Question

(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 5 liters per minute. The solution is mixed well and drained at 5 liters per minute. a. Let x be the amount of salt, in grams, in the fish tank after t minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c. dx grams/minute b. Find a formula for the amount of salt, in grams, after t minutes have elapsed. Your answer should be in terms of c and t. x(t)= grams c. In 10 minutes there are 25 grams of salt in the fish tank. What is the concentration of salt in the incoming brine? C = g/L

Answer 1

The rate of change in the amount of salt in the fish tank can be determined by subtracting the rate of salt flowing out from the rate of salt flowing in. The formula for the rate of change is dx/dt = 5(c - x) grams/minute. The amount of salt in the tank after t minutes can be calculated using x(t) = x(0) + (c - x(0)) * t.

Step-by-step explanation:

To find the rate of change in the amount of salt, dx/dt, in the fish tank after t minutes have elapsed, we need to consider the rate of salt flowing in and out of the tank. The rate of change is equal to the rate of salt flowing in minus the rate of salt flowing out.

The rate of salt flowing in is the concentration of salt in the incoming brine, c, multiplied by the rate at which the brine is flowing in, which is 5 liters per minute.

The rate of salt flowing out is the concentration of salt in the tank, x, multiplied by the rate at which the solution is being drained, which is also 5 liters per minute.

So, dx/dt = c * 5 - x * 5 = 5(c - x) grams/minute. This is the formula for the rate of change in the amount of salt in the fish tank.

For the amount of salt, x, after t minutes have elapsed, we can use the formula x(t) = x(0) + (c - x(0)) * t. Here, x(0) is the initial amount of salt in the tank, which is 0 grams, and c is the concentration of salt in the incoming brine.

Answer 2

a.
`(dx)/(dt) = 6(2)/(3) \cdot c`

b.
`x(t) = 6(2)/(3) \cdot c \cdot t`

c.
`c = (3)/(8) \ g/L`

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx = (5·c - 5·c/3)×dt = 20/3·c =
`6(2)/(3) \cdot c \cdot dt`

`(dx)/(dt) = 6(2)/(3) \cdot c`

b. The amount of salt, x after t minutes is given by the relation

`(dx)/(dt) = 6(2)/(3) \cdot c`

`dx = 6(2)/(3) \cdot c \cdot dt`

`x(t) = \int\limits \, dx = \int\limits \left ( 6(2)/(3) \cdot c \right) \cdot dt`

`x(t) = 6(2)/(3) \cdot c \cdot t`

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

`x(10) = 25 \ grams(15 \ in \ liters) = 6(2)/(3) * c * 10`

`6(2)/(3) * c =(25 \ grams )/(10)`

`c =(25 \ g/L )/(10 * 6(2)/(3) ) = (25 \ g/L)/(10 * (20)/(3) ) =(3)/(200) * 25 \ g/L= (75)/(200) \ g/L = (3)/(8) \ g/L`

`c = (3)/(8) \ g/L`