Results 1 to 4 of 4

Math Help - Differentiation

  1. #1
    Junior Member
    Joined
    Nov 2006
    Posts
    59

    Differentiation

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by camherokid View Post
    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?
    Very Important: the first thing we want to do is draw a diagram so we can see what's going on. Fill in everything on the diagram that we know, don't know, and want to know. (See the diagram I have attached before reading the solution).


    Let x be the distance between ship A and the initial position of ship B (B1).
    Let y be the distance ship B has traveled.
    Let z be the distance between the two ships.

    After 4 hours:

    A travels 140 km (35km/h *4 hrs). So the distance between A and B1 is 10 km. It is travelling at a speed of 35 km/h so dx/dt = - 35 km/h (note, the rate is negative since the distance is decreasing).

    B travels 100 km from its initial position.

    By pythagoras then, after 4 hours, z^2 = 10^2 + 100^2
    => z = sqrt(100 + 10000) = 100.50 km

    Now to solve the problem, we again use Pythagoras' Theorem:

    z^2 = x^2 + y^2 ..........now differentiate implicitely
    => 2z dz/dt = 2x dx/dt + 2y dy/dt
    => dz/dt = (2x dx/dt + 2y dy/dt)/2z

    after 4 hours: x = 10, y = 100, dx/dt = -35, dy/dt = 25, z = 100.50

    => dz/dt = [2(10)(-35) + 2(100)(25)]/2(100.5)
    => dz/dt = 21.39 km/h ....the rate at which the distance between the two ships are changing
    Attached Thumbnails Attached Thumbnails Differentiation-ship.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, camherokid!

    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?
    Code:
                                |
                                * B
                         x   *  |
                          *     | 25t
                       *        |
          * - - - - * - - - - - * Q
          P   35t   A  150-35t

    Ship A starts at point P, 150 km west of point Q.
    . . In t hours, it has moved 35t km east to point A.
    . . Hence: .AQ = 150 - 35t

    Ship B starts at point Q.
    . . In t hours, it has moved 25t km north to point B.

    Let x = AB
    Then: .x .= .(25t) + (150 - 35t)
    . . and we have: .x .= .1850t - 10,500t + 22,500 . [1]


    Differentiate with respect to time: .2x(dx/dt) .= .3700t - 10,500

    . . and we have: .dx/dt .= .(1850t - 5250)/x . [2]


    When t = 4, we have: .[1] .x .= .18504 - 10,5004 + 22,500 .= .10,100

    . . Hence: .x .= .√10,100 .= .10√101


    Substitute into [2]: .dx/dt .= .(18504 - 5350)/(10√101) .= .21.39329959


    Therefore, the ships are separating at about 21.4 km/hr.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Soroban View Post
    Hello, camherokid!

    Code:
                                |
                                * B
                         x   *  |
                          *     | 25t
                       *        |
          * - - - - * - - - - - * Q
          P   35t   A  150-35t

    Ship A starts at point P, 150 km west of point Q.
    . . In t hours, it has moved 35t km east to point A.
    . . Hence: .AQ = 150 - 35t

    Ship B starts at point Q.
    . . In t hours, it has moved 25t km north to point B.

    Let x = AB
    Then: .x .= .(25t) + (150 - 35t)
    . . and we have: .x .= .1850t - 10,500t + 22,500 . [1]


    Differentiate with respect to time: .2x(dx/dt) .= .3700t - 10,500

    . . and we have: .dx/dt .= .(1850t - 5250)/x . [2]


    When t = 4, we have: .[1] .x .= .18504 - 10,5004 + 22,500 .= .10,100

    . . Hence: .x .= .√10,100 .= .10√101


    Substitute into [2]: .dx/dt .= .(18504 - 5350)/(10√101) .= .21.39329959


    Therefore, the ships are separating at about 21.4 km/hr.
    Sorry again Soroban, i had no idea you were working on this problem. I checked the "Who's Online" Forum to see if anyone was working on it and saw no one. It's good to verify that I'm right though, so your work is not in vain.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. Differentiation and partial differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 30th 2010, 10:16 PM
  3. Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 4th 2010, 10:45 AM
  4. Differentiation Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 25th 2010, 06:20 PM
  5. Differentiation and Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 6th 2009, 04:07 AM

Search Tags


/mathhelpforum @mathhelpforum