Salt Water Flows
There are three 50-gallon drums. Initially,
- Drum 1 contains 40 gallons of water with 3 pounds of dissolved salt,
- Drum 2 contains 10 gallons of water with 2 pounds of dissolved salt, and
- Drum 3 is empty.
While it still has water, Drum 1 empties into Drum 2 at a rate of 5 gallons per minute. While it still has water, Drum 2 empties into Drum 3 at a rate of 4 gallons per minute.
When Drum 2 reaches its maximum water level (though still not completely full), how many pounds of dissolved salt does it contain?
Assume that the salt water solutions are perfectly mixed at all times.
The answer is 1.469.
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3 solutions
How did you integrate d/dt[(10+t)^4S]
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I have shown that the derivative of ( 1 0 + t ) 4 S is equal to 8 3 ( 1 0 + t ) 4 . Therefore ( 1 0 + t ) 4 S = 4 0 3 ( 1 0 + t ) 5 + c and I choose c to ensure that S = 2 when t = 0 . Then I divide by ( 1 0 + t ) 4 .
The trick was to note that multiplying the previous equation by ( 1 0 + t ) 3 turned the LHS into an exact differential. This process is known as finding the integrating factor.
What is wrong with defining S ( t ) in the following way? S ( t ) = 2 + 8 3 t − 1 0 + t 4 S ( t ) t Solving this at t = 8 gets me 1 . 8 , which is wrong, but I don’t see the problem with defining S ( t ) in this way. Thanks!
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Well, you are saying that the rate at which salt is flowing into the third drum is 1 0 + t 4 S , so that d t d S = 8 3 − 1 0 + t 4 S This is correct. However, you cannot just multiply that by t to get the amount of salt at time t : S = 2 + t × d t d S because d t d S is not constant. That is why we need to solve the differential equation.
I would have liked solving this riddle, but unfortunately I refuse to use imperial units. Upload it again with SI-units... :D
Derived and then solved with WolframAlpha the following ODE:
d(A)/dt = (3/40) 5 - [4/{10+t(5-4)}] A(t), A(0)=2,
find, solution as:
A(t)=(5/4)(10^4)/{(t+10)^4}+(3*t/40)+(3/4),
In "8" minutes Drum1 gets empty, thus, A(8) gives maximum salt content in Drum2,
A(8)=(5/4){(10/18)^4} + (24/40) +(3/4)
Answer = 1.46907
Using the same reasoning of Mark Hennings I reached the equation [ d t d s = 8 3 - t + 1 0 4 s ] but could not go any further. I feel that he skipped many steps from this point on because I failed to understand how did he reach [ d t d [(10+t)^4 s] = 8 3 (10+t)^4 ]
I did solve the problem though using java (which is ugly I know). Here is the code FYR:
void Solve() { for(int j = 0; j < 5; j++) { double interval = 1 / Math.pow(10.0, j); double totalTime = 8.0 * 60.0; double n = totalTime / interval;
double s = 2.0;
double inS = 3.0 / n;
double v = 10;
double outV = (4.0 /60.0) * interval;
double vIncrease = (1.0 / 60.0) * interval;
for(int i = 0; i < n; i++)
{
double outS = s * outV / v;
v +=vIncrease;
s += inS - outS;
}
System.out.print(interval + " ");
System.out.println(s);
} } //Output
// 1.0 1.468192639116571
// 0.1 1.4689866309394366
// 0.01 1.4690660157786068
// 0.001 1.4690739541307423
// 1.0E-4 1.4690747477871104
All I did was multiply the original equation by ( 1 0 + t ) 4 . It is then a matter of spotting that ( 1 0 + t ) 4 d t d S + 4 ( 1 0 + t ) 3 S = d t d [ ( 1 0 + t ) 4 S ]
Suppose that the second drum contains W gallons of water and S pounds of dissolved salt. While the first drum contains water, we see that d t d W = 5 − 4 = 1 where time t is measured in minutes. Thus the amount of water in the second drum increases steadily for the first 8 minutes, with W = 1 0 + t . At time t = 8 , the second drum contains 1 8 gallons of water, and the first drum is empty. After this the second drum empties steadily into the third drum until it is empty. We want to know how much salt is in the second drum at time t = 8 .
During the first 8 minutes, the second drum receives 8 3 pounds of salt per minute from the first drum, and loses an amount of salt to the third drum that is proportional to the current salt concentration in the second drum. Thus d t d S ( 1 0 + t ) d t d S + 4 S d t d [ ( 1 0 + t ) 4 S ] = 8 3 − W 4 S = 8 3 ( 1 0 + t ) = 8 3 ( 1 0 + t ) 4 With the initial condition that S = 2 when t = 0 , we deduce that S = 4 0 3 ( 1 0 + t ) + 4 ( 1 0 + t ) 4 5 0 0 0 0 making the required answer S ( 8 ) = 3 2 8 0 5 4 8 1 9 3 = 1 . 4 6 9 0 7 pounds of salt.