Question

What is the equation of the line normal to `f(x)=sin(x/2+pi)` at `x=pi/3`?

Answer 1

The equation is `y = -sqrt(3)/4x + (sqrt(3)pi - 6)/12`

Explanation:

Start by finding the y-coordinate of tangency.

`f(pi/3) = sin((pi/3)/2 + pi)`

`f(pi/3) = sin(pi/6 + pi)`

`f(pi/3) = sin((7pi)/6)`

`f(pi/3) = -1/2`

Now we find the derivative of `f(x)`, using the chain rule. We let `y = sinu` and `u = x/2 + pi`. Then `dy/(du) = cosu` and `(du)/dx = 1/2` (since `π` is only a constant).

The derivative therefore is

`dy/dx = dy/(du) * (du)/dx`

`dy/dx = cosu * 1/2`

`dy/dx = 1/2cos(x/2 + pi)`

We now find the slope of the tangent.

`dy/dx|_(x = pi/3) = 1/2cos((pi/3)/2 + pi)`

`dy/dx|_(x = pi/3) = 1/2cos((7pi)/6)`

`dy/dx|_(x = pi/3) = 1/2(-sqrt(3)/2)`

`dy/dx|_(x= pi/3) = -sqrt(3)/4`

We now have all the information we need to find the equation of the line.

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

`y - (-1/2) = -sqrt(3)/4(x - pi/3)`

`y + 1/2 = -sqrt(3)/4x + (sqrt(3)pi)/12`

`y = -sqrt(3)/4x + (sqrt(3)pi - 6)/12`

Hopefully this helps!