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Cartesian to polar unit vectors + Linear Combination

  1. Mar 20, 2015 #1
    I've been trying to solve this question all day. If somebody could point me in the right direction I would really appreciate it!

    (ii) A particle’s motion is described by the following position vector r(t) = 4txˆ + (10t − t)ˆy Determine the polar coordinate unit vectors ˆr and ˆθ for r. [4]

    v(t) is given as: v(t) = 4xˆ + (10t − 1)ˆy
    (iv) Using the ˆr and ˆθ you found in (ii) above, write v(t) as a linear combination of rˆ and ˆθ. [4]

    (v) Differentiate the expression for r(t) you got in part (ii) (in terms of ˆr and ˆθ, and using the expressions ˙rˆ = ˙θ ˆθ , ˙ˆθ = − ˙θrˆ derived in the lectures, show that you obtain the same answer as in part (iv)

    I understand that for θ^ = -sinθ + cosθ and r^=sinθ + cos θ I'm just unsure how to do part (ii) without it being really messy e.g with cos(tan^-1(10t-1/4)) where tan^-1(10t-1/4)=θ especially knowing I have to find a linear combination afterwords.

    The attempt at a solution
    My current answer for part (ii):
    r^= cos(tan^-1(10t-1/4))x^ + sin(tan^-1(10t-1/4))y^
    θ^= -sin(tan^-1(10t-1/4))x^ + cos(tan^-1(10t-1/4))y^

    And for part (iv) am I am right in saying that I need to find, a and b scalars for the following:
    4 = a*cos(tan^-1(10t-1/4)) +b*-sin(tan^-1(10t-1/4))
    10t-1 = a*sin(tan^-1(10t-1/4))+b*cos(tan^-1(10t-1/4))

    I'm really not sure about this one. It appears like it's going to difficult to eliminate the t value.
    Anyway, thanks in advance, I'm going to keep working on it now.
     
  2. jcsd
  3. Mar 20, 2015 #2

    Svein

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    I am pretty sure there's at least one typo here. Why would anybody specify (10t - t)y?
     
  4. Mar 20, 2015 #3
    Yes that was a typo, sorry.
    But when I say y^ -'y' hat (Or ^y - the typo) I mean it as the component, and it's the same with the others, it was just a formatting error
     
  5. Mar 20, 2015 #4
    Hi. This all seems a bit confusing, partly because of notation and possible typos... (What's the correct r(t) you are given, by the way?)
    Now regardless of that, you seem to have taken a wrong track in the beginning. So to avoid the hats i'm going to define: r-hat = er ; θ-hat = eθ , and: x-hat = ex ; y-hat = ey

    You know that in polar coordinates (because you mention it at some point):
    er = cosθ ex + sinθ ey , eθ = –sinθ ex + cosθ ey , and this is always true.
    You also know that: cosθ = x/r = x⋅(x2+ y2)–1/2, and: sinθ = y/r = y⋅(x2+ y2)–1/2 ,
    And this is also always true.
    Now you are given an expression: r(t) = x(t) ex + y(t) ey,
    So what does all that make of er and eθ?
     
  6. Mar 21, 2015 #5

    vela

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    I think the typo referred to was writing the y-component as ##10t-t##. Why not simply write it as ##9t##? Plus it's inconsistent with what you later said was the y-component of the velocity, ##10t-1##. That's not the derivative of ##10t-t##.
     
  7. Mar 22, 2015 #6

    BvU

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    Chris, can you clear this up ?

    Either

    ##\vec r(t) = 4t\hat x + (10 t^2 - t )\hat y\ \ ##, but then ##\vec v(t) = 4\hat x + (20 t - 1)\hat y##,

    or

    ##\vec v(t) = 4\hat x + (10 t - 1)\hat y\ \ ## and then ##\vec r(t) = 4t\hat x + (5 t^2 - t )\hat y##.


    Unless, of course, there is more than one typo harassing us...
     
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