Q.Sand is pouring from a pipe at the rate of $12 \text{ cm}^3/\text{s}$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4 \text{ cm}$?

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

Step 1: Write the given relation.

Height $h = \frac{1}{6}r$, so $r = 6h$. Volume of a cone:

Step 2: Differentiate with respect to time $t$.

Step 3: Substitute known values.

Given $\frac{dV}{dt} = 12$ cm$^3$/s and $h = 4$ cm: …

Why related rates works here

When a quantity changes over time (here, sand volume), any other quantity geometrically tied to it also changes. The trick is to eliminate the intermediate variable (radius) using the given constraint, so that volume becomes a function of height alone. Then differentiating both sides with respect to time gives a direct relation between $dV/dt$ and $dh/dt$.

Step-by-step solution

1. Write down what you know

• Volume flow rate: $\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}$ (positive because sand is accumulating).
• Cone geometry: $h = \frac{1}{6}r$, so $r = 6h$.
• We want $\frac{dh}{dt}$ when $h = 4$ cm.

2. Express volume in terms of height only

The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. Substitute $r = 6h$:

This is the key relation: volume depends only on the cube of the height.

3. Differentiate with respect to time

Both $V$ and $h$ are functions of time $t$. Differentiate implicitly:

4. Plug in known values

We know $\frac{dV}{dt} = 12$ and $h = 4$: …

Method: Eliminating a Variable via a Geometric Constraint Before Differentiating

This method applies to related-rates problems with a cone (or any shape with two dimensions, like height and radius, tied together by a fixed ratio) — the trick is to reduce the volume formula to a single variable BEFORE differentiating, rather than juggling two changing quantities.

Steps

Step 1: Write down the volume formula and the geometric constraint separately

For a cone, $V = \frac{1}{3}\pi r^2 h$, and the problem gives a fixed relationship between $h$ and $r$ (e.g. $h$ is always some fraction of $r$) that holds throughout the process.

Step 2: Use the constraint to eliminate one variable from the volume formula

Solve the constraint for the variable you do NOT need (commonly $r$, if the question asks about $h$), and substitute it into $V$. This leaves volume as a function of a single variable — here, $h$ alone — which makes the differentiation in the next step far simpler than trying to differentiate a two-variable product with an implicit constraint attached.

Step 3: Differentiate this single-variable volume expression with respect to time

Since the remaining variable (e.g. $h$) is itself a function of $t$, use the chain rule — a term in $h^3$ differentiates to $3h^2\frac{dh}{dt}$, and so on. …

Common Mistakes

Mistake 1: Inverting the given ratio between height and radius

Why it's wrong: the problem states the height is one-sixth of the radius, i.e. $h=\frac{r}{6}$, so $r=6h$ — using $h=6r$ (or $r=\frac{h}{6}$) instead flips the relationship and produces a volume formula that's wrong by a large factor after cubing. Correct approach: carefully translate "height is one-sixth of the radius" as $h=\frac{1}{6}r$, then solve for $r$ in terms of $h$ (not the other way around) before substituting into the volume formula.

Mistake 2: Substituting $h=4$ before differentiating with respect to time …