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Figure 1 shows a metal cube which is expanding uniformly as it is heated - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 7
Question 4
Figure 1 shows a metal cube which is expanding uniformly as it is heated. At time t seconds, the length of each edge of the cube is x cm, and the volume of the cube ... show full transcript
Worked Solution & Example Answer:Figure 1 shows a metal cube which is expanding uniformly as it is heated - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 7
Step 1
Step 2
find \( \frac{dx}{dt} \), when x = 8
Answer
Given that ( \frac{dV}{dt} = 0.048 ) cm³s⁻¹, we can use the chain rule to relate these rates:
[ \frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} ]
At x = 8:
[ \frac{dV}{dx} = 3(8^2) = 192 \text{ cm}^2 ]
Plugging in the values:
[ 0.048 = 192 \cdot \frac{dx}{dt} ]
Solving for ( \frac{dx}{dt} ):
[ \frac{dx}{dt} = \frac{0.048}{192} \approx 0.00025 \text{ cm/s} ]
Step 3
find the rate of increase of the total surface area of the cube, in cm²s⁻¹, when x = 8
Answer
The surface area S of the cube is given by:
[ S = 6x^2 ]
Differentiating with respect to t:
[ \frac{dS}{dt} = \frac{dS}{dx} \cdot \frac{dx}{dt} ]
Where:
[ \frac{dS}{dx} = 12x ]
At x = 8:
[ \frac{dS}{dx} = 12(8) = 96 ]
Now substituting ( \frac{dx}{dt} ) from part (b):
[ \frac{dS}{dt} = 96 \cdot 0.00025 = 0.024 \text{ cm}^2s^{-1} ]