Edexcel IAL June 2024 P4 (WMA14/01) Q4 Related Rates of Change, Radian Measure
Figure 1 shows a sketch of a segment PQRP of a circle with centre O and radius 5 cm.
Given that
- angle POR is θ radians
- θ is increasing, from 0 to π, at a constant rate of 0.1 radians per second
- the area of the segment PQRP is A cm²
(a) show that
\(\frac{dA}{d\theta} = k (1 – \cos \theta)\)
where k is a constant to be found.(4)
(b) Find, in cm²s⁻¹, the rate of increase of the area of the segment when \(\theta = \frac{\pi}{3}\)(4)
Solution to Part a: Rate of Change of Area of a Circular Segment
Key Concepts Used:
- Geometry of a Circle: Understanding the area of a sector and a triangle within a circle.
- Differentiation: Applying differentiation to find the rate of change of area with respect to angle.
- Chain Rule: Using the chain rule to connect the rate of change of area with respect to time.
Step-by-Step Working:
-
Define the Areas:
Sector Area (POR):
The area of a sector with angle \(\theta\) radians and radius \(r = 5 \text{cm}\) is:
\(Area_{sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 25 \times \theta = \frac{25}{2} \theta\)Triangle Area (POR):
The area of triangle POR (using \(\frac{1}{2} \text{absinC}\)):
\(Area_{triangle} = \frac{1}{2} \times 5 \times 5 \times \sin\theta = \frac{25}{2} \sin\theta\) -
Segment Area (PORP):
The segment area is the sector area minus the triangle area:\(A = Area_{sector} – Area_{triangle} = \frac{25}{2} \theta – \frac{25}{2} \sin\theta\)Simplify:
\(A = \frac{25}{2} (\theta – \sin\theta)\) -
Differentiate \( A \) with respect to \(\theta\):
\(\frac{dA}{d\theta} = \frac{25}{2} (1 – \cos\theta)\)
Here, \( k = \frac{25}{2} \), so:
\(\frac{dA}{d\theta} = k(1 – \cos\theta)\)
Final Answer
\(\frac{dA}{d\theta} = k(1 – \cos\theta)\)
Solution to Part b: Rate of Change of Area When \(\theta = \frac{\pi}{3}\)
Step-by-Step Working:
-
Given Rates:
\(\frac{d\theta}{dt} = 0.1\) radians per second (rate of increase of angle).
From part (a), \(\frac{dA}{d\theta} = \frac{25}{2} (1 – \cos\theta)\). -
Apply the Chain Rule:
\(\frac{dA}{dt} = \frac{dA}{d\theta} \times \frac{d\theta}{dt} = \frac{25}{2} (1 – \cos\theta) \times 0.1\)
Simplify:
\(\frac{dA}{dt} = \frac{2.5}{2} (1 – \cos\theta) = 1.25(1 – \cos\theta)\) -
Substitute \(\theta = \frac{\pi}{3}\):
\(\cos \left( \frac{\pi}{3} \right) = \frac{1}{2}\)
So:
\(\frac{dA}{dt} = 1.25 \left( 1 – \frac{1}{2} \right) = 1.25 \times \frac{1}{2} = 0.625 \text{cm}^2 \text{per second}\)Alternatively, in fractional form:
\(\frac{dA}{dt} = \frac{5}{8} \text{cm}^2 \text{per second}\)
Final Answer
\(\frac{dA}{dt} = \frac{5}{8} \text{cm}^2 \text{per second}\)