A community for students.
Here's the question you clicked on:
 1 viewing
lgbasallote
 3 years ago
a tank contains 100 gallons of water. in error, 300 lbs of salt are poured into the tank instead of 200 lbs. to correct this condition, a stopper is removed from the bottom of the tank allowing 3 gal of brine to flow out per minute. at the same time, 3 gallons of fresh water per minute are pumped into the tank. if the mixtire is kept uniform by constant stirring, how long will it take for the brine to contain the desired amount of salt?
lgbasallote
 3 years ago
a tank contains 100 gallons of water. in error, 300 lbs of salt are poured into the tank instead of 200 lbs. to correct this condition, a stopper is removed from the bottom of the tank allowing 3 gal of brine to flow out per minute. at the same time, 3 gallons of fresh water per minute are pumped into the tank. if the mixtire is kept uniform by constant stirring, how long will it take for the brine to contain the desired amount of salt?

This Question is Closed

hellow
 3 years ago
Best ResponseYou've already chosen the best response.1You might want to look at Practice Problems 4. There is a mixing problem there. I know there have also been mixing problems in the lecture, though I am not sure exactly where. Perhaps you can find some just before Practice Problems 4. Here is one way to approach the question: v= 100 gallons, s=amount of salt in tank, t= time in min \[ds/dt = [s(t)/100 gallons]\times (3 gallons) \]Just solve the ODE. I won't show all the steps. \[\int\limits_{}^{}1/s ds = \int\limits_{}^{} 3/100 dt\]\[s=ce ^{3/100t}\]\[s=300e ^{3/100t}\]Then we want to solve \[200=300e ^{3/100t}\]. We find t=100/3ln(2/3). (Don't worry, t is not negative since ln(2/3)<0.)

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0okay thanks..i'll try this

hellow
 3 years ago
Best ResponseYou've already chosen the best response.1Okay, I hope it makes sense. The ODE ds/dt =s(t)/100x3 is just saying that the change in the amount of salt over time is, first of all, negative, because the amount of salt in the tank is decreasing. How much is the amount of salt in the tank decreasing by? Well, at time t, the concentration of salt in the tank is the amount of salt in the tank divided by the volume, i.e. s(t)/100gallons. And, in 1 minute, 3 gallons leave the tank. So, we want to multiply s(t)/100gallons times 3 gallons to get the total amount of salt leaving the tank per minute. Thus ds/dt = s(t)/100gallons x 3 gallons. If the ODE is right, I think it should make sense to you.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.