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Math Help - Help Needed Desperately

  1. #1
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    Still Need Help Please

    A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat.

    A-._
    -..... _
    - ........_
    - ..........._
    - ............._
    - ................_
    C-................ _B

    Thats a 90 degree right triangle, the boat is at B, the winch at A, and line AC is the 12 feet.
    The winch pulls in rope at a rate of 4 feet per second. there is 13 feet of rope out (line AB), find the acceleration of the boat.


    okay, so i solved for line CB using the pythagorean theorem, which is 5 feet. i know i have to implicitely differentiate, then a second time because it is acceleration. could anyone help me.
    Last edited by doctorgk; December 4th 2007 at 08:29 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by doctorgk View Post
    A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat.

    A-._
    -..... _
    - ........_
    - ..........._
    - ............._
    - ................_
    C-................ _B

    Thats a 90 degree right triangle, the boat is at B, the winch at A, and line AC is the 12 feet.
    The winch pulls in rope at a rate of 4 feet per second. there is 13 feet of rope out (line AB), find the acceleration of the boat.


    okay, so i solved for line CB using the pythagorean theorem, which is 5 feet. i know i have to implicitely differentiate, then a second time because it is acceleration. could anyone help me.
    tell us the eqution you got and we'll take it from there
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  3. #3
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    equation

    x^2 + y^2 =s^2
    13^2 = x^2 + 12^2
    x=5

    2x(dx/dt) + 2y(dy/dt) = 2s(ds/dt)

    so... 5(dx/dt) = s(ds/dt)
    ds/dt = (13*4)/5

    how would i get the second derivative to get acceleration??
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by doctorgk View Post
    x^2 + y^2 =s^2
    13^2 = x^2 + 12^2
    x=5

    2x(dx/dt) + 2y(dy/dt) = 2s(ds/dt)

    so... 5(dx/dt) = s(ds/dt)
    ds/dt = (13*4)/5

    how would i get the second derivative to get acceleration??
    first of all, you do not want 3 unknowns, that just makes the problem harder. we use variables to represent quantities that are changing. so your equation should be:

    If we let AC = 12, BC = x and AB = y, then:

    12^2 + x^2 = y^2 .......and our aim is to find \frac {d^2 x}{dt^2}

    \Rightarrow 2x ~\frac {dx}{dt} = 2y~\frac {dy}{dt}

    \Rightarrow \frac {dx}{dt} = \frac yx~\frac {dy}{dt}

    Now what?
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  5. #5
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    Anchorage, AK
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    Chain rule

    Let us call the length of rope out (AB in diagram) u, and the distance of the boat from the dock (CB in diagram) x. We want a = \frac{d^{2}x}{dt^2}.
    We can get the velocity v=\frac{dx}{dt} by the chain rule \frac{dx}{dt}=\frac{dx}{du}\frac{du}{dt} and that we know \frac{du}{dt} is -4 ft/s (from the problem). This gives velocity in terms of u. Applying the chain rule similarly to v gives us acceleration a in terms of u, and so you plug in u=13 ft for the moment in question to get the result.

    --Kevin C.
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  6. #6
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    equation 2

    so u would differentiate again leaving only the constant y/x, which is just going to be 5/144 ?
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  7. #7
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    answer?

    is 2.6 ft/sec/sec right
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