Edexcel IAL October 2023 P4 (WMA14/01) Q8 Parametric Equations, Differentiation
Figure 3 shows a sketch of the curve C with parametric equations
\[x = 6t – 3 \sin 2t \quad y = 2 \cos t \quad 0 \leq t \leq \frac{\pi}{2}\]
The curve meets the y-axis at 2 and the x-axis at k, where k is a constant.
(a) State the value of k.(1)
(b) Use parametric differentiation to show that
\[ \frac{dy}{dx} = \frac{1}{c} \cdot \frac{\sin t}{\cos 2t} \]
where c is a constant to be found.(4)
The point P with parameter \( t = \frac{\pi}{4} \) lies on C
The tangent to C at the point P cuts the y-axis at the point N.
(c) Find the exact y coordinate of N, giving your answer in simplest form.
(3)
Solution to Part a: State the value of k
Key Concepts Used:
- Parametric axis intercepts: Set y = 0 for x-axis, x = 0 for y-axis
- Trigonometric evaluation: cos(π/2) = 0, sinπ = 0
Step-by-Step Working:
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Find where curve meets x-axis On x-axis, \( y = 0 \):
\[ 2\cos t = 0 \Rightarrow \cos t = 0 \Rightarrow t = \frac{\pi}{2} \]
Substitute into \( x \):
\[ x = 6 \left( \frac{\pi}{2} \right) – 3\sin \left( 2 \cdot \frac{\pi}{2} \right) = 3\pi – 3\sin\pi = 3\pi – 0 = 3\pi \]
So \( k = 3\pi \).
Final Answer
\[ 3\pi \]
Solution to Part b: Show \(\frac{dy}{dx} = \lambda\) cosec \(t\)
Key Concepts Used:
- Parametric differentiation: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)
- Trigonometric identities: \(cos2t = 1 – 2sin^2t\)
- Cosecant definition: \(cosec \, t = \frac{1}{sin \, t}\)
Step-by-Step Working:
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Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
\[ \frac{dx}{dt} = 6 – 3 \cdot 2\cos2t = 6 – 6\cos2t \]
\[ \frac{dy}{dt} = -2\sin t \]
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Compute \(\frac{dy}{dx}\)
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2\sin t}{6 – 6\cos2t} \]
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Simplify using identity \(cos2t = 1 – 2sin^2t\)
\[ 6 – 6\cos2t = 6 – 6(1 – 2\sin^2t) = 12\sin^2t \]
\[ \frac{dy}{dx} = \frac{-2\sin t}{12\sin^2 t} = \frac{-1}{6\sin t} = -\frac{1}{6}\csc t \]
So \(\lambda = -\frac{1}{6}\)
Final Answer
\[ \frac{dy}{dx} = -\frac{1}{6}\csc t \]
Solution to Part c: Find the y-coordinate of N
Key Concepts Used:
- Tangent line equation: Point-slope form
- y-intercept: Set \(x = 0\) in tangent equation
- Exact form simplification: Keeping surds and \(\pi\) in final answer
Step-by-Step Working:
-
Coordinates and gradient at P
At \(t = \frac{\pi}{4}\):
\[ x = 6\left(\frac{\pi}{4}\right) – 3\sin\left(\frac{\pi}{2}\right) = \frac{3\pi}{2} – 3 \]
\[ y = 2\cos\left(\frac{\pi}{4}\right) = 2\cdot\frac{\sqrt{2}}{2} = \sqrt{2} \]
Gradient:
\[ \frac{dy}{dx} = -\frac{1}{6}\csc\left(\frac{\pi}{4}\right) = -\frac{1}{6}\cdot\frac{1}{\sin(\pi/4)} = -\frac{1}{6}\cdot\frac{1}{\sqrt{2}/2} = -\frac{\sqrt{2}}{6} \]
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Equation of tangent at P
\[ y – \sqrt{2} = -\frac{\sqrt{2}}{6}\left(x – \left(\frac{3\pi}{2} – 3\right)\right) \]
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Find y-intercept N
At \(N, x = 0\):
\[ y – \sqrt{2} = -\frac{\sqrt{2}}{6} \left( 0 – \left( \frac{3\pi}{2} – 3 \right) \right) \]
\[ y – \sqrt{2} = \frac{\sqrt{2}}{6} \left( \frac{3\pi}{2} – 3 \right) \]
\[ y = \sqrt{2} + \frac{\sqrt{2}}{6} \cdot \frac{3\pi – 6}{2} \]
\[ y = \sqrt{2} + \frac{\sqrt{2}}{12} (3\pi – 6) \]
\[ y = \sqrt{2} + \frac{\sqrt{2}}{4} \pi – \frac{\sqrt{2}}{2} \]
\[ y = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{4} \pi \]
\[ y = \frac{\sqrt{2}}{2} \left( 1 + \frac{\pi}{2} \right) \]
Final Answer
\[ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{4} \pi \]