Differentiation(P4)

Coach Name: Sir Muhammad Abdullah Shah

Edexcel IAL January 2023 P4 (WMA14/01) Q5, Implicit Differentiation
Figure 2: Curve

Figure 2 shows a sketch of the curve with equation

\[ y^{2} = 2x^{2} + 15x + 10y \]

(a) Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).(2)

The curve is not defined for values of \(x\) in the interval (p, q), as shown in Figure 2.

(b) Using your answer to part (a) or otherwise, find the value of p and the value of q.

(Solutions relying entirely on calculator technology are not acceptable.)

(3)

Solution to Question a: Finding \(\frac{dy}{dx}\)
Key Concepts Used:
  • Implicit Differentiation

Step-by-Step Working:
  1. Differentiate Both Sides with Respect to \(x\)
    Given:

    \[ y^{2} = 2x^{2} + 15x + 10y \]
    Differentiate term-by-term:
    Left Side (\(y^{2}\)):
    \[ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \]
    Right Side (\(2x^2 + 15x + 10y\)):
    \[ \frac{d}{dx}(2x^2) = 4x \]
    \[ \frac{d}{dx}(15x) = 15 \]
    \[ \frac{d}{dx}(10y) = 10 \frac{dy}{dx} \]
    Combined Result:
    \[ 2y \frac{dy}{dx} = 4x + 15 + 10 \frac{dy}{dx} \]
  2. Collect \(\frac{dy}{dx}\) Terms

    \[ 2y \frac{dy}{dx} – 10 \frac{dy}{dx} = 4x + 15 \]
    Factor out \(\frac{dy}{dx}\):
    \[ \frac{dy}{dx}(2y – 10) = 4x + 15 \]
  3. Solve for \(\frac{dy}{dx}\)

    \[ \frac{dy}{dx} = \frac{4x + 15}{2y – 10} \]
Final Answer
\[ \frac{dy}{dx} = \frac{4x + 15}{2y – 10} \]
Solution to Part b: Finding the Interval \((p,q)\) Where the Curve is Not Defined
Key Concepts Used:
  • Undefined Derivative
  • Quadratic Solutions

Step-by-Step Working:
  1. Identify Undefined Conditions
    The derivative \(\frac{dy}{dx}\) is undefined when the denominator \(2y – 10 = 0\):

    \[ 2y – 10 = 0 \Rightarrow y = 5 \]
  2. Substitute \(y = 5\) into the Original Equation

    \[ (5)^2 = 2x^2 + 15x + 10(5) \]
    \[ 25 = 2x^2 + 15x + 50 \]
    \[ 2x^2 + 15x + 25 = 0 \]
  3. Solve the Quadratic Equation

    \[ x = \frac{-15 \pm \sqrt{15^2 – 4 \cdot 2 \cdot 25}}{2 \cdot 2} \]
    \[ x = \frac{-15 \pm \sqrt{225 – 200}}{4} \]
    \[ x = \frac{-15 \pm \sqrt{25}}{4} \]
    \[ x = \frac{-15 \pm 5}{4} \]
    Solutions:
    – \(x = \frac{-15+5}{4} = \frac{-10}{4} = -\frac{5}{2}\)
    – \(x = \frac{-15-5}{4} = \frac{-20}{4} = -5\)
  4. Determine the Interval \((p,q)\)
    The curve is not defined between these two x-values:

    \[ p = -5, \quad q = -\frac{5}{2} \]
Final Answer
\[ p = -5, \quad q = -\frac{5}{2} \]