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Math Help - tangent to the curve with a given slope help

  1. #1
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    tangent to the curve with a given slope help

    The question is:

    Find x, such that the slope of the tangent to the curve y=tanx is equal to 4. (over [0,2\pi))

    I've got:

    \frac{d}{dx}tanx=sec^2x

    I think I should set that equal to 4, but I'm not sure on how I'm supposed solve it. Any help would be appreciated.
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  2. #2
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    Re: tangent to the curve with a given slope help

    I presume you know that sec(x)= \frac{1}{cos(x)}. So that if sec^2(x)= 4, sec(x)= \frac{1}{cos(x)}= \pm 2. That is, cos(x)= \pm\frac{1}{2}. Can you solve that?
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: tangent to the curve with a given slope help

    Quote Originally Posted by amthomasjr View Post
    The question is:

    Find x, such that the slope of the tangent to the curve y=tanx is equal to 4. (over [0,2\pi))

    I've got:

    \frac{d}{dx}tanx=sec^2x

    I think I should set that equal to 4, but I'm not sure on how I'm supposed solve it. Any help would be appreciated.
    Good so far. So you know that sec^2x = 4 . Can you solve that for x?
    (Hint: sec(x) = 1/cos(x).)

    -Dan
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  4. #4
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    Re: tangent to the curve with a given slope help

    x=\arccos(\frac{1}{2}) correct?

    Then I just find the other values that fit on the interval right?
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: tangent to the curve with a given slope help

    I wouldn't bother with using inverse trig functions:

    Use:

    \cos\left(\frac{\pi}{3} \right)=\frac{1}{2}

    \cos\left(\frac{2\pi}{3} \right)=-\frac{1}{2}

    Now use the identity \cos(2\pi-x)=\cos(x) to find the other 2 roots, i.e., the quadrant III and IV roots.
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