00:01
Hi there, so for this problem, we have initial and we have a 500 gallons tan.
00:10
And it initially contains 200 gallons of brine.
00:16
And this contains 100 pounds, okay, of salt.
00:22
Now, brine, so the rate in in this is that one pound of, yes, one pound of salt per gallon.
00:33
Enters at a rate that is also given of 4 gallons per minute.
00:42
And then with that said, we are also given that the rate out is that the well -stir mixture flows out of the tan at a rate of 1 gallon, 1 pounds per minute.
00:58
Then we need to set up a differential equation for the amount of salt in the 10 at any given time, and also answer the question how much shall in this detail when it is full.
01:09
Okay, so the different initial, so the root of change of the amount of salt at any given time is equal to the rate in minus the rate out.
01:24
So with that said, the rate in is just a product between the two values that we are given for the rate in, so that will give us four, and this minus the rate out, so that will be one, which is the rate out, and this times the amount of salt at any, given time and this divided by the volume but the volume is changing because the rate out is different to the rate in so the difference between those two values is three so we will have the initial amount that is 200 and then this plus three times that um three times the time okay so that is the differential equation that we need to solve and the initial condition that we're given for this is that the initial amount of salt is equal to 100 pounds okay so to solve this equation, the first thing that we're going to do is to pass this term to the left.
02:16
So we'll have the rate of change of the amount of salt with respect to time plus a, this divided by 200 plus 3 times the time is equal to 4.
02:25
Now we need to multiply both sides of this by the integrating factor, and that integrating factor will be the integral of the differential in time divided by 200 plus 3 times the time.
02:38
So that will be simplifying this.
02:43
This integrating factor becomes just simply the exponential of the neparian logarithm of 200 plus 3 times a time.
02:52
And we know that the exponential function is the inverse function of the neparian logarithm, so we obtain just simply 200 plus 3 times the time.
03:01
So with that said, we can write out this expression as a word of change with respect to time of the period between the amount of salt, and 200 plus three times the time.
03:13
And then this is equal to 4 times 200 plus three times the time.
03:22
Then we now need to simply, well, separate the variables in here.
03:32
And then we can integrate both sides of this.
03:36
So for the left side of this, we obtain just simply the amount of sugar, of salt, sorry.
03:41
And this times 200 plus 3 times the time.
03:46
Then this is equal to, we will have 4 divided by 3, and this times 200.
03:58
Well, we will have that this is equal to, okay, so 200 plus 3 times the time, and that to the square, and then also this to the square in here.
04:17
So that will be 6 in here, okay? because remember that what we have done is to set an expression that that is u is equal to 200 plus 3 times the time.
04:32
So the differential in u is 3 times the differential in time...