Thread: Applications of differential equations questions

Salt is dissolved in a tank filled with 120 litres of water. Salt water containing 20g of salt per litre is poured in at a constant rate of 3 litres per minute and the mixture flows out at a constant rate of 3 litres per minute. The contents of the tank are kept well mixed at all times. Let the amount of salt in the tank (in grams) be denoted by S and the time (in minutes) be denoted by t.

i)Show that dS/dt = 1/40(2400-S)
ii) Given that 400g of salt was dissolved in the tank initially, find the amount of salt in the tank after 1 hour, giving your answer to the nearest gram.

I'm having trouble getting an equation to work with.



also another question. When Jenny retired in 2006, she put a sum of $5000 into a fund that has a constant rate of return of 5% per annum. Starting in 2006, she withdraws $400 each year and gives the money to her granddaughter as a birthday gift. Denote the amount of money Jenny has at time t years by $x.

i) The differential equation relating x and t is in the form dx/dt =kx +c, state the values of k and c.
ii) Solve the d.e and find the amount of money Jenny has after 15 years. giving your answer to the nearest integer.
iii)In which year will the granddaughter receive her last $400
iv) Comment on whether the model can be regarded as a good model of the situation in the real world.

again.. im having trouble getting the initial equation

dS/dt = salt in - salt out

dS/dt = (concentration)(flow rate in) - (concentration)(flow rate out)

dS/dt = (20g/l)(3l/min) - (Sg/120l)(3l/min)

dS/dt = 60 - 1/40S = 1/40( 2400 -S)

Separate and solve note equillibrium solution is S =2400 since S(0) = 400

2400 - S > 0