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Application Of Derivatives

Question
CBSEENMA12035095
Wired Faculty App

Find the equation of tangent to the curve:
straight x space equals space 1 plus cos space straight theta comma space space space space straight y equals space straight theta plus sinθ space space at space straight theta space equals space straight pi over 4.



Solution

The equations of the curve are straight x equals 1 plus cos space straight theta comma space space space space straight y space equals space straight theta plus sinθ
At straight theta space equals space straight pi over 4 comma
      straight x space equals space 1 plus space cos space straight pi over 4 space equals space 1 plus fraction numerator 1 over denominator square root of 2 end fraction comma space space space space straight y space equals space straight pi over 4 plus sin straight pi over 4 space equals space straight pi over 4 plus 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 1 plus fraction numerator 1 over denominator square root of 2 end fraction comma space space straight pi over 4 plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses Also comma space space space dx over dθ space equals space minus sin space straight theta comma space space space dy over dθ space equals space 1 plus space cos space straight theta therefore space space space space space space space 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 1 plus cos space straight theta over denominator negative sin space straight theta end fraction
At space straight theta space equals space straight pi over 4 comma space space space space space space dy over dx space equals space minus fraction numerator 1 plus cos space begin display style straight pi over 4 end style over denominator sin space begin display style straight pi over 4 end style end fraction space equals negative fraction numerator 1 plus begin display style fraction numerator 1 over denominator square root of 2 end fraction end style over denominator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style end fraction space equals space minus left parenthesis square root of 2 plus 1 right parenthesis space space therefore space space space space at space straight theta space equals space straight pi over 4 comma space space space slope space of space tangent space space equals space minus left parenthesis square root of 2 plus 1 right parenthesis space space space therefore space space space equation space of space tangent space is space space space straight y minus open parentheses straight pi over 4 plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses space equals space minus left parenthesis square root of 2 plus 1 right parenthesis space open square brackets straight x minus open parentheses 1 plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses close square brackets

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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