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PLS HELP: Application of Derivatives

  1. Jun 25, 2007 #1
    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
    [tex] dz/dt = 20 m/s [/tex]
    [tex] dx/dt = 30 m/s [/tex]

    [tex] h^2 = x^2 + y^2 + z^2 [/tex]
    [tex] h^2 = 40^2 + 30^2 + 20^2 [/tex]
    [tex] h^2 = 54 dh/dt [/tex]

    [tex] h^2 = x^2 + 30^2 + z^2 [/tex]
    [tex] 2h(dh/dt) = 2x(dx/dt) + 0 + 2z(dz/dt) [/tex]
    [tex] 2(54) dh/dt = 2(40) dx/dt + 0 + 2(20) dz/dt [/tex]
    [tex] 108 dh/dt = 80 dx/dt + 40 dz/dt [/tex]
    *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. jcsd
  3. Jun 25, 2007 #2
    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.
     
  4. Jun 25, 2007 #3

    Dick

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    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.
     
  5. Jun 25, 2007 #4
    I'm sorry, I don't quite understand. What do I need to change? Is the [tex] 54 dh/dt [/tex] not correct (Should it just be [tex] 30^2 + 20^2 [/tex] ?)
     
    Last edited: Jun 25, 2007
  6. Jun 25, 2007 #5
    it's not 3 variables
    it's
    [tex]h^2 = x^2 + y^2[/tex]

    Ok well that gives you 2 variables. I'm gonna assume x is the train's direction, z is the car's direction.
    [tex] 2(54) dh/dt = 2(40) dx/dt + 0 + 2(20) dz/dt [/tex] 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
  7. Jun 25, 2007 #6

    So...

    [tex]h^2 = 30^2 + 20^2 [/tex]
    [tex] h = 36.1 dh/dt[/tex]

    and then...

    [tex] 2h(dh/dt) = 2x(dx/dt) + 2y(dy/dt) [/tex]

    I'm getting confused as to what the x and y is now..still 40(x) & 20(y)?
    Ugh, I don't know! :confused:
     
  8. Jun 25, 2007 #7
    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
     
  9. Jun 25, 2007 #8
    Okay, well thanks for all your help i'll give that a try and post it again.
     
  10. Jun 25, 2007 #9
    I know this is wrong...i'm getting 1.38 m/s later.

    Can you please tell me your answer so I can at least compare with mine?
     
  11. Jun 25, 2007 #10
    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...
     
  12. Jun 26, 2007 #11

    Dick

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    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.
     
  13. Jun 26, 2007 #12
    I'm not sure when I should be plugging in the "2.00 s"

    Anyway, this is what I did.

    [tex] 2 (31.6) dh/dt = 2(30) + 2 (20) [/tex]
    [tex] dh/dt = 1.38 [/tex]

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