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Math Help - Derivatives of Inverse Trig. Functions

  1. #1
    Junior Member Coco87's Avatar
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    Derivatives of Inverse Trig. Functions

    Hey, I'm currently in Calc 2 on a section dealing with the Derivatives/Integrals of Inverse Trig. Functions. I ran across this problem, and it's just kind of completely out of the blue, and I'm not 100% sure what to do with it, but I think I have an idea.



    q) Where should the point P be chosen on the line segment AB so as to maximize the angle \theta

    My questions are: Could I assume that the triangles are right angles? I'm not sure I can. And, can anyone give me any hints on how to work the problem? I'd just like some guidance, don't want the answer

    Thanks!
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  2. #2
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    Quote Originally Posted by Coco87 View Post
    Hey, I'm currently in Calc 2 on a section dealing with the Derivatives/Integrals of Inverse Trig. Functions. I ran across this problem, and it's just kind of completely out of the blue, and I'm not 100% sure what to do with it, but I think I have an idea.



    q) Where should the point P be chosen on the line segment AB so as to maximize the angle \theta

    My questions are: Could I assume that the triangles are right angles? I'm not sure I can. Mr F says: Yes you can.

    And, can anyone give me any hints on how to work the problem? I'd just like some guidance, don't want the answer

    Thanks!
    It is sufficient to maximise \tan \theta (why?).

    \theta = \pi - (\alpha + \beta) \Rightarrow \tan \theta = - \tan (\alpha + \beta) (keep reading and it will be clear where the angles \alpha and \beta are).

    Let BP = x \Rightarrow PA = 3 - x.

    Then \tan \alpha = \frac{2}{x} and \tan \beta = \frac{5}{3-x}.
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  3. #3
    Junior Member Coco87's Avatar
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    Well, I did set it up the way you have, but I couldn't figure out the \tan\theta=-\tan(\alpha+\beta) part, so I just set:

    \tan\alpha = \frac{2}{x} \Rightarrow \alpha = \arctan\frac{2}{x}

    \tan\beta = \frac{5}{3-x} \Rightarrow \beta = \arctan\frac{5}{3-x}

    \theta = \pi-\arctan\frac{2}{x}-\arctan\frac{5}{3-x}

    then:

    \frac{d\theta}{dx}=\frac{2}{x^2+4}-\frac{5}{x^2-6x+34}

    solve \frac{d\theta}{dx} for 0

    x=-2+2\sqrt{5}

    |AP|=3-x\Rightarrow3-x=-2+2\sqrt{5}

    |AP|=x\Rightarrow x=5-2\sqrt{5}

    So \theta is maxed when P is \approx0.5279 from A

    That's the answer in the back of the book

    Thanks for the help!
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  4. #4
    Junior Member Coco87's Avatar
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    Sorry, I also forgot to ask, why would you use tangent? I know it looks to be more convenient, but just curious if that is really why.
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    Quote Originally Posted by Coco87 View Post
    Sorry, I also forgot to ask, why would you use tangent? I know it looks to be more convenient, but just curious if that is really why.
    Since the angle you want lies between 0 and pi/2 degrees and tan is strictly increasing over that interval, the value of x that maximises \theta will also maximise \tan \theta.

    And well done getting the right answer.
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  6. #6
    Junior Member Coco87's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Since the angle you want lies between 0 and pi/2 degrees and tan is strictly increasing over that interval, the value of x that maximises \theta will also maximise \tan \theta.

    And well done getting the right answer.
    Ah, I see. Thanks again!
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