Webwork 9.2 Hints

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Math 2300 Section 005 – Calculus II – Fall 2010

Hints for Webwork Section 9.2

Most of the Webwork problems for this section simply use the notions of geometric series that we developed in class. There are a few, however, that are challenging and require you to put a lot of thought in to setting up the series.

5. Consider a similar example. Us math folk really like coffee. Assume that at the start of each day I drink 300mg of caffeine. Also assume that the “halflife” of this caffeine in my system is 7.5 hours. So, in 7.5 hours I will effectively have 150mg of caffeine in my system, and in 15 hours that will be halved again and I will now have 75mg of caffeine in my system. Notice that at 24 hours (a full day after my last sip of coffee) I still have some caffeine in my system. How much? The answer: I take my initial amount and multiply it by half for each period of 7.5 hours. And since 24/7.5 of those periods happen in 24 hours, we find that I have

300\cdot\frac{1}{2}^{{24/7.5}}\approx 32.64\,\text{mg}

of caffeine in my system after 24 hours. At that instant, I drink another 300mg of caffeine and I now have

300+300\cdot\frac{1}{2}^{{24/7.5}}\,\text{mg}

in my system. This new amount is cut in half after 7.5 hours, and so on and so on. If we form a series with

a=300\qquad\text{and}\qquad x=\frac{1}{2}^{{24/7.5}}.

This tells me that upon drinking my coffee on day 1 I have 300mg of caffeine in my system. Upon drinking my coffee on day 2, I have

300+300\cdot\frac{1}{2}^{{24/7.5}}\,\text{mg}

in my system. Upon drinking my coffee on day 3, I have

300+300\cdot\frac{1}{2}^{{24/7.5}}+300\cdot\frac{1}{2}^{{2\cdot 24/7.5}}\,\text{mg}

and so on, and so on. The main idea here is that the quantity farthest to the left represents the 300mg of caffeine I just digested, while the quantity farthest to the right represents the caffeine I consumed way back on the first day.

6. Make sure for part (b) that you consider the fact that the vertical distance covered by the ball before it bounces for the first time is counted only once (since the ball only falls), while the vertical distance covered by the ball after that point is counted twice (since the ball rises and then falls). I.e., you may want to write the geometric series as if you were counting the first drop twice (it makes the geometric series easier to deal with) and then subtract anything that has been overcounted.

7. The same comment for 6 holds in this situation too. This problem is a bit on the challeneging side, but know that algebraic simplification can help.

© 2011 Jason B. Hill. All Rights Reserved.