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Math Help - related rates

  1. #1
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    related rates

    1. A weather balloon rises straight up in the air at a constant speed of 5 km/h. An observer 2 km from the balloon’s starting point tracks the balloon with a telescope. The angle between the telescope’s line of sight and the horizontal ground is a.Determine the rate at which a is changing when a=

    <br />
\frac{\pi}{3}<br />
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by santatrue View Post
    1. A weather balloon rises straight up in the air at a constant speed of 5 km/h. An observer 2 km from the balloon’s starting point tracks the balloon with a telescope. The angle between the telescope’s line of sight and the horizontal ground is a.Determine the rate at which a is changing when a=

    <br />
\frac{\pi}{3}<br />
    did you draw a diagram? you should always do this with related rates problems, where possible. see the diagram of the situation below. where would you go from here? what equation do you know that can relate a and x? it is a right-triangle, so you should immediately be thinking of things like Pythagoras' theorem and trig ratios
    Attached Thumbnails Attached Thumbnails related rates-balloon.gif  
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  3. #3
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    We can use tangent given that \frac{dy}{dt}=5, \;\ {\theta}=\frac{\pi}{3}, \;\ x=2

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

    y=2tan{\theta}

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

    5=2sec^{2}(\frac{\pi}{3})\frac{d{\theta}}{dt}
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by santatrue View Post
    1. A weather balloon rises straight up in the air at a constant speed of 5 km/h. An observer 2 km from the balloon’s starting point tracks the balloon with a telescope. The angle between the telescope’s line of sight and the horizontal ground is a.Determine the rate at which a is changing when a=

    <br />
\frac{\pi}{3}<br />
    tan a = \frac{y}{x}

    y = x \ tan \ a = 2 \ tan \ a

    Differentiate.

    y' = (2 sec^2 a)(a')

    5 = 2 sec^2 (\frac{\pi}{3})(a')

    a' = \frac{5}{2 sec^2 (\frac{\pi}{3})}
    Last edited by janvdl; June 22nd 2008 at 11:14 AM. Reason: Oops, bit too slow :D
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