1. Feb 4, 2008

### AquaGlass

1. The problem statement, all variables and given/known data

Two cars start moving from teh same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing two hours later?"

2. Relevant equations

So far I have the equation..
(z^2) = (x^2) + (y^2)

I know dy/dt = 60 mi/h and dx/dt = 25 mi/h.

3. The attempt at a solution

I implicity differentiated the first equation and ended up with:
dz/dt = (50x +120y)*(1/2z)

Since I do not know what x or y is I am unable to find the rate the distance between the two cars is increasing.

Thank you so much for your help!

2. Feb 4, 2008

### rocomath

$$x^2+y^2=z^2$$

taking the derivative

$$x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}$$

You're told that one travels 60 miles per hour and the other, 25 miles per hour. How far would they each have gone after 2 hours?

3. Feb 4, 2008

### AquaGlass

do I multiply the whole equation by 2 then?

4. Feb 4, 2008

### rocomath

Not the equation. Read my last sentence!

5. Feb 4, 2008

### AquaGlass

After two hours, they would have gone 120 miles and 50 miles?

6. Feb 4, 2008

### rocomath

Correct, now plug those values in. And the rate doesn't change, just the distance.

7. Feb 4, 2008

### PFStudent

Hey,

Welcome to the forum AquaGlass.

Exactly. But how did you get that? Once you figure out that, you will see how to re-express $x$ and $y$.

Or in other words what are the following proportional to,

$${\frac{dx}{dt}} = {\frac{?}{?}}$$

$${\frac{dy}{dt}} = {\frac{?}{?}}$$

Best,

-PFStudent

8. Feb 4, 2008

### AquaGlass

well then you get Zdz/dt = some number, but what about the Z then.. aren't we trying to find dz/dt? how do you cancel out the z?

9. Feb 4, 2008

### rocomath

Use the Pythagorean theorem to find your missing length.

$$x^2+y^2=z^2$$

10. Feb 4, 2008

### AquaGlass

ohhh ok i see! thank you so much!!