Now, the solution of this, the function that solves this equation.

The slope is equal to, the, the slope of this function is equal to the value of

the function itself, where the function itself we know is e to the t.

And I could multiply by some constant a, and it would be carried along on both

sides. So it's perfectly fine to multiply by

this constant. So this is the general solution of this

differential equation. Now, I have to make this general solution

satisfy my particular initial condition, which is at t equals 0, I have two

rabbits. So if I plug t equals 0 in here, e to the

0, any number to the 0 power, is unity. And so that means that a has to be 2.

So the solution that we're looking for for the number of rabbits as a function

of time t is 2 times e to the t. And if I plot that, I start off initially

with two rabbits. And over time, the number of rabbits

increases and over the course of ten years, I have close to 50,000 rabbits.

So this is an example of exponential increase of a population.

And so, the whole point of this is that this number e comes about in certain

types of problems. This is a very naturally occurring

problem. Another example is like radioactive

decay. Where the number of decaying atoms at

any, or decaying nuclei at any particular time is proportional to the number of

nuclei that I have at that time. And so, that would give me a decreasing

exponential function. but e pops up all the time from certain

naturally occurring problems. Let's take a minute now and look at the

derivatives of sine and cosine. So I'm going to start by plotting just

one cycle of sin wave here. And we said before that the derivative of

sine of t is cosine t. So that says that the slope of the

tangent line to every point on the sine curve is equal to the cosine at that

point. So here's a plot of one cycle of a cosine

wave, and let's just take a look at what the, the slope of the tangent lines would

be. So here, at t equals 0, if I draw the

tangent line, it is a positive slope, and it turns out to be maximum at this point.

As I go up the hill, the slope starts to decrease.

And when I get to the top of the hill, the slope is 0, so it's a horizontal

tangent line up here. Now, the function starts heading down the

hill, and the slope of the tangent line is negative.

And it becomes maximum, its maximum negative valueaAt the crossing point.

And then it starts to level out again, and goes back to 0.

So the slope here is 0. And now the slope with the tangent line

is positive. And it reaches it's maximum value here at

the crossing point. Now, if I take the derivative of cosine.

That is minus sine of t. So let's do the same kind of

construction. And say, okay here on the cosine curve at

t equals 0. The slope is 0, the slope at the tangent

line is 0. So that's right there.

And now, I'm heading down the hill, the slope is negative.

So, the and it becomes maximum negative at the crossing point.

So here's the crossing point. And as I keep going down into the valley

of cosine curve, the slope is always negative, but it's coming back towards 0.

So the slope of the tangent line at the bottom of the valley is zero.

And now, the slope turns positive and I'm heading up the hill and the slope becomes

maximum at the crossing point. And then after I go through the crossing

point, it starts to level off and the slope goes back to 0.

Now, the last if I take the derivative of this minus sign t.

Then that's just minus cosign t because if I just multiply, I'm just multiplying

both sides of this equation by negative 1.

So that derivative is then minus cosine t.

Now, if I take the derivative of minus cosine t.