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Math Help - I don't even know where to start...

  1. #16
    MHF Contributor MarkFL's Avatar
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    Re: I don't even know where to start...

    I would write:

    \tan(\theta)=\frac{x}{50}

    Differentiating with respect time time t, we find:

    \sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{50}\cdot  \frac{dx}{dt}

    \frac{dx}{dt}=50\sec^2(\theta)\frac{d\theta}{dt}

    Now plug in the given values:

    \frac{dx}{dt}=50\sec^2\left(\frac{\pi}{6} \right)\left(-\frac{9}{10} \right)

    So what do we get?
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  2. #17
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    Re: I don't even know where to start...

    I don't know how to plug in secant (squared) into my calculator... :S

    EDIT: -59.99?
    Last edited by theunforgiven; November 20th 2012 at 01:22 AM.
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  3. #18
    MHF Contributor MarkFL's Avatar
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    Re: I don't even know where to start...

    You shouldn't need to use you calculator. The given angle is a "special angle" for which we should know all of the trig functions at this angle.

    What is \cos\left(\frac{\pi}{6} \right) ?

    If you know this, then since \sec(\theta)\equiv\frac{1}{\cos(\theta)} you can invert, then square to get the value you want.
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  4. #19
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    Re: I don't even know where to start...

    Quote Originally Posted by MarkFL2 View Post
    You shouldn't need to use you calculator. The given angle is a "special angle" for which we should know all of the trig functions at this angle.

    What is \cos\left(\frac{\pi}{6} \right) ?

    If you know this, then since \sec(\theta)\equiv\frac{1}{\cos(\theta)} you can invert, then square to get the value you want.
    And that's exactly what I did. I ended up with -59.99 (or -60). Oh and \cos\left(\frac{\pi}{6} \right) = square root of 3 over 2.
    Last edited by theunforgiven; November 20th 2012 at 01:30 AM.
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  5. #20
    MHF Contributor MarkFL's Avatar
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    Re: I don't even know where to start...

    Yes, so what do you get when you invert and square?
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  6. #21
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    Re: I don't even know where to start...

    Quote Originally Posted by markfl2 View Post
    yes, so what do you get when you invert and square?
    -60. Hope that's what you are asking for.
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  7. #22
    MHF Contributor MarkFL's Avatar
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    Re: I don't even know where to start...

    That is the correct final answer, but I was asking what you got when you invert and square the value you found for the cosine function.
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  8. #23
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    Re: I don't even know where to start...

    Oh my bad. I got 0.66666666667
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  9. #24
    MHF Contributor MarkFL's Avatar
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    Re: I don't even know where to start...

    What I mean is:

    invert: \frac{2}{\sqrt{3}}

    square: \left(\frac{2}{\sqrt{3}} \right)^2=\frac{4}{3}

    Thus, we know:

    \sec^2\left(\frac{\pi}{6} \right)=\frac{4}{3}
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  10. #25
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    Re: I don't even know where to start...

    Stupid me! But thank you, sir, I truly appreciate all your help. You are a true math master! *bows down*
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  11. #26
    MHF Contributor MarkFL's Avatar
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    Re: I don't even know where to start...

    Glad to be of assistance!
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