00:01
Hi there, so for this problem, we are told that oil is pumped continuously from a well at a rate.
00:08
So let's say that the amount at a times t is a proportional to this.
00:17
Well, that means a concept of proportionality that we did determine times a.
00:21
Now we know that initially there is 3 million barrels.
00:30
So let's put that at 3 times 10 to the 6.
00:34
And after six years, the total amount is 1 .5 times 10 to the 6.
00:43
So with that said for part a of this problem, the question is, at what rate was the amount of oil in the well decreasing when there were 1 .8 times 10 to the 6 barrels remaining? okay.
01:03
So the question is the amount, the rate of a, or respect to time, when there were a that is equal to 1 .8 times 10 to the 6.
01:39
Okay.
01:43
So with that said, what we need to do first is to solve the expression that we are given.
01:53
So by separating the variable, so we will have this.
01:59
Then if we integrate both sides of this, for the left side, we obtain the neferial logarithm of a.
02:08
And for the right side, we obtain a constant of proportionality times d and this plus, it comes a constant of integration since.
02:17
Plan the exponential function in both sides of this.
02:19
We obtain the following.
02:21
Now with this we need to determine the values for this.
02:24
Now c, if we evaluate this at zero, we will obtain immediately that c is equal to three times 10 to the six, which is the initial value for this.
02:34
So with that, we will now have that the amount at any given time is equal to three times 10 to the six times the exponential of the comps in proportionality times the time.
02:44
Now, now we use the other condition of that after six years, there is 1 .5 times 10 to the 6 barrels.
02:52
So that will be this is equal to 3 times 10 to the 6 times the exponential of k times 6.
03:02
Then if we pass this to divide to the right side, we obtain 1 divided by 2 that will have the neparial logarithm of this is equal to this.
03:12
So the constant k is equal to 1 divided by 6 times the neparion logarithm of 1 divided by 2.
03:21
Then we can also write as just simply minus the neparion logarithm of 2 divided by 6.
03:29
So with this, we can have then that, yes, in fact, we can write that this is equal to the neparian logarithm of 2, elevated to minus 1 divided by 6.
03:54
So when we substitute that into the function that we have from before, so that will give us, the exponential function cancels with the neferian logarithm...