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Related Rates and particles

  1. Mar 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A particle is moving along the curve [itex]y=\sqrt{x}[/itex]. As the particle passes through the point (4,2), its x-coordinate increases at a rate of 3 cm/s. How fast is the distance from the particle to the origin changing at this instant?

    2. Relevant equations



    3. The attempt at a solution
    I made a diagram of the curve, connected the point (4,2), x=4, and the origin by a right triangle with z being the hypotenuse, x = 4, and y = 2.

    so [itex]z^2=x^2+y^2[/itex], after differentiating I arrive at
    [tex]\frac{dz}{dt}=\frac{1}{z}(x\frac{dx}{dt}+\frac{dy}{dt})[/tex]
    i know dx/dt 3 cm/s, but I have no idea how to find dy/dt, I have a feeling that I have to use the equation of the curve but I'm not sure at all
     
    Last edited: Mar 25, 2007
  2. jcsd
  3. Mar 25, 2007 #2
    You're given the acceleration (3cm/s2) at the X-ordinates.

    Maybe you should be intergrating to find the velocity.
     
  4. Mar 25, 2007 #3
    The equation is given, and you havent employed it yet
     
  5. Mar 25, 2007 #4
    Sorry I dont get what anyone is saying, so far the book hasn't taught integration so I dont think I should use it.

    sry I was doing this in the middle of the night ><

    so dy/dt = derivative of [itex]y=\sqrt{x}[/itex] with respect to t
    but how would the chain rule work there? would it be like
    [tex]\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}[/tex]
    right? so then I'd arrive to
    [tex]\frac{dy}{dt}=\frac{1}{2\sqrt{x}}\frac{dx}{dt}[/tex]
    so I'd plug in x=4, dx/dt = 3 cm/s, and obtain dy/dt
    ok i think Im getting this now thanks
     
    Last edited: Mar 25, 2007
  6. Mar 25, 2007 #5

    HallsofIvy

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    Yes, exactly! You were told that the particle is moving along the curve [itex]y= \sqrt{x}= x^{1/2}[/itex] so, using the chain rule,
    [tex]\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}= (1/2)x^{-1/2}\frac{dx}{dt}[/tex].
     
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