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# Match the vector equations with the graphs (a)-(h) given here. (GRAPH CANT COPY)$$\mathbf{r}(t)=t \mathbf{j}+(2-2 t) \mathbf{k}, \quad 0 \leq t \leq 1$$

Integrals

Vectors

Vector Functions

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### Video Transcript

Okay, folks. So in this video, we're gonna take a look at this factory equation, and we want to match this equation with one of the graphs. Okay, So I'm direct down this vector equation for you we have are as a function of t is given by t is given by first of all, I'm gonna rewrite this vector equation in a slightly different form. I'm gonna write as as as a vector. Okay, so we have zero for the x component, and we have tea for the white component, and we have tu minus two t through the Z component. Okay, So I'm just I'm just rewriting it in the in an actual vector form, okay? And were restricted with with the following restriction which says that zero has to be less. There you go to tea, and unless than you're gonna want. Okay, so we have this restriction here, So, first of all, I want you to notice it's the fact What? I want you what I want you to know this is the fact that we don't have an X component at all times. So whatever our final graphs and it looks like a better not have the next component that our graph better lies. It better lie in the wide Z component. I mean, I'm sorry. Let me rephrase that Our graph better lie in the Y Z plane. So if we have this three dimensional space here we have This is the X component. That is why component and this is the Z component paragraph. Can't look something like this. It cannot be. It cannot have a non zero x component and cannot be like this because at this point, obviously has a non zero x value and say so. So does this point. So what I'm saying is, it's not gonna have It's not gonna have an x component. So that's one thing to notice. Another thing to notice is that there is a simple relationship between y and Z because why is T and Z is two minus two t so there's a relationship there, and we could write that relationship as Z equals two minus two t. But T is why? So I'm gonna write that as two minus two. Why? So this you probably can recognize as just a linear function, So this is a linear function, and if you if you graph it out. And if you graph the horizontal axes as the Y axis and Z as the as the and the vertical axis at the Z axis, then this is a very simple graph to draw, because all you need to do is to recognize the Z intercept as to and why intercept as as one, you just connect them. This is the This is the two dimensional representation of our vector valued function. But now we want to upgrade this two dimensional Rafto a three dimensional graph. So how do we do that? It's very simple to do because all we need to do is go back to this wrath and to say that this is too. And this is one and we connect them. Oh, and missed it. But that's okay. I'm going to say that this is to me instead. So we have our three dimensional graph, and it looks like this. And if you go, if you look through all of the options from age and match it with one, you'll see that there's only one option that matches the graph that we have drawn here. And that option is let me see what's going on here? I can't find it. Where is it? Um, yeah, So, Part B. So be is the only graph given to us that matches with our with our own graph. So the answer for this question is part B, and we are done. Thank you. Bye.

University of California, Berkeley

Integrals

Vectors

Vector Functions

Lectures

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