Sponsor Area

Application Of Derivatives

Question
CBSEENMA12035116

Find the equation of the tangent and normal to the given curves at the points given:
y = x3 at (1, 1)

Solution

The equation of curve is y = x3
therefore space space space space space dy over dx space equals space 3 straight x squared
At (1, 1),   dy over dx space equals space 3 left parenthesis 1 right parenthesis squared space equals space 3
therefore space space space space straight m space equals space 3 comma space space where space straight m space is space slope space of space tangent.
therefore space space space space space the space equation space of space tangent space at space left parenthesis 1 comma space 1 right parenthesis space is
space space space space space space space space space space straight y minus 1 space equals space 3 left parenthesis straight x minus 1 right parenthesis semicolon space space space space space space or space space space straight y minus 1 space equals space 3 straight x minus 3 space space space space or space space space 3 straight x minus straight y minus 2 space equals space 0
Slope space of space normal space space equals space 1 third
therefore space space space space equation space of space normal space at space left parenthesis 1 comma space 1 right parenthesis space is
space space space space space space space space space space space space space space straight y minus 1 space equals space minus 1 third left parenthesis straight x minus 1 right parenthesis comma space space space space or space space space 3 straight y minus 3 space equals space minus straight x plus 1 space space space space or space space straight x plus 3 straight y minus 4 space equals space 0