Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Some useful review. The derivative of X to the north is n * X to the north -1. The derivative of cosine X is negative sine X. The derivative of sine X is cosine X. The derivative of tangent X is secant squared X. The derivative of natural log X 1 / X. The derivative of E to the X is E to the X. The Pythagorean identities cosine squared X plus sine squared X equal 11 plus tangent squared X equals secant squared X cotangent squared X + 1 equal cosecant squared XE to the natural log of U equal U natural log of E to EU equal U. The product rule. The derivative of U * v where U&V are functions would be U prime v + v prime U. The quotient rule. The derivative of two functions U / v derivative of the top times the bottom minus the derivative of the bottom times the top all over the bottom squared or U prime v -, v prime U all over v ^2. And then the chain rule. The derivative of a function inside of a function. We're going to take that inside function and find its derivative V prime, but then it's going to get multiplied by the derivative of EU substituting into the V equation. So we have V prime U prime of V. That might have sounded a little confusing. Let's do that again. We're going to take the derivative of the inside, IE the V prime here. Then we're going to take the derivative of U. But we're not going to change the inside. So it's basically the inside outside rule. We're going to take the derivative of the inside times the derivative of the outside, leaving the function alone on the inside when we do the derivative. So this U prime V is all one thing together, mathematical models solving real world problems using mathematics. So we're going to start by looking at some verify equations. But eventually we're going to lead up to if we're given a situation and we know some initial conditions, what's the original condition or equation that fits those real world situations. So we're going to verify if we have Y prime plus two y = 0, given Y equal 3 E to the -2 X, the first thing we're going to do is find the derivative. So E to the -2 X. We're going to use the chain rule here. So we're going to take the derivative of the -2 X, which is -2 and then we're going to times it by the derivative of E to something. And the derivative of E to something is just E to something. And we still have that constant of three there. So Y prime is just going to be -6 E to the -2 X. So when we're trying to verify this equation here, we're going to stick in Y prime -6 E to the negative two X + 2 * y. Well, the Y was given as 3E to the -2 X. We can see that that does indeed prove to be true. Looking at another one, this time we're going to do it in two parts. We have AY one, and we're going to do it all the way through. And then we're going to go back and we're going to do the verify again for Y2. And basically what we're trying to see in this equation is the realization that there can be many different equations that fit a given situation. So just because we find one that fits doesn't mean that we have found all of them. If we look at Y sub one equal E to the -2 X, we can have Y prime here being -2 E to the -2 X&Y double prime would be 4 E to the -2 X. So when we want to verify this, we get 4E to the negative two X + 4 * y prime, which was -2 E to the -2 X plus 4Y and our original Y was E to the -2 X and we want to see if that equals 0. So if we have 4 E to the -2 X -8 E to the negative two X + 4 E to the -2 X, we can see that that does indeed equal 0. So if we come to this other one and we have Y prime here, we've got to use the product rule. So the derivative of plain old X is 1. So we get E to the negative 2X plus the derivative of E to the new -2 X is going to be -2 E to the -2 X. And then that's still going to get multiplied by the X. So our Y prime, if we simplify it up, we get E to the -2 X -2 xe to the -2 X. When we do our Y double prime, E to the -2 X, we're going to bring down the -2 because the derivative of the exponent. So we're going to get -2 E to the -2 X. And for this next term, we have to do the product rule again. The derivative of X is just one. So we'd have -2 E to the -2 X. But then the derivative of the E to the -2 X derivative of that exponents -2 so -2 * -2 is going to give us +4. The X is going to come along for the ride. E to the -2 X. Oh, where did it go? 4 xe to the -2 X. So if we simplify this up, we get -4 E to the negative two X + 4 xe to the -2 X. So when we go back to our verify equation, we would get -4 E to the negative two X + 4 xe to the negative two X + 4 * y prime. Well, Y prime was E to the -2 X -2 xe to the -2 X plus 4Y and our Y was xe to the -2 X and we want to show that equals 0. So here we get -4 E to the negative two X + 4 xe to the negative two X + 4 E to the -2 X -8 xe to the negative two X + 4 xe to the -2 X. When we look here and combine our like terms, there's a -4 and there's a +4 E to the -2 XS, and here's a +4 and -8 and a +4 of XE to the -2 X. So this does indeed work. What we've really done is we've shown that multiple equations can work in an original initial condition. So Y prime plus 4Y prime plus four y = 0. There's more than one equation that would make that condition true. We want to solve for R given Y equal E to the RX and for Y prime double prime equal Y. So if we know that Y equal E to the RX, we know that Y prime equal. The derivative of the exponent is going to be R because R is a constant. So we get RE to the RX and Y double prime is going to be r ^2 e to the RX. So we want to know when does 4 R-squared E to the RX equal E to the RX. Well, we could think about taking everything to one side and having it set equal to 0. We can see both those terms have an E to the RX in them, so we could factor that out. We'd get four r ^2 -, 1 equaling 0 * E to the RX. So E to the RX equals 0 OR OR and four r ^2 - 1 = 0. E to the RX can't ever equal 0. If we think about taking the natural log of each side, we get RX equaling the natural log of 0, which doesn't exist. But over here if we thought about adding one, dividing by 4, and square rooting, we can see that R could equal positive or negative 1/2. Looking at another example, we're going to solve for C. In this case, we're given that Y prime equal to YY of X equals CE to the 2X and Y of 0 equal three. Well, if we know Y of 0 equal 3, we know every time X IS0Y is 3, so 3 is going to equal CE 2 * 0. Anything to the 0 power with the exception of 0 to the zero is 1. So in this case we can see that C is 3. So now we have a new equation and our equation says Y equal 3 E to the 2X. We want to show that this is true for this equation. So Y prime would equal 6 E to the 2X. And does 6E to the 2X really equal 2 * 3 E to the 2X? And the answer is that is a true statement, which is what we were trying to show. We're trying to solve for C, but we needed it to be true for this condition. If we had solved for C and when we substituted it into here it wasn't true, there would be no solution and it would mean that we couldn't have all three of those conditions being true at the same time. Solve for C given Y prime plus Y tangent X equal cosine cosine XY of X equal X + C cosine XY π = 0. Once again, this is saying every time we see the X, the Y is 0. So we're going to start with saying 0 = π + C times cosine of π and we know that cosine of π is -1. We could divide each side by a -1 so we would get 0 equal π + C or C is going to equal negative π as C is negative π. Now we actually have an equation of Y equaling X - π times cosine X so Y prime. We're going to have to use the product rule. The derivative of this first term is going to be one times the second, plus the derivative of cosine X is negative sine X times the 1st. If we simplify this up, we can see that this is cosine X -, X sine X + π sine X. So now we want to see if this condition here is going to be true when we substitute. So we're going to have cosine X -, X sine X + π sine X + y, which in this case was X - π cosine X times tangent X equaling cosine X. So tangent, we know a sine over cosine, and we can see that those cosines, one in the numerator, one in the denominator, are going to cancel. So we're going to end up with cosine X -, X sine X + π sine X. Go ahead and do the distribution. So plus X sine X minus π sine X equaling cosine X. Remember there was this X - π times that term. So when those cosine X is cancelled, we still had X times sine X and negative π times sine X. When we look here we can see that the negative X sine X and the positive X sine X are going to cancel. We can also see that π sine X and negative π sine X are going to cancel, so the left side and the right side equal each other. So all three of those conditions were met. The time rate of change of a population P is proportional to the square root of P. We're trying to figure out how to set up this equation and then to solve it. So when we look at this one, we have a proportionality. So we're going to have the time rate of change of population DP, DT. The population change in terms of time is proportional to proportionality means there's always some constant. In this case, we're going to use K to the square root of P. Now we can actually solve this through integration. What we're going to do is we're going to take everything with AP to one side and everything without AP to the other. So we get DP over square root of P equaling KDT. We could think of the square root of P as P to the -1 half DP equaling KDT. And then we're going to integrate each side. If I integrate the left side, I'm going to integrate the right side. Here we want to figure out how to integrate. So we're going to add 1 to the exponent and divide by the new exponent. If I divide by 1/2, it's the same thing as multiplying by two. Over here on the right side, T is our variable. K is just a constant, so the right side is going to turn into KT plus some constant C So the relationship would be 2P to the half equaling KT plus C. So the time rate of change of a population is proportional to the square root of P. This is the original equation based off of the derivative. We want to find at least one solution that will make this true. Now we're trying to come up with things that we already know. What's an equation where the derivative would equal the original function? Well, the derivative would equal the original function if we started with Y equal E to the X because the derivative of E to the X is just E to the X. A more simplistic case would be if we had Y equaling 0 because the derivative of a constant is always 0. So those are two different examples where the first derivative equals the original function. This next one a little harder. We need Y prime squared plus y ^2 to equal 1. Well, if we thought about our Y prime squared equaling zero, we could think about what squared would give us one. If y = 1, Y prime derivative of a constant is always zero, and we can see that 0 ^2 + 1 ^2 would equal 1. This would also hold true if we'd let Y equal -1 because the derivative of a constant is 0 and we'd get 0 ^2 + -1 ^2 would equal 1. And those are our examples. Thank you and have a wonderful day.