# Math Help - differentiate

1. 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.

$y = \frac{230}{tan\alpha}$

2. alpha equalling x?

3. 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.

4. 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?

5. 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

6. okay thank you, and just double checking, but tan can only be used on right angle triangles...

7. just double checking, but the derivitive of is y=1/(cos^[2][230\x])

8. 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)
$

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

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

9. oh my gosh i love you so so much!!!

10. remember ... it helps a great deal if one knows the original problem statement.

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