00:01
We usually set up the rate of change as a difference between a rate in and a rate out.
00:08
Here we've got a simple situation where there is nothing going into this barrel of oil, and v we are using as the symbol for volume in the tank.
00:25
What we would like to do is solve for that volume as a function of time, and we will write the rate at which it is being pumped out.
00:35
So usually you have to be told something about that outlet rate.
00:40
And here we're told it's being pumped out at a rate proportional to the volume that's in the tank.
00:50
So i'll use k as a constant of proportionality.
00:55
And because it is coming out, we are going to use a negative sign.
01:00
If something were being pumped in, we'd have to know something about that rate in, and we'd include it with a positive on the right -hand side.
01:12
Anyway, this is a simple equation that we can use the process, separate and integrate for.
01:24
So what we want to do with separate is to get the volume with the dv and get the dt with anything, time -related.
01:37
And then once we do that, we can integrate.
01:45
So we're going to cross -multify and get dv over v is minus k -t -t.
01:56
Now it's time to integrate.
01:59
Don't forget when you integrate.
02:01
You want to include an arbitrary constant that's going to be determined by your initial conditions.
02:11
In this case, just one initial condition.
02:16
In fact, what we are told is a couple of things that we'll get to, but i'll go ahead and list them.
02:24
We are told that the initial volume at t -equal zero volume of the oil in the tank is 10 to the 6th.
02:37
Let's see what units we are looking at just for fun.
02:45
Barrels, that's what i was hoping.
02:48
Yes, barrels.
02:57
And then we are told that six years later, so we have a time frame.
03:05
That b of 6 is 5 times 10 to the fifth barrels.
03:17
So yeah, we have enough information to work with, i believe.
03:24
So anyway, back to the integral.
03:26
A fairly easy one on the left -hand side, that's the logarithm, natural logarithm, of volume.
03:32
On the right -hand side even easier.
03:36
The time just comes out, and we still have that constant.
03:44
So to invert this and get the volume by itself, we want to take both sides and raise them as powers of e, the natural base.
03:58
And what will happen on the left is because raising something to the e, is the inverse function of logarithm.
04:09
Those two things cancel...