Edexcel IAL June 2022 P4 (WMA14/01) Q3, Differentiation, Related Rates of Change
A tablet is dissolving in water.
The tablet is modelled as a cylinder, shown in Figure 1.
At seconds after the tablet is dropped into the water, the radius of the tablet is xmm and the length of the tablet is 3x mm.
The cross-sectional area of the tablet is decreasing at a constant rate of \(0.5 \text{mm}^2 \text{s}^{-1}\)
(a) Find \(\frac{dx}{dt}\) when \(x = 7\)(4)
(b) Find, according to the model, the rate of decrease of the volume of the tablet when \(x = 4\)(4)
Solution to Part a: Rate of Change of Radius \(\left(\frac{dx}{dt}\right)\) when \(x = 7\)
Key Concepts Used:
- Surface Area
Step-by-Step Working:
Here we to determine how fast the radius is decreasing when \(x = 7 \text{mm}\).
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Express Cross-Sectional Area
The tablet is cylindrical with radius \(x \text{mm}\) and height \(3x \text{mm}\).
Where cross-sectional area (circular):
\( A = \pi r^2 \)
\( A = \pi x^2 \)
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Differentiate Area with Respect to Time Given:
\( \frac{dA}{dt} = -0.5 \text{mm}^2/\text{s} \)
Differentiate \( A = \pi x^2 \):
\( \frac{dA}{dt} = 2\pi x \frac{dx}{dt} \)
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Solve for \(\frac{dx}{dt}\)
\( -0.5 = 2\pi x \frac{dx}{dt} \)
\( \frac{dx}{dt} = \frac{-0.5}{2\pi x} = \frac{-1}{4\pi x} \)
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Evaluate at \( x = 7 \text{mm} \)
\( \frac{dx}{dt} = \frac{-1}{4\pi \times 7} \approx \frac{-1}{87.96} \approx -0.0114 \text{mm/s} \)
Hence, the radius decreases at approximately 0.0114 mm/s when \( x = 7 \text{mm} \).
Final Answer:
\( -0.0114 \text{mm/s} \)
(Negative sign indicates decrease)
(Negative sign indicates decrease)
Solution to Part b: Rate of decrease of the volume of the tablet when \( x = 4 \)
Key Concepts Used:
- Volume Differentiation
Step-by-Step Working:
To find how fast the volume is decreasing when \( x = 4 \text{mm} \), calculate in the following way.
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Express Volume
Volume of cylinder:
\( V = \pi x^2 h = \pi x^2 (3x) = 3\pi x^3 \)
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Differentiate Volume with Respect to Time
\( \frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} \)
First, find \(\frac{dV}{dx}\):
\( \frac{dV}{dx} = 9\pi x^2 \)
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From part (a), we have:
\( \frac{dx}{dt} = \frac{-1}{4\pi x} \)
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Combine Derivatives
\( \frac{dV}{dt} = 9\pi x^2 \times \left( \frac{-1}{4\pi x} \right) = -\frac{9x}{4} \)
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Evaluate at \(x = 4\) mm
\( \frac{dV}{dt} = -\frac{9 \times 4}{4} = -9 \text{mm}^3/\text{s} \)
Thus, the volume decreases at \(9 \text{mm}^3/\text{s}\) when \(x = 4\) mm.
Final Answer:
\( -9 \text{mm}^3/\text{s} \)
(Negative sign indicates decrease)
(Negative sign indicates decrease)