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Third question

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. HOw fast is the distance between the ships changing at 4:00pm.


    2. Relevant equations

    ??

    3. The attempt at a solution

    I was able to draw a picture of the problem, but I don't know where to go from there, and I also understand that I am trying to solve for dz/dt, but I have no clue how to do that, please help :).
     
  2. jcsd
  3. Oct 12, 2009 #2

    berkeman

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    Staff: Mentor

    Show us your work. Then we can offer tutorial hints. You need to do the work here at the PF.
     
  4. Oct 12, 2009 #3
    What? You mean to tell me you won't do my work for me? :O

    Here's what I got so far.



    dx/dt = 35km/h,,, dy/dt = 25km/h,,, dz/dt = ???

    Then we'll use the z2 = x2 + y2

    Then we implicit differentiate it in regards to time.

    2z(dz/dt) = dx(dx/dt) + 2y(dy/dt)

    Solve for dz/dt

    (dz/dt) = (1/z)(x*(dx/dt) + y*(dy/dt))

    This is where I start to get a little confused.... I know we plug in dy/dt and dx/dt into the equation, but what's z? and is dz/dt what I'm looking for?
     
  5. Oct 12, 2009 #4
    So then I have (dz/dt) = (1/z) * (35x + 25y), so would the xyz be the distance? Like 150 as x and.....?
     
  6. Oct 13, 2009 #5
    Calculus1:Related rates

    I'll give this a better title so people know what I'm talking about :).
    1. The problem statement, all variables and given/known data
    At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. HOw fast is the distance between the ships changing at 4:00pm.


    2. Relevant equations

    x2+y2 = z2

    3. The attempt at a solution

    Here's what I got so far.



    dx/dt = 35km/h,,, dy/dt = 25km/h,,, dz/dt = ???

    Then we'll use the z2 = x2 + y2

    Then we implicit differentiate it in regards to time.

    2z(dz/dt) = dx(dx/dt) + 2y(dy/dt)

    Solve for dz/dt

    (dz/dt) = (1/z)(x*(dx/dt) + y*(dy/dt))

    This is where I start to get a little confused.... I know we plug in dy/dt and dx/dt into the equation, but what's z? and is dz/dt what I'm looking for?
     
  7. Oct 13, 2009 #6

    Mark44

    Staff: Mentor

    Re: Calculus1:Related rates

    You need to take into account the positions of both ships at time t. Let At and Bt be the coordinates of the two ships at time t. Let's suppose that ship A is at the origin at noon, so the coordinates of A0 are (0, 0) and the coordinates of B0 are (150, 0) at noon.

    Ship A is moving east (to the right) at 35 km/hr, so what are its coordinates at an arbitrary time t? Ship B is moving north (upward) at 25 km/hr, so what are its coordinates at time t?

    Get an equation for the distance between points At and Bt, then differentiate both sides of that equation, and evaluate it when t = 4 (4:00pm).
     
  8. Oct 13, 2009 #7
    You almost got it..bear in mind that noon means 12:00, so from noon to 4:00 the 2 ships travelled 4 hours already...continue your work from there.
     
  9. Oct 13, 2009 #8

    Mark44

    Staff: Mentor

  10. Oct 13, 2009 #9

    berkeman

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    Staff: Mentor

    Merged the two threads. Do not multiple post, Unknown9.
     
  11. Oct 14, 2009 #10
    So I got z = sqr( (150-x)^2 + y^2)

    Then I differentiated that and got.

    dz/dt = 1/2( (150-x)^2 + y^2)^-1/2 * (-2(dx/dt)(150-x) + 2y(dy/dt))

    when I plugged in the numbers, which I used x as 140 y as 100, and dx/dt as 35 and dy/dt as 25, I got

    dz/dt = 1/(2*sqrt(10,100)) * 4300

    am I close?
     
  12. Oct 14, 2009 #11
    I got the same result.
     
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