I hope this isn't a silly question. I'm learning single variable calc, and having lots of fun with optimization problems. This isn't exactly an optimization problem, but something that came up while working on one.

Let's say I have a small circular garden with a short brick border. This border is perhaps 1 foot tall, so that any sun or rain that reaches the flowers has to come from directly overhead. Suppose that the radius from the center of the circular flowerbed to the outermost edge of the circular brick border is $r.$ I plant a metal rod at the circle's center. At the top of the rod is a fan blade of sorts: it's flat, thin, parallel to the ground, and has the shape of a circular sector with radius $r.$ This blade is opaque, so it provides some shade for that part of the flowerbed beneath it.

I give the blade a spin: as it's spinning, all of the flowerbed receives some shade. Then I get an idea: I automate the spinning of the blade. I can control the angular velocity, $w,$ of the blade with a remote control. Let $l$ be the amount of light (or, if you want, rain) admitted to the flowerbed. My question is this: Is it the case that $$\lim_{w\to \infty}l=0?$$

I have reasons for thinking this is the case, and other reasons for thinking it's nonsense. And if it is the case, then it's true regardless of the value $\theta$ of the central angle of the circular sector, right?

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    $\begingroup$ Assuming I understand the question correctly, the result depends only on $\theta$, and the amount of light received by the flowerbed is completely independent of $w$. $\endgroup$ – Alex Becker Aug 15 '13 at 1:05
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    $\begingroup$ Physically, what you're going for is probably nonsense when it comes to light or even rain. Here's a formulation that might be more along the intuitive lines you're going for: let's say that instead of blocking sun, you're blocking an assault of arrows (as in the weapon bow and arrow) of length $L$, which pass the horizontal-spinning blade vertically at speed $v$. How fast would the blade need to spin to block all arrows from hitting the grass below (and yes, such a speed exists)? Answering this might give you a better intuition about the problem with light and rain. $\endgroup$ – Omnomnomnom Aug 15 '13 at 1:28
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    $\begingroup$ Ok. So say that $v$ is measured in feet per second, and $L$ is likewise measured in feet. And to keep things clean we'll let $L$ be less than the height of the blade. Then the rotating blade has $\frac {L}{v}$ seconds to hit an incoming arrow--and that's just to keep the blade from missing the arrow altogether. I guess I see what you mean. With rain, we're shrinking $L$ (though maybe we're slowing down $v$). With light, we're making $L$ extremely small and increasing $v$ to the greatest speed possible. $\endgroup$ – Ryan Aug 15 '13 at 1:43
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    $\begingroup$ @Ryan yep, you've pretty much got it. It might be physically possible for rain, in a sense. Light seems out of the question, though. Then again, in a physical setting, you might be able to fan the rain away. $\endgroup$ – Omnomnomnom Aug 15 '13 at 1:53
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    $\begingroup$ @Omnomnomnom Great answer, BTW! The arrows problem really helped me gain some intuition about this. $\endgroup$ – Ryan Aug 15 '13 at 2:11

If the altitude of the fan blade is essentially the same as that of the wall surrounding the garden, then no "side light" needs to be factored in to the final answer. If the fan speed is controlled at a speed fast enough to eliminate any possibility that the blade might hover longer over a particular area during greatest sunlight intensity, then we don't need to take that into account. Given these things, and given $\theta$ as the single parameter identifying the area of the fan blade by circular section, we can then say:

$$l = 1 - {\theta \over 2\pi}$$

where $l$ is the amount of light reaching the garden. This is achieved by noting that $\theta {r^2 \over 2}$ is the area of a circular section, and $\pi r^2$ is the area of the entire circle. Then the difference of these areas is the section allowing light though the fan. Then the final formula starts as:

$$l = {\pi r^2 - \theta {r^2 \over 2} \over \pi r^2}$$

and it is easy to reduce from there.


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