# Related Rates and particles

1. Mar 25, 2007

### 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

Last edited: Mar 25, 2007
2. Mar 25, 2007

### Nibbler

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

Maybe you should be intergrating to find the velocity.

3. Mar 25, 2007

### f(x)

The equation is given, and you havent employed it yet

4. Mar 25, 2007

### 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.

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

Last edited: Mar 25, 2007
5. Mar 25, 2007

### HallsofIvy

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