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Continuity And Differentiability

Question
CBSEENMA12035013

If space straight x equals straight a left parenthesis straight theta plus sinθ right parenthesis comma space straight y equals straight a left parenthesis 1 minus cosθ right parenthesis comma space fimd space dy over dx at space straight theta equals straight pi over 2.

Solution
space straight x equals straight a left parenthesis straight theta plus sinθ right parenthesis space rightwards double arrow space dx over xθ equals straight a left parenthesis 1 plus cosθ right parenthesis
straight y equals straight a left parenthesis 1 minus cosθ right parenthesis space rightwards double arrow space dy over dθ equals straight a space sinθ
Now space dy over dθ equals fraction numerator straight a space sinθ over denominator straight a left parenthesis 1 plus cosθ right parenthesis end fraction
At space straight theta equals straight pi over 2 comma space dy over dx equals fraction numerator sin begin display style straight pi over 2 end style over denominator 1 plus cos begin display style straight pi over 2 end style end fraction equals fraction numerator 1 over denominator 1 plus 0 end fraction equals 1

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