Related Rates and particles

  • Thread starter bob1182006
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  • #1
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Homework Statement


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?

Homework Equations





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:

Answers and Replies

  • #2
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You're given the acceleration (3cm/s2) at the X-ordinates.

Maybe you should be intergrating to find the velocity.
 
  • #3
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I have a feeling that I have to use the equation of the curve but I'm not sure at all
The equation is given, and you havent employed it yet
 
  • #4
492
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Sorry I dont get what anyone is saying, so far the book hasn't taught integration so I dont think I should use it.

The equation is given, and you havent employed it yet
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:
  • #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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