Edexcel IAL October 2023 P4 (WMA14/01) Q2, Differentiation, Related Rates of Change
Figure 1 shows a cube which is increasing in size.
At time t seconds,
- the length of each edge of the cube is x cm
- the surface area of the cube is S cm²
- the volume of the cube is V cm³
Given that the surface area of the cube is increasing at a constant rate of 4 cm² s⁻¹
(a) show that \(\frac{dx}{dt} = \frac{k}{x}\) where k is a constant to be found,
(4)
(b) show that \(\frac{dV}{dt} = V^p\) where p is a constant to be found.
(3)
Solution to Part a: Show \(\frac{dx}{dt} = \frac{k}{x}\)
Key Concepts Used:
- Geometric formulas: Surface area of cube = 6x²
- Chain rule: \(\frac{dS}{dt} = \frac{dS}{dx}\cdot\frac{dx}{dt}\)
- Related rates: Connecting rates of change through differentiation
Step-by-Step Working:
-
Write surface area formula for a cube
For a cube with side length x, the total surface area is:
\[ S = 6x^2 \]
-
Differentiate both sides with respect to time t
Using the chain rule:
\[ \frac{dS}{dt} = \frac{dS}{dx}\cdot\frac{dx}{dt} \]
First find \(\frac{dS}{dx}\):
\[ \frac{dS}{dx} = 12x \]
So:
\[ \frac{dS}{dt} = 12x\cdot\frac{dx}{dt} \]
-
Substitute given rate We are told \(\frac{dS}{dt}=4\) cm²/s:
\[ 4 = 12x\cdot\frac{dx}{dt} \]
-
Solve for \(\frac{dx}{dt}\)
\[ \frac{dx}{dt} = \frac{4}{12x} = \frac{1}{3x} \]
-
Identify constant \(k\) Comparing with \(\frac{dx}{dt}=\frac{k}{x}\) we have:
\[ k = \frac{1}{3} \]
Final Answer
\[ k = \frac{1}{3} \]
Solution to Part b: Show \(\frac{dV}{dt} = V^p\)
Key Concepts Used:
- Geometric formulas: Volume of cube = \(x^3\)
- Chain rule: \(\frac{dV}{dt} = \frac{dV}{dx}\cdot\frac{dx}{dt}\)
- Power relationships: Converting between \(x\) and \(V\) using \(x = V^{1/3}\)
Step-by-Step Working:
-
Write volume formula for a cube For a cube with side length \(x\), the volume is:
\[ V = x^3 \]
-
Differentiate both sides with respect to time \( t \)
Using the chain rule:
\[ \frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} \]
First find \(\frac{dV}{dx}\):
\[ \frac{dV}{dx} = 3x^2 \]
So:
\[ \frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt} \]
-
Substitute \(\frac{dx}{dt}\) from part (a)
From part (a): \(\frac{dx}{dt} = \frac{1}{3x}\)
\[ \frac{dV}{dt} = 3x^2 \cdot \frac{1}{3x} = x \]
-
Express in terms of \( V \)
Since \( V = x^3 \), we can write \( x = V^{1/3} \):
\[ \frac{dV}{dt} = V^{1/3} \]
-
Identify constant \( p \)
Comparing with \(\frac{dV}{dt} = V^p\), we have:
\[ p = \frac{1}{3} \]
Final Answer
\[ p = \frac{1}{3} \]