Question
A 200-gallon tank is currently half full of water that contains 50 pounds of salt. A solution containing 1 pounds of salt per gallon enters the tank at a rate of 6 gallons per minute, and the well-stirred mixture is withdrawn from the tank at a rate of 6 gallons per minute. How many pounds of salt are in the tank 10 minutes later? Round your answer to 2 decimal places.
I got 104.82 but it is wrong.
Please explain the answer and how to get it.
I got 104.82 but it is wrong.
Please explain the answer and how to get it.
Answers
Steve
Why don't you explain your answer, and we can see whether your math is sound.
Did you start (with s(t) being the amount of salt present)
ds/dt = 1 - 6/100 s
s(0) = 50
?
Surely you can see that 104.82 is way off. Even with no salt draining out, only 110 lb would be present after 10 minutes!
Did you start (with s(t) being the amount of salt present)
ds/dt = 1 - 6/100 s
s(0) = 50
?
Surely you can see that 104.82 is way off. Even with no salt draining out, only 110 lb would be present after 10 minutes!
Sarah
my equation was y=900+C/e^(1/150)t
Steve
yeah, but how did you get it?
Where did the 900 and 150 come from?
With only 50 lb of salt at t=0, I don't see how it works. What is y measuring?
I started out by reasoning that with no draining,
ds/dt = 6
since since each minute 6 lbs of salt are added. OUCH! Typo, since I said 1 (because of 1 lb/gal, but I ignored the fact that 6 gal/min are added)
But, at the same time 6 of the 100 gallons drain out, taking 6/100 of the salt present with them. Hence the differential equation.
Where did the 900 and 150 come from?
With only 50 lb of salt at t=0, I don't see how it works. What is y measuring?
I started out by reasoning that with no draining,
ds/dt = 6
since since each minute 6 lbs of salt are added. OUCH! Typo, since I said 1 (because of 1 lb/gal, but I ignored the fact that 6 gal/min are added)
But, at the same time 6 of the 100 gallons drain out, taking 6/100 of the salt present with them. Hence the differential equation.