Question

What is the slope of the line normal to the tangent line of `f(x) = sin(3x-pi)` at `x= pi/3`?

Answer 1

`y = -1/3x + pi/9`

Explanation:

We start by finding the corresponding `y` coordinate.

`f(pi/3) = sin(3(pi/3) - pi)`

`f(pi/3) = sin(pi - pi)`

`f(pi/3) = sin(0)`

`f(pi/3) = 0`

We now differentiate. We let `y = sinu` and `u = 3x- pi`.

Then `dy/(du)= cosu` and `(du)/dx = 3`, since `pi` is but a constant.

By the chain rule,

`f'(x) = dy/(du) xx (du)/dx`

`f'(x) = cosu xx 3`

`f'(x) = 3cos(3x - pi)`

We now determine the slope of the tangent.

`f'(pi/3) = 3cos(3(pi/3) - pi)`

`f'(pi/3) = 3cos(0)`

`f'(pi/3) = 3`

The normal line is perpendicular to the tangent, so the slope will be the negative reciprocal.

`m_"normal" = -1/3`

We can now determine the equation.

`y - y_1 = m(x- x_1)`

`y - 0 = -1/3(x- pi/3)`

`y = -1/3x + pi/9`

Hopefully this helps!