How do you find the equation for the lines that are tangent and normal to the curve y=1+cosx at the point (pi/2, 1)?

1 Answer
Sep 1, 2017

Start with the derivative

Explanation:

For y = 1 + cosx,
dy/dx = -sinx.
At pi/2, we have
dy/dx = -sin(pi/2) = -1.
This is the slope of the tangent line.

Use the point-slope form of a line to find its equation.
y - y_1 = m(x - x_1)

In this case
y - 1 = -1(x - pi/2)
Solve for y.
y = -x + pi/2 + 1

The normal is perpendicular to the tangent.
So, if m is the slope of the tangent line, then -1/m is the slope of the normal.

In this case, the normal slope is 1.

Use the same point, and the point-slope form of a line, to obtain the equation of the normal line.