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Wislar
 2 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)
Wislar
 2 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)

This Question is Closed

wio
 2 years ago
Best ResponseYou've already chosen the best response.1plug in \(x(0,0), x(1,1), y(0,0), y(1,1)\)

Wislar
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, that's the function. That's it?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1No, that's the function. The image is the range of values you get from the "domain" S.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Seems to me that \[ x(0,0) \leq x \leq x(1,1) \]

Wislar
 2 years ago
Best ResponseYou've already chosen the best response.0Alright, I have (0,0)>(0,0) (1,0)>(1,1) (0,1)>(0,0) (1,1)>(1,2) Is that it?

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Look at the highest and lowest of x and y.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1@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.

Wislar
 2 years ago
Best ResponseYou've already chosen the best response.0The 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

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Then your image is the set of all the transformed points.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Well, 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 \(S = \{(u, v)  u \in [0,1] \land v \in [0,1]\}\)

Wislar
 2 years ago
Best ResponseYou've already chosen the best response.0That's what I did. Thanks!
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