4.3.2

[STARTS AT 10:40]

Instructor: A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is 350 degrees Fahrenheit. The temperature of the kitchen is 75 degrees Fahrenheit. And after 5 minutes, the temperature of the pizza is 340 degrees Fahrenheit. 

We would like to wait until the temperature of the pizza reaches 300 degrees before cutting and serving it. How much longer will we have to wait? 

All right. Well, our initial temperature is 350. We're talking about the temperature, and it is going to become the ambient temperature. It's going to start at some T0, the initial temperature-- in our case, 350 degrees-- and it will slowly approach 75. Slowly approach 75 in a rate that is proportional to the difference between its own temperature and the ambient temperature. 

And if we let T of t represent the temperature, then dT of dt, capital T being our temperature, lowercase being our time, and that tells the rate of change. That's this right here. And again, this is a solution, or the solution, general solution to this. 

But let's first start by writing this out as an initial value problem. We've got dT over dt, temperature over time, equals our proportionality constant, which we don't know at this point-- we will find us out-- times T minus our TS. This actually represents the ambient temperature. So here Ta equals TS-- or I used a capital. Those two are actually synonymous. 

So here our ambient temperature is 300, no, it's 75, that's at 75. And we also know that T0, or T of 0, is equal to 350. That is our initial temperature. So there's our initial value problem. 

So setting this up in such a way, we can write this as the integral dT divided by T minus 75 equals-- I'm going through the antiderivative in here-- the integral of k dt, so that we have natural log absolute value of T minus 75 equals kt plus c. 

Now, if we exponentiate at this point, this should begin to take shape, begin looking like that formula from 2.8. So we have natural-- no. No more natural log, just the absolute value of T minus 75. And I'm going to drop the absolute value, because we're going to deal with that with the constant. e to the kt. I'm going to make that c1 where c1 equals e to the c. And adding 75 to both sides, we find that T is equal to 75 plus c1 e to the kt. 

Now let's use our value of T of 0 being 350 to solve for c1. Again, evaluating 0 for T, this becomes c1, 75 equals 350. Subtracting 75, we have 275 equals c1. So that our specific, our particular solution, is T equals 75 plus 275e to the kt. 

Now, we slide back to this equation. T equals T0 minus T ambient, which would be 350 minus 75. Hey, that's 275e to the kt plus T0, which is just the 75. So that formula does in fact work. 

Now we don't know the k value yet. But we were given another piece of information here. We were told that after five minutes, the temperature is 340, which tells us that T of 5 is 340. 

So let's take this equation, apply it to that. T of 5 equals 340. So this is going to be 340 equals 75 plus 275e to the k times 5. 

All right, let's see. Subtracting 75 from both sides, we get 265 e the 5k. So that was minus 75. We'll go ahead and divide by 275, which reduces to 53/55. And that is e to the 5k. We will apply the natural log-- 

Woman: These are the top results. 

Instructor: --to get, let's see, 5k equals natural log 53/55 and then dividing by 5. So k is equal to natural log 53/55 divided by 5, making our particular solution. Now we have a complete picture T equals 75 plus to 275 e to the, let's go ahead and make that approximate value and calculate, we can approximate that as negative 0.007408. So that's e to the negative 0.007408 t. 

All right. Now to actually answer the question, at what point will the pizza be 300 degrees Fahrenheit? When will that be 300 degrees Fahrenheit. Well, now that we have our complete particular solution, we can actually do some algebra. That's all we have left at this point. 

300 equals 75 plus 275e to the negative 0.007408t. We will subtract 75 to make this 225. e to the negative 0.007408t dividing by 275 subtracting 75 there. And that reduces to 9/11. 

So 9/11 equals e to the negative 0.007408t. Taking our natural log, as we did in the last, to find k. Natural log of 9/11 equals negative 0.007408t. 

And finally, dividing by that value of k, 7408, we get that t is approximately 27.09 minutes. 

Now at this point we have waited already, 5 minutes. So that means that we need to wait 22.09 more minutes.