# The Differential Equation Model for Exponential Growth - Problem 4

The solution to a differential equation dy/dx = ky is y = ce^{kx}. This can be used to solve problems involving rates of exponential growth and decay, such as the function for the temperature of a cooling object.

The population of a group of animals is given by a function of time, p(t). So, the rate of growth of the population is p'(t). If the rate of growth is proportional to the population, p'(t) = kp(t), where k is a constant. Now, this is of the form dy/dx = ky, so this differential equation can be solved to find that p(t) = ce^{kx}. Use the initial condition to find the value of c.

We could look at another example. This is actually a pretty famous example of Newton’s Law of Cooling. The problem is; a pie fresh out of the oven is cooling in a 70 degree Fahrenheit room according to the differential equation dh/dt equals k times the quantity h minus 70. This rule is called Newton’s Law of cooling.

Basically the way it works is, the rate of change of temperature, the rate that the temperature is going to decrease, because it just came out of the oven is proportional to the difference in temperatures of the room and the pie. So the smaller that difference in temperature gets, the smaller the rate of change of the temperature gets. Let’s use this differential equation to find the temperature function for the pie and we have actually have this formula up here.

This formula looks very similar to the differential equation that we’re starting with. We have the k, we have a difference, we have a derivative here my y has become h, my x has become t but that’s really the only difference. So my solution is going to be y minus a, or in this case, h minus 70, equals Ce to the kx, kt, because x is t in this problem. And k, again I don’t know the value of k, or c, but I can maybe use this fact; the pie was 150 degrees right from the oven. I can use that fact maybe to find one of the values.

Let me plug that in. Right from the oven, that means that t equals zero, so when I plug zero in for t, I could get 350 for h. H is the temperature of the pie. H(0) equals 350, gives me 350 minus 70 equals Ce to the k times zero. This is nice because k times zero is zero, e to the zero is 1, and that gives me that C equals 350 minus 70. C equals 280.

At least I have one of the parameters, C. So I have h minus 70 equals 280 e to the kt. Let me write that up here. 280 e to the kt. Now let’s make sure that I didn’t miss any other information. Cooling in a 70 degree room, that’s covered in the h minus 70. Pie was 350, I used that to get the 280, we’re good.

The last thing I need to do is to find k using the fact that after 30 minutes the pie is 250 degrees Fahrenheit. So you’ll notice that I haven’t got the value of k yet. So that’s what I’m going to do now. I’m going to use the fact that when t is 30, h is 200.

From this equation 200 minus 70 equals 280 e to the k times t, in this case would be 30. 200 minus 70 is 130. 130 equals 280 e to the 30k. I divide both sides by 280, that gives me 130, over 280 equals e to the 30k. And then I have to take the natural log of both sides. Natural log of 130 over 280 equals 30k. And finally the value of k is going to be 1/30 of this. Let me take that over here. K equals 1/30 natural log of 130 over 280. I’m going to do this on my calculator, get an approximate value and then I’ll be done. I’ll have my value of k.

Natural log of 130 divided by 280 divided by 30 equals -0.02557. I’m going to call it -0.0256. That’s my k value. So even though they didn’t ask for it, I’m going to write out the final function. I’ll move the 70 over. I get h equals 70 plus 280e to the -0.0256t. This is a function for the temperature of the pie after it comes out of the oven. You can see that over time, this term is going to disappear, it's exponential to k, this term is going to dwindle all the way to zero and the pie’s temperature is going to approach 70.

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