Thread: Find equation of the tangent line

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. c(t)=cost, cost + sin^2(t) t = pi

Well right off there is a problem because the curve is at an endpoint when t = pi, so the slope isn't defined. However, I suppose we can take a limit as t approaches pi.

What is the location of the point? We have:

x =
y =

Then for t = we have:
x = -1
y = -1

So what is the slope of the curve at that point? We have

c'(t) = ,

So the slope of a line passing through that point would be:



For t = we have:

Slope =


Oooookay, so whats the equation of a line passing through (-1,-1) with slope 3?

y = mx + b ---> -1 = -3 + b ----> b = 2

Then our equation of the line is:

y = 3x + 2