FPM(2023) January Paper 02 que 9 and solution

Problem Statement

A cube has edges of length $x$ cm. The total surface area, $A \, \text{cm}^2$, of the cube is increasing at a constant rate of $0.45 \, \text{cm}^2/\text{s}$.

Find the rate of increase, in $\text{cm}^3/\text{s}$, of the volume of the cube at the instant when the total surface area of the cube is $384 \, \text{cm}^2$.

Solution

Step 1: Relationship between Surface Area and Edge Length

The surface area of the cube is given by:

$$ A = 6x^2 $$

Differentiating with respect to time $t$, we have:

$$ \frac{dA}{dt} = 12x \frac{dx}{dt} $$

Step 2: Relationship between Volume and Edge Length

The volume of the cube is given by:

$$ V = x^3 $$

Differentiating with respect to $t$, we have:

$$ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} $$

Step 3: Determine $x$ when $A = 384 \, \text{cm}^2$

Given $A = 384 \, \text{cm}^2$, solve for $x$:

$$ 6x^2 = 384 \implies x^2 = 64 \implies x = 8 \, \text{cm} $$

Step 4: Calculate $ \frac{dx}{dt} $

Given $ \frac{dA}{dt} = 0.45 \, \text{cm}^2/\text{s} $:

$$ \frac{dA}{dt} = 12x \frac{dx}{dt} \implies 0.45 = 12(8)\frac{dx}{dt} \implies \frac{dx}{dt} = \frac{0.45}{96} = 0.0046875 \, \text{cm}/\text{s} $$

Step 5: Calculate $ \frac{dV}{dt} $

Substitute $x = 8 \, \text{cm}$ and $ \frac{dx}{dt} = 0.0046875 \, \text{cm}/\text{s} $ into:

$$ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} $$

$$ \frac{dV}{dt} = 3(8^2)(0.0046875) = 3(64)(0.0046875) = 0.9 \, \text{cm}^3/\text{s} $$

Final Answer

The rate of increase of the volume of the cube is:

$$ \boxed{0.9 \, \text{cm}^3/\text{s}} $$