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Math Help - word problem help

  1. #1
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    word problem help

    A weather balloon is released 60 feet from an observer with a camera. if the balloon is rising at 10 ft. per second how fast is the angle of the camera changing when the balloon is 100 ft. high?

    I did the problem and got 14.29 radians per second, but I'm not too sure that i did it right. Thanks in advance
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  2. #2
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    Quote Originally Posted by wisezeta View Post
    A weather balloon is released 60 feet from an observer with a camera. if the balloon is rising at 10 ft. per second how fast is the angle of the camera changing when the balloon is 100 ft. high?

    I did the problem and got 14.29 radians per second, but I'm not too sure that i did it right. Thanks in advance

    your solution is incorrect ... what did you do to get it?
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  3. #3
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    First I made a right triangle with th ebase as 60, the right leg as 100, the hypotnuse as s, and the angle as theta
    then i made the equation tanx=(116.62/60)10
    116.62 was my hyponuse
    I took the derivative of the equation and got sec^2 x (dx/dt)=(116.62/60)10
    I then solved for dx/dt and got my answer
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  4. #4
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    Quote Originally Posted by wisezeta View Post
    First I made a right triangle with th ebase as 60, the right leg as 100, the hypotnuse as s, and the angle as theta
    then i made the equation tanx=(116.62/60)10
    116.62 was my hyponuse
    I took the derivative of the equation and got sec^2 x (dx/dt)=(116.62/60)10
    I then solved for dx/dt and got my answer
    the side that you made 100 (height of the balloon) is variable ...

    \frac{d}{dt}\left(\tan{\theta} = \frac{h}{60}\right)

    \sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{60} \cdot \frac{dh}{dt}

    now, try again to find \frac{d\theta}{dt} when h = 100
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    I'm still confused. Where do I place 100 and where do i place 10?
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  6. #6
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    Quote Originally Posted by wisezeta View Post
    I'm still confused. Where do I place 100 and where do i place 10?
    h = 100 ... you'll need that to figure out \sec{\theta}

    \frac{dh}{dt} = 10 ft/s
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  7. #7
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    so is the answer .126 radians per second?
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  8. #8
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    [quote=skeeter;428168]the side that you made 100 (height of the balloon) is variable ...

    \sec^2{\theta} \cdot \frac{d\theta}{dt} = \frac{1}{60} \cdot \frac{dh}{dt}

    \frac{d\theta}{dt} = \cos^2{\theta} \cdot \frac{1}{60} \cdot \frac{dh}{dt}

    using right triangle trig ... \cos{\theta} = \frac{60}{\sqrt{60^2+100^2}}

    \frac{d\theta}{dt} = \left(\frac{60}{\sqrt{60^2+100^2}}\right)^2 \cdot \frac{1}{60} \cdot 10

    try again
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  9. #9
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    Oh i see what i did. I took the actual cosine of the ratio. Ok so is the answer .0044?
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  10. #10
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    Quote Originally Posted by wisezeta View Post
    Oh i see what i did. I took the actual cosine of the ratio. Ok so is the answer .0044?
    you're off by a factor of 10 ...

    \frac{d\theta}{dt} \approx 0.044
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  11. #11
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    Ok i just wrote it down wrong on the computer. Thank you very much
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