Sunday, May 13, 2012

GeekDad Puzzle of the Week Answer: Coffee Conundrum II

Coffee Conundrum II:
Q: Each cup of coffee gives me a jolt and then the jolt decays across time according to the following equation (t in minutes): Jitters=10-[(t-10)^2]/10
So at minute 10 after consuming a cup of coffee (which, for the purposes of this puzzle happens instantly), I reach a maximum of 10 jitters. At a combined 20 jitters, I go catatonic. With what frequency can I instantly consume coffee without the combined jitters passing this important tipping point?
The diagram shows the exact tipping point situation. Focus on the red curve which is the second cup of coffee. Right at the middle of the parabola, the second cup is at its maximum (10 jitters). At this point, the effects of the first cup are diminishing and the effects of the third cup are increasing. Notice, because of symmetry, they have the exact same value when they cross. The tipping point will be when cup 1 is contributing 5 jitters, cup 2 is at its maximum of 10 jitters and cup 3 is also contributing 5 jitters (5+10+5 = 20). (Note you can extend this diagram out to 4 cups, 5 cups, etc. but you'll never have more than 3 cups contributing to the total jitters at once. If you did, you'd definitely be over the tipping point of 20.)

As stated in the puzzle, the height of one parabolic curve is given by the formula:
Jitters = 10 - [ ( t - 10 ) ^ 2 ] / 10

Solving this for a Jitters value of 5 we have:
5 = 10 - [ ( t - 10 ) ^ 2] / 10
-5 = -[ ( t - 10 ) ^ 2] / 10
5 = [ ( t - 10 ) ^ 2] / 10
50 = ( t - 10 )^2
t - 10 = √50
t = 10 ± √50
t = 10 ± 5√2

That means that the coffee Jitters are at a height of 5 either 5√2 seconds before the peak or 5√2 seconds after the peak. This also happens to be the minimum frequency between cups (5√2 seconds or approximately 7.071068 seconds apart) to avoid jitters.

If you can handle exactly 20 jitters without going catatonic, then you could handle drinking cups of coffee every 5√2 seconds (≈7.071068 seconds)

If you must stay under 20 jitters, then you would have to pick a frequency just over every 5√2 seconds.
A: Frequency of 5√2 seconds (or approximately every 7.0710678118654752440084436210484903928483593768847403 seconds)

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