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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)

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  1. wio
    • 2 years ago
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    plug in \(x(0,0), x(1,1), y(0,0), y(1,1)\)

  2. Wislar
    • 2 years ago
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    Yes, that's the function. That's it?

  3. vf321
    • 2 years ago
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    No, that's the function. The image is the range of values you get from the "domain" S.

  4. wio
    • 2 years ago
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    Seems to me that \[ x(0,0) \leq x \leq x(1,1) \]

  5. Wislar
    • 2 years ago
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    Alright, I have (0,0)->(0,0) (1,0)->(1,1) (0,1)->(0,0) (1,1)->(1,2) Is that it?

  6. wio
    • 2 years ago
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    Look at the highest and lowest of x and y.

  7. vf321
    • 2 years ago
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    @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
    • 2 years ago
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    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
    • 2 years ago
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    Then your image is the set of all the transformed points.

  10. Wislar
    • 2 years ago
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    So I had to correctly?

  11. Wislar
    • 2 years ago
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    it**

  12. vf321
    • 2 years ago
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    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
    • 2 years ago
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    That's what I did. Thanks!

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