Edexcel IAL October 2021 P4 (WMA14/01) Q5 Parametric Equations, Differentiation, Cartesian Conversion
Figure 1 shows a sketch of the curve C with parametric equations
\( x = 5 + 2 \tan t \quad y = 8 \sec^2 t \quad -\frac{\pi}{3} \leq t \leq \frac{\pi}{4} \)
(a) Use parametric differentiation to find the gradient of C at x=3(4)
The curve C has equation \(y = f(x)\), where f is a quadratic function.
(b) Find \(f(x)\) in the form \(a(x+b)^2 + c\), where a, b and c are constants to be found.(3)
(b) Find \(f(x)\) in the form \(a(x+b)^2 + c\), where a, b and c are constants to be found.(3)
(c) Find the range of f.(2)
Solution to Part a: Gradient at \(x = 3\) for Parametric Curve \(x = 5 + 2\tan t\), \(y = 8\sec^2 t\)
Key Concepts Used:
- Parametric differentiation
- Trigonometric identities
Step-by-Step Working:
Find \(\frac{dy}{dx}\) when \(x = 3\), using parametric differentiation.
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Recall Parametric Differentiation Formula or Chain Rule
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \)
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Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
\( x = 5 + 2\tan t \)
\( \frac{dx}{dt} = 2\sec^2 t \)
Now, differentiate y w.r.t t.
\( y = 8\sec^2 t \)
\( \frac{dy}{dt} = 16\sec t\cdot\sec t\tan t \)
\( \frac{dy}{dt} = 16\sec^2 t\tan t \)
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Find \(\frac{dy}{dx}\)
\( \frac{dy}{dx} = \frac{16\sec^2 t \times \tan t}{2\sec^2 t} = 8\tan t \)
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Determine t when \(x = 3\)
Set \(x = 3\) in equation of variable of x.
\( x = 5 + 2\tan t \)
\( 3 = 5 + 2\tan t \)
\( 2\tan t = -2 \)
\( \tan t = -1 \)
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Compute gradient by substituting \(\tan t = -1\) into \(\frac{dy}{dx}\).
\( \frac{dy}{dx} = 8(-1) = -8 \)
Final Answer:
-8
Solution to Part b: Express \( f(x) \) in the form of \( a(x + b)^2 + c \)
Key Concepts Used:
- Trigonometric Identities
Step-by-Step Working:
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Convert parametric equations to Cartesian form \( y = a(x + b)^2 + c \). To do so use Trigonometric Identity.
Recall:
\( 1 + \tan^2 t = \sec^2 t \)
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Express \( \sec^2 t \) in Terms of \( x \)
From \( x = 5 + 2\tan t \)
\( \tan t = \frac{x – 5}{2} \)
Thus,
\( \sec^2 t = 1 + \left( \frac{x – 5}{2} \right)^2 \)
\( \sec^2 t = 1 + \left( \frac{(x – 5)^2}{4} \right) \)
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Substitute into \( y \)
\( y = 8\sec^2 t \)
\( y = 8\left( 1 + \frac{(x – 5)^2}{4} \right) \)
\( y = 8 + 2(x – 5)^2 \)
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Rewrite in Required Form
\( y = 2(x – 5)^2 + 8 \)
Here, \( a = 2, \, b = -5, \, c = 8 \).
Final Answer:
\( y = 2(x – 5)^2 + 8 \)
Solution to Part c: Range of \( f(x) \)
Key Concepts Used:
- Range of cartesian equation y
Step-by-Step Working:
We need to understand the given Quadratic Function. The function found in part b is a parabola opening upwards (since \( a = 2 > 0 \), where \( a \) the coefficient of \( (x – 5)^2 \)). The vertex is at (5,8), which is the minimum point.
However, considering parametric constraints given that \( t \in \left[ -\frac{\pi}{3}, \frac{\pi}{4} \right] \), to find the range, we need to find the respective values of \( y \) at \( -\frac{\pi}{3} \) and \( \frac{\pi}{4} \).
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So, at \( t = -\frac{\pi}{3} \):
\( y = 8 \sec^2 t \)
\( y = 8 \sec^2 \left( -\frac{\pi}{3} \right) = 8 \left( \frac{1}{\cos^2 \left( -\frac{\pi}{3} \right)} \right) = 8 \times 4 = 32 \, (\text{maximum value of } y) \)
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Similarly, let’s do it for \( t = \frac{\pi}{4} \):
\( y = 8 \sec^2 t \)
\( \tan t = 1 \implies x = 5 + 2(1) = 7 \)
\( y = 8 \sec^2 \left( \frac{\pi}{4} \right) = 8 \left( \frac{1}{\cos^2 \frac{\pi}{4}} \right) = 8 \times 2 = 16 \)
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Now, to determine range, we already know that
– The minimum value of \( y \) occurs at the vertex: \( y = 8 \) (from the quadratic equation found in part b).
– The maximum value occurs at \( t = -\frac{\pi}{3} \): \( y = 32 \).
– So, the range of \( f \) is
\( 8 \leq y \leq 32 \)
Final Answer:
\( 8 \leq y \leq 32 \)