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Application Of Derivatives

Question
CBSEENMA12035086
Wired Faculty App

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

Solution

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