Edexcel IAL October 2023 P4 (WMA14/01) Q5 Implicit Differentiation
Figure 2 shows a sketch of the curve \(C\) with equation
\[ y^{3} – x^{2} + 4x^{2}y = k \]
where \(k\) is a positive constant greater than 1
(a) Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).(5)
The point P lies on \(C\). Given that the normal to \(C\) at P has equation \(y=x\), as shown in Figure 2,
(b) find the value of \(k\).
(5)
Solution to Part a: Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\)
Key Concepts Used:
- Implicit differentiation: Differentiating both sides with respect to \(x\)
- Product rule: For term \(4x^2y\)
- Algebraic manipulation: Isolating \(\frac{dy}{dx}\)
Step-by-Step Working:
-
Differentiate implicitly with respect to \(x\)
\[ \frac{d}{dx}(y^3) – \frac{d}{dx}(x^2) + \frac{d}{dx}(4x^2y) = 0 \]
-
Apply differentiation rules
– \(\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}\)
– \(\frac{d}{dx}(x^2) = 2x\)
– \(\frac{d}{dx}(4x^2y) = 8xy + 4x^2\frac{dy}{dx}\) (using product rule)
-
Combine terms
\[ 3y^2\frac{dy}{dx} – 2x + 8xy + 4x^2\frac{dy}{dx} = 0 \]
-
Collect \(\frac{dy}{dx}\) terms
\[ \frac{dy}{dx}(3y^2 + 4x^2) + 8xy – 2x = 0 \]
\[ \frac{dy}{dx}(3y^2 + 4x^2) = 2x – 8xy \]
\[ \frac{dy}{dx} = \frac{2x – 8xy}{3y^2 + 4x^2} \]
Final Answer
\[ \frac{dy}{dx} = \frac{2x – 8xy}{3y^2 + 4x^2} \]
Solution to Part b: Find the value of \( k \)
Key Concepts Used:
- Normal line properties: Gradient relationship between tangent and normal
- Substitution method: Using \( y = x \) to simplify equations
- Domain consideration: Eliminating invalid solutions based on \( k > 1 \)
Step-by-Step Working:
-
Use information about the normal
The normal at \( P \) has equation \( y = x \), so:
– Gradient of normal = 1
– Therefore, gradient of tangent = -1 (negative reciprocal)
-
Set \(\frac{dy}{dx} = -1\)
\[ \frac{2x – 8xy}{3y^2 + 4x^2} = -1 \]
\[ 2x – 8xy = -3y^2 – 4x^2 \]
\[ 4x^2 + 3y^2 – 8xy + 2x = 0 \]
-
Since \( P \) lies on both \( C \) and \( y = x \)
Substitute \( y = x \) into both equations:
From Step 2:
\[ 4x^2 + 3x^2 – 8x^2 + 2x = 0 \]
\[ -x^2 + 2x = 0 \]
\[ x(-x + 2) = 0 \]
Solutions: \( x = 0 \) or \( x = 2 \)From original curve equation:
\[ y^3 – x^2 + 4x^2y = k \]
Substitute \( y = x \):
\[ x^3 – x^2 + 4x^3 = k \]
\[ 5x^3 – x^2 = k \]
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Determine valid solution
If \( x = 0 \), then \( k = 0 \) (invalid since \( k > 1 \))
If \( x = 2 \), then \( k = 5(8) – 4 = 40 – 4 = 36 \)
Final Answer
\[ 36 \]