Related Rates and particles

[bob1182006]

Homework Statement

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?

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 $z^2=x^2+y^2$, 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 haven't employed it yet

 

[bob1182006]

Sorry I don't get what anyone is saying, so far the book hasn't taught integration so I don't think I should use it.

The equation is given, and you haven't 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 I am 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,
$$\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}= (1/2)x^{-1/2}\frac{dx}{dt}$$.