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 = $\displaystyle \cos (t)$
y = $\displaystyle \cos (t) + \sin ^2(t)$

Then for t = $\displaystyle \pi$ we have:
x = -1
y = -1

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

c'(t) = $\displaystyle - \sin (t)$, $\displaystyle - \sin (t) + 2 \cos (t) \sin (t)$

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

$\displaystyle \frac{- \sin (t) + 2 \cos (t) \sin (t)}{- \sin (t)}$

For t = $\displaystyle \pi$ we have:

Slope = $\displaystyle \lim_{t \rightarrow \pi } \frac{- \sin (t) + 2 \cos (t) \sin (t)}{- \sin (t)} = 3$


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