Sponsor Area

Application Of Derivatives

Question
CBSEENMA12035133

Find the equation of the tangent and normal to the given curves at the points given:
y = x4 – 6x3 + 13x2 – 10x + 5  at ( 0 , 5 )

Solution

The equations of curve is y = x4 – 6x3 + 13x2 – 10x + 5
therefore space space space space dy over dx space equals space 4 straight x cubed minus 3 bx squared plus 26 straight x minus 10
At (0, 5). dy over dx space equals space 0 minus 0 plus 0 minus 10 space equals space minus 10 comma space space which space is space slope space of space tangent
therefore space space space space space space equation space of space tangent space at thin space left parenthesis 0 comma space 5 right parenthesis space is space
space space space space space space straight y space minus space 5 space space equals space minus 10 left parenthesis straight x minus 0 right parenthesis comma space space space space or space space space space straight y minus 5 space equals space minus 10 space straight x space space space space space or space space space 10 straight x plus straight y minus 5 space equals space 0
Slope space of space normal space equals space 1 over 10
therefore space space space space space space space equation space of space normal space at space left parenthesis 0 comma space 5 right parenthesis space is
space space space space space space space space space space straight y minus 5 space equals space 1 over 10 left parenthesis straight x minus 0 right parenthesis
or space space space space 10 straight y minus 50 space equals space straight x space space space space space space or space space space straight x minus 10 straight y plus 50 space equals space 0