## Wislar 3 years ago Find the image of the set S under the given transformation: u=0, u=1, v=0, v=1, x=v, y=u(1+v^2)

1. wio

plug in $$x(0,0), x(1,1), y(0,0), y(1,1)$$

2. Wislar

Yes, that's the function. That's it?

3. vf321

No, that's the function. The image is the range of values you get from the "domain" S.

4. wio

Seems to me that $x(0,0) \leq x \leq x(1,1)$

5. Wislar

Alright, I have (0,0)->(0,0) (1,0)->(1,1) (0,1)->(0,0) (1,1)->(1,2) Is that it?

6. wio

Look at the highest and lowest of x and y.

7. vf321

@Wislar are you implying that $$S = \{0, 1\} \times \{0, 1\}$$? From your question, it seems to just be $$S = \{ (0,0), (1,1)\}$$. Please clarify.

8. Wislar

The points on the first image are (0,0), (1,0), (1,1), (0,1) which is a square and plugging those into the equations for x=v and y=u(1+v^2) to get those points

9. vf321

Then your image is the set of all the transformed points.

10. Wislar

11. Wislar

it**

12. vf321

Well, I'm not going to check your arithmetic, but it should be the four points on the right side of your arrows in your above comment. Also make sure that $$S$$ only relates to those four points, and is NOT defined by $$S = \{(u, v) | u \in [0,1] \land v \in [0,1]\}$$

13. Wislar

That's what I did. Thanks!