A tank originally contains 100 gal of fresh water. Then water containing .5 Ib of salt water per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at a same rate.After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving ar the same rate. Find the amount of salt in the tank at the end of an additional 10 min.
The answer in the book is 7.42 Ib
how do you work it out?
First model:
\(\frac{d y}{d t}=y_{i n}-y_{o u t}\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=\left(0.5 \frac{\mathrm{lb}}{\mathrm{gal}}\right)\left(2 \frac{\mathrm{gal}}{\mathrm{min}}\right)-\left(\frac{y}{100} \frac{\mathrm{lb}}{\mathrm{gal}}\right)\left(2 \frac{\mathrm{gal}}{\mathrm{min}}\right)\)
\(\frac{d y}{d t}=1-\frac{y}{50}\)
\(\frac{\mathrm{dy}}{1-\frac{y}{50}}=\mathrm{dt}\)
\(-50 \ln \left|1-\frac{y}{50}\right|=t+C\)
\(1-\frac{y}{50}=e^{-\frac{t}{50}} e^{C_{2}}\)
\(y=50-k e^{-\frac{t}{50}}\)
\(y(0)=0\)
\(0=50-k ; k=50\)
\(y(t)=50\left(1-e^{-\frac{t}{50}}\right)\)
At \(t=10\) minutes:
\(y(10)=9.063 \mathrm{lbs}\)
We need this because it will become our initial condition for the second model.
Second model: There is no salt going in, so:
\(\frac{\mathrm{dy}}{\mathrm{d} t}=y_{\mathrm{in}}-y_{\mathrm{out}}\)
\(\frac{\mathrm{dy}}{\mathrm{dt}}=\left(0 \frac{\mathrm{lb}}{\mathrm{gal}}\right)\left(2 \frac{\mathrm{gal}}{\mathrm{min}}\right)-\left(\frac{y}{100} \frac{\mathrm{lb}}{\mathrm{gal}}\right)\left(2 \frac{\mathrm{gal}}{\mathrm{min}}\right)\)
\(\frac{d y}{d t}=-\frac{y}{50}\)
\(\frac{d y}{y}=-\frac{d t}{50}\)
\(\ln |y|=-\frac{t}{50}+C\)
$$ y=k e^{-\frac{t}{50}} $$
Now be careful here. The origin of this differential equation occurs when \(t=10\) for the OLD differential equation. Which means \(y(10)\) in the OLD MODEL is \(y(0)\) in the NEW MODEL because in the eyes of the new model, the previous 10 minutes didn't even happen because salt was going into the tank.
So we say:
\(y(0)=9.063\)
\(9.063=k\)
\(y(t)=9.063 e^{-\frac{t}{50}}\)
\(y(10)=7.42 \mathrm{lbs}\)
A tank originally contains 100 gal of fresh water. Then watercontaining .5 Ib of salt...
A tank originally contains 100 gal of fresh water. Then water containing to of salt per gallon is poured into the tank at a rate of 2 gamin, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 9 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end...
A tank originally contains 140 gal of fresh water. Then water containing 1/2 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min. the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the...
Chapter 2, Section 2.3, Additional Question 02 A tank originally contains 100 gal of fresh water. Then water containinglb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 10 gal/min, with the mixture again leaving at the same rate. Find the amount of salt...
Chapter 2, Section 2.3, Additional Question 02 A tank originally contains 10 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to gal of fresh water. Then water containing leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 9 gal/min, with the mixture again leaving at the same rate. Find the amount of...
Chapter 2, Section 2.3, Additional Question 02 A tank originally contains 100 gal of fresh water. Then water containing = lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 10 gal/min, with the mixture again leaving at the same rate. Find the amount...
A tank originally contains 400 gal of fresh water. Then water containing 0.9 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 12 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the...
Please show your work (and reasons!) clearly and explain it like you think I'm dumb. I have tried this problem 3 times by myself and gotten 3 different (wrong) answers! I need your help! Thanks! 1 3. [2.3 3 A tank originally contains 100 gallons of fresh water. Then water containing lb of salt per gallon is poured into the tank at a rate of 2 gal/min and the mixture is allowed to leave at the same rate. After 10...
A tank originally contains 100 gallons of fresh water. Water containing lb of salt per gallon is poured into the tank at a rate of 2 gallons per minute, and the mixture is allowed to leave at the same rate. After 10 minutes the salt water solution flowing into the tank suddenly switches to fresh water flowing in at a rate of 2 gallons per minute, while the solution continues to leave the tank at the same rate (a) Write...
A tank with capacity of 700 gal of water originally contains 300 gal of water with 50 lb of salt in solution Water containing 1 lb of salt per gallon is entering at a rate of 4 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Let Q(t) (in pounds) be the amount of salt in the tank and V(t) (in gallons) be the volume of water in the tank. a) Find...
A tank with capacity of 600 gal of water originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Let Qct) Ib be the amount of salt in the tank, Vt) gal be the volume of water in the tank. Find the amount of...