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Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12035439
Wired Faculty App

If space space space straight x equals 2 space cos space straight ө minus cos space 2 straight ө space and space straight y equals 2 space sin space straight ө minus sin space 2 straight ө comma space find space open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses subscript straight theta equals straight pi over 2 end subscript

Solution
space space space space space space space space space space space space space space space space space space space straight x equals 2 space cos space straight ө minus cos space 2 straight ө comma space space space straight y equals 2 space sin space straight ө minus sin space 2 straight ө therefore space space space space space space space space dx over dθ equals negative 2 space sin space straight theta plus 2 space sin space straight theta comma space dy over dθ equals 2 space cos space straight theta minus 2 space cos space 2 straight theta Now space space space space dy over dx equals fraction numerator begin display style dy over dθ end style over denominator begin display style dx over dθ end style end fraction equals fraction numerator 2 space cos space straight theta minus 2 space cos space 2 straight theta over denominator negative 2 sin space straight theta plus 2 space sin space 2 straight theta end fraction equals fraction numerator cos space straight theta minus cos space 2 straight theta over denominator sin space 2 straight theta minus sin space straight theta end fraction space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 2 space sin begin display style fraction numerator 3 straight theta over denominator 2 end fraction end style sin begin display style straight theta over 2 end style over denominator 2 space cos space begin display style fraction numerator 3 straight theta over denominator 2 end fraction end style sin begin display style straight theta over 2 end style end fraction equals tan fraction numerator 3 straight theta over denominator 2 end fraction therefore space space space space space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction equals open parentheses sec squared fraction numerator 3 straight theta over denominator 2 end fraction close parentheses. open parentheses 3 over 2 close parentheses dθ over dx equals 3 over 2 sec squared straight theta over 2 open parentheses fraction numerator 1 over denominator negative 2 sin space straight theta plus 2 space sin space 2 straight theta end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 3 over denominator 4 space cos squared begin display style straight theta over 2 end style end fraction cross times fraction numerator 1 over denominator sin space 2 straight theta minus sin space straight theta end fraction open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction close parentheses subscript straight theta equals straight pi over 2 end subscript equals fraction numerator 3 over denominator 4 space cos squared begin display style straight theta over 2 end style end fraction cross times fraction numerator 1 over denominator sin space straight pi minus sin begin display style straight pi over 2 end style end fraction equals fraction numerator 3 over denominator 4 cross times begin display style 1 half end style end fraction cross times fraction numerator 1 over denominator 0 minus 1 end fraction equals 3 over 2 cross times left parenthesis negative 1 right parenthesis equals negative 3 over 2

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