Edexcel IAL January 2024 P4 (WMA14/01) Q4 Differentiation, Related Rates of Change
A cone, shown in Figure 2, has
- fixed height 5 cm
- base radius r cm
- slant height l cm
(a) Find an expression for l in terms of r(1)
Given that the base radius is increasing at a constant rate of 3cm per minute,
(b) find the rate at which the total surface area of the cone is changing when the radius of the cone is 1.5 cm. Give your answer in cm per minute to one decimal place.
[The total surface area, S, of a cone is given by the formula \( S = \pi r^2 + \pi rl \)]
(4)
Solution to Part a: Expression for l in terms of r
Key Concepts Used:
- Pythagoras’ Theorem: Relating height, radius, and slant height in a right circular cone
Step-by-Step Working:
-
Use Pythagoras’ Theorem
In a right-angled triangle formed by height, base radius, and slant height:
\[ l^2 = r^2 + h^2 \]
Given \( h = 5 \, \text{cm} \):
\[ l^2 = r^2 + 25 \]
\[ l = \sqrt{r^2 + 25} \]
Final Answer
\[ l = \sqrt{r^2 + 25} \]
Solution to Part b: Rate of change of surface area when \( r = 1.5 \, \text{cm} \)
Key Concepts Used:
- Surface area formula: Total surface area = base area + curved surface area
- Differentiation techniques: Product rule for \(\pi r\sqrt{r^2 + 25}\)
- Related rates: Chain rule \(\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt}\)
- Numerical evaluation: Substitution and approximation
Step-by-Step Working:
-
Write surface area formula
\[ S = \pi r^2 + \pi r l \]
Substitute \( l = \sqrt{r^2 + 25} \):
\[ S = \pi r^2 + \pi r \sqrt{r^2 + 25} \]
-
Differentiate \( S \) with respect to \( r \)
\[ \frac{dS}{dr} = \frac{d}{dr} [\pi r^2] + \frac{d}{dr} [\pi r (r^2 + 25)^{1/2}] \]
First term:
\[ \frac{d}{dr} [\pi r^2] = 2\pi r \]
Second term (using product rule):
Let \( u = \pi r, v = (r^2 + 25)^{1/2} \)
\[ \frac{du}{dr} = \pi, \quad \frac{dv}{dr} = \frac{1}{2} (r^2 + 25)^{-1/2} \cdot 2r = \frac{r}{\sqrt{r^2 + 25}} \]
\[ \frac{d}{dr} [\pi r (r^2 + 25)^{1/2}] = \pi (r^2 + 25)^{1/2} + \pi r \cdot \frac{r}{\sqrt{r^2 + 25}} \]
\[ = \pi \sqrt{r^2 + 25} + \frac{\pi r^2}{\sqrt{r^2 + 25}} \]
-
Combine terms
\[ \frac{dS}{dr} = 2\pi r + \pi \sqrt{r^2 + 25} + \frac{\pi r^2}{\sqrt{r^2 + 25}} \]
\[ = \pi \left[ 2r + \sqrt{r^2 + 25} + \frac{r^2}{\sqrt{r^2 + 25}} \right] \]
-
Use chain rule for \(\frac{dS}{dt}\)
Given \(\frac{dr}{dt} = 3 \, \text{cm/min}\):
\[ \frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt} \]
\[ = 3\pi \left[ 2r + \sqrt{r^2 + 25} + \frac{r^2}{\sqrt{r^2 + 25}} \right] \]
-
Evaluate at \( r = 1.5 \)
\[ \sqrt{r^2 + 25} = \sqrt{(1.5)^2 + 25} = \sqrt{2.25 + 25} = \sqrt{27.25} \]
\[ \frac{dS}{dr} = \pi \left[ 2(1.5) + \sqrt{27.25} + \frac{2.25}{\sqrt{27.25}} \right] \]
\[ = \pi \left[ 3 + \sqrt{27.25} + \frac{2.25}{\sqrt{27.25}} \right] \]
Numerically:
\[ \sqrt{27.25} \approx 5.220 \]
\[ \frac{2.25}{\sqrt{27.25}} \approx 0.431 \]
\[ \frac{dS}{dr} \approx \pi [3 + 5.220 + 0.431] = \pi \times 8.651 \approx 27.175 \]
\[ \frac{dS}{dt} \approx 27.175 \times 3 \approx 81.525 \]
Final Answer
81.5 cm²/min