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Eliminating Parameter for <-rsinθ,rcosθ> ≠ assumed function

  1. Jun 5, 2015 #1
    Hello, I am attempting to eliminate the parameter from r'(θ) = r'(θ)=<-10sinθ,10cosθ> , but when I do, I get back to the same equation as I would for eliminating the parameter for r(θ)=<10cosθ,10sinθ>

    (1) [itex]x^2 + y^2 = 10[/itex]

    [itex]y = \sqrt{10-x^2}[/itex]

    (2) [itex]y' = \frac{-x}{\sqrt{10-x^2}}[/itex]

    then,

    since [itex]x^2 + y^2 = 10[/itex], if we let

    [itex]x=\sqrt{10}cosθ[/itex]
    [itex]y=\sqrt{10}sinθ[/itex], then

    (3) r(θ)=<10cosθ,10sinθ>

    (4)r'(θ)=<-10sinθ,10cosθ>

    then how come if I let,

    [itex]x=-\sqrt{10}sinθ[/itex]
    [itex]y=\sqrt{10}cosθ[/itex],

    I get

    [itex] \frac{-x}{\sqrt10}=sinθ[/itex]
    [itex] \frac{y}{\sqrt10}=cosθ[/itex]

    [itex] (\frac{-x}{\sqrt{10}})^2 + (\frac{y}{\sqrt{10}})^2= 1 [/itex]

    [itex]x^2 + y^2 = 10[/itex]

    But, my question is:

    Why does eliminating the parameter in

    r'(θ)=<-10sinθ,10cosθ> ⇒ [itex]x^2 + y^2 = 10[/itex]

    instead of what seems more intuitive:

    r'(θ)=<-10sinθ,10cosθ> ⇒[itex]y' = \frac{-x}{\sqrt{10-x^2}}[/itex]


    It just feel like it would make more sense for a y(x) ⇒ r(θ) and y'(x) ⇒ r'(θ)

    So I wonder why y(x) ⇒ r(θ) and y'(x) ⇒ r(θ)

    In words, what I'm asking, is when I introduce the theta parameter for the semicircle (1), I get the paremetric/vector function (3). Then if I take the derivative of the semicircle I get (2), and when I introduce the theta parameter, I get (4). But if I eliminate the parameter in (3), I go back to (1), but when I eliminate the parameter of (4), I don't go back to (2), but instead go back to (1) as well.

    Why does this occur?
     
  2. jcsd
  3. Jun 5, 2015 #2

    Mark44

    Staff: Mentor

    Because r'(θ) is perpendicular to r(θ), and both vectors have the same magnitudes. This means that r'(θ) is ##\pi/2## "ahead" of r(θ), but is tracing out the same path.
    Not to me it doesn't. The two derivatives are different rates of change. y' measures how y changes relative to changes in x (which runs horizontally), while r' measures how r changes relative to changes in θ. To me this is apples and oranges.
     
  4. Jun 5, 2015 #3
    Thank you, this makes sense to me. I just came across the property r(θ)⋅r'(θ) = 0 if the magnitudes are equal in chapter 13 of my calculus book last week. I don't fully understand the proof just yet. I will work through it though. May I ask, are r(θ) perpendicular r'(θ) because they have equal magnitudes or do they have equal magnitudes because they are perpendicular. This is probably a chicken and the egg question that doesn't even makes sense. If the question is not even valid then don't worry about answering it.

    Also, would this property only hold true for circles?
     
  5. Jun 5, 2015 #4

    Mark44

    Staff: Mentor

    They are perpendicular because the angle between them is 90°; that is, one of them is the rotation of the other, with no other change, which makes their magnitudes equal.
     
  6. Jun 7, 2015 #5
    Mark44, I will have to ponder this a bit more. I will return to this thread soon.
     
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