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Thread: Trig and Calculus:

  1. #1
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    Question Trig and Calculus:

    Trig and Calculus:-problem.jpgTrig and Calculus:-2.jpg

    The textbook I used didn't provide a detailed enough explanation.
    How should I approach this problem?


    Thank you in advance!
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  2. #2
    MHF Contributor
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    Re: Trig and Calculus:

    let $\arctan2(x,y)$ be the quadrant adjusted arc tangent of point $(x,y)$

    $\theta = \arctan2(x,400)$

    $\beta = \arctan2(x,225)$

    $\alpha(x) = \theta-\beta = \arctan2(x,400)-\arctan2(x,225)$

    $\dfrac {d \alpha}{dx}= \dfrac{225}{x^2+50625}-\dfrac{400}{x^2+160000}$

    Now solve

    $\dfrac {d \alpha}{dx} = 0$ for $x$
    Thanks from Nour2712
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