Edexcel IAL October 2022 P4 (WMA14/01) Q6 Parametric Equations, Differentiation, Domain & Range
Figure 3 shows a sketch of the curve \( C \) with parametric equations
\( x = 1 + 3 \tan t \quad y = 2 \cos 2t \quad -\frac{\pi}{6} \leq t \leq \frac{\pi}{3} \)
The curve crosses the x-axis at point \( P \), as shown in Figure 3.
(a) Find the equation of the tangent to \( C \) at \( P \), writing your answer in the form \( y = mx + c \), where \( m \) and \( c \) are constants to be found.(5)
The curve \( C \) has equation \( y = f(x) \), where \( f \) is a function with domain \(\left[ k, 1 + 3\sqrt{3} \right]\)
(b) Find the exact value of the constant \( k \).(1)
(b) Find the exact value of the constant \( k \).(1)
(c) Find the range of \( f \).(2)
Solution to Question 16(a): Equation of the Tangent at \( P \)
Key Concepts Used:
- Equation of Tangent
- Point – Slope Formula
Step-by-Step Working:
Find the tangent line where C intersects the x-axis (y = 0).
-
Find t at Point P (where y = 0)
Set y = \(2cos2t\) = 0:
\( cos2t = 0 \Rightarrow 2t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{4} \)
-
Find Coordinates of P
Substitute \(t = \frac{\pi}{4}\) into x:
\( x = 1 + 3tan\left(\frac{\pi}{4}\right) = 1 + 3(1) = 4 \)
Thus, \(P = (4,0)\).
-
Compute \(\frac{dy}{dx}\) at \(t = \frac{\pi}{4}\)
First, find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):
\( \frac{dx}{dt} = 3sec^2t \)
\( \frac{dy}{dt} = -4sin2t \)
Now,
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-4sin2t}{3sec^2t} \)
At \(t = \frac{\pi}{4}\):
\( \frac{dy}{dx} = \frac{-4(1)}{3(\sqrt{2})^2} = \frac{-4}{6} = -\frac{2}{3} \)
-
Write Tangent Equation
Using point-slope form:
\( y – 0 = -\frac{2}{3}(x – 4) \)
\( y = -\frac{2}{3}x + \frac{8}{3} \)
Final Answer:
\( y = -\frac{2}{3}x + \frac{8}{3} \)
Solution to Question 16(b): Exact Value of \(k\) (Domain Lower Bound)
Key Concepts Used:
- Parametric Equation
Step-by-Step Working:
Find the minimum x-value in the domain \([k, 1 + 3\sqrt{3}]\).
-
Analyze \(x = 1 + 3\tan t\) for \(t \in \left[ -\frac{\pi}{6}, \frac{\pi}{3} \right]\)
And since we need the lower bound of the domain, the minimum x occurs at \(t = \frac{\pi}{6}\) and the upper bound of the domain at \(t = \frac{\pi}{6}\):
\( x = 1 + 3\tan \left( \frac{\pi}{6} \right) = 1 + 3 \left( \frac{1}{\sqrt{3}} \right) = 1 + \sqrt{3} \)
- Thus, \(k = 1 + \sqrt{3}\).
Verification:
We can also verify that the value we found is the lower bound of the domain by substituting the other value of \(t\) in the parametric equation of variable \(x\).
At \(t = \frac{\pi}{3}\):
\( x = 1 + 3\tan \left( \frac{\pi}{3} \right) = 1 + 3\sqrt{3} \)
\( x = 1 + 3\tan \left( \frac{\pi}{3} \right) = 1 + 3\sqrt{3} \)
Final Answer:
\( 1 + \sqrt{3} \)
Solution to Question 16(c): Range of \(f\)
Key Concepts Used:
- Range of Cartesian Equation
Step-by-Step Working:
To determine all possible y-values for \(t \in \left[ -\frac{\pi}{6}, \frac{\pi}{3} \right]\), we need to compute the upper and lower value of \(y\) of curve.
-
Let’s analyze and compute range
\( y = 2\cos 2t \)
\(\cos 2t\) is increasing on \(\left[ -\frac{\pi}{6}, 0 \right]\) and decreasing on \(\left[ 0, \frac{\pi}{3} \right]\). -
At \( t = \frac{\pi}{6} \):
\( y = 2\cos 2t = 2\cos 2 \times \frac{\pi}{6} = 1 \)
-
At \( t = \frac{\pi}{3} \):
\( y = 2\cos 2t = 2\cos 2 \times \frac{\pi}{3} = -1 \)
-
So,
Maximum \( y: 2 \) (at \( x=0 \))
Minimum \( y: -1 \) (at \( t=\frac{\pi}{3} \))
Hence,
\( -1 \leq y \leq 2 \)
Final Answer:
\( -1 \leq y \leq 2 \)