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