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Part C, find the particular solution– W as a function of t– to the differential equations, derivative of W with respect to t is equal to 1 over 25 times w minus 300, with initial condition, at time 0, W is equal to 1,400. And the units in this problem, it was 1,400 tons they told us.

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## 2011 Calculus AB free response #5c. AP Calculus AB

So it might seem strange to see a differential equations problem on an AP exam. You say, wait, this isn’t fair. This isn’t a differential equations class.

And the thing to keep in mind is if, on an AP exam, you see differential equations, they’re really asking you to solve a differential equation that can be solved using the skills that you would learn in an AP class. And the main class of differential equations that you could solve are the ones that are separable differential equations. The ones where you can separate the independent and the dependent variables. And if you don’t have any idea what I’m talking about, I’m about to show you.

So let me rewrite this differential equation. We have dw, dt is equal to 1 over 25 times W minus 300. Now this differential equation, we have W here. And we have this dt.

But we don’t have t anywhere else. So what I want to do is I want to get the parts that involve W onto the left-hand side, and I want to get anything that deals with dt on the right-hand side. So let’s divide both sides of this equation by w minus 300.

So we would have 1 over W minus 300 dw dt– the derivative of W with respect to t– is equal to– I divide this side by W minus 300, so it’s equal to 1 over 25. Now I want to get the dt’s on this side. So let’s multiply both sides by dt.

[INAUDIBLE] the differential is just a very, very small change in t. And so we get– and I’m a little bit imprecise with my notation here. But this is one way that you can think about it when you are dealing with the differential equations like this, separable differential equations. You have 1 over W minus 300 dw– let me make sure that the dw doesn’t look like it’s in the denominator– dw is equal to 1 over 25 dt.

And now we can integrate both sides, which is essentially just saying that we’re taking– since this is equal to this– that each little increment of W times this is equal to each little increment of dt times 1 over 25. We can then say, well, if we sum up all of the increments of W times this quantity– if we sum up the infinite number of infinitesimally small increments– that would be the same thing as summing up all of these characters. Because each of these characters is really the same as each of these characters. So let’s do that sum.

We’re essentially just integrating both sides of this. And here, you might be able to do this in your head. Or we could do the formal u substitution, just to make it clear what we’re doing. If you said u is equal to W minus 300, then what is du? If you wanted the differential.

So if you want to take the derivative of u with respect to W, you would say du with respect to W is equal to what? It is equal to 1. The derivative of this with respect to W is 1. Or if you multiply both sides times dw, you have du is equal to dw.

And so this left-hand side right over here– this integral– could be rewritten as the integral of 1 over u du. We define this as u. And we just saw that du is the same thing as dw, so this could be rewritten as du. And this is one of the basic integrals, so hopefully we’ve learned.

And this is equal to the natural log of the absolute value of u. Or if you unwind the substitution, this is equal to the natural log of the absolute value of W minus 300. So let me write this down.

So the left-hand side right over here. And you might be able do that in your head. You say, hey, look.

You have W minus 300. Its derivative is 1, so it’s essentially there. So I can just take this as if this was 1 over– I could just take the antiderivative here and treat this whole thing as a variable. So it would just be the natural log of this entire thing, or the absolute value of that entire thing.

So the integral of the left-hand side is the natural log of the absolute value of W minus 300. And the derivative of the right-hand side– and this is a little bit more straightforward– or sorry, the antiderivative with respect to t on the right-hand side is 1 over 25 t, or we could say t over 25. And then these are both indefinite integrals.

You could have a constant on both sides or one side. We could just put it on one side right over here. You could have put plus some constant. You could have added some C1 over here.

And then you could’ve had C2 over here. But then you could’ve subtracted C1 from both sides, and then just have a C3 constant over here. But to simplify it, you could just say, look, this is going to be equal to this plus some constant. You could’ve put the constant on that side as well.

It doesn’t matter. All the constants would merge into one. Now we’ve gotten pretty far. We want to solve the particular solution.

Right now, we only have this constant. And we also haven’t written W as a function of t just right yet. They’re implicitly expressed. It is more of a relationship than a function right now.

So the first thing we can do is try to solve for C. We’re told in the problem that W of 0 is equal to 1400 tons. They told us that. That’s one of our initial conditions.

So we know that the natural log, that when t is equal to 0, W is equal to 1400. So W is equal to 1,400 when t is equal to 0 plus C. Or another way to think about it is, the natural log of the absolute value of 1,100 is equal to C.

And obviously, this is a positive value. So we could just say that C is equal to the natural log of 1,100. We can drop the absolute value sign because the absolute value of 1,100 is 1,100.

So now we can rewrite this part right over here. Let me do it in yellow. Let me start– that’s not yellow.

Let me start right over here. So we could write the natural log of the absolute value of W minus 300. And actually, we know that W is an increasing function.

W starts at 1,400 and only goes up from there as t gets larger and larger. So if we’re starting at 1,400 and only getting larger and larger values for W, this expression right here is never going to get negative for positive t values. And that’s all we care about. And so we can actually drop the absolute value sign.

This will always be a positive value. So we could say the absolute value of W minus 300 is equal to t over 25– and we’ve just solved for our constant– plus the natural log of 1,100. And now we can just solve this explicitly for W. We can raise e to both sides.

This value is equal to this value. The left hand side is equal to the right hand side. So e raised to this power is the same thing as e raised to that power.

So we have e to the natural log of W minus 300 is going to be equal to e to the t over 25 plus the natural log of 1,100. Let me take this– actually, just to get the real estate, let me delete that. All right. Now I can keep working right below that.

And so e to the natural log of something is just going to be equal to that something. The natural log is what power do I have to raise e to to get this thing. Well I’m raising e to that power, so I’m going to get this thing. So the left-hand side just becomes W minus 300, and the right-hand side, we could rewrite this as e to the t over 25 times e to the natural log of 1,100.

We have the same base, different exponents. If you wanted to simplify this, you would just add the exponents, which is exactly what we did up here. So these two things are equivalent. e to the natural log of 1,100, that is 1,100.

The natural log of 1,100 is a power that you raise e to to get to 1,100. So if you raise e to that power, you get 1,100. So that is 1,100.

So we get W minus 300 is equal to 1,100 e to the t over 25. And then you just add 300 to both sides. W is equal to 1,100 e to the t over 25 plus 300.

And we’re done. We have solved the differential equation. We found the particular solution. We had to solve for C to find that particular solution.

We’re done. And we did it using just basic integration. We didn’t have to use any of the fancy differential equation solving tools that you might learn in a more advanced class.