Well, it seems this node's author has no intention of providing a
solution to this problem of extreme importance to
frat boys worldwide, here I go:
a)
Firstly, we start with 12 ounces of beer:
b(t) = 12
Then we add 2 ounces per second:
b(t) = 12 + 2t
Then, we need to add the basic rate of Jim's drinking, being 1.5:
b(t) = 12 + 2t - 1.5t
Lastly, we add the rate at which he slows drinking, being c ounces per second per second. However, since this is a continual slide, not a series of steps down in speed, the actual figure we must insert is the integral of ct (constant * time), which is (ct2/2):
b(t) = 12 + 2t - 1.5t + (ct2/2)
This holds up to the point that Jim stops drinking the beer, but it does have a problem in that once t exceeds the number of seconds Jim spends chugging, his rate of drinking becomes negative (backwash!). However, for the purposes of this question I take it that the exercise ends once Jim stops drinking, so this shouldn't be a problem. Therefore the solution is:
b(t) = 12 + t/2 + ct2/2
b) It doesn't take a genius to see that if he starts drinking the beer at 1.5 ounces per second, and slows at .1 per second, he will stop after 15 seconds. So we simply substitute t = 15 into our function and see if it exceeds 22:
b(15) = 12 + 15/2 + 0.1*152/2
b(15) = 12 + 7.5 + 11.25
b(15) = 30.75
Therefore, as you can see, the beer bong will overflow before Jim finishes.
The exact time at which the beer bong overflows can be found by putting the equation into general form as a quadratic equation:
22 = 12 + t/2 + 0.1t2/2
0 = 0.05t2 + 0.5t - 10
this allows the use of the quadratic formula:
-0.5 +- sqrt((0.5)2 - 4*0.05*(-10))
t = ------------------------------------------------
0.1
This resolves down to:
t = 10, -20
Since the negative value is meaningless for the purposes of this question, we can safely say the beer bong will overflow after 10 seconds.
c) This question has no answer. Seeing as the rate of drinking is always slower than the rate of filling, it is clearly impossible for the beer bong ever to empty if the rate of drinking only falls. Of course, it would be possible for the beer bong to empty if the rate of drinking were to be increasing, but if that were the case, then Jim would never finish.
Indeed, even if Jim's drinking rate started higher than the rate of slowing, it would still be impossible for the beer bong to be empty when finished, as finishing requires he reduce his drinking to 0 ounces per second, while the filling rate is always constant. As such, unless Jim were to go from a drinking rate above 2 ounces per second down to zero instantly (which he can't do under the circumstances of the question), the rate of filling has to be higher than the rate of drinking for at least some period of time prior to finishing. In other words, no matter what the constants in this question, this problem has no solution.
However, just to prove it mathematically, here is the simultaneous equation for this problem:
0 = 12 + t/2 + ct2/2
t = 1.5/c (since that is how long it takes for Jim to finish)
which can be substituted in to form:
0 = 12 + 0.75c + 0.5c(1.5/c)2
0 = 12 + 0.75c + 0.5c*2.25/c2
0 = 12 + 0.75c + 1.125/c
0 = 0.75c2 + 12c + 1.125
which is another quadratic formula, so we again use the quadratic equation:
-12 +- sqrt(122 - 4*0.75*1.125)
c = -----------------------------------------
1.5
which resolves to:
c = -0.0943058, -15.9057
Both of these answers are totally meaningless for the purposes of the problem we have been given since they give forever increasing rates of drinking. Just in case you're curious, they are "valid" answers if we could accept negative time values, however this problem does not allow that. As such, it has been proven that there is no value of the constant which will result in Jim finishing drinking the beer bong at the same time that it empties.
So there you have it frat boys, next time you're chugging from a beer bong, you know how to calculate if it is going to overflow before you finish. After all, it's not that hard to remember and apply the quadratic formula while drunk is it?