Application of Derivatives

Question

Find the equation of the tangent and normal to the given curves at the points given:
y = x⁴ –6x³+ 13x²– 10x + 5 at (1 ,3)

Solution

The equation of curve is
y = x⁴ –6x³+ 13x²– 10x + 5
$therefore space space space space space space dy over dx space equals space 4 straight x cubed minus 18 straight x squared plus 26 straight x minus 10$
$At space left parenthesis 1 comma space 3 right parenthesis comma space space space dy over dx space equals space 4 left parenthesis 1 right parenthesis cubed minus 18 left parenthesis 1 right parenthesis squared plus 26 left parenthesis 1 right parenthesis space minus 10 space space space space space space space space space space space space space space space space space space space space space space equals 4 minus 18 plus 26 minus 10 space equals space 2 comma space space which space is space slope space of space tangent therefore space space space space space space the space equation space of space tangent space at space left parenthesis 1 comma space 3 right parenthesis space is space space space space space space space space space space space space straight y minus 3 space equals space 2 left parenthesis straight x minus 1 right parenthesis space space space or space space space space straight y minus 3 space equals space 2 straight x minus 2 space space space or space space 2 straight x minus straight y plus 1 space space equals space 0 Slope space of space normal space equals space minus 1 half therefore space space space space space straight y minus 3 space equals space minus 1 half left parenthesis straight x minus 1 right parenthesis comma space space space space space or space space space space 2 straight y minus 6 space equals space minus straight x plus 1 space space space space or space space straight x plus 2 straight y minus 7 space equals space 0$

For Daily Free Study Material Join wiredfaculty