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Math Help - Related rates problem

  1. #1
    Newbie Exotique's Avatar
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    Related rates problem

    I just want to check if my reasoning and answer are correct. The question is:
    A rocket is launched straight up, and it's altitude is given by h=10t^2 after t seconds. You are on the ground 300m from the launch site watching the rocket going up. The line of sight from you to the rocket makes and angle \theta with the horizontal. By how many radians per second is \theta changing 10 seconds after the launch?

    My reasoning is as follows:
    Need to find \frac{d\theta}{dt}

    h=10t^2, so \frac{dh}{dt}=20t.

    \tan\theta=\frac{h}{300}, so \frac{d\theta}{dh}=\frac{300}{h^2+90000}.

    Remembering t=10, \frac{dh}{dt}=20t=200

    Also, since h=10t^2, h=1000 when t=10.

    Therefore, \frac{d\theta}{dh}=\frac{300}{h^2+90000}=\frac{300  }{91000}


    So using Chain Rule:
    \frac{d\theta}{dt}=\frac{d\theta}{dh}\cdot\frac{dh  }{dt}

    =\frac{300}{91000}\cdot200=\frac{60}{91}

    Any help will be much appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Exotique View Post
    I just want to check if my reasoning and answer are correct. The question is:
    A rocket is launched straight up, and it's altitude is given by h=10t^2 after t seconds. You are on the ground 300m from the launch site watching the rocket going up. The line of sight from you to the rocket makes and angle \theta with the horizontal. By how many radians per second is \theta changing 10 seconds after the launch?

    My reasoning is as follows:
    Need to find \frac{d\theta}{dt}

    h=10t^2, so \frac{dh}{dt}=20t.

    \tan\theta=\frac{h}{300}, so \frac{d\theta}{dh}=\frac{300}{h^2+90000}.

    Remembering t=10, \frac{dh}{dt}=20t=200

    Also, since h=10t^2, h=1000 when t=10.

    Therefore, \frac{d\theta}{dh}=\frac{300}{h^2+90000}=\frac{300  }{91000}


    So using Chain Rule:
    \frac{d\theta}{dt}=\frac{d\theta}{dh}\cdot\frac{dh  }{dt}

    =\frac{300}{91000}\cdot200=\frac{60}{91}

    Any help will be much appreciated.
    \tan(\theta)=\frac{h}{300}

    Now differentiate with respect to t:

     <br />
\sec^2(\theta) \frac{d\theta}{dt}=\frac{1}{300}\frac{dh}{dt}<br />

    Now at t=10, h=1000, \tan(\theta)=10/3 so \cos(\theta)=3/\sqrt{109} and \frac{dh}{dt}=200

    CB
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  3. #3
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by Exotique View Post
    Any help will be much appreciated.
    Not what I got. It's basically a right triange with the the adjacent side to \theta being 300m and the opposite side is the height, h = 10t^2.

    To relate all of them, you use TOA.

     tan {\theta} = \frac{opposite}{adjacent}

     tan {\theta} = \frac{10t^2}{300}

     tan {\theta} = \frac{t^2}{30}<< simplifying it a bit..

    now we take the derivative of both sides with respect to time. Since \theta is a variable, we have to use the chain rule.

     \frac{d}{dt}\left[tan {\theta}\right] =\frac{d}{dt}\left[\frac{t^2}{30}\right]

     sec^2{\theta} \frac{d\theta }{dt} = \frac{t}{15}

    \frac{d\theta }{dt} = \frac{t}{15sec^2{\theta}}

    now we plug in our numbers; make sure to use radians on your calculator; not degrees.

    At t = 10,  h = 1000 m and using some trigonometry we find that \theta \approx 1.279339532. Use the memory function on your calculator!

    \frac{d\theta }{dt} = \frac{10}{15sec^2(1.279339532)}

    \frac{d\theta }{dt} = \frac{6}{109} \approx 0.055 rad/s
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  4. #4
    Newbie Exotique's Avatar
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    Quote Originally Posted by eddie2042 View Post
    Not what I got. It's basically a right triange with the the adjacent side to \theta being 300m and the opposite side is the height, h = 10t^2.

    To relate all of them, you use TOA.

     tan {\theta} = \frac{opposite}{adjacent}

     tan {\theta} = \frac{10t^2}{300}

     tan {\theta} = \frac{t^2}{30}<< simplifying it a bit..

    now we take the derivative of both sides with respect to time. Since \theta is a variable, we have to use the chain rule.

     \frac{d}{dt}\left[tan {\theta}\right] =\frac{d}{dt}\left[\frac{t^2}{30}\right]

     sec^2{\theta} \frac{d\theta }{dt} = \frac{t}{15}

    \frac{d\theta }{dt} = \frac{t}{15sec^2{\theta}}

    now we plug in our numbers; make sure to use radians on your calculator; not degrees.

    At t = 10,  h = 1000 m and using some trigonometry we find that \theta \approx 1.279339532. Use the memory function on your calculator!

    \frac{d\theta }{dt} = \frac{10}{15sec^2(1.279339532)}

    \frac{d\theta }{dt} = \frac{6}{109} \approx 0.055 rad/s
    Can you please explain what might be wrong with my reasoning?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Exotique View Post
    Can you please explain what might be wrong with my reasoning?
    if h=1000 then h^2=1000000

    CB
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  6. #6
    Newbie Exotique's Avatar
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    Seems I failed to spot that. Peer review works wonders, thank you! But if I had squared h after substituting, we would have arrived at the same answer, so my logic is correct, I assume.
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