00:01
Okay, very well.
00:02
So here we have our parameterization, r of r comma theta, which is r cosine of theta, comma, theta, and the z coordinate is going to be r only.
00:25
Okay, so here we need to find the two partial derivatives of this vector function.
00:33
That is, okay, the partial derivative of our vector function with respect to r.
00:39
This is easy.
00:40
This one is just cosine of theta, comma, sine of theta, comma one.
00:48
And now we need to evaluate this guy here at the point 2 comma pi.
00:55
Okay so at the point 2 comma pi are the partial derivative of r with respect to the radius at the point 2 comma pi well this one is going to be negative 1 01 perfect now let's do the same for the partial derivative with respect to theta okay this one is going to be r multiplied by negative sign of theta comma consign of theta, comma, zero.
01:30
Okay, perfect.
01:31
So the evaluation of this guy, the evaluation of this guy at the point 2 .5, well, this one is just 0.
01:44
Okay, so 2 .5, so it's gonna be 0, negative 2 and 0.
01:52
Okay, perfect.
01:53
So at this point we can find the question of our tangent plan.
02:01
This is easy.
02:02
In order to do this, let's find the normal vector at 2.
02:07
Pi first.
02:09
Well, this normal vector is just the determinant of the matrix i, j, k.
02:16
Okay, here we are going to have negative 1, 01, that is this vector here.
02:21
And 0 negative to 0.
02:25
Okay, so the first coordinate of this guy is going to be 2.
02:30
The second coordinate of this guy is going to be 0...