00:01
Hi there, so for this problem, we are told that a tan contains 100 gallons of pure water.
00:07
So this is the initial volume, 100 gallons of pure water.
00:16
And at the time equals to zero, a sugar water solution.
00:20
And then we know that the solution coming in is 0 .2 pounds of sugar per gallon.
00:30
And it enters at a rate that is also given, that is three columns per minute, okay? so once we are given that, we are told that simultaneously a drain is opened, so the rate out is at three columns per minute.
00:54
So assume that the solution in the 10 is kept perfectly midst of all time.
00:59
So for part a of this problem, the question is, what will be the sugar content in the 10 after 20 minutes? now, first of all, we're going to set that adds of tea is the amount of sugar.
01:19
This, at any given time, tea in the 10, okay? so now as we are given this, we know that the rate in is just the product between the 3.
01:38
Gallons per minute times the 0 .2 pounds per gallon.
01:45
So in here we will have that this is 0 .6 pounds per minute.
01:51
So that's the rate in.
01:53
Now the rate out is equal to the three gallons per minute, this times the amount at any given time, and this divided by the initial volume that is 100.
02:07
So we can simplify this as just simply 0 .0 .0.
02:11
Times the amount at any given time.
02:15
So with that said, we know that the rate of change of the amount of sugar at any given time is equal to the rate in minus the rate out.
02:27
Okay.
02:29
So then this will be equal to 0 .6 minus 0 .03 times the amount of sugar at any given time.
02:39
Okay.
02:40
Once we have that, then we know that we can write this in the following way, so that will be the derivative of the amount of sugar with respect to time.
02:51
This plus 0 .03 times the amount of sugar at any given time is equal to 0 .6.
03:01
Now we're going to multiply both sides by the integrating factor.
03:05
And in this case, that integrating factor is just the exponential of this term in here, 0 .03, and this times the time.
03:16
So then we can write this in the following way.
03:18
We can write this as the derivative with respect to time of the amount s times the exponential of 0 .03, this times the time.
03:28
And then this is equal to 0 .6 times the exponential of 0 .03 times the time.
03:36
Once we have this, what we can do is to separate the variables.
03:41
Let me just move this differential to this side in here.
03:46
And then we can integrate both sides of this.
03:51
So the solution for the left side is just the amount of sugar at any given time, times the exponential of 0 .03 times the time is equal to.
04:01
Then we will have 0 .6 divided by 0 .03.
04:07
This times the s -penumtional of 0 .03 times the time plus a constant of integration that we're going to label as c.
04:16
Now we can divide everything by this exponential factor in here...