Edexcel IAL June 2023 P4 (WMA14/01) Q2 Implicit Differentiation
The point \(P\) lies on \(C\) and has \(x\) coordinate 2
(a) Find the \(y\) coordinate of \(P\).
(3)
(b) Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).
(4)
(c) Find the \(x\) coordinate of \(Q\) giving your answer in the form \(\frac{a\ln 2+b}{c\ln 2+d}\) where \(a\), \(b\), \(c\) and \(d\) are integers to be found.
(3)
Solution to Part a: Find the \(y\)-coordinate of \(P\)
Key Concepts Used:
- Substitution: Plugging given x-value into equation
- Quadratic solving: Using quadratic formula
- Domain consideration: Selecting valid solution based on \(y\geq 0\)
Step-by-Step Working:
-
Substitute \(x=2\) into the equation
\[
\begin{cases}
2^2 – 4(2)y + y^2 = 13 \\
4 – 8y + y^2 = 13
\end{cases}
\]
-
Rearrange to form quadratic equation
\[ y^2 – 8y + 4 – 13 = 0 \]
\[ y^2 – 8y – 9 = 0 \]
-
Solve the quadratic equation
Using the quadratic formula:
\[ y = \frac{8 \pm \sqrt{(-8)^2 – 4(1)(-9)}}{2(1)} \]
\[ y = \frac{8 \pm \sqrt{64 + 36}}{2} \]
\[ y = \frac{8 \pm \sqrt{100}}{2} \]
\[ y = \frac{8 \pm 10}{2} \]
-
Identify valid solution
\[ y = \frac{18}{2} = 9 \quad or \quad y = -\frac{2}{2} = -1 \]
Since \( y \geq 0 \), we take \( y = 9 \).
Final Answer
\[ 9 \]
Solution to Part b: Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\)
Key Concepts Used:
- Implicit differentiation: Differentiating both sides with respect to \(x\)
- Exponential derivative: \(\frac{d}{dx}(a^x) = a^x \ln a\)
- Product rule: For term \(4xy\)
- Algebraic manipulation: Isolating \(\frac{dy}{dx}\)
Step-by-Step Working:
-
Differentiate implicitly with respect to \(x\)
\[ \frac{d}{dx}(2^x) – \frac{d}{dx}(4xy) + \frac{d}{dx}(y^2) = 0 \]
-
Apply differentiation rules
– \(\frac{d}{dx}(2^x) = 2^x \ln 2\)
– \(\frac{d}{dx}(4xy) = 4y + 4x \frac{dy}{dx}\) (using product rule)
– \(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\)
-
Combine terms
\[ 2^x \ln 2 – (4y + 4x \frac{dy}{dx}) + 2y \frac{dy}{dx} = 0 \]
\[ 2^x \ln 2 – 4y – 4x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \]
-
Collect \(\frac{dy}{dx}\) terms
\[ -4x \frac{dy}{dx} + 2y \frac{dy}{dx} = 4y – 2^x \ln 2 \]
\[ \frac{dy}{dx} (2y – 4x) = 4y – 2^x \ln 2 \]
\[ \frac{dy}{dx} = \frac{4y – 2^x \ln 2}{2y – 4x} \]
Final Answer
\[ \frac{dy}{dx} = \frac{4y – 2^x \ln 2}{2y – 4x} \]
Solution to Part c: Find the x-coordinate of Q
Key Concepts Used:
- Tangent line equation: Point-slope form
- x-intercept: Setting \( y = 0 \) in line equation
- Algebraic simplification: Rearranging to required form
Step-by-Step Working:
-
Find \(\frac{dy}{dx}\) at P (2,9)
\[ \frac{dy}{dx} = \frac{4(9) – 2^2 \ln 2}{2(9) – 4(2)} = \frac{36 – 4 \ln 2}{18 – 8} = \frac{36 – 4 \ln 2}{10} = \frac{18 – 2 \ln 2}{5} \]
-
Equation of tangent at P Using point-slope form:
\[ y – 9 = \frac{18 – 2 \ln 2}{5} (x – 2) \]
-
Find \(x\)-intercept \(Q\)
At \(Q\), \(y = 0\):
\[ -9 = \frac{18 – 2 \ln 2}{5} (x – 2) \]
\[ x – 2 = \frac{-45}{18 – 2 \ln 2} \]
\[ x = 2 – \frac{45}{18 – 2 \ln 2} \]
-
Simplify to required form
\[ x = \frac{2(18 – 2 \ln 2) – 45}{18 – 2 \ln 2} \]
\[ x = \frac{36 – 4 \ln 2 – 45}{18 – 2 \ln 2} \]
\[ x = \frac{-9 – 4 \ln 2}{18 – 2 \ln 2} \]
Multiply numerator and denominator by \(-1\):
\[ x = \frac{9 + 4 \ln 2}{-18 + 2 \ln 2} \]
\[ x = \frac{4 \ln 2 + 9}{2 \ln 2 – 18} \]
Final Answer
\[ \frac{4 \ln 2 + 9}{2 \ln 2 – 18} \]