Related Rates

[bob1182006]

1. The problem statement, all variables and given/known data
A particle is moving along the curve y=\sqrt{x}. 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 z^2=x^2+y^2, after differentiating I arrive at
\frac{dz}{dt}=\frac{1}{z}(x\frac{dx}{dt}+\frac{dy} {dt})
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

[Nibbler]

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

Maybe you should be intergrating to find the velocity.

[f(x)]

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

[bob1182006]

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 y=\sqrt{x} with respect to t
but how would the chain rule work there? would it be like
\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}
right? so then I'd arrive to
\frac{dy}{dt}=\frac{1}{2\sqrt{x}}\frac{dx}{dt}
so I'd plug in x=4, dx/dt = 3 cm/s, and obtain dy/dt
ok i think Im getting this now thanks

[HallsofIvy]

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