Question

What is the equation of the normal line of `f(x)=x^3/sqrt(16x^2-x)` at `x=2`?

Answer 1

Equation of normal at `x=2` is `y-1.016=-0.992253(x-2)`

Explanation:

For a function `y=f(x)`, the slope of tangent at `x=x_0` i.e. at `(x_0.f(x_0))` is given by `f'(x_0)`, where `f'(x)=(dy)/(dx)`

and hence slope of normal is `-1/(f'(x_0))` as normal is perpendicular to tangent.

and while equation of tangent is `y-f(x_0)=f'(x_0)(x-x_0)`

and equation of normal is `y-f(x_0)=-1/(f'(x_0))(x-x_0)`

Here we have `f(x)=x^3/sqrt(16x^2-x)` and at `x_0=2`. As `f(x_0)=8/sqrt62`, hence we are seeking normal at point `(2,8/sqrt62)` or `(2,1.016)`

Further using quotient rule `f'(x)=(3x^2sqrt(16x^2-x)-x^3((32x-1)/(2sqrt(16x^2-x))))/(16x^2-x)`

and `f'(2)=(12sqrt62-8(63/(2sqrt62)))/62`

and slope of normal is `-62/(12sqrt62-8(63/(2sqrt62)))`

= `-sqrt62/(12-504/124)=-sqrt62/(12-126/31)`

= `-(31sqrt62)/(372-126)=-(31sqrt62)/246=-0.992253`

and equation of normal is `y-1.016=-0.992253(x-2)`

graph{(y-1.016+0.992253(x-2))(x^3/sqrt(16x^2-x)-y)=0 [-10, 10, -5, 5]}