Edexcel IAL January 2022 P4 (WMA14/01) Q4 Differentiation, Rates of Change
A regular icosahedron of side length xcm, shown in Figure 1, is expanding uniformly.
The icosahedron consists of 20 congruent equilateral triangular faces of side length xcm.
(a) Show that the surface area, A cm\(^2\), of the icosahedron is given by
\( A = 5\sqrt{3}x \)
Given that the volume, Vcm\(^3\), of the icosahedron is given by
\( V = \frac{5}{12} \left( 3 + \sqrt{5} \right)x^3 \)
(b) show that
\( \frac{dV}{dA} = \frac{(3+\sqrt{5})x}{8\sqrt{3}} \)
(3)
The surface area of the icosahedron is increasing at a constant rate of 0.025 cm\(^2\)s\(^{-1}\).
(c) Find the rate of change of the volume of the icosahedron when x=2, giving your answer to 2 significant figures.(3)
(c) Find the rate of change of the volume of the icosahedron when x=2, giving your answer to 2 significant figures.(3)
Solution to Part a: Area of Icosahedron
Key Concepts Used:
- Surface Area
Step-by-Step Working:
Derive the formula for the surface area \( A \) of a regular icosahedron with side length \( x \).
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Understand the Structure
A regular icosahedron has 20 identical equilateral triangular faces. Area of one equilateral triangle with side \( x \):
\( A = \frac{1}{2} ab \sin \theta \)
\( A = \frac{1}{2} \times x^2 \times \sin 60 \)
\( Area = \frac{\sqrt{3}}{4} x^2 \)
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Let’s calculate Total Surface Area now. Multiply the area of one face by 20:
\( A = 20 \times \frac{\sqrt{3}}{4} x^2 = 5 \sqrt{3} x^2 \)
Final Answer:
\( A = 5 \sqrt{3} x^2 \)
Solution to Part b: Rate of Volume Change for Icosahedron
Key Concepts Used:
- Chain rule
- Surface area and volume derivatives
Step-by-Step Working:
In this part, we have to show the relationship between \(\frac{dV}{dA}\) and \(x\).
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Differentiate Volume V with Respect to \(x\) Given:
\( V = \frac{5}{12} (3 + \sqrt{5}) x^3 \)
\( \frac{dV}{dx} = \frac{5}{12} (3 + \sqrt{5}) \times 3 x^2 = \frac{5}{4} (3 + \sqrt{5}) x^2 \)
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Differentiate Surface Area A with Respect to \(x\) From part (a):
\( A = 5 \sqrt{3} x^2 \)
\( \frac{dA}{dx} = 10 \sqrt{3} x \)
-
Apply the Chain Rule
\( \frac{dV}{dA} = \frac{dV / dx}{dA / dx} = \frac{5}{4} (3 + \sqrt{5}) x^2 \)
Simplify:
\( \frac{dV}{dA} = \frac{(3 + \sqrt{5}) x}{8 \sqrt{3}} \)
Final Answer:
\( \frac{dV}{dA} = \frac{(3 + \sqrt{5}) x}{8 \sqrt{3}} \)
Solution to Part c: Rate of Change of Volume
Key Concepts Used:
- Chain Rule
- Related Rates
Step-by-Step Working:
We have to find \(\frac{dV}{dt}\) when \(x = 2\) cm and \(\frac{dA}{dt} = 0.025 \text{cm}^2/\text{s}\).
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Use the Chain Rule
From part (b):
\( \frac{dV}{dA} = \frac{dV}{dt} \times \frac{dt}{dA} \)
\( \frac{dV}{dt} = \frac{dV}{dA} \times \frac{dA}{dt} \)
Substitute \(\frac{dV}{dA}\) and \(\frac{dA}{dt} = 0.025\):
\( \frac{dV}{dt} = \frac{(3 + \sqrt{5})x}{8\sqrt{3}} \times 0.025 \)
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Evaluate at \(x = 2\)
\( \frac{dV}{dt} = \frac{(3 + \sqrt{5}) \times 2}{8\sqrt{3}} \times 0.025 \)
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Numerical calculation:
\( 3 + \sqrt{5} \approx 5.236 \)
\( \frac{dV}{dt} = \frac{5.236 \times 2}{8\sqrt{3}} \times 0.025 \)
\( \frac{dV}{dt} \approx \frac{10.472}{13.856} \times 0.025 \)
\( \frac{dV}{dt} \approx 0.0189 \text{cm}^3/\text{s} \)
Final Answer:
\( 0.019 \text{cm}^3/\text{s} \)