Mathematics, 16.07.2020 23:01 hadilalhjajih

A tank contains 4,000 L of brine with 18 kg of dissolved salt. Pure water enters the tank at a rate of 40 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. a. How much salt is in the tank after t minutes?
b. How much salt is in the tank after 30 minutes? (Round your answer to one decimal place.

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Answer from: wrightstephanie193

In this exercise we have to use the integral to calculate the salt concentration:

(a) $A(t)=18e^{-\frac{t}{100} }$

(b) 13.3 kg

Knowing that the volume of brine in the tank = 4000L, the initial Amount of salt, A(0)=18 kg. The rate of change in the amount of salt in the tank at any time t is represented by the equation:

$\frac{dA}{dt} = Rate \ in - Rate \ out$

Rate In = (concentration of salt in inflow)(input rate of brine). Since pure water enters the tank, concentration of salt in inflow =0.

Rate In = 0

Rate Out=(concentration of salt in outflow)(output rate of brine)

$\frac{A(t)}{4000}*(40)$

$= \frac{A(t)}{100}$

Therefore:

$\frac{dA}{dt} = \frac{A(t)}{100}\\\frac{dA}{dt} + \frac{A(t)}{100} = 0$

This is a linear D.E. which we can then solve for A(t). Integrating Factor: $e^{\int\limits {\frac{t}{100} } \, dt\\e^{t/100}$ . Multiplying all through by the Integrating Factor:

$\frac{dA}{dt} = e^{t/100}+\frac{A(t)}{100}e^{t/100}\\(Ae^{1/100})'=0$

Taking integral of both sides:

$Ae^{t/100}=C\\A(t)=Ce^{-t/100}$

Recall our initial condition:

$A(0)=18 kg\\18=Ce^{0}\\C=18$

Therefore, the amount of salt in the tank after t minutes is:

$A(t)=18e^{-t/100}$

(b)When t=30 mins

$A(30)=18e^{-30/100}\\=18e^{-0.3}\\=13.3$

The amount of salt in the tank after 30 minutes is 13.3kg.

See more about concentration at link

Answer from: lrnnwb

(a) $A(t)=18e^{ -\frac{t}{100}}$

(b)13.3 kg

Step-by-step explanation:

The volume of brine in the tank = 4000L

Initial Amount of salt, A(0)=18 kg

The rate of change in the amount of salt in the tank at any time t is represented by the equation:

$\dfrac{dA}{dt}=$Rate In$-$Rate Out$

Rate In = (concentration of salt in inflow)(input rate of brine)

Since pure water enters the tank, concentration of salt in inflow =0

Rate In = 0

Rate Out=(concentration of salt in outflow)(output rate of brine)

$=\frac{A(t)}{4000}\times 40\\ =\frac{A(t)}{100}$

Therefore:

$\dfrac{dA}{dt}=-\dfrac{A(t)}{100}\\\dfrac{dA}{dt}+\dfrac{A(t)}{100}=0$

This is a linear D.E. which we can then solve for A(t).

Integrating Factor: $e^{\int \frac{1}{100}d}t =e^{ \frac{t}{100}$

Multiplying all through by the I.F.

$\dfrac{dA}{dt}e^{ \frac{t}{100}}+\dfrac{A(t)}{100}e^{ \frac{t}{100}}=0e^{ \frac{t}{100}}\\(Ae^{ \frac{t}{100}})'=0$

Taking integral of both sides

$Ae^{ \frac{t}{100}}=C\\A(t)=Ce^{ -\frac{t}{100}}$

Recall our initial condition

A(0)=18 kg

$18=Ce^{ -\frac{0}{100}}\\C=18$

Therefore, the amount of salt in the tank after t minutes is:

$A(t)=18e^{ -\frac{t}{100}}$

(b)When t=30 mins

$A(30)=18e^{ -\frac{30}{100}}\\=18e^{ -0.3}\\=13.3 $kg(correct to 1 decimal place)$

The amount of salt in the tank after 30 minutes is 13.3kg

Answer from: Quest

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )

Answer from: Quest

r = 78539458390583095803458309458039485093485 or just 3

step-by-step explanation:

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A tank contains 4,000 L of brine with 18 kg of dissolved salt. Pure water enters the tank at a rate...