Question

fish tank initially contains 35 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. Let xx be the amount of salt, in grams, in the fish tank after tt minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dtdx/dt, in terms of the amount of salt in the solution

Answer 1

Answer:

Therefore, the rate of change in the amount of salt is $\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$

$\frac{grams }{min}$

Explanation:

Given:

Initial volume of water $V = 35$ lit

Flowing rate = 5 $\frac{Lit}{min}$

The rate of change in the amount of salt is given by,

   $\frac{dx}{dt} =$ ( Rate of salt enters tank - rate of sat leaves tank )

Since tank is initially filled with water so we write that,

$x(0) = 0$

Let amount of salt in the solution is $c$,

  $\frac{dx}{dt} = \frac{5c}{1 } - \frac{x(t) \times 5}{100}$

  $\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$ $\frac{grams}{min}$

Therefore, the rate of change in the amount of salt is $\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$

$\frac{grams }{min}$