Answer:
Therefore, the rate of change in the amount of salt is [tex]\frac{dx}{dt} =( 5c}{ - \frac{x }{20})[/tex]
[tex]\frac{grams }{min}[/tex]
Explanation:
Given:
Initial volume of water [tex]V = 35[/tex] lit
Flowing rate = 5 [tex]\frac{Lit}{min}[/tex]
The rate of change in the amount of salt is given by,
[tex]\frac{dx}{dt} =[/tex] ( Rate of salt enters tank - rate of sat leaves tank )
Since tank is initially filled with water so we write that,
[tex]x(0) = 0[/tex]
Let amount of salt in the solution is [tex]c[/tex],
[tex]\frac{dx}{dt} = \frac{5c}{1 } - \frac{x(t) \times 5}{100}[/tex]
[tex]\frac{dx}{dt} =( 5c}{ - \frac{x }{20})[/tex] [tex]\frac{grams}{min}[/tex]
Therefore, the rate of change in the amount of salt is [tex]\frac{dx}{dt} =( 5c}{ - \frac{x }{20})[/tex]
[tex]\frac{grams }{min}[/tex]
