. Differentiating:

This is the engine of the problem: the rate of change of volume equals the (constant) cross-sectional area times the rate of change of height.

3. Plug in what we know. We are told $\frac{dV}{dt} = 314$ m³/h. So:

4. Solve for $\frac{dh}{dt}$. $\frac{dh}{dt} = \frac{314}{100\pi} = \frac{3.14}{\pi}.$ …

Method: Related Rates When One Dimension is Fixed

This method handles related-rates problems where a container's shape has one dimension held constant (here, the tank's radius) — recognising this collapses the volume formula to a single active variable, making the related-rates step much simpler than the general two-variable case.

Steps

Step 1: Identify which dimensions are constant and which change with time

For a cylindrical tank being filled, the radius $r$ is fixed by the tank's shape; only the depth $h$ and the volume $V$ change as wheat is poured in. Spotting this before differentiating tells you exactly which symbol needs a $\frac{d}{dt}$ and which is just a number.

Step 2: Write the volume formula with the constant substituted in

With $r=10$ m fixed, this becomes $V = 100\pi h$ — a formula in a single active variable, $h$.

Step 3: Differentiate with respect to time

Because $r$ never changes, no $\frac{dr}{dt}$ term appears at all — this is what makes a "fixed cross-section" problem simpler than a general two-variable related-rates problem.

Step 4: Substitute the given rate and solve

$\frac{dh}{dt} = \frac{1}{100\pi}\,\frac{dV}{dt}.$ …

Common Mistakes

Mistake 1: Differentiating $r$ as though it were also changing

A student might write $\frac{dV}{dt}=2\pi r h\frac{dr}{dt}+\pi r^2\frac{dh}{dt}$ out of habit from the general product-rule form for related rates. Why it's wrong: the tank's radius is fixed by its shape — only the wheat's depth changes as it's poured in, so $\frac{dr}{dt}=0$ and that whole term must vanish; including it wrongly introduces an unknown that can't be solved for. Correct approach: substitute the constant $r=10$ into the volume formula before differentiating, since $r$ never varies here.

Mistake 2: Forgetting to use $\pi\approx3.14$ consistently and mismatching units …