A solid metal cube is heated. At time \(t\) seconds each edge of the cube has length \(x\) cm, the volume of the cube is \(V\) cm\(^{3}\) and the total surf...

(a) Value of \(\dfrac{dx}{dt}\) [1]

The edge length increases at the constant rate \(0.02\) cm s\(^{-1}\), and a rate of change of \(x\) with respect to time is precisely \(\dfrac{dx}{dt}\), so

\[\frac{dx}{dt} = 0.02 \quad \textbf{[B1]}\]

A B mark: it is awarded for the correct value alone and is independent of every other part.

(b) \(V\) in terms of \(x\), and \(\dfrac{dV}{dx}\) [2]

A cube of edge \(x\) has volume

\[V = x^{3}, \qquad \text{so} \qquad \frac{dV}{dx} = 3x^{2} \quad \textbf{[B1][B1]}\]

Two independent B marks, one for each result.

(c) Rate of increase of the volume when \(x = 5\) [3]

The question links a rate with respect to time to a rate with respect to length, which is exactly what the chain rule is for. Nothing gives \(V\) as a function of \(t\) directly, so connect the two derivatives already found:

\[\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} = 3x^{2} \times 0.02 = 0.06x^{2} \quad \textbf{[M1][A1]}\] \[\text{At } x = 5: \quad \frac{dV}{dt} = 0.06 \times 25 = 1.5 \ \mathrm{cm^{3}\,s^{-1}} \quad \textbf{[A1]}\]

The M mark is for a correct chain-rule statement and survives an arithmetic slip; the two accuracy marks require the expression and then the value with its unit. Writing the chain the wrong way round, as \(\dfrac{dV}{dx} \div \dfrac{dx}{dt}\), is the standard error, and a unit check catches it: cm\(^{2}\) multiplied by cm s\(^{-1}\) gives cm\(^{3}\) s\(^{-1}\), which is a volume rate, whereas dividing does not.

(d) Show that \(\dfrac{dS}{dt} = 0.24x\) [4]

A cube has six congruent square faces, each of area \(x^{2}\), so

\[S = 6x^{2} \quad \textbf{[B1]}\] \[\frac{dS}{dx} = 12x \quad \textbf{[B1]}\]

Then apply the chain rule again:

\[\frac{dS}{dt} = \frac{dS}{dx} \times \frac{dx}{dt} \quad \textbf{[M1]}\] \[= 12x \times 0.02 = 0.24x \quad \textbf{[A1]}\]

The result is given, so all four marks are for the derivation; a candidate who writes \(\dfrac{dS}{dt} = 0.24x\) and stops earns nothing. The two B marks are independent, the M mark is for the chain-rule structure, and the final A mark requires the exact constant. Using \(S = x^{2}\) or \(S = 4x^{2}\) rather than \(6x^{2}\) is the wrong turn here: a solid cube has a top and a bottom as well as four sides.

(e) Rate of increase of the surface area when \(V = 216\) cm\(^{3}\) [3]

The condition is given as a volume, but the formula from part (d) needs \(x\), so convert first:

\[V = 216 \ \Rightarrow \ x^{3} = 216 \ \Rightarrow \ x = 6 \ \mathrm{cm} \quad \textbf{[M1][A1]}\] \[\frac{dS}{dt} = 0.24 \times 6 = 1.44 \ \mathrm{cm^{2}\,s^{-1}} \quad \textbf{[A1]}\]

The M mark is for recognising that \(x^{3} = 216\) must be solved, and it survives a slip in the cube root; the final accuracy mark requires the value with its unit.

The wrong turn this part catches. Substituting \(216\) straight into \(\dfrac{dS}{dt} = 0.24x\) to get \(51.84\). The variable in that formula is the edge length, not the volume, and mixing the two is the single most common loss on this question. A magnitude check settles it: the edge of the cube is only \(6\) cm and it grows at \(0.02\) cm s\(^{-1}\), so a surface-area rate of \(51.84\) cm\(^{2}\) s\(^{-1}\) is far too large for so slow a growth.

Further check. At \(x = 6\) the volume rate is \(0.06 \times 36 = 2.16\) cm\(^{3}\) s\(^{-1}\), which exceeds the value \(1.5\) cm\(^{3}\) s\(^{-1}\) found at \(x = 5\); both rates should increase with \(x\), and they do.