Welcome to calculus. I'm Professor Greist. We're about to begin lecture 23. Bonus material. In our main lesson, we gave a collection of substitutions involving trigonometric or hyperbolic trigonometric functions. These work great for computing some integrals. But at times, you've got to be careful. Let's consider a simple example of a differential equation mode that is amenable to a trigonometric substitution. Let's say that a financial model projects that your marginal profits Are going to be a constant plus some term that is proportional to the square of the net profits. What do you think about such a model? Does that sound like a good investment? Well let's translate what that model says. Into a differential equation. If P as a function of t is your net profit. Then what do we mean when we say that the marginal profits are equal to a constant plus a term proportional to the square of the net. Well the marginal profit. Is the derivative, dP dt, to say that that marginal profit is equal to a constant, let's make sure that it's a positive constant by calling b squared. Plus some term proportional to the square of the net profits, let's call that term a squared p squared, so that the constant of proportionality here is also positive as it must be. Then how do we solve this differential equation. If we perform substitution. Then, on the left, we need to compute the integral of dP over b squared plus a squared P squared. On the right, we have the integral of dt. Now, which trigonometric substitution is appropriate? Well, clearly, using a tangent is what we're going to want. So, let us substitute for P, B over A times tangent of theta. DP is therefore a secant squared. Now, when we plug that in, well, we get B over A secant squared theta for DP. And in the denominator b squared secant squared Theta. The secant squareds cancel, the b's cancel, and we're left with d Theta over a times b. Now that integrates to simply Theta over ab. Setting that equal to the integral of a dt, that is, t plus constant, allows us to very easily substitute back again for theta the arctangent. We therefore have t plus a constant equals one over ab Times the arctan of a over, times P. Solving for P as a function of t, after a little algebra, yields b over a times tangent of quantity, a b t plus a constant. We can show based on an initial condition that that constant is say, zero. Let's say we're starting off with no profit at all, times zero, very good. We've got our answer, and we've done the mathematics correctly... But what do you think about this model for the net profit? Is it realistic? Well, what is happening to P as T increases from zero? Well, we're getting a tangent function which means that at some finite time, t and the function is going to get a large, a very large, it's going to blow up to infinity. In fact, these are called blow-up solutions, or singularities, and they arise, sometimes, when performing a trigonometric substitution. Based on the nature of the tangent, cotangent, secant, and cosecant functions. Now you may think that such a solution to a differential equation is inherently nonphysical. We don't have things going to infinity and finite time, however there are examples of physical settings where blow ups, or singularities in the differential equation are important to consider. For example in certain areas of physics one has a resonance phenomenon. Think of a vibration that is forced at a certain critical frequency that leads to feedback. This corresponds to a blow-up solution to the relevant differential equation. But it is not something ignorable. Indeed one often has to engineer around such resonances, for safety purposes. Other examples include vortices, in fluid dynamics, or electromagnetics. These can arise as singular solutions. To the relevant differential equations. And nobody would argue that things like tornadoes or hurricanes are unimportant or uninteresting. In fact, the differential equation that models three dimensional fluid dynamics, that is the Navier Stokes equation. Is a very fascinating set of equations, unfortunately we don't have enough calculus down as yet for us to be able to appreciate these equations fully in this class. However, it is an open problem in mathematics, whether or not the solutions do these differential equations have finite time blowups or singularities in them? This is one example of a very simple sounding problem which is nevertheless open and unknown to the best of mathematicians. Many very good mathematicians are currently, working on this problem. After you've had a little bit of multi variable calculus, you might be able to work it too.