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Math Help - differentiate

  1. #16
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    Without knowing the viewing capacity of the eye, the only answer you can give is a function of the optimal viewing distance y as the maximal viewing angle \alpha changes.

    y = \frac{230}{tan\alpha}
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  2. #17
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    alpha equalling x?
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  3. #18
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    You can call it whatever you like as long as you know what it stands for. In this case, \alpha or x is the maximal viewing angle.
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  4. #19
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    okay, so how does knowing the viewing capacity affect the equation? i think the equation is ment for people with 20/20 perfect vision, as it hasnt been specified that the view has imperfections. does this help?
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  5. #20
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    Well if you know what the maximum viewing angle \alpha is, then you simply input it into the function and you'll get the optimal viewing distance y.

    So if \alpha = \frac{\pi}{4}, then y = \frac{230}{tan\frac{\pi}{4}} = 230 cm
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  6. #21
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    okay thank you, and just double checking, but tan can only be used on right angle triangles...
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  7. #22
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    just double checking, but the derivitive of is y=1/(cos^[2][230\x])
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  8. #23
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    let \alpha be the angle of elevation from eye level to the bottom of the painting.

    let \theta be the angle from the bottom of the painting to the top of the painting, i.e. the viewing angle.

    let x be the horizontal distance to the wall in meters.

    then ...

    \alpha = \arctan\left(\frac{0.3}{x}\right)

    \theta + \alpha = \arctan\left(\frac{2.3}{x}\right)

    so ...

    \theta = \arctan\left(\frac{2.3}{x}\right) - \arctan\left(\frac{0.3}{x}\right)<br />

    \frac{d\theta}{dx} = \frac{\frac{-2.3}{x^2}}{1 + \frac{5.29}{x^2}} - \frac{\frac{-0.3}{x^2}}{1 + \frac{0.09}{x^2}} = \frac{-2.3}{x^2+5.29} + \frac{0.3}{x^2 + 0.09}

    setting \frac{d\theta}{dx} = 0 yields the equation ...

    \frac{2.3}{x^2+5.29} = \frac{0.3}{x^2 + 0.09}

    solve for x ...

    x = 0.83 meters
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  9. #24
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    oh my gosh i love you so so much!!!
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  10. #25
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    remember ... it helps a great deal if one knows the original problem statement.
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