A tank initially contains 200 gallons of water in which 50 lbs of salt are dissolved. A salt solution containing 0.5 lb of salt per gallon is poured into the tank at a rate of 1 gal/min. The mixture in the tank is stirred and drained off at the rate of 2 gal/min.
a) Find the amount of salt in the tank until the tank is empty.
b) Find the concentration of salt in the tank until the tank is empty.
The correct answer and explanation is :
Alright, let’s dive in carefully.
We are given:
- Initial volume ( V(0) = 200 ) gal, initial salt ( S(0) = 50 ) lb
- Inflow: 1 gal/min, salt concentration = 0.5 lb/gal
- Outflow: 2 gal/min
- So net volume decreases at ( 2 – 1 = 1 ) gal/min
Thus, volume at time ( t ) is:
[
V(t) = 200 – t
]
The tank will be empty when ( V(t) = 0 ), so at ( t = 200 ) minutes.
Step 1: Set up the differential equation
The rate of change of salt ( S(t) ) is:
[
\frac{dS}{dt} = \text{rate in} – \text{rate out}
]
Rate in = (inflow rate) × (inflow concentration)
= ( (1 \, \text{gal/min})(0.5 \, \text{lb/gal}) = 0.5 \, \text{lb/min} )
Rate out = (outflow rate) × (current concentration in tank)
= ( (2 \, \text{gal/min})\left(\frac{S(t)}{V(t)}\right) = 2 \times \frac{S(t)}{200-t} )
Thus:
[
\frac{dS}{dt} = 0.5 – 2\frac{S(t)}{200-t}
]
Step 2: Solve the differential equation
Rearranging:
[
\frac{dS}{dt} + \frac{2}{200-t} S = 0.5
]
This is a linear differential equation. The integrating factor ( \mu(t) ) is:
[
\mu(t) = e^{\int \frac{2}{200-t} \, dt} = e^{-2\ln(200-t)} = (200-t)^{-2}
]
Multiplying through by ( (200-t)^{-2} ):
[
(200-t)^{-2} \frac{dS}{dt} + \frac{2}{(200-t)^3} S = 0.5(200-t)^{-2}
]
The left-hand side is:
[
\frac{d}{dt}\left( (200-t)^{-2} S \right)
]
Thus:
[
\frac{d}{dt}\left( (200-t)^{-2} S \right) = 0.5(200-t)^{-2}
]
Integrating both sides:
[
(200-t)^{-2} S = 0.5 \int (200-t)^{-2} \, dt
]
We know:
[
\int (200-t)^{-2} \, dt = \frac{(200-t)^{-1}}{1}
]
Thus:
[
(200-t)^{-2} S = 0.5 (200-t)^{-1} + C
]
Multiplying through by ( (200-t)^2 ):
[
S(t) = 0.5 (200-t) + C(200-t)^2
]
Step 3: Apply initial condition
At ( t=0 ), ( S(0) = 50 ):
[
50 = 0.5(200) + C(200)^2
]
[
50 = 100 + 40000C
]
[
40000C = -50
]
[
C = -\frac{1}{800}
]
Thus:
[
S(t) = 0.5(200-t) – \frac{1}{800}(200-t)^2
]
Final answers:
a) Amount of salt in the tank:
[
\boxed{S(t) = 0.5(200-t) – \frac{1}{800}(200-t)^2}
]
valid for ( 0 \leq t \leq 200 ).
b) Concentration (salt per gallon):
[
\text{Concentration} = \frac{S(t)}{V(t)} = \frac{S(t)}{200-t}
]
300-word Explanation:
In this mixing problem, the key is setting up a balance between the salt entering and leaving the tank. Initially, there are 50 lbs of salt in 200 gallons. Salt solution flows in at 1 gal/min, adding 0.5 lbs of salt each minute. However, the tank drains faster, at 2 gal/min, causing the volume to drop by 1 gallon per minute.
Since the volume depends linearly on time as ( V(t) = 200 – t ), the concentration inside the tank is changing. We built a differential equation based on this rate of change: the amount of salt added minus the amount leaving. After rearranging the terms, we recognized the form of a linear differential equation and used an integrating factor to solve it.
The solution involves finding the integrating factor, multiplying through, integrating, and then applying the initial condition to find the specific solution. In this case, solving gives:
[
S(t) = 0.5(200-t) – \frac{1}{800}(200-t)^2
]
for the amount of salt.
Since the volume at any time ( t ) is ( V(t) = 200 – t ), the concentration is the ratio ( S(t)/(200-t) ).
Notice, as time progresses, the volume decreases, and although salt is being added, the faster draining causes the concentration to behave nonlinearly. Eventually, when the tank empties at 200 minutes, both the volume and the amount of salt drop to zero.
This problem nicely shows how mixing, flow rates, and concentration interact dynamically over time.