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Math Help - Related rates, finding an angle.

  1. #1
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    Related rates, finding an angle.

    I was trying to solve this problem:

    A ladder 13ft long is leaning against a wall. It suddenly starts to slide away at the base of the ladder. When it is 12ft away from the wall it is sliding away at 5ft/s

    if it is 12 feet away then at that moment the other side is 5ft high
    I found that at that moment the side on the wall is coming down at -12ft/s

    What I can't find is how much the angle between the base and the ground is changing at that moment. I know it is -1 rad/s but I can't get there.
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  2. #2
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    Quote Originally Posted by david13579 View Post
    I was trying to solve this problem:

    A ladder 13ft long is leaning against a wall. It suddenly starts to slide away at the base of the ladder. When it is 12ft away from the wall it is sliding away at 5ft/s

    if it is 12 feet away then at that moment the other side is 5ft high
    I found that at that moment the side on the wall is coming down at -12ft/s

    What I can't find is how much the angle between the base and the ground is changing at that moment. I know it is -1 rad/s but I can't get there.
    I assume you mean the angle between the ladder and the ground ... the rate of change is not -1 rad/s , it's -13/12 rad/s
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  3. #3
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    Quote Originally Posted by skeeter View Post
    I assume you mean the angle between the ladder and the ground ... the rate of change is not -1 rad/s , it's -13/12 rad/s
    -1 is the answer my books gives. And yes, I meant the base of the ladder and the ground.
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  4. #4
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    If \theta is the angle between the ladder and the ground then tan(\theta)= \frac{h}{x} where h is the height of the ladder on the wall and x is the distance from the base of the ladder to the wall.

    \frac{1}{1+ \theta^2}\frac{d\theta}{dt}= \frac{1}{x}\frac{dh}{dt}- \frac{h}{x^2}\frac{dx}{dt}.

    We are told that, at this point, x= 12, h= 5, \frac{dx}{dt}= 5, and you have calculated that \frac{dh}{dt}= -12. Also tan(\theta)= \frac{5}{12} so \theta= 0.3948 radians.

    Putting all of that into the formula:
    \frac{1}{1+ 0.1559}\frac{d\theta}{dt}= \frac{1}{12}(-12)- \frac{5}{144}(5}= \frac{12+ 25}{144}= \frac{169}{144}
    = -1.0187 which is very close to -1.
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  5. #5
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    well dip me in dog ...

    ... sure does work out to be -1

    \theta = \arctan\left(\dfrac{h}{x}\right)

    \dfrac{d\theta}{dt} = \dfrac{x \cdot \frac{dh}{dt} - h \cdot \frac{dx}{dt}}{x^2} \cdot \dfrac{1}{1 + \left(\frac{h}{x}\right)^2}

    \dfrac{d\theta}{dt} = \dfrac{12 \cdot (-12) - 5 \cdot 5}{12^2} \cdot \dfrac{1}{1 + \left(\frac{5}{12}\right)^2}

    \dfrac{d\theta}{dt} = \dfrac{-169}{12^2} \cdot \dfrac{1}{1 + \left(\frac{5}{12}\right)^2}

    \dfrac{d\theta}{dt} = \dfrac{-169}{12^2\left[1 + \left(\frac{5^2}{12^2}\right)\right]} = -1

    need to watch my arithmetic.
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  6. #6
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    Quote Originally Posted by skeeter View Post
    well dip me in dog ...

    ... sure does work out to be -1

    \theta = \arctan\left(\dfrac{h}{x}\right)

    \dfrac{d\theta}{dt} = \dfrac{x \cdot \frac{dh}{dt} - h \cdot \frac{dx}{dt}}{x^2} \cdot \dfrac{1}{1 + \left(\frac{h}{x}\right)^2}

    \dfrac{d\theta}{dt} = \dfrac{12 \cdot (-12) - 5 \cdot 5}{12^2} \cdot \dfrac{1}{1 + \left(\frac{5}{12}\right)^2}

    \dfrac{d\theta}{dt} = \dfrac{-169}{12^2} \cdot \dfrac{1}{1 + \left(\frac{5}{12}\right)^2}

    \dfrac{d\theta}{dt} = \dfrac{-169}{12^2\left[1 + \left(\frac{5^2}{12^2}\right)\right]} = -1

    need to watch my arithmetic.
    Thanks . Though I had already worked it out with the help of my professor.
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