Math  /  Calculus

Question5. The tank in Fig. 28 contains 80 lb of salt dissolved in 500 gal of water. The inflow per minute is 20 lb of salt dissolved in 20 gal of water. The outflow is 20gal/min20 \mathrm{gal} / \mathrm{min} of the uniform mixture. Find the time when the salt content y(t)y(t) in the tank reaches 95%95 \% of its limiting value (as tt \rightarrow )\infty).
Fig. 28. Tank in Problem 28

Studdy Solution

STEP 1

1. The tank initially contains 80 lb of salt in 500 gallons of water.
2. The inflow rate is 20 lb of salt in 20 gallons of water per minute.
3. The outflow rate is 20 gallons per minute.
4. The mixture in the tank is uniform at all times.

STEP 2

1. Set up the differential equation for the salt content in the tank.
2. Solve the differential equation to find the function y(t) y(t) .
3. Determine the limiting value of y(t) y(t) as t t \rightarrow \infty .
4. Calculate the time when y(t) y(t) reaches 95% of its limiting value.

STEP 3

Set up the differential equation for the salt content in the tank:
Let y(t) y(t) be the amount of salt in the tank at time t t .
The rate of change of salt in the tank is given by the difference between the inflow and outflow rates of salt:
dydt=Inflow rate of saltOutflow rate of salt\frac{dy}{dt} = \text{Inflow rate of salt} - \text{Outflow rate of salt}
The inflow rate of salt is 20lb/min 20 \, \text{lb/min} .
The outflow rate of salt is proportional to the concentration of salt in the tank, which is y(t)500lb/gal×20gal/min \frac{y(t)}{500} \, \text{lb/gal} \times 20 \, \text{gal/min} .
Thus, the differential equation becomes:
dydt=2020y500\frac{dy}{dt} = 20 - \frac{20y}{500}
Simplify the equation:
dydt=20y25\frac{dy}{dt} = 20 - \frac{y}{25}

STEP 4

Solve the differential equation to find y(t) y(t) :
Rearrange the equation:
dydt+y25=20\frac{dy}{dt} + \frac{y}{25} = 20
This is a linear first-order differential equation. The integrating factor is:
e125dt=et25e^{\int \frac{1}{25} \, dt} = e^{\frac{t}{25}}
Multiply through by the integrating factor:
et25dydt+125et25y=20et25e^{\frac{t}{25}} \frac{dy}{dt} + \frac{1}{25} e^{\frac{t}{25}} y = 20 e^{\frac{t}{25}}
Integrate both sides:
(ddt(et25y))dt=20et25dt\int \left( \frac{d}{dt} \left( e^{\frac{t}{25}} y \right) \right) \, dt = \int 20 e^{\frac{t}{25}} \, dt
et25y=500et25+Ce^{\frac{t}{25}} y = 500 e^{\frac{t}{25}} + C
Solve for y(t) y(t) :
y(t)=500+Cet25y(t) = 500 + Ce^{-\frac{t}{25}}
Apply the initial condition y(0)=80 y(0) = 80 :
80=500+C80 = 500 + C
C=420C = -420
Thus, the solution is:
y(t)=500420et25y(t) = 500 - 420e^{-\frac{t}{25}}

STEP 5

Determine the limiting value of y(t) y(t) as t t \rightarrow \infty :
As t t \rightarrow \infty , et250 e^{-\frac{t}{25}} \rightarrow 0 .
Thus, the limiting value of y(t) y(t) is:
ylim=500y_{\text{lim}} = 500

STEP 6

Calculate the time when y(t) y(t) reaches 95% of its limiting value:
95% of the limiting value is:
0.95×500=4750.95 \times 500 = 475
Set y(t)=475 y(t) = 475 and solve for t t :
475=500420et25475 = 500 - 420e^{-\frac{t}{25}}
420et25=25420e^{-\frac{t}{25}} = 25
et25=25420e^{-\frac{t}{25}} = \frac{25}{420}
Take the natural logarithm of both sides:
t25=ln(25420)-\frac{t}{25} = \ln\left(\frac{25}{420}\right)
Solve for t t :
t=25ln(25420)t = -25 \ln\left(\frac{25}{420}\right)
Calculate the value:
t75.54minutest \approx 75.54 \, \text{minutes}
The time when the salt content reaches 95% of its limiting value is approximately:
75.54minutes \boxed{75.54 \, \text{minutes}}

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