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Application Of Derivatives

Question
CBSEENMA12035093
Wired Faculty App

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


Solution

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