1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
    Thanks.
     
    Last edited: Jan 21, 2010
  2. jcsd
  3. Jan 21, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Gold Member

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Gold Member

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

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

    Char. Limit

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What does it mean to find dy/dt in this question?(parametrics)
Loading...