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Vector Algebra

Question
CBSEENMA12035677

If straight x equals straight a space sin space 2 straight t left parenthesis 1 plus cos space 2 straight t right parenthesis space and space straight y space equals straight b space cos space 2 straight t left parenthesis 1 minus cos space 2 straight t right parenthesis comma space then space find dy over dx space at space straight t space equals straight pi over 4.

Solution
straight x equals straight a space sin space 2 straight t left parenthesis 1 plus cos space 2 straight t right parenthesis comma
straight y equals space straight b space cos space 2 straight t left parenthesis 1 minus cos space 2 straight t right parenthesis
dx over dt space equals 2 straight a space cos space 2 straight t left parenthesis 1 plus cos space 2 straight t right parenthesis space plus space straight a space sin space 2 straight t left parenthesis negative 2 space sin space 2 straight t right parenthesis
space space space space equals space 2 straight a space cos 2 straight t space plus space 2 acos squared 2 straight t minus 2 straight a space sin squared 2 straight t
space space space space equals 2 acos 2 straight t space plus space 2 acos 4 straight t
space space space space dy over dt equals negative 2 bsin 2 straight t left parenthesis 1 minus cost space 2 straight t right parenthesis plus bcos 2 straight t left parenthesis 2 sin 2 straight t right parenthesis
space space space space space space space equals space minus 2 bsin 2 straight t space plus 2 bsin space 2 tcos 2 straight t space plus 2 bcos 2 straight t space sin 2 straight t
space space space space space space space equals negative 2 bsin 2 straight t plus 4 bsin 2 tcos 2 straight t
space space space space space space space equals negative 2 bsin 2 straight t plus 2 bsin 4 straight t
     fraction numerator begin display style dy over dt end style over denominator begin display style dx over dt end style end fraction equals fraction numerator negative 2 bsin 2 straight t plus 2 bsin 4 straight t over denominator 2 acos 2 straight t plus 2 acos 4 straight t end fraction
space space space rightwards double arrow space space space dy over dx equals fraction numerator negative 2 bsin 2 straight t plus 2 bsin 4 straight t over denominator 2 acos 2 straight t plus 2 acos 4 straight t end fraction
space space space space space rightwards double arrow space open vertical bar dy over dx close vertical bar subscript straight t equals straight pi over 4 end subscript space equals space fraction numerator negative 2 bsin begin display style fraction numerator 2 straight pi over denominator 4 end fraction end style plus 2 bsin begin display style fraction numerator 4 straight pi over denominator 4 end fraction end style over denominator 2 acos begin display style fraction numerator 2 straight pi over denominator 4 end fraction end style plus 2 acos begin display style fraction numerator 4 straight pi over denominator 4 end fraction end style end fraction
space space space space space rightwards double arrow open vertical bar dy over dx close vertical bar subscript straight t equals straight pi over 4 end subscript space equals fraction numerator negative 2 bsin begin display style straight pi over 2 end style plus 2 bsinπ over denominator 2 acos begin display style straight pi over 2 end style plus 2 acosπ end fraction
space space space space space rightwards double arrow open vertical bar dy over dx close vertical bar subscript straight t equals straight pi over 4 end subscript space equals fraction numerator negative 2 straight b over denominator negative 2 straight a end fraction equals straight b over straight a
space space space space space space space space therefore space space space open vertical bar dy over dx close vertical bar subscript straight t equals straight pi over 4 end subscript space equals straight b over straight a
space space space space space space space space space