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Math Help - what does it mean to find dy/dt in this question?(parametrics)

  1. #1
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    what does it mean to find dy/dt in this question?(parametrics)

    Original question:


    A particle moves in the xy-plane so that at any time t 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 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







    I need your help.
    Thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by hangainlover View Post
    Original question:


    A particle moves in the xy-plane so that at any time t 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 0.

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

    (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
    ....
    actually, \frac {dy/dt}{dx/dt} = \frac {dy}{dx} = \frac {e^t~{\color{red} +}~ e^{-t}}{e^t ~{\color {red}-}~ e^{-t}}

    i will tell you the answer, the limit is 1. how would you get there?

    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 )
    take the hint, it's a hyperbola. consider what x^2 - y^2 would be (it would actually be nicer to think of what \left( \frac x2 \right)^2 - \left( \frac y2 \right)^2 is. do those expressions look familiar? think hyperbolic functions)

    (d) On the axes provided, sketch the path of the particle showing the velocity vector
    after doing part (c), this should be ok
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  3. #3
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    how part b is applicable in reality

    Quote Originally Posted by Jhevon View Post
    ok

    actually, \frac {dy/dt}{dx/dt} = \frac {dy}{dx} = \frac {e^t~{\color{red} +}~ e^{-t}}{e^t ~{\color {red}-}~ e^{-t}}

    i will tell you the answer, the limit is 1. how would you get there?

    take the hint, it's a hyperbola. consider what x^2 - y^2 would be (it would actually be nicer to think of what \left( \frac x2 \right)^2 - \left( \frac y2 \right)^2 is. do those expressions look familiar? think hyperbolic functions)

    after doing part (c), this should be ok

    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?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by hangainlover View Post
    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?
    the magnitude of velocity vector is speed. dy/dx gives you the slope of the velocity vector. of course there are reasons for finding dy/dx. it is another way of thinking about direction
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  5. #5
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    acceleration at that t

    Quote Originally Posted by Jhevon View Post
    the magnitude of velocity vector is speed. dy/dx gives you the slope of the velocity vector. of course there are reasons for finding dy/dx. it is another way of thinking about direction

    Oh dy/dx is the acceleration at that specific t.
    Is this correct?
    I believe it is..
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by hangainlover View Post
    Oh dy/dx is the acceleration at that specific t.
    Is this correct?
    I believe it is..
    no. acceleration is the second derivative: \frac {d^2y}{dx^2}
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  7. #7
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    (dy/dt)/(dx/dt) is...

    Quote Originally Posted by Jhevon View Post
    no. acceleration is the second derivative: \frac {d^2y}{dx^2}

    (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)
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  8. #8
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    (dy/dt)/(dx/dt) being unitless seems problematic....

    Quote Originally Posted by hangainlover View Post
    (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)

    Unlike slope of a tangent line in v(t) function, where the unit of slope is m/s^2, (dy/dt)/(dx/dt) is unitless. what am i doing wrong. is (dy/dt)/(dx/dt) denoting something other?
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