00:01
The problem is find an equation of the tangent to the curve at the given point, then graph the curve and the tangent.
00:10
X is equal to sine pi t, y is equal to t squared plus t, and the point is 0 .2.
00:19
First we compute the slope of the tangent, dy, x.
00:25
This is equal to dy d t over d x dx.
00:31
D .y d t is 2 t plus 1 and d x d t is pi times cosine high t and 1 xy is at the point 02 we have sine high t is equal to 0 and t squared plus t is equal to 2 is equal to 2 then we have t is equal to 1 or t is equal to negative 2.
01:33
And when we plug in t equal to 1 and t equal to negative 2 to d, y, dx, we can find both of those terms are equal to negative 3 over pi.
01:50
So, for example, one t is equal to 1...