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Class 12 Continuity And Differentiability · Long Answer3 Marksparametric_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #72

For x=\(a \left(\cos{\left(2 t \right)} + 1\right) \sin{\left(2 t \right)}\), y=\(b \left(1 - \cos{\left(2 t \right)}\right) \cos{\left(2 t \right)}\), find dy/dx at t=\(\frac{\pi}{4}\). Assume a≠0.

✅ Step-by-Step Solution

Final Answer: \(\frac{b}{a}\)
  1. Differentiate in t: dx/dt=\(2 a \left(\cos{\left(2 t \right)} + 1\right) \cos{\left(2 t \right)} - 2 a \sin^{2}{\left(2 t \right)}\) and dy/dt=\(- 2 b \left(1 - \cos{\left(2 t \right)}\right) \sin{\left(2 t \right)} + 2 b \sin{\left(2 t \right)} \cos{\left(2 t \right)}\).
  2. Thus dy/dx=(dy/dt)/(dx/dt)=\(\frac{b \tan{\left(t \right)}}{a}\).
  3. Evaluate at the specified t to obtain \(\frac{b}{a}\).
  4. Therefore dy/dx=\(\frac{b}{a}\).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Differentiate in t: dx/dt=\(2 a \left(\cos{\left(2 t \right)} + 1\right) \cos{\left(2 t \right)} - 2 a \sin^{2}{\left(2 t \right)}\) and dy/dt=\(- 2 b \left(1 - \cos{\left(2 t \right)}\right) \sin{\left(2 t \right)} + 2 b \sin{\left(2 t \right)} \cos{\left(2 t \right)}\).
  • [1 mark] Thus dy/dx=(dy/dt)/(dx/dt)=\(\frac{b \tan{\left(t \right)}}{a}\).
  • [1 mark] Evaluate at the specified t to obtain \(\frac{b}{a}\).
  • [0.5 marks] Therefore dy/dx=\(\frac{b}{a}\).

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