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

Answer:

a. $\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. $x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. $c = \dfrac{3}{8} \ g/L$

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 = $6\frac{2}{3} \cdot c \cdot dt$

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

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

$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

$dx = 6\frac{2}{3} \cdot c \cdot dt$

$x(t) = \int\limits \, dx = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt$

$x(t) = 6\frac{2}{3} \cdot c \cdot t$

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

$x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10$

$6\frac{2}{3} \times c =\dfrac{25 \ grams }{10}$

$c =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} } = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200} \ g/L = \dfrac{3}{8} \ g/L$

$c = \dfrac{3}{8} \ g/L$