NCERT Class 12 · Ch 5 Miscellaneous Exercise · Q15
NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Miscellaneous Exercise Question 15
If (x−a)²+(y−b)²=c², c>0, examine [1+(y′)²]^(3/2)/y″ and its independence of a,b.
💡 Concept Hints
- Identify the domain and the inner and outer functions, or the relevant one-sided limits.
- Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.
✅ Step-by-Step Solution
Final Answer: −c on the upper semicircle; +c on the lower semicircle. Its absolute value is c.
- Implicit differentiation gives y′=−(x−a)/(y−b). Differentiating again gives y″=−c²/(y−b)³, for y≠b.
- The numerator is [c²/(y−b)²]^(3/2)=c³/|y−b|³.
- The signed quotient is −c·(y−b)³/|y−b|³=−c sign(y−b). It is constant on each differentiable semicircle and independent of a,b.
- The two branches have opposite signs; a single signed constant does not hold across both. At the vertical tangencies y=b the derivative expression is undefined.
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