A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 4 liters per minute. The solution is mixed well and drained at 4 liters per minute. Let x be the amount of salt, in grams, in the fish tank after t minutes have elapsed.
a. 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/dt =

Step-by-step explanation:

Data provided in the question:

Initial volume of water = 15 liters

Concentration of salt = 4 liters per minute

x be the amount of salt, in grams, in the fish tank after t minutes have elapsed

The unknown concentration of incoming brine c.

Now,

Salt concentration per minute =

For x amount of salt =  

Therefore,

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}