0:00
There.
00:01
So for this problem, we are told that initially a tan contains 1 ,000 liters of brine with 50 kilograms of dissolved salt.
00:11
So that is the initial value for the amount of salt at zero is equal to 50 kilograms.
00:21
And we are told that brine containing 10 grams of salt per liter is flowing into the tank at a constant rate of 10 liters per minute.
00:33
So we have 10 grams per liter that are flowing at a rate of 10 liters per minute.
00:42
And we are also told that if the contents of the 10 are kept throughout limits at all times, and if the solution also floats out at a rate of 10 liters per minute, so the flow rate flow out is also 10 liters per minute.
01:05
So what we need to ask for this question is how much salt remains in the tank at the end of the 40 minutes.
01:13
So we need to find the amount of salt after 40 minutes.
01:21
Now, with that said, the first thing that we need to write is the we know that s of d, s of d, is going to representate the amount of salt, amount of salt in the water.
01:41
And then the derivative of this with respect to time is equal to the rate in, minus the rate out.
02:02
And the rate is going to be equal to, well, the rate in is the values that we have in here, this product right here, this number with this number.
02:14
Now, of course, you need to pass this from grams to kilograms.
02:19
So we just need to divide 10 by 100 that we know is going to be 0 .01 kilograms per liter.
02:30
And we multiplied this by the rate of flowing, which is 10 liters per minute.
02:38
And this minus the rate out.
02:40
Now, the rate out in this case is a value that we don't know.
02:44
That is x kilograms.
02:54
And this times, well, in this, we have x kilograms divided by the volume, which is 1 ,000 liters.
03:07
And this times the rate.
03:11
We know that also the flow out of this tank is 10 liters per minute.
03:16
So we have 10 liters per minute for the flow out.
03:21
So if we simplify this expression, we will find that this is equal.
03:26
We are going to leave the units for now.
03:33
But we are going to have 0 .1 minus 3 .1.
03:37
Adds divided by 100.
03:40
And we can write this by the following form where you can write this as 10 minus adds and all of this divided by 100.
03:48
In here, what we have done is just multiply that 0 .1 by 100 and divided by 100.
03:56
So we can write it in this way.
03:58
Now, to solve for this equation and what we need to do is to write it in the following form.
04:07
Take the integral of this so that we pass this to the other side and this 100 will leave it in there and this is going to be equal to 1 over 100 times the differential in time so what we are going to do in here is to take the integral in both sides of this equation now the one at the left is going to give us minus the neparion logarithm of the absolute value of 10 minus s because that's the integral, the integral of that expression.
04:46
And this is going to give it in the right side, we're going to produce time divided by 100.
04:54
And of course, because we are integrated, this will produce a constant of integration...