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Math Help - Rate of Change With Chief Sohcahtoa

  1. #1
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    Rate of Change With Chief Sohcahtoa

    A hot air balloon is rising straight up from a level field and is tracked by a range finder 500ft from the lift point. At the moment the range finder's elevation angle is pi/4, the angle is increasing by .14 radians per minute. How fast is the balloon rising at that instant.

    So using the the trigonometric laws I know that the current height is equal to tan θ = op/adj which for this problem would be tan pi/4 = opp/500. I can use that to get the height, but can I also use that for the derivative to calculate the height?

    I guess I'm not really sure how to relate the change in angle towards the change in height necessarily.

    I'm also a little thrown off by the use of radians. Would it be advantageous to convert to degrees and then convert back to radians at the end?

    So far the only guess I have at a derivative would be sec^2 θ * dθ/dt = dh/dt / 500, but I'm pretty sure that's pretty wrong.
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  2. #2
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    You're on the right track.

    \frac{y}{500}=tan{\theta}

    y=500tan{\theta}

    \frac{dy}{dt}=500sec^{2}{\theta}\frac{d{\theta}}{d  t}

    \frac{dy}{dt}=500sec^{2}(\frac{\pi}{4})\cdot\frac{  7}{50}
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    Quote Originally Posted by Menudo View Post
    A hot air balloon is rising straight up from a level field and is tracked by a range finder 500ft from the lift point. At the moment the range finder's elevation angle is pi/4, the angle is increasing by .14 radians per minute. How fast is the balloon rising at that instant.
    Why do you doubt yourself?

    Code:
                     .
                    /|
                  /  |
                /    |
              /      h
            /        |
          /          |
        /  θ         |
    --o-------500----+
    As you observed, we have

    h = 500\tan\theta

    So, we differentiate implicitly with respect to time t,

    \frac{dh}{dt} = 500\sec^2\theta\frac{d\theta}{dt}

    When the angle is \frac\pi4, we have \frac{d\theta}{dt} = 0.14, so we substitute:

    \frac{dh}{dt} = 500\cdot0.14\sec^2\left(\frac\pi4\right)

    = 500\left(\frac{14}{100}\right)\left[\frac1{\cos^2\left(\frac\pi4\right)}\right]

    = 5\cdot14\cdot\left(\sqrt2\right)^2 = 70\cdot2 = 140

    If you carry through the units, you should see that we have feet per minute, so the balloon is rising at a rate of 140\text{ ft/min}.
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  4. #4
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    Wow, thank you so much. I'm glad to know I was on the right track. I'll try to trust my instincts a little better when I suspect they're well founded.
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