Respuesta :
Answer:
a. [tex]\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c[/tex]
b. [tex]x(t) = 6\frac{2}{3} \cdot c \cdot t[/tex]
c. [tex]c = \dfrac{3}{8} \ g/L[/tex]
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 = [tex]6\frac{2}{3} \cdot c \cdot dt[/tex]
[tex]\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c[/tex]
b. The amount of salt, x after t minutes is given by the relation
[tex]\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c[/tex]
[tex]dx = 6\frac{2}{3} \cdot c \cdot dt[/tex]
[tex]x(t) = \int\limits \, dx = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt[/tex]
[tex]x(t) = 6\frac{2}{3} \cdot c \cdot t[/tex]
c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;
[tex]x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10[/tex]
[tex]6\frac{2}{3} \times c =\dfrac{25 \ grams }{10}[/tex]
[tex]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[/tex]
[tex]c = \dfrac{3}{8} \ g/L[/tex]
