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Class 12 Applications Of Derivatives · Short Answer2 Marksrelated_rates

Class 12 Maths · Applications Of Derivatives · Short Answer #48

Two squares have positive side lengths x and y=x−x² (so 0<x<1). Find the rate of change of the second area with respect to the first.

✅ Step-by-Step Solution

Final Answer: dA₂/dA₁=2x²−3x+1
  1. A₁=x² and A₂=(x−x²)².
  2. dA₂/dA₁=[2(x−x²)(1−2x)]/(2x)=(1−x)(1−2x).
  3. Thus dA₂/dA₁=2x²−3x+1, for 0<x<1.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] A₁=x² and A₂=(x−x²)².
  • [1 mark] dA₂/dA₁=[2(x−x²)(1−2x)]/(2x)=(1−x)(1−2x).
  • [0.5 marks] Thus dA₂/dA₁=2x²−3x+1, for 0<x<1.

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