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Math Help - Applications of trig

  1. #1
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    [SOLVED] Applications of trig

    A rotating light L is situated at sea 180 metres from the point P on a straight shoreline. The light rotates through one revolution every 10 seconds. Show that the rate at which a ray of light moves along the shore at a point 300 metres from P is 136pi m/s.

    Could someone please help with this question?

    Thanks!
    Last edited by ipokeyou; April 10th 2010 at 03:23 PM.
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  2. #2
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    Related rates nearly always depend on the chain rule, so you might want to try filling up this pattern...



    ... where straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case time), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    Assuming the shore is perpendicular to the line PL, we can use basic trigonometry to express x, the distance of the beam along the shore from L, in terms of theta the angle of the beam away from the line PL...



    So differentiate with respect to the inner function, and the inner function with respect to t...

    Spoiler:


    Use a right-triangle diagram to see that sec(arctan(t)) = \sqrt{1 + t^2}

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    Last edited by tom@ballooncalculus; April 9th 2010 at 02:52 PM.
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  3. #3
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    Quote Originally Posted by ipokeyou View Post
    A rotating light L is situated at sea 180 metres from the point P on a straight shoreline. The light rotates through one revolution every 10 seconds. Show that the rate at which a ray of light moves along the shore at a point 300 metres from P is 136pi m/s.

    Could someone please help with this question?

    Thanks!
    given ...

    \frac{d\theta}{dt} = \frac{2\pi}{10} rad/s

    x = 300 m

    relationship between x and \theta ...

     <br />
\theta = \arctan\left(\frac{x}{180}\right)<br />

    take the time derivative of the above equation, sub in your given values and solve for \frac{dx}{dt}
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    Quote Originally Posted by skeeter View Post
    given ...

    \frac{d\theta}{dt} = \frac{2\pi}{10} rad/s

    x = 300 m

    relationship between x and \theta ...

     <br />
\theta = \arctan\left(\frac{x}{180}\right)<br />

    take the time derivative of the above equation, sub in your given values and solve for \frac{dx}{dt}
    how do i take the time derivative? because you cant differentiate
     <br />
\theta = \arctan\left(\frac{x}{180}\right)<br />
    with respect to t?

    and also if x = 300 then cant we just sub this into the above equation?

    sorry, i dont think i'm on the right track here..
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    Quote Originally Posted by ipokeyou View Post
    how do i take the time derivative? because you cant differentiate
     <br />
\theta = \arctan\left(\frac{x}{180}\right)<br />
    with respect to t?

    and also if x = 300 then cant we just sub this into the above equation?

    sorry, i dont think i'm on the right track here..
    Post #2 said to use the chain rule and then gave you all you needed. Have you been given examples or done similar questions so that you understand what this means?

    From the chain rule: \frac{dx}{dt} = \frac{dx}{d \theta} \cdot \frac{d \theta}{dt}.
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    Quote Originally Posted by mr fantastic View Post
    Post #2 said to use the chain rule and then gave you all you needed. Have you been given examples or done similar questions so that you understand what this means?

    From the chain rule: \frac{dx}{dt} = \frac{dx}{d \theta} \cdot \frac{d \theta}{dt}.
    yes i have done similar questions. but i dont really understand what the diagram means...
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  7. #7
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    Quote Originally Posted by ipokeyou View Post
    how do i take the time derivative? because you cant differentiate
     <br />
\theta = \arctan\left(\frac{x}{180}\right)<br />
    with respect to t?
    You can, using the chain rule - since x and theta are both functions of t.

    Quote Originally Posted by ipokeyou View Post
    and also if x = 300 then cant we just sub this into the above equation?
    Yes, to find theta for x = 300. Sub into the differentiated equation to find dx/dt for x = 300 and d theta / dt = 2 pi / 10.

    Quote Originally Posted by ipokeyou View Post
    yes i have done similar questions. but i dont really understand what the diagram means...
    Use whatever you can...
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  8. #8
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    Quote Originally Posted by ipokeyou View Post
    yes i have done similar questions. but i dont really understand what the diagram means...
    If you understand problems involving related rates, then you ought to understand how the chain rule is used. So substitute the things told to you in this thread into the 'formula' I gave in my previous post.
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    Quote Originally Posted by ipokeyou View Post
    how do i take the time derivative? because you cant differentiate
     <br />
\theta = \arctan\left(\frac{x}{180}\right)<br />
    with respect to t?
    sure you can ...

    \frac{d}{dt}\left[\theta = \arctan\left(\frac{x}{180}\right)\right]

    \frac{d\theta}{dt} = \frac{\frac{1}{180}}{1 + \left(\frac{x}{180}\right)^2} \cdot \frac{dx}{dt}
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  10. #10
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    Quote Originally Posted by skeeter View Post
    sure you can ...

    \frac{d}{dt}\left[\theta = \arctan\left(\frac{x}{180}\right)\right]

    \frac{d\theta}{dt} = \frac{\frac{1}{180}}{1 + \left(\frac{x}{180}\right)^2} \cdot \frac{dx}{dt}
    oh. that's what i did. but i thought it was wrong... sorry about that. but thanks!
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