00:01
So we are given the vector valued function, r of t, equal to 2 plus cosine of t, 3 plus sine of 2t, and t.
00:18
And we are asked to find the tangent line, the equation of the tangent line, at the point t0 equals pi over 2.
00:31
To do this, we'll need two things.
00:32
The first thing we'll need is to evaluate the vector at the point pi over 2.
00:38
So let's do that first.
00:40
We have 2 plus cosine of pi over 2, and then 3 plus sign of 2 times pi over 2 is just pi.
00:51
And then we just plug in pi over 2 at the end.
00:53
And last component.
00:55
Cosine of pi over 2 is 0, so this leaves me with 2.
00:58
Sign of pi is 0, so that leaves with 3, and then the last 1 is just pi over 2...