Edexcel IAL January 2024 P4 (WMA14/01) Q3 Implicit Differentiation, Stationary Points
The curve C, shown in Figure 1, has equation
\[ y^2 x + 3 y = 4 x^2 + k \quad y > 0 \]
where k is a constant.
(a) Find \(\frac{dy}{dx}\) in terms of x and y(5)
The point P (p, 2), where p is a constant, lies on C.
Given that P is the minimum turning point on C,
(b) Find
(i) the value of p
(ii) the value of k
(4)
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 \(y^2\) x
- Algebraic manipulation: Isolating \(\frac{dy}{dx}\)
Step-by-Step Working:
-
Differentiate implicitly with respect to x
\[ \frac{d}{dx} (y^2 x) + \frac{d}{dx} (3 y) = \frac{d}{dx} (4 x^2) + \frac{d}{dx} (k) \]
-
Apply differentiation rules
– For \( y^2 x \) (product rule):
\[ \frac{d}{dx} (y^2 x) = y^2 + 2xy \frac{dy}{dx} \]
– For \( 3y \):
\[ \frac{d}{dx} (3y) = 3 \frac{dy}{dx} \]
– For \( 4x^2 \):
\[ \frac{d}{dx} (4x^2) = 8x \]
– For \( k \):
\[ \frac{d}{dx} (k) = 0 \]
-
Combine terms
\[ y^2 + 2xy \frac{dy}{dx} + 3 \frac{dy}{dx} = 8x \]
-
Collect \(\frac{dy}{dx}\) terms
\[ \frac{dy}{dx} (2xy + 3) = 8x – y^2 \]
\[ \frac{dy}{dx} = \frac{8x – y^2}{2xy + 3} \]
Final Answer
\[ \frac{dy}{dx} = \frac{8x – y^2}{2xy + 3} \]
Solution to Part b-i: Find the value of \( p \)
Key Concepts Used:
- Turning point condition: \(\frac{dy}{dx} = 0\)
- Substitution: Using coordinates of point \( P \)
Step-by-Step Working:
-
Use turning point condition At a turning point, \(\frac{dy}{dx} = 0\). From part (a), numerator must be zero:
\[ 8x – y^2 = 0 \]
-
Substitute coordinates of \( P(p, 2) \)
\[ 8p – (2)^2 = 0 \]
\[ 8p – 4 = 0 \]
\[ 8p = 4 \]
\[ p = \frac{1}{2} \]
Final Answer
\[ \frac{1}{2} \]
Solution to Part b-ii: Find the value of k
Key Concepts Used:
- Point on curve: Substituting coordinates into original equation
- Solving for constant: Using given information to find unknown parameter
Step-by-Step Working:
-
Substitute \( P \left( \frac{1}{2}, 2 \right) \) into original equation
\[ y^2 x + 3y = 4x^2 + k \]
\[ (2)^2 \left( \frac{1}{2} \right) + 3(2) = 4 \left( \frac{1}{2} \right)^2 + k \]
-
Simplify both sides Left side:
\[ 4 \times \frac{1}{2} + 6 = 2 + 6 = 8 \]
Right side:
\[ 4 \times \frac{1}{4} + k = 1 + k \]
-
Solve for k
\[ 8 = 1 + k \]
\[ k = 7 \]
Final Answer
\[ 7 \]