Question

What is the equation of the normal line of `f(x)= tanx` at `x = pi/8`?

Answer 1

Equation of normal line is `x+ 1.17y =0.87`

Explanation:

`x=pi/8 ~~0.39 ; f(x)=tanx = or f(x)= tan (pi/8) ~~ 0.41`

So at `(0.39,0.41)` tangent and normal is drawn.

Slope of the tangent is `f'(x)= sec^2x`. at `x=pi/8`

`sec^2x=1/cos^2x= 1/(cos (pi/8))^2 ~~ 1.17` . Slope of tangent

is `m_1=1.17` Normal line is perpendicular to tangent , so

slope of the normal line `m_2= -1/m_1 = -1/1.17`

Equation of normal line at point `(0.39,0.41) ; m_2= -1/1.17`

is `y-y_1=m_2(x-x_1) or (y- 0.41) = -1/1.17(x-0.39)` or

`1.17 (y- 0.41) = -(x-0.39)` or

`x+1.17y=0.39+(1.17*0.41)` or

`x+1.17y=0.39+(1.17*0.41) or x+1.17y =0.87`

Equation of normal line is `x+ 1.17y =0.87` [Ans]