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: