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Math Help - Related Rates - Angle rate

  1. #1
    Member nautica17's Avatar
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    Related Rates - Angle rate

    I'm not sure weather this is correct or not, but I gave it my best shot and it looks correct... but I'm not sure if it is.

    A 10 ft plank is leaning against a wall. If at a certain instant the bottom of the plank is 2 ft from the wall and is pushed toward the wall at a rate of 6 in/sec, how fast is the acute angle that the plank makes with ground increasing?
    So here is my try:

    cos(theta) = 2/10 = 2/10
    -sin(theta) d(theta)/dt = 2/10 dx/dt
    -sin(theta) d(theta)/dt = 2/10 * .5ft/sec
    -sin(theta) d(theta)/dt = 1/10
    d(theta)/dt = 1/10 / -sin(theta)
    d(theta)/dt = (1/10) / (sqrt(96) / 10)
    d(theta)/dt = 1/sqrt(96) <---- answer?

    I'm not sure weather or not to convert inches to feet. But when I don't convert then I get 12/sqrt(96) as my answer. :| Am I on the right track here?
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  2. #2
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    Quote Originally Posted by nautica17 View Post
    I'm not sure weather this is correct or not, but I gave it my best shot and it looks correct... but I'm not sure if it is.



    So here is my try:

    cos(theta) = 2/10 = 2/10
    -sin(theta) d(theta)/dt = 2/10 dx/dt
    -sin(theta) d(theta)/dt = 2/10 * .5ft/sec
    -sin(theta) d(theta)/dt = 1/10
    d(theta)/dt = 1/10 / -sin(theta)
    d(theta)/dt = (1/10) / (sqrt(96) / 10)
    d(theta)/dt = 1/sqrt(96) <---- answer?

    I'm not sure weather or not to convert inches to feet. But when I don't convert then I get 12/sqrt(96) as my answer. :| Am I on the right track here?
    your original equation should be

    \cos{\theta} = \frac{x}{10}

    take the time derivative ...

    -\sin{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{10} \cdot \frac{dx}{dt}

    note that \frac{dx}{dt} = -\frac{1}{2} \, ft/sec because the variable distance x is decreasing.

    -\frac{\sqrt{96}}{10} \cdot \frac{d\theta}{dt} = \frac{1}{10} \cdot  \left(-\frac{1}{2}\right)

    \frac{d\theta}{dt} = \frac{1}{10} \cdot  \left(-\frac{1}{2}\right) \cdot \left(-\frac{10}{\sqrt{96}}\right)

    \frac{d\theta}{dt} = \frac{1}{8\sqrt{6}} \, rad/sec
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