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Numerade Educator



Problem 25 Medium Difficulty

For the vector field $\mathbf{F}$ and curve $C$, complete the following:
a. Determine the points (if any) along the curve C at which the vector field $\mathbf{F}$ is tangent to $C$.
b. Determine the points (if any) along the curve C at which the vector field $\mathbf{F}$ is normal to $C$
c. Sketch $C$ and a few representative vectors of $\mathbf{F}$ on $C$.
$$\mathbf{F}=\left\langle\frac{1}{2}, 0\right\rangle ; C=\left\{(x, y): y-x^{2}=1\right\}$$


a. $(\frac{1}{2},0)$
b. 0
c. graph


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Video Transcript

This is a three part question given a vector field and curb or wondering the points where this vector is tangent to the curve, where it's normal to the curb and then we'll sketch the curb and some vectors f on seat. So what do we have here? We have the vector field f is going to be a constant vector field, half common zero. The curve C is going to be equal to X comma. Why such that? Why minus X squared is equal to one. So to start off before we get to our A, B and C parts, we need to be able to get a parameter ization for our curve. So we have Why is equal to X squared plus one? We can just make that by moving this X Where to the other side. So an easy parameter ization we could do is t for X and then t squared, plus one for why, since this is what is going to since we have a necklace, that formula for why it allows us to do this, making it an easy parameter ization. All right, so now we have what our sea of T is for part A. We want to find any points if they exist Where f is tangent to see This would mean that our curves tangent sector is the same It's how do you find the tangent Vector? You just take the derivative of the both parts of respect t of our privatisation. So we have one comment to t So are there any point X comma y which would originate from values of t such that this vector is equal to this sector? Well, this answer is going to be no since we have 1.5 and they don't ever change So there's no way we're ever gonna have a tangent vector, So we have no point. But we have attention. Vector part B. We wanna find point where f is normal to see. This is where our f is going to be orthogonal to the pain defector. So we're wondering what points we have where half on the one dot dot trying with one comma two t is equal to zero. Well, we would just take the stop product and see what value did. He will give us what we want. We have half times one plus zero times to t which, unfortunately, doesn't give us any solutions here because this zero causes this time to go away. And then we have half is recorded zero, which is clearly not true. So we don't have any points for this part either. So this vector field and this curb aren't really very much in line with each other to see that we're gonna go ahead and draw it. So the function why equals X squared plus one? We've dealt with a lot before and passed courses typically. So we start here at this 10.1 the Y intercept, and then we draw X squared. Looks like about this. So then we would have the point if X is one, then once where plus two is too. So this should should be to its scaled. Right? So at this point, the tangent vectors going something like this our vector half comma zero would just be going a little bit like that. And so we would have it are aspect er going several places along it and we never really increase in this direction. It doesn't really make sense. So that's why we don't have a tangent or necessarily a normal vector at

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