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Thread: rate of change with a rising balloon

  1. #1
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    rate of change with a rising balloon

    A weather balloon that is rising vertically is being observed from a point on the ground 300 ft from the spot directly beneath the balloon. At what rate is the balloon rising when the angle between the ground and the observers line of sight is 45 degrees, and is increasing 1 degree per second.

    What i did is Tan b = y / 300 I did implicit diferentiation so 300 sec^2b (db/dt) = dy/dt.

    300 (2.83) (1) = dy/dt
    848 ft

    The other problem Im having is:

    Two straight roads intersect at right angles. At 10 am a car passes through the intersection headed due east at 30 mph. At 11 am a truck heading due north at 40 mph passes through the intersection. Assume that the 2 vehicles maintain the given speeds and directions. At what rate are they separating at 1 pm?

    For this one I used the pythagoream theorem and s^2 = X^2 + Y^2

    Then I did implicit diferentiation so 2s (ds/dt) = 2x (dx/dt) + 2Y (dy/dt)

    Simplified ds/dt = ( X (dx/dt) + Y (dy/dt) ) / S

    So after i plug all the values in is ds/ dt = ( 80 (40) + 90 (30) ) / (120.42)

    ds/ dt = 48.99 mph

  2. #2
    Senior Member skeeter's Avatar
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    your set-up on the balloon problem is o.k., but one big mistake ...
    you have to use radians for angle measurements.

    y = 300*tan(b)

    dy/dt = 300*sec<sup>2</sup>(b)*(db/dt)

    dy/dt = 300*2*(pi/180) = approx 10.5 ft/sec


    in your second problem, you have to take into account that both vehicles did not pass through the intersection at the same time.

    to make things easy, let 11am be t = 0

    position of eastbound car is x = 30 + 30t

    position of northbound truck is y = 40t

    s<sup>2</sup> = x<sup>2</sup> + y<sup>2</sup>

    2s(ds/dt) = 2x(dx/dt) + 2y(dy/dt)

    ds/dt = [x(dx/dt) + y(dy/dt)]/s

    now find ds/dt at 1pm, which is at time t = 2.

  3. #3
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    Re: rate of change with a rising balloon

    A weather balloon that is rising vertically is being observed from a point on the ground 300 ft from the spot directly beneath the balloon. At what rate is the balloon rising when the angle between the ground and the observers line of sight is 45 degrees, and is increasing 1 degree per second.

    What i did is Tan b = y / 300 I did implicit diferentiation so 300 sec^2b (db/dt) = dy/dt.

    300 (2.83) (1) = dy/dt
    848 ft
    That is one mighty fast balloon you have there.

    Make sure you have consistent units.

    [tex]\L\\\frac{d{\theta}}{dt}=\frac{\pi}{180} \;\ radians, \;\ {\theta}=\frac{\pi}{4}, \;\ x=300, \;\ \frac{dy}{dt}=?[/tex]

    [tex]\L\\tan({\theta})=\frac{y}{300}[/tex]

    [tex]\L\\sec^{2}({\theta})\frac{d{\theta}}{dt}=\frac{1} {300}\frac{dy}{dt}[/tex]

    [tex]\L\\\frac{dy}{dt}=300sec^{2}(\frac{\pi}{4})(\frac{ \pi}{180})=\frac{10\pi}{3} \;\ ft/sec.[/tex]

  4. #4
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    Thank youy very much for all your help!!

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