Edexcel IAL June 2024 P4 (WMA14/01) Q3 Implicit Differentiation, Equation of Normal
The curve C is defined by the equation
\(8x^3 – 3y^2 + 2xy = 9\)
Find an equation of the normal to C at the point (2, 5), giving your answer in the form ax + by + c = 0, where a, b and c are integers.(7)
Solution to Question: Rate of Change of Area of a Circular Segment
Key Concepts Used:
- Implicit differentiation: Differentiate both sides with respect to x
- Substitution into derivative: Evaluate \(\frac{dy}{dx}\) at specific point
- Normal gradient: \(m_{normal} = -\frac{1}{m_{tangent}}\) & use the point-slope form: \(y – y_1 = m(x – x_1)\)
- Linear equation conversion: Rearrange to ax + by + c = 0 with integer coefficients
Step-by-Step Working:
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Find \(\frac{dy}{dx}\) using implicit differentiation
Differentiate both sides with respect to x:\(\frac{d}{dx}(8x^3) – \frac{d}{dx}(3y^2) + \frac{d}{dx}(2xy) = \frac{d}{dx}(9)\)\(24x^2 – 6y\frac{dy}{dx} + \left(2x\frac{dy}{dx} + 2y\right) = 0\) -
Collect \(\frac{dy}{dx}\) terms
\(\frac{dy}{dx} (2x – 6y) + 24x^2 + 2y = 0\)\(\frac{dy}{dx} (2x – 6y) = -24x^2 – 2y\)\(\frac{dy}{dx} = \frac{-24x^2 – 2y}{2x – 6y}\)\(\frac{dy}{dx} = \frac{24x^2 + 2y}{6y – 2x}\)
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Evaluate \(\frac{dy}{dx}\) at (2,5)
\(\frac{dy}{dx} = \frac{24(2)^2 + 2(5)}{6(5) – 2(2)} = \frac{96 + 10}{30 – 4} = \frac{106}{26} = \frac{53}{13}\)
So the gradient of the tangent at (2,5) is \(\frac{53}{13}\).
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Find gradient of the normal
The normal is perpendicular to the tangent, so:\(m_{normal} = -\frac{1}{m_{tangent}} = -\frac{13}{53}\) -
Equation of the normal line
Using point-slope form at (2,5):\(y – 5 = -\frac{13}{53}(x – 2)\)Multiply through by 53:
\(53(y – 5) = -13(x – 2)\)\(53y – 265 = -13x + 26\)\(13x + 53y – 291 = 0\)
Final Answer
\(13x + 53y – 291 = 0\)