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?