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Application Of Derivatives

Question
CBSEENMA12035069

Find the slope of the normal to the curve straight x space equals space 1 minus straight a space sin space straight theta comma space space space straight y space equals space space straight b space cos squared straight theta space space at space straight theta space equals space straight pi over 2.

Solution
straight x space equals space 1 minus straight a space sin space straight theta comma space space space straight y space equals space space straight b space cos squared straight theta space space at space straight theta space equals space straight pi over 2.
  The equations of the curve are straight x space equals space space 1 minus straight a space sin space straight theta comma space space space space straight y space equals space straight b space cos squared straight theta
therefore space space space space space dx over dθ space equals space minus straight a space cos space straight theta comma space space dy over dθ space equals space minus 2 straight b space cos space straight theta space sin space straight theta
Now dy over dx space equals space fraction numerator begin display style dy over dθ end style over denominator begin display style dx over dθ end style end fraction space equals space fraction numerator negative 2 straight b space sinθ space cosθ over denominator negative straight a space cosθ end fraction space equals space fraction numerator 2 straight b over denominator straight a end fraction sin space straight theta
At straight theta space equals space straight pi over 2 comma space  dy over dx space equals space fraction numerator 2 straight b over denominator straight a end fraction sin space straight pi over 2 space equals space fraction numerator 2 straight b over denominator straight a end fraction cross times 1 space equals space fraction numerator 2 straight b over denominator straight a end fraction
This is slope of tangent to the curve
therefore space space space space slope space of space normal space to space the space curve space equals space minus fraction numerator 1 over denominator begin display style fraction numerator 2 straight b over denominator straight a end fraction end style end fraction space equals space minus fraction numerator straight a over denominator 2 straight b end fraction