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Homework Help: Intuition Behind Particular Related Rates Question

  1. Mar 1, 2014 #1


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    1. The problem statement, all variables and given/known data
    A particle is moving along the graph of y=sqrt(x). At what point on the curve are the x-coordinate and the y-coordinate of the particle changing at the same rate?

    2. Relevant equations
    y = sqrt(x)
    y' = 1/(2sqrt(x))

    3. The attempt at a solution
    The solution to the problem is to differentiate both sides of the eqn with respect to time and then determine when dy/dt = dx/dt.

    But I don't understand how the x-coordinate has a rate of change since it's an independent variable? Can someone explain how to comprehensively understand the solution and I suppose, wording of this problem?

  2. jcsd
  3. Mar 1, 2014 #2
    Edit: As shown in later posts, this is wrong and I need sleep.

    I think we need to make the assumption that the particle is moving at a constant speed, and that the independent variable is time (not x).

    That changes the equation for y:

    y = sqrt(t)

    and gives us an equation for x:

    x = t^2

    I admit, though, the question is very vague and I can't be sure that this is what it means.
    Last edited: Mar 2, 2014
  4. Mar 1, 2014 #3


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    If x = t^2, then sqrt(x) = t, which simplifies the function to:

    y = sqrt(sqrt(x))

    Not sure I understand.
  5. Mar 1, 2014 #4
    You are thinking of y as a function of x. Think of them as two independent functions:

    x(t) = t^2

    y(t) = sqrt(t)

    This allows you to differentiate both x and y with respect to t, rather than with respect to one another. Once you have both derivatives, you can find where they have equal rates of change.

    Note that t^2 is just a reflection of sqrt(t). It would have been equivalent to say t = sqrt(x), which would make the x and y plots identical, but with different axis labels.

    Does that make sense? I'm not sure that I am explaining how I got the equation for x very well.
  6. Mar 1, 2014 #5


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    Use the chain rule. dy/dt=(dy/dx)*(dx/dt). If dy/dt=dx/dt then what is dy/dx?
  7. Mar 1, 2014 #6


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    Well, dy/dx must equal 1. So I guess we have to assume y and x are functions of another variable (time I suppose)? It's rather unintuitive to do so since x is generally the independent variable.
  8. Mar 1, 2014 #7
    I don't think we can assume constant speed -- I think all that's implied by the problem is that the independent variable is t, and then velocity is going to depend on dx/dt and dy/dt.

    There's an error in your substitution here. And I think you have the same error too Nick -- eliminating t would also give you sqrt(sqrt(x)) instead of just sqrt(x).

    For these parameterization problems, it's usually easiest to assign t to x, and then find what y as a function for t would be. For this problem, you would get:

    [itex]x = t[/itex] (by arbitrary assignment) and
    [itex]y = \sqrt(t)[/itex]

    Then you want to find when [itex]\frac{dx}{dt}[/itex] and [itex]\frac{dy}{dt}[/itex] are equal, and then what point that time corresponds to.

    Hope this was clear enough.
  9. Mar 1, 2014 #8
    Gah, you're right. In a post I made an hour or so ago I said I should get some sleep... I really should do that before trying to do any more math.
  10. Mar 1, 2014 #9


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    Yes, time is the independent variable but dy/dx must equal 1. So at what (x,y) point is dy/dx=1? That doesn't depend on the time.
  11. Mar 1, 2014 #10


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    Thanks jackarms, your explanation was great. If dy/dx = 1 then 1/(2sqrt(x)) = 1 and thus x = 1/4.
  12. Mar 2, 2014 #11

    Ray Vickson

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    To answer the question there is no need to assume a constant speed; it will work perfectly well if you assume the position is <x(t),y(t)>, with y(t) = sqrt(x(t)) and x(t), y(t) differentiable functions of t. That's all you need.
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