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Wednesday, November 19, 2008

5-foot Girl and her Shadow...

Okay so today in class was a pretty fun one. We started with a problem: A light is at the top of a 16 foot pole. A girl, 5 feet tall walks away from the pole at a rate of 4 ft/sec. At what rate is the tip of her shadow moving when she is 18 ft. from the pole? At what rate is the length of shadow increasing?

Okay so we tackled this question together. First we wrote out everything we had, and everything we need, thanks to the question we already had a rate to work from. Which was the rate at which the girl walked away from the pole, it was 4 ft./sec. We had the height of the pole: 16ft, and we had the girls height: 5ft. We need to find the rate when the shadow is moving and the rate of the length of the shadow as it increases. Let's draw out a diagram!

As you can see, that is a triangle, not just a regular triangle, it's a right angle triangle. Okay, but there are 2 triangles, and they happen to be similar, because they are proportional, just one is smaller, with smaller sides. Okay, we'll get back to that later, but for now we should write down our rates.

Our rates are in relation to time, because if there was no time, there would be no movement going on, and well we wouldn't have a moving shadow or girl without time.. right? We're given the rate of the distance covered by the girl which is 4ft./sec, so our first rate is equal to 4. Our second rate is the rate at which the shadow is moving and our final rate is the rate of the length of her shadow as it increases. Now, all the letters such as "d", "s" and "a" are given by our diagram. From our diagram, we got "a" from adding together d and s (d + s = a, pay attention to this equation, we're going to use it later on)
. We should find some ratios, because the triangles in the diagram are similar (refer to diagram), we know that side "s" is to 5 ft. as side "a" is to 16 ft. s/5 = a/16, from this we know that 16/5 = a/s. We know that a is equal to d+s, now we just throw it in.
First things first, substitute a for d+s and rewrite the equation, then put like values on one side and solve by finding the lowest common denominator, etc. Substitute dd/dt with 4
, because that's what we were originally given. simplify 4/16 to 1/4. Divide out 11/80 to finally find ds/dt, which is the answer to one of out questions! The rate at which the tip of the shadow is moving when the girl is at 18ft. away from the lamp post.

Now that we have ds/dt and dd/dt, we can solve for da/dt.

That was the answer to our last question, the rate the length of her shadow is increasing.

We also learned a little jig for the Quotient Rule, we rehearsed it a few times so feel free to practice. It goes like this: lo - di- hi minus hi - di - lo all over lo - lo. Where hi is the numerator of the given function, lo is the denominator and di is short for derivative.

Last thing we did in class was similar to the question we got in the beginning, but I'm not sure what it was about, because we didn't have much time to work on it. I'm sure we'll go over it thoroughly next class though. But for now, my post is done! Next scribe will be Joyce.

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