Match the vector equations with the graphs (a)-(h) given here. (GRAPH CANT COPY)

$$\mathbf{r}(t)=(2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{k}, \quad 0 \leq t \leq \pi$$

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Campbell University

Oregon State University

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University of Michigan - Ann Arbor

Okay, folks. So in this video, we're gonna take a look at this vector equation. We have the following victory equation that we would like to match with one of the graphs that there were given. So let me write it down for you. We have are as a function of t can be written as to co sign of tea and that's for the X component and applies to sign of T. And that's where the Z component I would like to remind you that that I in care the same thing as X and Z Okay? No. And we also have a restriction. We we have that zero has to be less than were equal to t, which is less than or equal to pi. So this is our restriction and disserve vector function. We want to graph it out. The way we're gonna graph it out is by first noticing the fact that we don't have a white component for this function. That means our graph is going to be in the xz plane. So whatever your graph end up looking like a better not have a white component. If it does, it's wrong. OK, so that's That's a nice thing to notice. Another thing to notice is that Is that if you remember, polar coordinates the X component of you know, if this is the X axis, that's why acts that the X component of the point can be written as our co sign data and why component of a point can be written as r sine theta. So what I'm saying is, whenever you see something like this, you can be certain that you have something like a polar cornet, something like a polar equation going on here. But if you look here, this is exactly in the same form as as this equation, because we have a scaler multiplied by a co sign and we have a scaler here multiplied by a side. That means we have a polar equation going on here. So how about how do we graft polar equations that they're very simple graph? Well, we're gonna grab it out and see what we see what happens. Um, if I say that this is the X axis and this is the Z axis, you know, it's gonna look something like a circle because that's what polar acquaintance usually do whenever you have a polar equation that's gonna look something like a circle, especially if it's in the same form as what we are given here. So when t is zero, which is within our range of teas with T zero X is too and why? I mean, Z is zero and when t is pie, what were given here is co sign of pies, negative ones. We have negative to wear here and t sign both. Pie is zero. So we have see again zero. And when t is pie half, we have for the X component zero and we have for the Z component once I mean two, because we're scaling it by two. So this is our half circle. This is 1/2 circle and of course we're ignoring. We have been ignoring the y component beat because we don't have a wide component. But now, because this is a three dimensional graph, let's just add in the three the third axis, which is the y axis like this. And this is the positive side of the of the Y axis because, you know, right and rule. And now we're gonna look because this is the answer when you look for it. We're gonna look for I'm the one graph from the list of options that were given. You can see that the only option that matches our description is is is H because, first of all, it's still only half circle that we have out of all of these options, the only other circle is a full circle. We want 1/2 circle. So So this is all we need in order to know that the answer is the age. But I just want to say a little bit more. I want to say that that our graphics and look exactly like the graph given in part H. But that's okay, because, as you remember, you can, you know, randomly reorient your three dimensional axes. Your three axes could be randomly reoriented. There's nothing superior about the way that you oriented Drax is, or the way I orient my access. They're the same thing you could match. First of all, for example, how do you know that my graft is the same thing as the graph given in part H. You can say this. You can. You can say that that when Why is zero X has to be equal. You can say that that paragraph has to intercept with the X axis at X equals two and negative too. And that matches the graph given in part age. Because Partridge has a point where X is to y zero z zero and has a point at the polar opposite end of that. And and it has this point as well. So So the graph given given in part H has all of the graph that we need. So that's our answer. Even though they look, they look a little bit differently. They're the same thing. All right. I think that's it for this video. Thank you. Until next time. Bye bye.

University of California, Berkeley