Class 12 Continuity And Differentiability · Long Answer3 Marksparametric_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #154

For x=\(a \left(2 t \sin{\left(2 t \right)} + \cos{\left(2 t \right)}\right)\), y=\(a \left(- 2 t \cos{\left(2 t \right)} + \sin{\left(2 t \right)}\right)\), find d²y/dx². Assume a≠0,t≠0 and cos2t≠0.

✅ Step-by-Step Solution

Final Answer: \(\frac{1}{2 a t \cos^{3}{\left(2 t \right)}}\)

Differentiate in t: dx/dt=\(4 a t \cos{\left(2 t \right)}\) and dy/dt=\(4 a t \sin{\left(2 t \right)}\).
Thus dy/dx=(dy/dt)/(dx/dt)=\(\tan{\left(2 t \right)}\).
Use d²y/dx²=[d(dy/dx)/dt]/(dx/dt)=\(\frac{1}{2 a t \cos^{3}{\left(2 t \right)}}\).
Therefore d²y/dx²=\(\frac{1}{2 a t \cos^{3}{\left(2 t \right)}}\).

📋 CBSE Stepwise Marking Scheme

[0.5 marks] Differentiate in t: dx/dt=\(4 a t \cos{\left(2 t \right)}\) and dy/dt=\(4 a t \sin{\left(2 t \right)}\).
[1 mark] Thus dy/dx=(dy/dt)/(dx/dt)=\(\tan{\left(2 t \right)}\).
[1 mark] Use d²y/dx²=[d(dy/dx)/dt]/(dx/dt)=\(\frac{1}{2 a t \cos^{3}{\left(2 t \right)}}\).
[0.5 marks] Therefore d²y/dx²=\(\frac{1}{2 a t \cos^{3}{\left(2 t \right)}}\).

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Class 12 Maths · Continuity And Differentiability · Long Answer #154

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 5: Continuity And Differentiability

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