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Math Help - trig function - proof

  1. #1
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    Cool trig function - proof

    i have the equation

    r(t) = sin(2t)i + 2cos(t)j where 0<t<2pi

    I have to prove that using this equation that the object it describes passes through the origin twice in the time limit. Does this just mean I have to show the values of t that give r(t) = 0? i.e. pi/2 and 3pi/2?

    cheers for the help
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  2. #2
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    Quote Originally Posted by thermalwarrior View Post
    i have the equation

    r(t) = sin(2t)i + 2cos(t)j where 0<t<2pi

    I have to prove that using this equation that the object it describes passes through the origin twice in the time limit. Does this just mean I have to show the values of t that give r(t) = 0? i.e. pi/2 and 3pi/2?

    cheers for the help
    Yes.

    Having said that, I hope you got those values in the following way:

    sin(2t) = 0 .... (1)
    2 cos(t) = 0 .... (2)

    Solve (1) and (2) simultaneously over the domain 0 < t < 2 pi: t = pi/2, 3pi/2.
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  3. #3
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    i pretty much used that way of doing it but question is only 2 marks so i imagine very little detail is needed. Thanks for the post!!!
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    Question

    a further part of this question is asking when are the times that the displacement is perpendicular to the velocity. Clearly one time would be t=0 as the i and j directions are mutually perpendicular to each other. How would i find the others? Is there some way involving dot products?
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    Quote Originally Posted by thermalwarrior View Post
    a further part of this question is asking when are the times that the displacement is perpendicular to the velocity. Clearly one time would be t=0 as the i and j directions are mutually perpendicular to each other. How would i find the others? Is there some way involving dot products?
    Solve \frac{d\vec{r}}{dt} \cdot \vec{r} = 0 (over the domain 0 < t < 2 pi ?)

    This will boil down to solving 2 \sin (2t) (\cos (2t) - 1) = 0 (over the domain 0 < t < 2 pi ?).....

    Note: Some of the resulting solutions might not be wanted - why?
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  6. #6
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    how did you reach that final equation to solve?? have you just multiplied the velocity with the displacement???
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  7. #7
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    Quote Originally Posted by thermalwarrior View Post
    how did you reach that final equation to solve?? have you just multiplied the velocity with the displacement???
    I took the dot product.

    If you're stuck, show the working you've got and I'll get you over the hump.
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  8. #8
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    Question

    okay,

    r(t) = sin(2t) + 2cos(t)

    r(t)' = 2cos(2t) - 2sin(t)

    r(t)' * r(t) =

    2sin cos(2t)^2 - 4 sin cos(t)^2

    this is clearly the bit thats wrong but not sure why.....
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  9. #9
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    Quote Originally Posted by thermalwarrior View Post
    okay,

    r(t) = sin(2t) i + 2cos(t) j

    r(t)' = 2cos(2t) i - 2sin(t) j

    r(t)' * r(t) =

    2sin cos(2t)^2 - 4 sin cos(t)^2 Mr F says: Here is the error.

    this is clearly the bit thats wrong but not sure why.....
    Note my edits in red ....

    When you take the dot product:

    sin(2t) [2 cos(2t)] - [2 cos t] [2 sin (t)] = 0

    and you know that 2 sin(t) cos(t) is sin(2t), right?

    Have you learned about the dot product (also called the scalar product)? Do you understand that it's very different to multiplying two scalars ......?
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  10. #10
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    im okay with trig identities but the dot product hasnt really been explained very well to me.
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  11. #11
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    Quote Originally Posted by thermalwarrior View Post
    im okay with trig identities but the dot product hasnt really been explained very well to me.
    Your survival guide:

    i.i = j.j = k.k = 1.

    i.j = i.k = j.k = 0.
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  12. #12
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    ive got that bit written down but not sure how to interpret in relation to this quesiton.

    with regards dot product

    i now have that in the case of these equations

    a*b=a1b1 + a2b2

    taking a to be the velocity components and b to be displacement components

    which leads to 2cos(2t)*sin(2t) and -2sin(t)*2cps(t)


    is this correct??
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  13. #13
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    if i follow this through I get

    2cos(2t)sin(2t) - 4sin(t)cos(t)

    cos(2t)sin(2t) - 2sin(t)cos(t)

    cos(2t)sin(2t) - sin(t)

    any better???
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  14. #14
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    Quote Originally Posted by thermalwarrior View Post
    if i follow this through I get

    2cos(2t)sin(2t) - 4sin(t)cos(t) = 0

    cos(2t)sin(2t) - 2sin(t)cos(t) = 0

    cos(2t)sin(2t) - sin(2t) = 0

    any better???
    Note the red edits (especially in the last line).
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  15. #15
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    okay so now i understand how you got to the final part apart from one thing - where does the -1 come from???

    thanks
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