If the surface area of a cube is increasing at a rate of 3.6 cm²/sec, retaining its shape; then the

Q: Wait, why do we need to find $\frac{da}{dt}$? Can't we just relate $\frac{dV}{dt}$ directly to $\frac{dA}{dt}$?

That is a brilliant intuition! You absolutely can relate them directly by expressing $V$ in terms of $A$ (since $V = a^3$ and $A = 6a^2$, then $a = \sqrt{A/6}$, so $V = (A/6)^{3/2}$). However, differentiating that expression is often more prone to algebraic errors. Using $\frac{da}{dt}$ as an intermediate 'bridge' is the standard, safest, and most elegant way to handle related rates in JEE problems.

Q: Why does the side length $a$ cancel out in the final step? Is that a coincidence?

It is not a coincidence at all! It is the beautiful geometry of the cube. Notice that the rate of change of surface area is proportional to $a \cdot \frac{da}{dt}$, and the rate of change of volume is proportional to $a^2 \cdot \frac{da}{dt}$. When you divide them, the $a$ terms align perfectly. It is a sign that your derivation is correct.

Q: What if the cube wasn't retaining its shape? Could I still solve it?

If the cube didn't retain its shape, it wouldn't be a cube! The problem explicitly states 'retaining its shape,' which is the key constraint that allows us to use a single variable $a$. If it were a general rectangular prism, you would need multiple variables and partial derivatives, which is a much more advanced topic. Stick to the symmetry here!

Question

If the surface area of a cube is increasing at a rate of $3.6	ext{cm}^2/	ext{sec}$, retaining its shape; then the rate of change of its volume (in $	ext{cm}^3/	ext{sec}$), when the length of a side of the cube is $10	ext{cm}$, is:

Sigma Percentile · Accepted Answer

### The Geometry of Growth: An Expanding Cube

Imagine you are standing in a laboratory, watching a perfect, crystalline cube grow. It is not just getting bigger; it is expanding uniformly, maintaining its perfect cubic symmetry at every single microsecond.

This is the essence of a 'related rates' problem in calculus. We are not just looking at static shapes; we are looking at the dynamic, living pulse of geometry.

### The Foundation

To understand this expansion, we must first define the 'DNA' of our cube. Let $a$ be the side length of the cube at any given moment $t$.

The surface area $A$ is the sum of the areas of its six square faces, giving us the elegant relation:
$$A = 6a^2$$

Simultaneously, the volume $V$, the space contained within this expanding boundary, is defined by:
$$V = a^3$$

These two equations are the bedrock of our journey. They link the two-dimensional 'skin' of the cube to its three-dimensional 'heart.'

### The Chain Rule Bridge

We are given that the surface area is increasing at a constant rate: $\frac{dA}{dt} = 3.6 	ext{ cm}^2/	ext{sec}$. Our mission is to find the rate of change of the volume, $\frac{dV}{dt}$, at the exact instant when $a = 10 	ext{ cm}$.

To get from the surface area to the volume, we need a translator. That translator is the rate of change of the side length, $\frac{da}{dt}$.

We start by differentiating our surface area equation with respect to time $t$. Using the chain rule, we get:
$$\frac{dA}{dt} = \frac{d}{dt}(6a^2) = 12a \frac{da}{dt}$$

This is the moment where many students stumble. Do not forget that $a$ is a function of time! The derivative of $a^2$ is not just $2a$; it is $2a \cdot \frac{da}{dt}$. This is the heartbeat of the problem.

### The Connection

Now, we substitute our known rate, $\frac{dA}{dt} = 3.6$, into our derivative equation:
$$3.6 = 12a \frac{da}{dt}$$

Solving for $\frac{da}{dt}$, we find:
$$\frac{da}{dt} = \frac{3.6}{12a} = \frac{0.3}{a}$$

This expression, $\frac{da}{dt} = \frac{0.3}{a}$, is our golden key. It tells us exactly how fast the side length is growing at any size $a$.

### The Volume Synthesis

Now, we turn to the volume. We differentiate $V = a^3$ with respect to time $t$:
$$\frac{dV}{dt} = 3a^2 \frac{da}{dt}$$

Here is the magic. We can substitute our expression $\frac{da}{dt} = \frac{0.3}{a}$ directly into this equation:
$$\frac{dV}{dt} = 3a^2 \left( \frac{0.3}{a} ight)$$

Watch the algebra collapse beautifully. One $a$ in the denominator cancels one $a$ in the numerator, leaving us with:
$$\frac{dV}{dt} = 3a 	imes 0.3 = 0.9a$$

### The Final Climax

We have arrived at a simple, linear relationship: the rate of change of volume is exactly $0.9$ times the current side length. Now, we simply plug in the value given in the problem, $a = 10 	ext{ cm}$:
$$\frac{dV}{dt} = 0.9 	imes 10 = 9 	ext{ cm}^3/	ext{sec}$$

And there it is. At the moment the side length is $10 	ext{ cm}$, the volume is expanding at **$9 	ext{ cm}^3/	ext{sec}$**. You have successfully navigated the relationship between surface area and volume, using the chain rule to bridge the gap.

Animated Solution for Mathematics - Differentiation: If the surface area of a cube is increasing at a rate of $3.6 cm^{2} / sec$ , retaining its shape; then the rate of change of its volume (in $cm^{3} / sec$ ), when the length of a side of the cube is $10 cm$ , is:

Enter Numerical Value:

Visualized Solution

Visualizing the Expanding Cube

Identifying the Given Information

Differentiating Surface Area

Substituting the Known Rate

Isolating $\frac{d a}{d t}$

Differentiating Volume

Linking the Rates

Simplifying the Expression

Final Calculation

Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of Growth

The Foundation

The Chain Rule Bridge

The Connection

The Volume Synthesis

The Final Climax

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