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# Match the vector equations with the graphs (a)-(h) given here. (GRAPH CANT COPY)$$\mathbf{r}(t)=t \mathbf{i}+(1-t) \mathbf{j}, \quad 0 \leq t \leq 1$$

## Graph (c)

Integrals

Vectors

Vector Functions

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### Video Transcript

Okay, folks. So in this video, we're gonna talk about this problem where were given a vector equation, and we want to match it with one of the graphs that are given here, okay. And the graphs go from a to H. Okay, so we're going to write down this vector equation. So first of all we have are as a function of T. This is a vector function vector valued function parameter rised by the variable T. And the X component of this function is given by t and the y component of this. Some dysfunction is given by wanted minus t. And I just want you to know that the Z component is not mentioned. And when when the problem doesn't mention the Z component, that means zero. So the Z component of this of this vector value function is zero at least for the time interval of zero. Less than or equal to you less than or equal to one. Okay, so in this time interval, there's no Z component. What that means is that all other things aside, at least we know one thing that we that we definitely know is that this graph of this function is definitely going to lie in and the x Y axis. Okay, so that's what we know for sure. So So just by that, we can eliminate a lot of the choices. Um, so a lot of the graphs where the Z component is not zero is going to be eliminated. But now let's try and relate the X and y component of this some of this vector function. We know that the Y component or the J component is one minus t. So I'm gonna write Why equals one minus t. Um, and we know that the X component of this function is his team. So there is, As you can see, there's a little bit of a relationship between why Annex? And that relationship could be written as why equals one minus t. But T is X. So we have this. This is the relationship between why and x of this vector valued function so we could graph it out. I mean, I definitely hope that you know how to graft this simple, you know, line out. We have X axis year. And why access here? Then why does one minus X is just going to give you a line. Who's why? Intercept is one and X intercept is also one. Um, by because of the fact that we're only interested in the time interval between zero and one, we're gonna raise the interval. Um, outside of that, OK, we're only gonna keep you know, this portion of the graph so narrow. Now, the problem is like 90% done, because all we need to do now is to graph it out. So let me go ahead and grab you a three dimensional space. This ex Cornett, this Y cornet is the Z cup cornet, and we're gonna, you know, basically just match this thing when a map it to here. Ok, we're gonna upgrade our two dimensional graph to a three dimensional one, and that's very easy to do. Um, all we need to dio, for example, there's a couple of tricks here. We can describe Exeter's at first, which is located at one. And same as why, Right? So we have these x intercept and why intercepts and we connect them because it's the line, we can connect them and and that's it. This is the graph. So now we just we just look for look for the right graph from the list of options that's given to us, and we can see we can see that that the option C is clearly the same graph that that I have just drew right here. Okay, so we're gonna pick, um, the sea option. And that's it for this video. Thank you. Fine.

University of California, Berkeley

Integrals

Vectors

Vector Functions

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