00:01
This is a three -part question given a vector field and a specific curve.
00:04
We're wondering points where this vector is tangent to the curve, where it's normal to the curve, and then we'll sketch the curve and some vectors f on c.
00:14
So what do we have here? we have our vector field f.
00:17
It is going to be a constant vector field, half comma zero.
00:22
Our curve c is going to be equal to x comma y, such that y might be, minus x squared is equal to 1.
00:33
So to start off before we get to our a, b, and c parts, we need to be able to get a parameterization for our curve.
00:42
So we have y is equal to x squared plus 1.
00:46
We can just make that by moving this x squared to the other side.
00:50
So an easy parameterization we could do is t for x and then t squared plus 1 to y.
00:57
Since this is what is going to, since we have an explicit formula for y, it allows us to do this, making it an easy parameterization.
01:05
All right, so now we have what our c of t is.
01:09
For part a, we want to find any points, if they exist, where f is tangent to c.
01:15
This would mean that our curve's tangent vector is the same as f.
01:19
How do you find the tangent vector? you just take the derivative of both parts which respect to t of our parameterization.
01:26
So we have 1 comma 2t.
01:29
So are there any points x comma y, which would originate from values of t such that this vector is equal to this vector? well, this answer is going to be no since we have one in half and they don't ever change...