00:01
Hi there, so for this problem, we're told that a tank with a capacity of 100 gallons.
00:08
Okay, initially contains.
00:12
So initially we have the gallons with 10 pounds of salt.
00:20
Now, freshwater enters our rate.
00:23
So freshwater means that there are 0 pounds per gallon.
00:27
On this mixture that is entering of this and this had a rate of 2 gallons per minute.
00:34
Okay, so then the question for this is, well, and the well as their mixture is pumped out at the rate.
00:46
So this is the rate in the rate out is at 1 gallon per minute.
00:54
Okay, so then we need to compute the amount of salt in the tank at the 1st moment when the tank is filled.
01:03
So then we know that the rate of change of the amount of salt with respect to time is then the rate in minus the rate out.
01:11
Now, the rate in is the pro between this value and this value in here.
01:15
So we know that the 1st term in here is just 0.
01:22
Okay, and then this is minus the rate out.
01:24
The rate out is 1st of all, we have a, which is the amount of salt at any given time.
01:33
This divided by the initial volume for this that is 50 minus.
01:41
Then we do the difference between this value and this value in here.
01:45
The rate in minus the rate out.
01:47
So we will obtain this times the time.
01:50
Okay, we have that.
01:52
And then in here, we need to multiply this by the rate out, which is just 1.
02:01
So we can just leave it like this.
02:03
So this is the differential equation that we need to solve for this.
02:09
Okay, so what we can do is to make some separation of variables.
02:13
So we will have the differential in a divided by a is equal to minus the differential in the time divided by 50 minus the time.
02:23
And then we just need to integrate both sides of this.
02:27
For the left side, we obtain the napolean logarithm of a.
02:31
And for the other side, we will obtain the napolean logarithm and the integral of this...