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Math Help - Related Rates 2

  1. #1
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    Related Rates 2

    At noon, Ship A is 100km west of ship B. Ship A is sailing south at 35km/h and ship B is sailing at 25km/h.
    How fast is the distance between the ships changing at 4PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    At noon, Ship A is 100km west of ship B. Ship A is sailing south at 35km/h and ship B is sailing at 25km/h.
    How fast is the distance between the ships changing at 4PM.
    you forgot to state what direction ship B is going in
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  3. #3
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    B is sailing north
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    At noon, Ship A is 100km west of ship B. Ship A is sailing south at 35km/h and ship B is sailing north at 25km/h.
    How fast is the distance between the ships changing at 4PM.
    see diagram below

    let x be the distance ship B has travelled
    let y be the distance ship A has travelled
    let z be the distance between the ships at any time

    after 4 hours:

    ship A has travelled 35(4) = 140 km
    ship B has travelled 25(4) = 100 km

    by pythagoras, after 4 hours
    z^2 = 100^2 + (x + y)^2 = 100^2 + (140 + 100)^2 = 67600
    => z = 260 km

    Now, by pythagoras again:

    z^2 = 100^2 + (x + y)^2
    => 2z dz/dt = 2(x + y)(dx/dt + dy/dt)
    => dz/dt = [2(x + y)(dx/dt + dy/dt)]/2z = [(x + y)(dx/dt + dy/dt)]/z
    now plug in the values:
    => dz/dt = [(100 + 140)(25 + 35)]/260 = 240(60)/260
    => dz/dt = 55.38 km/h .........the rate at which the distance between the ships is changing
    Attached Thumbnails Attached Thumbnails Related Rates 2-ship.gif  
    Last edited by Jhevon; March 25th 2007 at 11:55 PM.
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  5. #5
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    Hello, camherokid!

    I have a slightly different approach to motion problems.
    I think it's neater ... see what you think.


    At noon, Ship A is 100km west of ship B.
    Ship A is sailing south at 35km/h
    and ship B is sailing north at 25km/h.
    How fast is the distance between the ships changing at 4 PM?
    Code:
                              * B
                            / ↑
                          /   ↑ 25t
                        /     ↑
        P * - - - - - / - - - * Q
          ↓       x /         :
          ↓       /           :
      35t ↓     /             : 35t
          ↓   /               :
          ↓ /                 :
        A * - - - - - - - - - * R
                   100

    Ship A starts at point P and sails south at 35 km/hr.
    . . In t hours, it has sailed 35t km south to point A.

    Ship B starts at point Q and sails north at 25 km/hr.
    . . In t hours, it has sailed 25t km north to point B.

    Let x = AB.

    In right triangle BRA, we have: .x .= .(60t) + 100 .= .3600t + 10,000

    Differentiate with respect to time: .2x(dx/dt) .= .7200t

    . . and we have: .dx/dt .= .3600t/x .[1]


    At 4 PM (t = 4), the triangle has: AR = 100, BR = 240. .Hence: x = 260.

    Substitute into [1]: .dx/dt .= .3600(4)/260 . .55.38 km/hr
    . . Their distance is increasing.

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