Question

$If space space straight x equals fraction numerator 3 at over denominator 1 plus straight t squared end fraction comma space straight y equals fraction numerator 3 at squared over denominator 1 plus straight t squared end fraction comma space then space find space dy over dt space at space straight t equals 2$

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Solution

straight x equals fraction numerator 3 at over denominator 1 plus straight t squared end fraction comma space straight y equals fraction numerator 3 at squared over denominator 1 plus straight t squared end fraction therefore space dx over dt equals 3 straight a straight d over dt open square brackets fraction numerator straight t over denominator 1 plus straight t squared end fraction close square brackets equals 3 straight a open square brackets fraction numerator left parenthesis 1 plus straight t squared right parenthesis.1 minus straight t. space 2 straight t over denominator left parenthesis 1 plus straight t squared right parenthesis squared end fraction close square brackets equals 3 straight a open square brackets fraction numerator 1 plus straight t squared minus 2 straight t squared over denominator left parenthesis 1 plus straight t squared right parenthesis squared end fraction close square brackets therefore space dx over dt equals fraction numerator 3 straight a left parenthesis 1 plus straight t squared right parenthesis over denominator left parenthesis 1 plus straight t squared right parenthesis squared end fraction Also space dy over dt equals 3 straight a straight d over dt open square brackets fraction numerator straight t squared over denominator 1 plus straight t squared end fraction close square brackets equals 3 straight a open square brackets fraction numerator left parenthesis 1 plus straight t squared right parenthesis.2 straight t minus straight t squared.2 straight t over denominator left parenthesis 1 plus straight t squared right parenthesis squared end fraction close square brackets space space space space space space space space space space space space space space space space space equals 3 straight a open square brackets fraction numerator 2 straight t plus 2 straight t cubed minus 2 straight t cubed over denominator left parenthesis 1 plus straight t squared right parenthesis squared end fraction close square brackets equals fraction numerator 3 straight a left parenthesis 2 straight t right parenthesis over denominator left parenthesis 1 plus straight t squared right parenthesis squared end fraction therefore space dy over dt equals fraction numerator 6 at over denominator left parenthesis 1 plus straight t squared right parenthesis squared end fraction Now space dy over dt equals 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 6 at over denominator left parenthesis 1 plus straight t squared right parenthesis squared end fraction cross times fraction numerator left parenthesis 1 plus straight t squared right parenthesis squared over denominator 3 straight a left parenthesis 1 plus straight t squared right parenthesis end fraction therefore space dy over dt equals fraction numerator 2 straight t over denominator 1 minus straight t squared end fraction At space straight t equals 2 comma space dy over dx equals fraction numerator 2 cross times 2 over denominator 1 minus 4 end fraction equals fraction numerator 4 over denominator negative 3 end fraction equals negative 4 over 3.

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