Consider a fish tank that initially contains 300 liters of water with 100 grams of salt dissolved in it. Water containing 4 grams of salt per liter is allowed to run into the tank at a rate of 2 liters per minute. The thoroughly stirred mixture in the tank is also allowed to drain out at a rate of 3 liters per minute.

(a) Find a differential equation for the amount of salt x(t) in the tank at time t.
(b) Solve the DE in (a).

Step-by-step explanation:

given that  a fish tank that initially contains 300 liters of water with 100 grams of salt dissolved in it. Water containing 4 grams of salt per liter is allowed to run into the tank at a rate of 2 liters per minute.

The thoroughly stirred mixture in the tank is also allowed to drain out at a rate of 3 liters per minute.

Let X(t) be the amount of salt in the tank at time t

X(0) = 100 gms.

Volume of tank at time t = initial volume+t(incoming rate-outgoing rate)

= 300+(2-3)t= 300 -t

Rate of change of X would be=incoming salt rate-outgoing salt rate

,X(0) = 100 gm

i.e.

this is linear DE

Solution woul dbe

Use initial condition to find t

X(0) = 100 = 1200+300^3C

C = -4.074(10^-5)

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

they are simplified

step-by-step explanation: