IBDP Maths AHL 5.14 Implicit differentiation AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Area formula: \( A = \frac{\sqrt{3}}{4} x^2 \). A1
Differentiate with respect to time: \( \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt} \). M1
Given: \( \frac{dx}{dt} = 4 \, \text{cm s}^{-1} \), \( x = 5\sqrt{3} \, \text{cm} \). A1
Substitute: \( \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2 \cdot 5\sqrt{3} \cdot 4 = \frac{\sqrt{3} \cdot 40 \sqrt{3}}{4} = \frac{120}{4} = 30 \, \text{cm}^2 \text{s}^{-1} \). A1
[4 marks]
▶️ Answer/Explanation
Implicit differentiation of \( x^2 + xy + y^2 = 3 \): \( 2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \). M1 A1
Rearrange: \( \frac{dy}{dx} (x + 2y) = -(2x + y) \), so \( \frac{dy}{dx} = -\frac{2x + y}{x + 2y} \). M1
Substitute \( x = -1 \), \( y = k \): \( \frac{dy}{dx} = -\frac{2(-1) + k}{-1 + 2k} = \frac{2 – k}{2k – 1} \). M1 A1
[5 marks]
Tangent parallel to \( x \)-axis: \( \frac{dy}{dx} = 0 \), so \( \frac{2 – k}{2k – 1} = 0 \), \( 2 – k = 0 \), \( k = 2 \). A1
[1 mark]
▶️ Answer/Explanation
Differentiate implicitly: \( 4x^2 + y^2 = 4 \), so \( 8x + 2y \frac{dy}{dx} = 0 \). M1 A1
Solve: \( \frac{dy}{dx} = -\frac{8x}{2y} = -\frac{4x}{y} \). A1
Note: Award M1A0 for \( 8x + 2y \frac{dy}{dx} = 4 \).
[3 marks]
At \( \left( \frac{2}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \), \( \frac{dy}{dx} = -\frac{4 \cdot \frac{2}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = -4 \). A1
[1 mark]
Volume: \( V = \pi \int y^2 \, dx \). M1
From \( 4x^2 + y^2 = 4 \), \( y^2 = 4 – 4x^2 \). Limits: \( x = 0 \) to \( x = 1 \). A1
So: \( V = \pi \int_0^1 (4 – 4x^2) \, dx = \pi \left[ 4x – \frac{4}{3} x^3 \right]_0^1 = \pi \left( 4 – \frac{4}{3} \right) = \pi \cdot \frac{8}{3} = \frac{8\pi}{3} \). A1 A1
Note: If correct except for omitting \( \pi \), award 2 marks.
[4 marks]