# Homework Help: PLS HELP: Application of Derivatives

1. Jun 25, 2007

### louloulou

1. The problem statement, all variables and given/known data
An elevated train on a track 30.0 m above the ground crosses a street (which is at right angles to the track) at the rate of 20.0 m/s. At that instant, an automobile, approaching at the rate of 30.0 m/s, is 40.0 m from a point directly beneath the track.

Find how fast the train and the automobile are separating 2.00 seconds later.

2. Relevant equations
No equation was given.

3. The attempt at a solution
$$dz/dt = 20 m/s$$
$$dx/dt = 30 m/s$$

$$h^2 = x^2 + y^2 + z^2$$
$$h^2 = 40^2 + 30^2 + 20^2$$
$$h^2 = 54 dh/dt$$

$$h^2 = x^2 + 30^2 + z^2$$
$$2h(dh/dt) = 2x(dx/dt) + 0 + 2z(dz/dt)$$
$$2(54) dh/dt = 2(40) dx/dt + 0 + 2(20) dz/dt$$
$$108 dh/dt = 80 dx/dt + 40 dz/dt$$
*don't know what to do next*

Am I doing this right so far? Any suggestions are really appreciated, thanks a lot!

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

### bob1182006

it's not 3 variables, you only have x/y, say.. x = the train going to the right, y = the car moving up

but your equation's right, also differentiate w/respect to t, you'll have to use the chain rule.

3. Jun 25, 2007

### Dick

Write out (x,y,z) for each vehicle explicitly as a function of t. Use that to write h explicitly as a function of t.

4. Jun 25, 2007

### louloulou

I'm sorry, I don't quite understand. What do I need to change? Is the $$54 dh/dt$$ not correct (Should it just be $$30^2 + 20^2$$ ?)

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

### bob1182006

it's not 3 variables
it's
$$h^2 = x^2 + y^2$$

Ok well that gives you 2 variables. I'm gonna assume x is the train's direction, z is the car's direction.
$$2(54) dh/dt = 2(40) dx/dt + 0 + 2(20) dz/dt$$ is wrong, you're given the locations (30m, 40m) but you're asked for 2.0 s later, so you need to add the distance that they traveled in those seconds, so the speeds *2.0 + location and you plug that in for the x,z,h. And what is dx/dt? dh/dt? dz/dt?

Last edited: Jun 25, 2007
6. Jun 25, 2007

### louloulou

So...

$$h^2 = 30^2 + 20^2$$
$$h = 36.1 dh/dt$$

and then...

$$2h(dh/dt) = 2x(dx/dt) + 2y(dy/dt)$$

I'm getting confused as to what the x and y is now..still 40(x) & 20(y)?
Ugh, I don't know!

7. Jun 25, 2007

### bob1182006

no you need to plug-in 2h, 2x, 2y, what's h,x,y? after 2.0 Seconds, you're given the initial location and the speed they travel so you can find how far they've traveled in 2.0 s and add that to the initial location

8. Jun 25, 2007

### louloulou

Okay, well thanks for all your help i'll give that a try and post it again.

9. Jun 25, 2007

### louloulou

I know this is wrong...i'm getting 1.38 m/s later.

10. Jun 25, 2007

### bob1182006

I got something like, the train/car are separating at a rate of 36.0 m/s.

Can you show your work? maybe you're write and I did some algebra mistake...

11. Jun 26, 2007

### Dick

Suppose the car is going in the +y direction and the train in the +x and the (x,y) paths cross at (0,0). The position of the car is (0,-40+30*t,0). That's the hard one, can you do the train? So then what's h(t)? You seem to just be getting confused about what are x,y,z.

12. Jun 26, 2007

### louloulou

I'm not sure when I should be plugging in the "2.00 s"

Anyway, this is what I did.

$$2 (31.6) dh/dt = 2(30) + 2 (20)$$
$$dh/dt = 1.38$$

But something is not right because I still need to do something with the 2.00 seconds:uhh: