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Limiting Value of a Vector Function

  1. Mar 17, 2016 #1
    1. The problem statement, all variables and given/known data
    The position vector of a particle is given in the terms of t, by,
    s = (e-t+3*cos(2t)i+2tj+(e-t+3*sin(2t)k)

    Find the limiting value of speed when t approaches positive infinity. The answer says "s = ..."

    3. The attempt at a solution
    I have evaluated the limits of all the components of this vector function individually:
    e-t = 0
    3*cos(2t) = -3 to 3
    2t = positive infinity
    3*sin(2t) = -3 to 3.

    I know that the magnitude of the velocity vector is equal to the speed of the particle, so I'm not sure if i should have done the limits on the velocity vector that i found by differentiating the position vector? but the answer is in the form of "s = ..."

    Any help would be appreciated, Thanks!
     
  2. jcsd
  3. Mar 17, 2016 #2

    blue_leaf77

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    That can be a typo because if the magnitude of the particle's position vector has a limit then it means the particle is gradually coming to a stop (which is clearly not judging from ##\mathbf{s}(t)##). You should instead calculate the magnitude of the velocity vector.
     
  4. Mar 17, 2016 #3
    This is the view of the question that I've got:
    157fbjc.png

    I'm not sure how I would go about calculating the magnitude of the velocity vector, given all I have is t approaching infinity. Given that this is the form of the velocity vector function:

    v = ds/dt = (-e-t-6*sin(2t)i+2j+(-e-t+6*cos(2t)k)

    If I were to do the limits of each component and then do the magnitude of that it'd make sense, but the limits of the cosine and sine fluctuate :eek::eek:
     
  5. Mar 17, 2016 #4

    Dick

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    The velocity vector does not have a limit, as you say. The SPEED (which is NOT a vector) does. Write down an expression for the speed and find it's limit.
     
  6. Mar 17, 2016 #5

    blue_leaf77

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    Like Dick says, first calculate ##|\mathbf{v}|##, then find its limit as ##t## approaches infinity.
     
  7. Mar 17, 2016 #6
    Here is my attempt at interpreting what you guys have told me thus far:

    s = (e-t+3*cos(2t)i+2tj+(e-t+3*sin(2t)k)
    v = ds/dt = (-e-t-6*sin(2t)i+2j-(e-t+6*cos(2t)k)

    Speed = length/magnitude of velocity vector, v:

    |v| = √((-e-t-6*sin(2t))2+(2)2+(-e-t+6*cos(2t))2)

    lim t → ∞ √((-e-t-6*sin(2t))2+(2)2+(-e-t+6*cos(2t))2)

    -e^-t = 0
    -6*sin(2t) = -6 to 6
    2 = 2
    6*cos(2t) = -6 to 6

    I get to here but am unsure how to interpret the '-6 to 6' limits for the trig functions? I'm assuming if I can evaluate each part of the limit of |v| then I'll be able to get a scalar number from it.
     
  8. Mar 17, 2016 #7

    blue_leaf77

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    In order to compute the limit easier, simplify the above equation by expanding the squares and then employ a trigonometric identity.
     
  9. Mar 17, 2016 #8
    So by expanding the squares I have:

    |v| = (-e^-2t-36*sin^2(2t))+(2^2)+(-e^-2t+36*cos^2(2t))^(0.5)

    Limits for t approaching infinity:
    -e^-2t = 0
    -36*sin^2(2t) = 0 to 36
    2^2 = 4
    36*cos^2(2t) = 0 to 36

    Still not sure if this is the correct way to evaluate this?
     
  10. Mar 17, 2016 #9

    blue_leaf77

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    Not so correct, you missed some terms as well as problem with signs. Let me ask you, how do you expand ##(a+b)^2##?
     
  11. Mar 17, 2016 #10
    Ah! My mistake, trying to reply as I woke up lol.

    |v| = (-e^-t-6*sin(2t))^2+(2)^2+(-e^-t+6*cos(2t))^2)^(0.5)
    Doing the expansion
    (-e^-t-6*sin(2t))(-e^-t-6*sin(2t)+4+(-e^-t+6*cos(2t))(-e^-t+6*cos(2t))
    => (e^-2t+6*e^-t*sin(2t)+6*e^-t*sin(2t)+36*sin^2(2t)+4+(e^-2t-6*e^-t*cos(2t)-6*e^-t*cos(2t)+36*cos^2(2t))
    => e^-2t+12*e^-t*sin(2t)+36*sin^2(2t)+4+e^-2t-12*e^-t*cos(2t)+36*cos^2(2t))

    Would this be the correct way of expanding the equation? I'm not that familiar with trigonometric identities but I'm thinking that the terms cos^2 and sin^2 may have another form?
     
  12. Mar 17, 2016 #11

    blue_leaf77

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    Yes they do, take a look at this.
    Then you should familiarize yourself.
     
  13. Mar 17, 2016 #12
    e^-2t+12*e^-t*sin(2t)+36*sin^2(2t)+4+e^-2t-12*e^-t*cos(2t)+36*cos^2(2t))

    => 2e^-2t+12e^-t*sin(2t)+4-12*e^-t*cos(2t)+36*(sin^2(2t)+cos^2(2t))
    => 2e^-2t+12e^-t*(sin(2t)-cos(2t))+4+36

    Lim t->infinity of 2e^-2t+12e^-t*(sin(2t)-cos(2t))+4+36

    limit of any e^-2t function is 0
    Therefore,
    the limit of the expansion is equal to 4+36 = 40

    Hence, |v| = √(40).

    Does this seem like the correct usage of the trigonometric identities? I'm not sure if there would be any use (if it is possible) to simplify the sin(2t)-cos(2t) given that the limit of e^-t is zero anyway.
     
  14. Mar 17, 2016 #13

    Dick

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    Yes, that looks ok.
     
  15. Mar 18, 2016 #14

    blue_leaf77

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    No, that's fine. As you know, there is no point simplifying these terms as they go to zero.
     
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