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Math Help - Related Rates Problem

  1. #1
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    Related Rates Problem

    Hi there,
    Would anyone be able to tell me if I'm doing this problem correctly? Any help/pointers will be greatly appreciated. Thanks! Ana


    A plane flying with a constant speed of 29 km/min passes over a ground radar station at an altitude of 16 km and climbs at an angle of 45 degrees. At what rate is the distance from the plane to the radar station increasing 5 minutes later?

    x = 29 cos(45)t = 20.5t

    y = 29 sin(45)t = 20.5t

    Overall distance from radar station at any given time t =

    d = sqrt(x^2 + (y + 16)^2)

    = sqrt(420.5t^2 + 420.5t^2 + 656t + 256)

    = (841t^2 + 656t + 256)^(1/2)

    d(d/dt) = (1/2)(841t^2 + 656t + 256)^(-1/2)(1682t + 656)

    for t = 5, answer = 28.924 km/min
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Related Rates Problem

    I would probably use the law of cosines to state (with D representing the distance from the plane to the station):

    D^2=16^2+(29t)^2+2\cdot16\cdot29t\cos(45^{\circ})

    D^2=16^2+(29t)^2+464\sqrt{2}t

    Implicitly differentiate with respect to time t:

    2D\frac{dD}{dt}=2\cdot29t\cdot29+464\sqrt{2}

    \frac{dD}{dt}=\frac{29^2t+232\sqrt{2}}{D}

    Hence:

    \frac{dD}{dt}\left|_{t=5}=\frac{29^2(5)+232 \sqrt{2}}{D(5)}=\frac{29(145+8 \sqrt{2})}{\sqrt{21281+2320 \sqrt{2}}}\approx28.924\,\frac{ \text{km}}{ \text{min}}
    Thanks from phenol
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  3. #3
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    Re: Related Rates Problem

    Hey MarkFL2,
    Thanks for your help. I tried doing it the way you described but kept coming up with the wrong answer. But now looking at the way you did it, I see what I was doing wrong. Thanks again for your help (as well as all the other times you've helped me in the past). I really appreciate it!! Ana
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Related Rates Problem

    Hey, glad to help!
    Thanks from phenol
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