6.8 Try It Problems [STARTS AT 4:40] INSTRUCTOR: Now, in our next example, according to experienced baristas, the optimal temperature to serve coffee is between 155 degrees Fahrenheit and 175 degrees Fahrenheit. Suppose coffee is poured at a temperature of 200 degrees Fahrenheit, and after 2 minutes in a 75-degree Fahrenheit room, it has cooled to 185 degrees Fahrenheit. When is the coffee first cool enough to serve. When is the coffee too cold to serve? Round answers to the nearest half of a minute. I'm going to solve this question differently than the last one, but we'll begin the same. So this is an exponential decay problem because the coffee begins hot and decreases. So the equation we'll use is y equals y0 e to the negative kt. We have an initial temperature of 200 degrees. This is y0 equals 200. And we have a point in time. We have after 2 minutes in a 75-degree Fahrenheit room, it is cooled to 185 degrees. So we have the ordered paired t, comma, y equal to 2, comma, 185 as the temperature after 2 minutes. All right. Now, the way I'm going to do this differently is I'm going to solve the equation, y equals y0, e to the negative kt for both k and t. Let's begin writing it as y equals 200e to the negative kt. We will divide by 200 on both sides. So we have y divided by 200 equals e to the negative kt. We will take the natural log of both sides. So we have the natural log of y divided by 200 equals natural log of e to the negative kt. Natural log of e is 1, so this means we'll have this equal to natural log of y over 200 equals negative kt. What this means is we can diverge or split in two directions. We can say that k is equal to negative 1 over t times natural log of y over 200, or we could say that t is equal to negative 1 over k natural log of y divided by 200. Both of these are useful because they're related. But when you first need to find what k is-- we'll use k equals negative 1 over t times natural log of y over 200. We will use the t value of 2 and a y value of 185 so that k is equal to negative 1 over 2 natural log of 185 divided by 200. Now, that actually simplifies to be 37/40, so this will be-- so k is equal to negative 1/2 times natural log of 37 over 40. Hold on to that. We're adding this to our bank of information. Now, the reason I have solved this equation also for t is because we are given, a couple of times, our temperatures to find. We're trying to find the time at two different temperatures. We have a temperature when it's 155 degrees Fahrenheit and 175 degrees Fahrenheit. So at 155 degrees Fahrenheit, those are y values. This means that our time is equal to negative 1 over k. So let's just write this as negative 1 divided by negative 1/2 natural log of 37 over 40 times natural log of 155 divided by 200. Let me just get my calculator and find out what that should be. OK. So that actually could simplify to be 2 over 2 over natural log of 37/40 times natural log of 155/200. In decimal, that is-- that is 6.5 minutes. 6.5 minutes. Now, replacing the value of 175, the second time that I'm looking for, is going to result in t equals-- I'll go ahead and write that as 2 over natural log of 37/40 times natural log of 175/200. And that will be that t is approximately 3.5 minutes. Now, I just realized that our question actually says round to the nearest half of a minute, which means these would have been 3.5 and 7, probably, or still 6 and 1/2. So 6 and 1/2 minutes when it is 155 degrees and 175 degrees in 3.4 minutes. [ENDS AT 11:33]