00:01
Hi there, so for this problem we have a 200 gallons tan.
00:06
And initially, it contains 100 columns of salt, of salt water at a concentration that is given of 1 5th hounds per gallon.
00:23
So with this, we can already obtain the initial amount because that will be this times this.
00:29
So that will give us that initial amount of salt in this 10 is equal to 20 pounds.
00:40
Now, with that said, we have a salt solution coming in.
00:46
It has 1 divided by 10 pounds of salt per gallon.
00:56
And this at a rate that is also given and that rate is equal to 4.
01:01
Columns per minute and in this rate out and yes it is drain out at a rate of two columns per minute so with that said we need to assume the perfect medicine throughout the process so we need to find the quantity of salt in the tan at the moment that it begins to overflow so that means at the moment where the volume of this tan is in yes when the volume of this salt water is equal to 200 gallons.
01:40
So with that said, first of all, we're going to consider that the amount of, and the rate of change of the amount of salt with respect to time is equal to the rate in minus the rate out.
01:50
So with that said, let's write first the rate in.
01:55
So the rate in is just the product between this and this, because when we do that product, we can cancel the gallons, go to gallons, and we will obtain pounds per minute.
02:05
So that will be, 4 divided by 10, so that will give us 0 .4 for the rate in.
02:13
This minus the rate out.
02:15
So that will be the amount of salt at a times t.
02:18
This times the rate out, which is 2.
02:21
And then this divided by the initial volume for that, which is 100.
02:32
And this plus, well, we need to do the difference between the rate in and the rate out, so that will be 4 minus 2.
02:41
So that will be that this volume is increasing at 2 times the time.
02:46
Okay, so then this is the expression, 0 .4, and then this minus the amount a.
02:54
We can divide everything by 2, so that will be 50 plus the time.
02:59
Okay? this is the differential equation that we need to solve, and also we consider that initial amount we know from before, is 20, 20 pounds, okay? so let's try to solve the differential equation.
03:15
So we first we can pass this term to the left.
03:18
So we will have the rate of change of a with respect to time plus a divided by 50 plus the time is equal to 0 .4.
03:28
Now we can multiply both sides of this by an integrating factor.
03:33
So that integrating factor is the exponential of the integral of the integral of this term right here, so that will be the differential in time, divided by 50 plus the time.
03:50
So we know that then this is just the exponential of the nebaryon logarithm of 50 plus the time.
03:57
The exponential cancels with the neferian logarithm, so we will obtain just simply 50 plus the time.
04:07
So that is the integrating factor that we need to multiply the bulk so with that integrating factor we can write this in the following way so that will be 50 plus the time times the amount of salt at any given time is equal to 0 .4 times 50 plus the time...