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Homework Help: What does it mean to find dy/dt in this question?(parametrics)

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle moves in the xy-plane so that at any time t ³ 0 its position (x,y) is given by
    x = e^t + e^-t and y = e^t - e^-t .
    (a) Find the velocity vector for any t ³ 0.

    I know how to do part a it is simply (derivative of x, derivative of y)

    (b) Find lim (dy/dt)/(dx/dt)
    (t approaching infinity)

    What does this (dy/dt)/(dx/dt) tell us?
    I know dy/dt is going to be y component of the velocity and dx/dt is the x component of the velocity. Yet, simply dividing y component of the velocity at certain t by x component of the velocity at certain t ... what does this mean?

    So, i just attempted to find the limit of (dy/dt)/(dx/dt) as t approaches infinity without actual understanding.

    (e^t - e^-t) /(e^t + e^-t) ;the first one, the derivative of y, the second, the derivative of x
    In this i run into the problem of the function being undefined. I fiddled with it a little but could not find a way to avoid the problem

    Help me deal with my stupidity,

    (c) The particle moves on a hyperbola. Find an equation for this hyperbola in
    terms of x and y.

    i do not know how to isolate t in either function x or function y (in order to subsitute that for t in the other function )

    (d) On the axes provided, sketch the path of the particle showing the velocity vector

    2. Relevant equations

    3. The attempt at a solution

    I need your help.
    Last edited: Jan 21, 2010
  2. jcsd
  3. Jan 21, 2010 #2


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    Hint: Think about the magnitude and direction representation of a vector.

    Try dividing both the top and bottom by [itex]e^t[/itex].

    Hint: Try squaring both x and y.
  4. Jan 22, 2010 #3

    Char. Limit

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    Another hint for then first part... Try writing it by multiplying a reciprocal.

    Also, I don't see why your function could ever be undefined. Want to know why? Well, graph the denominator and take a look at the range.
  5. Jan 22, 2010 #4
    so for part (b), I can find the limit.
    Yet, i still have problem understanding the concept.
    dy /dx is going to be relation between dy and dx
    Its not like it is going to give you the magnitude of the velocity
    Just for the sake of the argument, let's say the velocity, at t =3, is (4,5) in some other function.
    Then, the magnitude of the velocity is square root of 4^2 + 5^2
    but if you do dy/dx, it is simply the slope, 5/4.
    So, in practicality, is there any point doing dy/dx?
  6. Jan 22, 2010 #5


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    I was trying to get you to identify what the quantity [itex]v_y/v_x[/itex] corresponds to.
  7. Jan 22, 2010 #6

    Maybe im just missing something, but for part (b) my (dy/dt)/(dx/dt) is (e^t+e^-t)/(e^t-e^-t))
    at t=0, the denominator gets 0
    That is why we are doing manipulation of that to avoid it being undefined.
  8. Jan 22, 2010 #7
    SO, is it just the slope ? y component of the velocity over x component of the velocity?
    is that all?

    oh that would be the acceleration at that particular t
  9. Jan 22, 2010 #8

    Char. Limit

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    When I looked at your problem in the first post, the denominator was [tex]e^t+e^{-t}[/tex]. I looked up there again and it still is. At t=0, that denominator equals 2.
  10. Jan 22, 2010 #9
    Im terribly sorry, I typed the (dy/dt)/(dx/dt) wrong
    i turned it upside down.
    Actual (dy/dt)/(dx/dt) should be (e^t + e^-t)/(e^t - e^-t)
    now the denominator = 0 at t=0

    I apologize..
  11. Jan 22, 2010 #10


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    Is the problem asking about what the limit of (dy/dt)/(dx/dt) tells you?
  12. Jan 22, 2010 #11
    yes as
    lim ((dy/dt)/(dx/dt)
    as t approaches infinity.
  13. Jan 22, 2010 #12


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    No. For one thing, [itex]v_y/v_x[/itex] is unitless, unlike acceleration.
  14. Jan 22, 2010 #13
    (dy/dt)/(dx/dt) is the slope of (x component of velocity, y component of the velocity) at that specific t

    in velocity function, if you find a tangent line at one particular t, the slope of the tanget line is the acceleration. Isn't it?

    I don't see how (dy/dt)/(dx/dt) is different from that slopeof the tangent line in this case

    (i know the second derivative is usually acceleration in s(t). Maybe its parametrics that confuses me)

    (yeah, it is unitless, so it cannt be the acceleration at that t.
    in v(t) the unit of a slope is m/s^2)
    But im still confused. What about my logic here?
    Could you please disprove my idea. I think im wrong but i just cannot see it
  15. Jan 22, 2010 #14


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    Only on an a vs. t graph.
  16. Jan 22, 2010 #15

    Char. Limit

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    Well, for one thing, your x and y equations describe position, not velocity. Also, why not just factor out an e^t from top and bottom... or better yet, have you yet learned partial fraction expansion?

    Or would that even work here...
  17. Jan 22, 2010 #16
    right. Thanks for having me understand that part.
    Wow... that was dumb. it was possible to get acceleration by getting slope of tangent line in v(t) because the y axis is the total velocity (not being divided into x and y components) and x axis is the time.
    Also sqrt((dx/dt)^2 + (dy/dt)^2) will be the speed at a specific t.
    Then, now im just back to square one.
    what is the use of (dy/dt)/(dx/dt)
    and i still have problem of getting 0 for the denominator. I did all manipulation i could; yet nothing worked.
  18. Jan 22, 2010 #17

    Char. Limit

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    Well, without specifying specific functions for dy/dt and dx/dt, why not think about this:

    Division is multiplication by a reciprocal. For example, 4/2=4*2^-1=4*(1/2). Now try that same process with dx/dt.
  19. Jan 22, 2010 #18
    what i did was (i learned partial fraction but i don't know if it what i am doing is the method)

    (e^t + e^-t)/(e^t - e^-t) = 1 + (2e^-t)/(e^t-e^-t) = 1+ ((e^t)*(2e^-2t))/((e^t)*(1-e^-2t)) = 1+ (2e^-2t)/(1-e^-2t)

  20. Jan 22, 2010 #19

    so you mean like (dy/dt)/(dx/dt) = (dy/dt) * (dt/dx) ???
    and what do yu mean by specifying functions for dy/dt ??
    are you referring to how i substituted the derivatives of x(t) and y(t) for (dx/dt) and (dy/dt) ???
  21. Jan 22, 2010 #20

    Char. Limit

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    Yep. So now you have it written as (dy/dt)(dt/dx). Now let's distribute, and we get (dy*dt)/(dt*dx).

    See anything that can cancel out?

    Now, about the 2/0 thing, that just I suppose means this isn't defined at t=0. However, try factoring out and canceling e^t from top and bottom, if you can, and I am not sure if you can.

    Also, as a side note, try looking up the hyperbolic sine and cosine functions...
  22. Jan 22, 2010 #21
    so from (dy*dt)/(dt*dx), i end up getting dy/dx
    but i dont have anything to substitute for dy/dx because x and y are given in parametrics.
    if we can isolate t from x(t) and substitute that for t in y(t).. That would be wonderful.
    But, i do not see how i can do that.
    In Eastern time, its 2:23 and i ve gotta go to school tomorrow
    Thanks for your help.
    I will think about this.
    If you can, i hope you can help me again with this question tomorrow.
  23. Jan 22, 2010 #22

    Char. Limit

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    I can try, considering that I'm in Pacific Time...

    What this tells you is that the derivative of your y(t) equation over the derivative of your x(t) equation gives you an equation v(x)=s'(x)... and I think looking at the hyperbolic functions will help as well.
  24. Jan 22, 2010 #23
    I'm in a hurry to go somewhere, so I apologize if the answers are repeated in some cases:

    The vector representation of the position of this particle would look good enough to say everything about how to determine velocity so I asuume you've done this before.

    dy/dt, if is put into simple words, is the y-component of the velocity vector you've found from the first question. And dx/dt is the horizontal or x-component of velocity. So we have

    [tex](dy/dt)/(dx/dt)=\frac{e^t+e^{-t}}{ e^t - e^{-t}}[/tex].

    Just factor out e^t from both denominator and numerator which yields

    [tex](dy/dt)/(dx/dt)=\frac{1+e^{-2t}}{ 1 - e^{-2t}}[/tex]. Now take the limit of this function as t-->oo.

    Yes it does move on a hyperbola. This is so easy and need not be dragged to those boring calculations of extracting the common parameter of y and x. Just observe two functions:

    [tex]\frac{x}{2}= \frac{e^t + e^{-t}}{2}\equiv cosh(t)[/tex],
    [tex]\frac{y}{2}= \frac{e^t - e^{-t}}{2}\equiv sinh(t)[/tex].

    These two hold in the following very simple and fundamental relation:


    So substitute x and y into this relation from the pair of equations we provided above.

    And this is nothing catchy for us but you only.

    Last edited: Jan 22, 2010
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