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Math Help - Related Rates problem

  1. #1
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    Related Rates problem

    Searched through previous threads, cannot find one like this listed.

    A Weather Balloon that is rising vertically is being observed from a point on the ground 2 miles from the spot directly beneath the balloon. At what rate is the balloon rising when the angle between the ground and the observer's line of sight is pi/5 and is increasing at 0.2 radians per minute?

    x = theta
    h = height of balloon
    dx/dt = ?
    dh/dt = .2 radians per minute



    So distance between person and balloon is 2 miles. So...tan(x) = h/2
    differentiate: tan(x) = h/2 --> sec^2(x)(dx/dt)=(1/2)(dh/dt)

    2sec^2(dh/dt)=dx/dt ---> 2sec^2(.2) = dx/dt...

    this is where i get lost if i already didn't mess up bad to begin with...
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  2. #2
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    A Weather Balloon that is rising vertically is being observed from a point on the ground 2 miles from the spot directly beneath the balloon. At what rate is the balloon rising when the angle between the ground and the observer's line of sight is pi/5 and is increasing at 0.2 radians per minute?

    x = theta
    h = height of balloon
    dx/dt = ?
    dh/dt = .2 radians per minute
    sin(theta)=opp/adj

     <br />
\sin{\frac{\pi}{5}}=\frac{h}{2}<br />

     <br />
2\sin{\frac{\pi}{5}}=h<br />

     <br />
h=2\sin{\theta}<br />

    2\sin{\frac{\pi}{5}}=2\sin{\theta}

     <br />
2\sin{\frac{\pi}{5}}(\frac{dh}{dt})=\frac{-\cos{\theta}}{(\sin{\theta})^2}(\frac{dx}{dt})<br />

     <br />
2\sin{\frac{\pi}{5}}(.2)=\frac{-\cos{\frac{\pi}{5}}}{(\sin{\frac{\pi}{5}})^2}(\fra  c{dx}{dt})<br />

     <br />
\frac{2\sin{\frac{\pi}{5}}(.2)}{\frac{-\cos{\frac{\pi}{5}}}{(\sin{\frac{\pi}{5}})^2}}=\fr  ac{dx}{dt}<br />

     <br />
\frac{dx}{dt}\approx{.9367}<br />
miles/min


    im doing related rates in my calc class now too and i think this is how you can do this problem.hope i helped
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  3. #3
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    A Weather Balloon that is rising vertically is being observed from a point on the ground 2 miles from the spot directly beneath the balloon. At what rate is the balloon rising when the angle between the ground and the observer's line of sight is pi/5 and is increasing at 0.2 radians per minute?

    x = theta
    h = height of balloon
    dx/dt = ?
    dh/dt = .2 radians per minute
    You are solving for the rate of change of the height with respect to time, not the rate of change of the angle with respect to time. You already know what that is. I believe you are misreading what that last line in the problem is saying: ". . .and is increasing at 0.2 radians per minute" - you should know automatically that this refers to the rate of change of the angle of elevation, and not the height of the balloon (nevermind we already said distance was in terms of miles. . .and that only in special case - usually when dealing with circular motion - do we describe rate of change of position in terms of degree measure).

    tan(\theta) = \frac{H}{2}

    Taking the derivative with respect to time:

    sec^{2}(\theta)\frac{d\theta}{dt} = \frac{1}{2}\frac{dH}{dt}

    Solving for the rate of change of the the height of the baloon:

    2sec^{2}(\theta)\frac{d\theta}{dt} = \frac{dH}{dt}

    We are given that the rate of change of the angle at some time t is 0.2 radians and that the angle at this time t is \frac{\pi}{5}. Therefore:

    2sec^{2}(\frac{\pi}{5})(0.2) = \frac{dH}{dt}

    Drewbear there are some fundemental misunderstandings with your solution. If you are using sine to represent the ratio between the height of the balloon and the distance to the observer, you are saying that the magnitude of the line of sight is 2 miles. This is not the case. The observer, is two miles from a spot directly beneath the vertically rising balloon; this is your adjacent side. We then want an equation in terms of this 2 miles, and the height of the ballon - therefore we use tangent.
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  4. #4
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    Thank you very much ANDS!. I kept forgetting to plug in pi/5 back into equation.
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