Back To CourseMath 104: Calculus
14 chapters | 116 lessons | 11 flashcard sets
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Let's think for a minute about the mad scientist in some movie who's trying to take over the world. Have you ever noticed that his plans, his dastardly plans, always fail for some reason? Well often, they show some chemicals bubbling in a beaker somewhere, so I think his dastardly plans fail because he doesn't understand what chemists and chemical engineers understand, which is, he doesn't understand differential equations.
Remember that a differential equation relates a variable with its rate of change. So let's think about a chemical reaction. Now an engineer and chemist, they understand that you can write the amount of chemical that you have, let's call it A, how it changes as a function of time. So you've got dA/dt. dA/dt = -kA. So in this case, what you're doing is you're throwing chemical A into a pot, and you're letting it react. And it's going to react with a rate k times the concentration of A, where k can be any number you want. People who understand differential equations can take this differential equation, then, and determine from it the concentration of A as a function of time. So can you write A as a function of time?
Well let's go back, and let's think about equations that we're more familiar with first. Let's say that we're looking at Super C, the human cannonball. We know that Super C starts out at a height of 0 at time t=0. When you shoot him out of the cannon, he's shot out at 13 meters per second, straight up. This means that his velocity straight up is 13 meters per second. He's always pulled down by gravity, though, which is 9.8 meters per second, squared, downwards. So given all of this information, can we determine his height as a function of time? Well first he's pulled down by gravity, which is acceleration at -9.8 meters per second squared. Acceleration is nothing more than the change in velocity over time. So I could write his acceleration at all points along his flight as being dv/dt=-9.8. It's minus because it's always being pulled back toward Earth. Now if I want to use this equation to find his actual velocity as a function of time, I'm going to multiply both sides by dt and integrate. Well, I know what the integral of dv is; it's just v. And I know what the integral of -9.8dt is; it's -9.8t. I have to add a constant of integration here, because I'm not integrating over some set time, I'm just taking an indefinite integral. So I end up with velocity equals -9.8t, plus my constant, C.
Now if I actually want to know where he is at any given point in time, I need to determine what C is. So what do I know? I know that his velocity when he was shot out of the cannon, so his velocity at time = 0, was equal to 13. So if I plug in 13 for v and 0 for t, I find that this constant, C, must be equal to 13. Okay, so we've got his velocity as a function of time. Velocity = -9.8t + 13. Can I find his position from this? Well, his velocity is dx/dt - it's how fast his position is changing with respect to time. So if I set this equal to dx/dt and multiply both sides of the equation by dt, I can integrate and find the integral of dx equals the integral of (-9.8t + 13)dt. Well, x=-4.9t^2 + 13t, plus my constant of integration. Once again, if I want to know where he is at any given point in time, I can't actually leave this constant of integration here. I need to solve for C somehow. Well, I know that at time t=0, he was at 0 height - he was just about to be shot out of the cannon, or just starting to be shot out of the cannon - so at t=0, x=0, so C has to be 0. I know that his position, his height, as a function of time, equals -4.9t^2 + 13t. So what, exactly, did I do here, other than find out how high he was at a point in time?
Well, I solved a differential equation. In particular, I solved the equation, the second derivative of x with respect to time equals -9.8, and I solved that for x as a function of time subject to what I'm going to call the initial conditions x=0 and dx/dt=13. That is, x=0, and the velocity equals 13 at time equals 0; hence, it's the initial conditions. So what does all of this have to do with determining the concentration of chemical A to avoid the dastardly fate of our dastardly mad scientist? Well, here's my differential equation: dA/dt=-kA. To solve this, to find A as a function of time, I need to integrate. But in order to integrate, I need to have t on one side of the equation and A on the other. If I just multiply both sides by dt, I still have an A on this side of the equation.
So this point is known as the separation of variables. Separation of variables means that we're going to rewrite a differential equation, like dx/dt, so that x is only on one side of the equation, and t is only on the other. This is kind of like making an explicit equation. Not all differential equations can be separated. So not all equations can be solved explicitly; some equations are implicit. It's the same thing with differential equations. But when you can write them explicitly, with x on one side and t on the other, you can use this separation of variables concept.
We actually did that with Super C. We had dv/dt=-9.8. We got t on one side of the equation by multiplying both sides by dt. When we had our velocity, we again got t limited to this side of the equation by multiplying it by dt. So we ended up with x on one side and t on the other. Can we do this for our chemical equation dA/dt=-kA? Well, if I divide both sides by A and multiply both sides by dt, my equation becomes dA/A=-kdt. So here, A is limited to the left side, and t is limited to the right side. I can integrate this just as I did for Super C, so I've got the integral of 1/A da equals the integral of -kdt. Well, the integral of 1/A is the natural log of A, and the integral of -k is -kt. Here's my constant of integration that I have to include, because I'm not taking any limits on these integrals. So I end up with lnA=-kt + C. That's almost there, but I really want A as a function of t, not the natural log of A as a function of t, so I'm going to take both sides of this equation and take e to the power of the left side and e to the power of the right side, so that this left side becomes A, and this right side becomes e^(-kt) + C. Now, if I knew what A was for some given C, I could solve this and get rid of that C.
So let's say we have an initial value again. Let's say that at time t=0, the concentration of A was just equal to 1. So I can plug in 0 for t, and I get e^0 + C = 1, because the concentration is 1 at t=0. This means that e^C is equal to 1. Well, I could take the natural log of that, but let's instead use what I know about exponentials, and let's write e^(-kt+C) = e^(C) * e^(-kt). So I've just split up this exponential. Well e^C, we've just determined, is 1. But it's important to not that e^C is still just a constant number to a constant power. So e^C I could write as just a different constant variable. Let's call it C sub 2. So I know even before I use my initial value that the concentration A = C sub 2 * e^(-kt). Now if I use the concentration of A at time equals 0 is 1, then I get A=e^(-kt).
So let's review. Differential equations are absolutely everywhere in physics. They relate a variable with its rate of change, such as the position of Super C with his velocity and the concentration of our chemical with the rate that it's being depleted. Often, we can solve these differential equations using a separation of variables. In separation of variables, we split the independent and dependent variables to different sides of the equation. In the case of Super C, we split everything that depended on t to the right-hand side and the velocity, or his position, to the left-hand side, so all the x's were on the left and all the t's were on the right. In the case of concentration, we put all of the A's, that is, all of the concentration variables, on the left, and all the time variables, all the t's, on the right.
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Back To CourseMath 104: Calculus
14 chapters | 116 lessons | 11 flashcard sets