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# Match the vector equations with the graphs (a)-(h) given here. (GRAPH CANT COPY)$$\mathbf{r}(t)=\left(t^{2}-1\right) \mathbf{j}+2 t \mathbf{k}, \quad-1 \leq t \leq 1$$

## Graph (f)

Integrals

Vectors

Vector Functions

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### Video Transcript

Okay, folks. So in this video, we're gonna take a look at this factor equation. This'd is this is a problem. Number seven. Um, so we have this vector equation right here that says are as a function of t gives you t squared minus one in the UAE. Um, access. I'm the right. I just wrote J hat as as white hat, but they're the same thing. Plus to t in the in the Z axis, and we're restricting ourselves to negative one, less than or equal to t less thundery would want. So the first thing I want you to notice is the fact that we don't have a next component. That means our graph is gonna be in the Wide Z plane on Lee so that that's the first thing to know this. Another thing to notice is that we can There's a relationship between y and Z because they're both parameter rised by the same by the same variable. So we have the y component of all of her points. That's gonna be raft out. The y component is always gonna be t squared minus one. I have very horrible handwriting today, so and the Z component of that of all all of our points is is gonna be written as to t So we have this relationship here that we can that we can take advantage of. The way we're gonna take advantage of this is by saying, is by rewriting why as a function of Z, how do we do that? Well, he squared. I'm gonna massage this t squared into something else. I can say that he square is the same thing as fourty sward divided by four. Okay, so I don't do anything. I just multiplied by foreign divided by four. I could do that minus one. Okay, before it's you squared, the reason that multiplied by four is because if you recognize that Z squared is exactly 40 squared because these two teams so z squared is 40 squared. So now we have y as a function of Z. So why is Z squared over four minus one and this is very helpful. The reason I say this is very helpful is because we can now and draw it out. We can draw why it's a function of C. So if we draw, this is Thea, you know usually would put X here, but now it's Z, but it's the same thing. It doesn't matter. So we have. Why the function of Z. You recognize that this is just a problem, you know it's a problem, but like it's scaled differently and it's shifted downward. But still it's a perhaps less we can draw it out. So you know the I guess it's called Vertex of the crab HLA is shifted downward by ones We're here. We're going to start off here and we're gonna have a problem. That's it. That's our problem. All we need to do now is to shift or, like, map this problem into a three dimensional space. And right now it's in right now is living in two dimensions. We're gonna move this problem from two dimensions up to three dimensions. So how do we do that? Well, um, if I, uh if I graph my three dimensional space, this is the positive side of why and this is the positive side of Z. I'm going to leave the X axis for now. I'm gonna forget about it for now because X is always zero. So let's graph this. So the way we're gonna graph this is by taking, you know, this graph right here and just, you know, and just shake it around and, you know, you just you just kind of like manipulator in a way that that's gonna look something like this. You know, when, When, When? When Z is zero. Why is negative one? So I'm gonna start all right here and it's gonna look like a parabola. So that's one thing I know for certain. One thing we know for certain is the fact that the vertex of the problem is that negative one. It's the equals, negative one. I mean, why equals negative one and it's a problem. So these are the two things that we definitely know for sure we don't care. We don't care about the X axis. We don't care about anything else. The only two things that we need to know in order to do this problem or the facts of that. First of all, it's a problem. Second of all, it's ver taxes at y equals negative one. Now all we need to do is is to add the axe axe is there and now we look for it. We have a couple of options from a to B all the way to each. And we're gonna look for the option that matches our graph right here. And if you look through it, you'll see that that part F matches their description. Exactly what? This is Our graph. This is Arena, sir. That's it. Thank you.

University of California, Berkeley

Integrals

Vectors

Vector Functions

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