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anonymous

  • one year ago

x^2/4 + (y+2)^2 = 1 The task is; "Parametrize the curve" What does it mean? What is expected? I have tried looking for several examples, but none are similar. Both got two functions (x and y) and such. Is the task to find the two functions (x and y)? How is this solved (or rather, how is this expected to be solved)?

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  1. inkyvoyd
    • one year ago
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    hmm

  2. inkyvoyd
    • one year ago
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    a parametrization is accomplished with the introduction of a third variable. Say, t

  3. inkyvoyd
    • one year ago
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    the equation you've provided is one of a ellipse centered at (0,-2).

  4. inkyvoyd
    • one year ago
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    so, we can actually define t such that we get the same equation of the ellipse as we sweep t across values. to give you an idea, a parametrization of a circle x^2+y^2=25 would be x=5 sin t and y = 5 cos t. now, because x^2+y^2=5^2* sin^2t+5^2*cos^2t=25(1)=25, you know you ahve the equation of a circle in terms of a parameter t.

  5. inkyvoyd
    • one year ago
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    now, if you are familiar with vector valued functions you can actually express this parametrization as a single vector valued function. In this example that would be r(t)=<5sin(t),5cos(t)> note that r would normally in bold to represent that it is a vector. Another thing to notice is that parametrizations are not unique. q(s)=<5cos(s),5sin(s)> comes out to the same thing... it also draws a circle, though it starts at a different point when s=0 versus when t=0

  6. inkyvoyd
    • one year ago
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    Using my example with the circle, do you have any ideas as to how you would parametrize the equation of your ellipse?

  7. anonymous
    • one year ago
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    Not yet. Reviewing old material

  8. anonymous
    • one year ago
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    Slowly getting there.... !

  9. inkyvoyd
    • one year ago
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    May I ask, what is the class youa re in and is there a chapter name for this topic? there are many ways to approach this topic and I am not sure how your material expects you to.

  10. anonymous
    • one year ago
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    "Parametric Curves"

  11. inkyvoyd
    • one year ago
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    is this under calculus 2?

  12. anonymous
    • one year ago
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    I dont know what number; but it is Calculus

  13. inkyvoyd
    • one year ago
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    Ok. let's work through this. With the equation x^2/4 + (y+2)^2 = 1 you can tweak my example slightly to get the same thing why don't we rewrite the equation as the sum of squares to apply the identity sin^2 x+cos^2 x in our parametrization.

  14. anonymous
    • one year ago
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    Wondering if there is a formula for conversion from x^2/9 to 3cos t etc

  15. inkyvoyd
    • one year ago
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    there probably is, but I try to do so algebraically - generally speaking there are so many formulas in calculus that it does not make sense to memorize every one unless you will use it constantly... derivations are more important.

  16. inkyvoyd
    • one year ago
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    anyways, the first term is not in perfect square form... can you rewrite it in the form (blah)^2?

  17. anonymous
    • one year ago
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    first step for me is to find out "how" to convert it. Second step is to understand how (in order for me to learn). Everyone learns differently - I learn by examples and step by step solutions

  18. anonymous
    • one year ago
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    calculus 2 hard 5 me

  19. inkyvoyd
    • one year ago
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    Yes. I am working with you right now on this. there are a few more steps after rewriting the first term as a perfect square. As I have explained before, you'll want to write your equation in a more familiar form if you want any luck with it. I would agree that everyone learns differently, but I would also like to add that there are two ways to get through calculus, and math in general. One can struggle repeatedly with a problem and figure it out with just the principles involved, or, one can look in the book and follow the same exact problem with different numbers and go through it step by step. The former is mathematics. The second is following rules. Most people can get by with the second, but a word of warning: typically tests go after the former.

  20. anonymous
    • one year ago
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    I am somehow able to get A's and B's with this method - dont ask me how... xP

  21. inkyvoyd
    • one year ago
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    If that is all you want, I can give you the answer and let you be done with it. I hope for your sake you learn before upper level university classes.

  22. anonymous
    • one year ago
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    Actually, I dont need an answer - I just need to understand what is being asked by the question "Parametrize the curve" ... ! I do not understand the question. Is it asking me to rewrite it in order to get a single function of t?

  23. inkyvoyd
    • one year ago
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    the other problem with doing textbook examples step by step is that you forget them. if you have to remember EVERY single problem type, it would be anightmare. But if you understood the fundamentals deeply, it might take you a bit more time, but you would reach the conclusions. Now I am not saying it is practical to derive everything on spot. But I think this problem is doable with a bit of insight and a bit of thinking.

  24. anonymous
    • one year ago
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    Appreciate the help btw

  25. inkyvoyd
    • one year ago
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    Well, I tried to clarify that by making another example and solving it step by step

  26. inkyvoyd
    • one year ago
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    you can check the wiki article if you want an idea of what you are doing. https://en.wikipedia.org/wiki/Parametric_equation

  27. anonymous
    • one year ago
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    ... ! x = 2cost y = sint - 2 t€(0, 2pi) ?

  28. anonymous
    • one year ago
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    I think I am getting closer... !

  29. IrishBoy123
    • one year ago
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    " Is it asking me to rewrite it in order to get a single function of t?" yes, sort of but you will get x = x(t) and y = y(t)

  30. inkyvoyd
    • one year ago
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    I think you have it.

  31. anonymous
    • one year ago
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    Ellipse: \[\frac{ (x-x_0)^2 }{ a^2 } + \frac{ (y-y_0)^2 }{ a^2 } = 1\] Parametrized form(?): \[x = a*cost + x_0\]\[y = b*sint+ y_0\]

  32. inkyvoyd
    • one year ago
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    yes. the second a^2 should be a b^2 but otherwise yes. Can you visualize it?

  33. anonymous
    • one year ago
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    There is no edit button! x.x

  34. inkyvoyd
    • one year ago
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    Well, given that you've deduced what I think is the correct answer, I'm going to head off for the night. Feel free to let me know if you have any additional questions.

  35. anonymous
    • one year ago
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    I am wondering if it would be correct that \[f(t) = 2cost + sint - 2\]

  36. inkyvoyd
    • one year ago
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    no.

  37. anonymous
    • one year ago
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    Damn - Wondering if it is possible to create a single function of t

  38. inkyvoyd
    • one year ago
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    yes. but you need knowledge of vector valued functions.

  39. inkyvoyd
    • one year ago
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    it would be r(t)=<2cos t,sin t - 2>

  40. inkyvoyd
    • one year ago
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    note that you are actually returning multiple scalars (a vector) for a single scalar input. This isn't a function in the form taht you are used to seeing, because it maps R to R^2 rather than R to R

  41. inkyvoyd
    • one year ago
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    https://en.wikipedia.org/wiki/Vector-valued_function you can extend it to three dimensions. if we took r(t)=<2cos t,sin t - 2,t> we would have the equation of a elliptical helix.

  42. inkyvoyd
    • one year ago
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    anyways, I am off for the night. If you are curious, you can check paul's online math notes http://tutorial.math.lamar.edu/ or khan academy https://www.khanacademy.org/ for some particularly good tutorials on vector valued functions and the general geometry of space.

  43. anonymous
    • one year ago
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    Thanks - good night

  44. anonymous
    • one year ago
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    For reference; this is each and every step I took in order to get the x(t) and y(t) (which, I assume; is parametrizing the curve - PLEASE CORRECT ME IF I AM WRONG, OR MISSING INFORMATION). 1. Looking at trigometric formulas; I can see that \[\cos^2x + \sin^2x = 1\] 2. I set \[\cos^2x = x^2/4\], set root on both sides and get; \[cosx = x/2\] Then I put x on one side and get \[x(t) = 2cost\] 3. I set the other; \[\sin^2x = (y+2)^2\], set root on both sides and get; \[sinx = y+2\] Then I put y on one side and get \[y(t) = sinx - 2\] y = y(t) and x = x(t) Crossing fingers that this is all that was asked by the task!

  45. IrishBoy123
    • one year ago
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    for \(\cos^2x + \sin^2x = 1\) write \(\cos^2t + \sin^2t = 1\) and the same applies all the way down from these, ie introduced your new parameter, t so for \(\cos^2x = x^2/4\) write \(\cos^2t = x^2/4\)

  46. anonymous
    • one year ago
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    My bad, wrote X because it was easier than alpha, and I forgot x was already in use!

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