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Application Of Derivatives

Question
CBSEENMA12035089
Wired Faculty App

Find the equation of the tangent at straight t space equals space straight pi over 4 space to space the space curve space straight x space equals sin space 3 straight t comma space space space space space straight y space equals space cos 2 straight t.

Solution

The equations of curve are
                          straight x space equals sin space 3 space straight t comma space space space space straight y space equals space cos space 2 straight t
At   straight t space equals space straight pi over 4 comma space space space space space space space straight x space equals space sin space fraction numerator 3 straight pi over denominator 4 end fraction space equals space sin space open parentheses straight pi minus straight pi over 4 close parentheses space equals space sin 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 cos straight pi over 2 space equals space 0
therefore space space 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 0 close parentheses space space space space space space dx over dt space equals space 3 space cos 3 straight t comma space space dy over dt space equals space minus 2 space sin space 2 straight t space space space space space dy over dx space equals space fraction numerator begin display style dy over dt end style over denominator begin display style dx over dt end style end fraction space equals space fraction numerator negative 2 space sin space 2 straight t over denominator 3 space cos space 3 straight t end fraction
At space straight t space equals space straight pi over 4 comma space space space dy over dx space equals space minus 2 over 3 fraction numerator sin space begin display style straight pi over 2 end style over denominator cos space begin display style fraction numerator 3 straight pi over denominator 4 end fraction end style end fraction space equals space minus 2 over 3 fraction numerator 1 over denominator cos space open parentheses straight pi minus begin display style straight pi over 4 end style close parentheses end fraction space equals space fraction numerator 2 over denominator 3 cos space begin display style straight pi over 4 end style end fraction space equals space fraction numerator 2 over denominator 3 cross times begin display style fraction numerator 1 over denominator square root of 2 end fraction end style end fraction space equals space fraction numerator 2 square root of 2 over denominator 3 end fraction
therefore space space space space slope space of space tangent space space equals space fraction numerator 2 square root of 2 over denominator 3 end fraction therefore space space space space the space equation space of space tangent space at space open parentheses fraction numerator 1 over denominator square root of 2 end fraction comma space 0 close parentheses space is space space space space space space space straight y minus 0 space equals space fraction numerator 2 square root of 2 over denominator 3 end fraction 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 space space 3 straight y space equals space 2 square root of 2 straight x minus 2 space space space space or space space 2 square root of 2 straight x minus 3 straight y minus 2 space equals space 0

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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