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Application Of Derivatives

Question
CBSEENMA12035100

Find the equation of the tangent and normal to the curve
straight x space equals space cost comma space space space space space straight y space equals space sin space straight t space space space at space straight t space equals space straight pi over 4

Solution

The equations of the curve are straight x space equals space cos space straight t comma space space space space straight y space equals space sin space straight t  At space space space straight t space equals space straight pi over 4 comma space space space straight x space equals space cos space straight pi over 4 space equals space fraction numerator 1 over denominator square root of 2 end fraction comma space space space straight y space equals space sin space straight pi over 4 space equals space fraction numerator 1 over denominator square root of 2 end fraction
therefore space space space space point space of space contact space is space open parentheses fraction numerator 1 over denominator square root of 2 end fraction comma space space space space fraction numerator 1 over denominator square root of 2 end fraction close parentheses
Also,     dx over dt space equals space minus sin space straight t comma space space space dy over dt space equals space cot space space space space space space space space space space space rightwards double arrow space space space space dy over dx space equals space fraction numerator begin display style dy over dx end style over denominator begin display style dx over dt end style end fraction space equals space fraction numerator cos space straight t over denominator negative sin space straight t end fraction space equals space minus cot space straight t
 At space straight t space equals space straight pi over 4 comma space space space dy over dx space equals space minus cot straight pi over 4 space equals space minus 1
therefore space space space space space space at space straight t space equals space straight pi over 4 comma space space slope space of space tangent space space equals space minus 1
therefore space space space space the space equation space of space tangent space at space straight t space equals space straight pi over 4 space is space
                   straight y minus fraction numerator 1 over denominator square root of 2 end fraction space equals space minus 1 open parentheses straight x minus fraction numerator 1 over denominator square root of 2 end fraction close parentheses space space space space space or space space space space space straight y minus fraction numerator 1 over denominator square root of 2 end fraction space equals space minus straight x plus fraction numerator 1 over denominator square root of 2 end fraction
or     straight x plus straight y minus square root of 2 space equals space 0
Also slope of normal  = negative fraction numerator 1 over denominator negative 1 end fraction space equals space 1
therefore space space space space space equation space of space normal space at space straight t space equals space straight pi over 4 space is
                             straight y minus fraction numerator 1 over denominator square root of 2 end fraction space equals space 1 open parentheses straight x minus fraction numerator 1 over denominator square root of 2 end fraction close parentheses space space space space or space space space space straight x minus straight y space equals space 0

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