Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Exponential change and separable differential functions. The law of exponential change states Y equal Y not E to the KT, where K greater than 0 means growth and K less than 0 is DK and K is the rate constant of the equation. Y not is the initial amount and Y is the final amount after time T differential equation. If we have dy DT equal KY, we're going to put all of the YS on one side and everything else on the other. So we're going to end up with D, y / y equaling K DT and now we're going to integrate both sides. So the integral of 1 / y DY is going to be the natural log of the absolute value of Y, the integral of KDT. Remember K is a constant, so that's just going to be KT plus some constant C. Now that C could come on either side, we want to have it on the right so that when we take E the each side to the base E. So if I have natural log of the absolute value of Y equaling KT plus C, if I take each side to the E power, that E to the L1 is going to cancel leaving us the absolute value of Y that's going to equal E to the KT plus C. Now a different way to think of that would be just to change that logarithmic form into exponential form. So whenever we have two things in the exponent that are added, we can think about separating those as each of the bases being multiplied together. So we get Y equal absolute value of Y equaling E to the kte to the C where E to the C is some constant. We call that constant A. So we get Y equaling some constant AE to the KT. Now we can drop the absolute value because E to any number, E to any C is going to be positive, and E to the KT is also going to be positive. Next we're going to think about what happens when t = 0. So if t = 0, we're going to realize that we have the final value and the original value is going to be the same at times 0. So Y not equal AE to the K0. Anything to the 0 power is 1. So our Y not is really just A. So Y equal Y not E to the KT. Sometimes you'll see it as Y equal AE to the KT, where A is that initial value. Number of cases of disease is reduced by 20% per year. If 10,000 cases today, how long for 1000 cases? So we're going to use this Y equal Y not E to the KT equation. If we're reduced by 20%, it means we have 80% left. So .80 of our 10,000 original equals our 10,000 E to the K and this is per year. So one is our time. We're going to find our constant, so we're going to start by dividing each side by 10,000, reducing it, and we get 4/5. Change the exponential into logarithmic, so we get K equaling the natural log of 4/5. Now once we find K, it's going to be true for every single problem that uses this information. So now we want to know how long. So we're trying to find the time for 1000 cases. So if we have 1000 equaling the original amount of 10,000 E to the K, we found the KA moment ago, Lane 4/5 * t we're going to do the same steps. We're going to divide by 10,000. We're going to get 110th. We get Ellen 4/5 * t, but that T could go up into the exponent by the power rule for logarithms so that this E to the Ln would cancel. So we get 110th equaling 4 fifths T If 110th equals 4/5 T, take the natural log you each side natural log and 110th equal natural log of 4/5. Bring the T down in front and divide each side by lane 4/5. So lane 110th divided by lane 4/5 is about 10.32 years. We can do problems that are involving interest also. So the interest paid by compounding continuously A equal A not E to the RT or sometimes it's thought of P equal P not E to the RT. Pert shampoo is how we frequently have students try to help themselves. Remember this Pert, the Pert shampoo radioactive DK, The K is going to be a negative when we have DK, so we could think of it as a negative of a positive. So if K was greater than 0, some books talk about Y equal Y, not E to the KT where K is less than zero. I just want you to be familiar that it could be written either way and it's dependent on whether we're putting a stipulation K is greater than or K is less than 0. If we're doing a half life problem, half life would mean that we end with half of what we started with. So 1/2 of Y not would equal Y, not E to the KT. the Y nots would divide out. We'd get 1/2 equal E to the KT putting taking the natural log of each side. So natural log of 1/2 natural log of E to the KT. The natural log of E are going to cancel so we get KT on the right side. 1/2 we could think of as Ln to the two to the -1 power and then we could bring that -1 down in front. So negative Ln 2 is going to equal KT. Solving for K, we can get the constant is negative Ln 2 / T, or solving for time, time equal negative Ln 2 / K So we either need the constant or the time to be able to figure out the relationship on a half life. Newton's law of cooling H is temperature of an object, T is time, HS is the temperature in the surrounding area. So DHDT equal negative KH minus HS. So we're going to start by letting Y equal H minus HS. If Y equals that, then we take the derivative in terms of time. So we get DYDT equal DDT of H minus HS or DYDT equal DHD T -, D HSDT. Now this HS is the temperature of the surrounding, so it's a constant. The temperature of the surrounding is not changing, and we know that the derivative of many constants just zero. So DYD T = D HDT. So DYDT from up here is just negative KH minus HS because we decided that or we showed that DHDT was really just DYDT. So by our transitive property, those three things have to all equal each other. Now we're going to take and put all the YS on one side. That H minus HS we said was Y. So we're going to have D, y / y equaling negative KDT taking the integral of both sides just should look what like what we did in the beginning. So we'd end up with the natural log of the absolute value of Y equaling negative KT plus some constant. Changing that index potential form, we get Y equal E to the negative KT plus C separating that up. So we have E to the C, E to the C We're just going to call some constant a. So when t = 0, we know that the original and the ending is the same amount. So Y would equal Y not. When time is 0. Solving for that, we get a equal Y not. This is the equation we had before. So Y equal Y, not E to the negative KT. We basically have just proven it twice now. Since H minus HS equaled Y, we can see that H minus HS equal H naughty minus HSE to the negative KT. That's Newton's law of cooling. Thank you and have a wonderful day.