Class 12 Continuity And Differentiability · Long Answer3 Marksparametric_derivatives
Class 12 Maths · Continuity And Differentiability · Long Answer #89
For x=a cosθ+b sinθ, y=a sinθ−b cosθ, prove y²y″−xy′+y=0 wherever y≠0.
✅ Step-by-Step Solution
Final Answer: Identity proved.
- Differentiate the parametric coordinates: dx/dθ=−a sinθ+b cosθ=−y and dy/dθ=a cosθ+b sinθ=x.
- Thus y′=(dy/dθ)/(dx/dθ)=−x/y for y≠0.
- Differentiate −x/y with respect to x: y″=(−y+xy′)/y². Multiplying by y² gives y²y″=−y+xy′.
- Therefore y²y″−xy′+y=0.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Differentiate the parametric coordinates: dx/dθ=−a sinθ+b cosθ=−y and dy/dθ=a cosθ+b sinθ=x.
- [1 mark] Thus y′=(dy/dθ)/(dx/dθ)=−x/y for y≠0.
- [1 mark] Differentiate −x/y with respect to x: y″=(−y+xy′)/y². Multiplying by y² gives y²y″=−y+xy′.
- [0.5 marks] Therefore y²y″−xy′+y=0.
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