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
namralos
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.

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