00:01
Now we have to find the unit vector parallel to the tangent line to the curve y is equal to 2 sine x at pi by 6 1.
00:13
Now we first find d .y by d x.
00:15
So dy by d x comes out to be equal to 2 cause x.
00:19
Slope at point pi by 6 1 is equal to 2 cos pi by 6 which is equal to root 3.
00:26
So now we find the equation of tangent.
00:29
So equation of tangent at this point pi by 6 1 will be y minus y 1 equal to root 3 that is slope x minus x1.
00:40
So this comes out to be equal to y is equal to root 3x minus root 3 by 6 pi plus 1.
00:46
So let us mark this as 1.
00:49
Now we have to find a unit vector parallel to this tangent.
00:53
So now y is equal to root 3x is the equation of the line which is parallel to.
00:59
To the tangent line because the slopes are equal.
01:03
Now over here also the slope is root 3 and over here also the slope is root 3.
01:08
So the slope, so this equation if line is parallel to the tangent at pi by 6 1 and passing through the origin because this is of the form y is equal to mx...