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Thread: 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 $\displaystyle h=10t^2$ after $\displaystyle t$ 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 $\displaystyle \theta$ with the horizontal. By how many radians per second is $\displaystyle \theta$ changing 10 seconds after the launch?

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

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

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

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

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

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


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

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

    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 $\displaystyle h=10t^2$ after $\displaystyle t$ 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 $\displaystyle \theta$ with the horizontal. By how many radians per second is $\displaystyle \theta$ changing 10 seconds after the launch?

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

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

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

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

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

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


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

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

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

    Now differentiate with respect to $\displaystyle t$:

    $\displaystyle
    \sec^2(\theta) \frac{d\theta}{dt}=\frac{1}{300}\frac{dh}{dt}
    $

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

    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 $\displaystyle \theta$ being 300m and the opposite side is the height, $\displaystyle h = 10t^2$.

    To relate all of them, you use TOA.

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

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

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

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

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

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

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

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

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

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

    $\displaystyle \frac{d\theta }{dt} = \frac{6}{109} \approx 0.055 rad/s$
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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 $\displaystyle \theta$ being 300m and the opposite side is the height, $\displaystyle h = 10t^2$.

    To relate all of them, you use TOA.

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

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

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

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

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

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

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

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

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

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

    $\displaystyle \frac{d\theta }{dt} = \frac{6}{109} \approx 0.055 rad/s$
    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 $\displaystyle h=1000$ then $\displaystyle h^2=1000000$

    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 $\displaystyle h$ after substituting, we would have arrived at the same answer, so my logic is correct, I assume.
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