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zonazoo Group Title

How do we graph this ellipse? 2x^2 + (y^2)/2 = 1

  • one year ago
  • one year ago

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  1. zepdrix Group Title
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    Let's get it into standard form.\[\large \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1\] Hmm, so the x has a 2 on it. Multiplying by 2 is the same as dividing by 1/2.\[\large 2x^2 \qquad \rightarrow \qquad \frac{x^2}{\left(\frac{1}{2}\right)}\]Then let's apply the square to both the top and bottom term.\[\large \frac{x^2}{\left(\frac{1}{2}\right)} \qquad \rightarrow \qquad \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2\]Now we can clearly see what our \(a\) value is.

    • one year ago
  2. zepdrix Group Title
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    Doing similar to the y gives us something like this I believe, \[\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=1\] And then to graph it.. ummm trying to remember. If x=0, y extends out to \(\pm \sqrt2\). if y=0, x extends out to \(\pm \dfrac{1}{\sqrt2}\)

    • one year ago
  3. zepdrix Group Title
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    |dw:1359930596041:dw|

    • one year ago
  4. zepdrix Group Title
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    Hmm lemme throw this equation into a grapher real quick. I wanna make sure I didn't miss anything.

    • one year ago
  5. zepdrix Group Title
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    Yah it looks good.

    • one year ago
  6. zonazoo Group Title
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    thank you, very well put explanation ... can you graph these with a ti-83 or anything like that? and if that equation was set equal to 2 instead of 1, you would just set each variable = to zero and then solve for your new x and y

    • one year ago
  7. zepdrix Group Title
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    \[\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=2\]Divide both sides by 2,\[\large \left(\frac{x}{\frac{\sqrt2}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2\cdot \sqrt2}\right)^2=1\]Which simplifies to,\[\large \left(\frac{x}{1}\right)^2+\left(\frac{y}{2}\right)^2=1\]These would be our new \(a\) and \(b\) values.

    • one year ago
  8. zepdrix Group Title
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    Calculator? :o hmm thinking.

    • one year ago
  9. zepdrix Group Title
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    If you wanted to put it into your calculator, you would probably want to avoid the fancy ellipse standard form. From the start of the problem, \[\large 2x^2 + (y^2)/2 = 1\]Solving for y gives us,\[\large y=\pm 2\sqrt{1-2x^2}\]And you'd probably have to put it in as 2 separate functions,\[\large y_1=2\sqrt{1-2x^2} \qquad \qquad y_2=-2\sqrt{1-2x^2}\]

    • one year ago
  10. zepdrix Group Title
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    There's probably another way to do it, but I'm not too familiar with calculators :\

    • one year ago
  11. zonazoo Group Title
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    so we always want it in the form of =1 then... the next problem i got was -x + ln y = C how would we graph that as C changes?

    • one year ago
  12. zepdrix Group Title
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    If we solved for y, we would see that this is an exponential function.\[\large \ln y=C+x\]Exponentiating both sides,\[\large e^{\ln y}=e^{C+x} \qquad \rightarrow \qquad y=e^Ce^x\] As C changes? :O hmm it looks like it's just changing the scaling on our function. It just grows a lot faster as C increases.

    • one year ago
  13. zepdrix Group Title
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    We can't graph an unknown constant :D I guess you could just graph a few different functions and compare. With different coeeficients in front, e^2, e^3 and so on.

    • one year ago
  14. zonazoo Group Title
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    these problems ask me to "graph several members of each family", and i assume they mean as C changes right

    • one year ago
  15. zonazoo Group Title
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    like just pick random C's and graph ??

    • one year ago
  16. zepdrix Group Title
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    When C=0, we have the graph \(y=e^x.\) When C=1, we have the graph \(y=e\cdot e^x\), so this one will grow about 3 times as fast as e^x. When C=2, we have, \(y=e^2\cdot e^x\), so this one will grow like 7 times faster than e^x. Yah just plug in random C's. Graph a few members of the family! :)

    • one year ago
  17. zepdrix Group Title
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    Oh it's not just growing faster, it's also moving the y-intercept :O I didn't think about that.

    • one year ago
  18. zepdrix Group Title
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    https://www.desmos.com/calculator/2saf2toes8

    • one year ago
  19. zonazoo Group Title
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    does the members also include the original equation do you think?

    • one year ago
  20. zepdrix Group Title
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    The equation we're graphing `is` the original equation, we've just moved some things around so it's easier to graph.

    • one year ago
  21. zepdrix Group Title
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    https://www.desmos.com/calculator/agvkcwjylj

    • one year ago
  22. zonazoo Group Title
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    For example... the first one... the problem asks us to find the orthogonal trajectories for the family of curves and graph several members of each family. x = Cy^4 was the original problem and after we find the orthogonal trajectories our equation becomes 2x^2 + y^2/2 = C so I am unsure if I should just graph random C's of the resulting equation or both.

    • one year ago
  23. zonazoo Group Title
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    or what "several members of each family" actually means

    • one year ago
  24. zepdrix Group Title
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    I'm not exactly sure about the orthogonal trajectories, But here are several members of each family graphed, The equations should be displayed on the left. https://www.desmos.com/calculator/mqoozlfnq2

    • one year ago
  25. zepdrix Group Title
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    I guessing that's probably what they want you to do, to graph a few of those.

    • one year ago
  26. zonazoo Group Title
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    That is a cool looking graphing calculator... but i see that both equations are orthogonal... so I guess we did the problem right... so I would just graph both then since they probably want to see them orthogonal

    • one year ago
  27. zepdrix Group Title
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    Oh I didn't notice that! Yah they intersect at right angles! Hah! That's cool :D

    • one year ago
  28. zonazoo Group Title
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    can anyone use that calculator... thats just on a website for everyone to use?

    • one year ago
  29. zepdrix Group Title
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    Yah I can't remember who showed it to me. Maybe Satellite or someone.. But I had to bookmark it immediately, very useful tool! :) Yah it's just a free resource I think.

    • one year ago
  30. zonazoo Group Title
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    well thank you again. for all your help.

    • one year ago
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