00:01
Hi there, so for this problem, we are told that a tank contains 50 kilograms of salt and 2 ,000 liters of water.
00:12
A solution of a concentration that is given of 0 .0 125 kilograms of salt per liter enters a tank at the rate of 9 liters per minute.
00:27
Now the solution is mixed and drains from the tank at the same rate.
00:32
So the first question in here, question a, is about what is the concentration of our solution in the tan initially? so we know that the initial amount of salt is 50 kilograms and the initial volume is 2 ,000 liters of water.
00:54
So from this we can immediately obtain the value for the initial concentration.
01:01
So that will be the value that you put in there, 0 .025 kilograms per liter.
01:10
Now, for part b of this problem, we are asked to find the amount of salt in the 10 after 4 .5 hours.
01:21
So the time that we are given in this case is 4 .5 hours.
01:29
We need to pass this into minutes because the expression that we are going to find is going to be in minutes.
01:40
So we know that 1 hour is equal to 16 minutes.
01:46
So this times in minutes is equal to a value of 270 minutes.
01:56
Now, we're going to let y of t represents the amount of salt in the time.
02:07
Amount of salt and after 10 minutes.
02:17
Now, we know that the derivative of that amount would result.
02:21
Respect to time is equal to the rate in minus the rate out.
02:34
So the rate in, so let me pick it in here, the rate in is the value that we are given and 0 .0 125 kilograms that enters.
02:59
This, this is per liter, sorry per liter.
03:09
This times at the rate that is entering, that is 9 liters per minute.
03:22
And this minus the rate out.
03:24
The rate out is an amount of salt at a given times t.
03:30
This divided by the...
03:36
The volume that is 2 ,000, this in kilograms per liter.
03:45
And this times the rate out.
03:47
And again, is, in this case, we are told that it runs from the 10 at the same rate.
03:52
So we need to multiply this by, again, nine letters per minute.
04:00
So simplifying this term in here, we will have that the differential equation is the right 0 .0125 times 9 so we obtained 0 .1125 kilograms per minute this minus um okay so let me see if we can simplify this 2000 we got a good well we're gonna leave it just like that so that will be 9 y divided by 2000.
04:53
We're going to leave everything without units.
04:55
So to just simplify things in here.
04:58
So that will be minus 9 times y divided by 2000.
05:06
Now, with that set, and let me just, what we can do in this case to put everything in the same is to just simply multiply 2 ,000 times 0 .11 .25.
05:29
So we obtain 225 minus 9 times y, this divided by 2000.
05:42
Okay, so now what we can do is to separate the variables.
05:46
So we can pass the 2 ,000...