Question

DIFFERENTIAL EQUATIONS WEBWORK PROBLEM

(1 point) Consider two interconnected tanks as shoun in the tigure above. Tank1 initial contains 10 L(iters of water and 215 g ot salt, while tank 2 initally contains 90 L ot water and 35g ot salt Water containing 50 gL ot salt is poured into tank1 at a rate of 4 Limin while the mixture to ing into tank 2 contains a salt concentrationof45L ot salt and is tlowing at the rate of 1 Limin. The two connecting tubes have a tow rate oLmin trom tank 1 to tank 2; and ot 3 Lmin trom tank 2 back to tank1 2 ls dralined at the rate lank ot b L/min You may assume hat he solutions in each ank are oroughly mixed so at the concentration of he mi any tank along any C the tubes has the same concentration of salt as e ank as a whole. This is not completely realistic. but as re leaving in real physics, we are going to work with the approximate, rather than exact description. The 'real' equations of physics are often too complicated to even write down precisely, much less solve.) How does the water In each tank change over time? Let pii) and bethe amount of salt in g at timet in tanks 1 and 2 respectively. Write differential ecuations tor and y. (As usual, use the symbols p and g rather than p and au). Give the initial values: p/O) q(0) to the differential Show the equation that needs to be solved to ind a constant solution to the differential equation: A constant solution is obtained ifpi | for all time t and q(r) = for all time t

If you can't read the picture:

Consider two interconnected tanks as shown in the figure above. Tank 1 initial contains 10 L (liters) of water and 215 g of salt, while tank 2 initially contains 90 L of water and 355 g of salt. Water containing 50 g/L of salt is poured into tank1 at a rate of 4 L/min while the mixture flowing into tank 2 contains a salt concentration of 45 g/L of salt and is flowing at the rate of 1 L/min. The two connecting tubes have a flow rate of 7 L/min from tank 1 to tank 2; and of 3 L/min from tank 2 back to tank 1. Tank 2 is drained at the rate of 5 L/min.

You may assume that the solutions in each tank are thoroughly mixed so that the concentration of the mixture leaving any tank along any of the tubes has the same concentration of salt as the tank as a whole. (This is not completely realistic, but as in real physics, we are going to work with the approximate, rather than exact description. The 'real' equations of physics are often too complicated to even write down precisely, much less solve.)

How does the water in each tank change over time?

Let p(t) and q(t) be the amount of salt in g at time t in tanks 1 and 2 respectively. Write differential equations for p and q. (As usual, use the symbols p and q rather than p(t) and q(t).

Answer 1

Consider two interconnected tanks as shown in the figure above. Tank 1 initial contains 10 L (liters) of water and 215 g of salt, while tank 2 initiallv contains 90 L of water and 355 g of salt. Water containing 50 g/L of salt is poured into tankl at a rate of 4 L/min while the mixture flowing into tank 2 contains a salt concentration of 45 g/L of salt and is flowing at the rate of 1 L/min. The two connecting tubes have a flow rate of 7 L/min from tank 1 to tank 2; and of 3 L/min from tank 2 back to tank 1. Tank 2 is drained at the rate of 5 L/min. How does the water in each tank change over time? Let p(t) and q(t) be the amount of salt in g at time t in tanks 1 and 2 respectively. Write differential equations for p and q. (As usual, use the symbols p and q rather than p(t) and q(t) Ans Let Inital Volume of Tank 1 10 L Initial Volume of Tank 2 90L Consider the inflow and outflow of water for each tank Tank 1: Inflow = 4 L/min + 1 L/min = 5 L/min Outflow = 7 L/min Tank 2: Inflow = 1L/min-TL/min = 5 Lmim Outflow = 5 L/min + 3L/min = 8 L/min Now let p(t) = amount of salt, in grams, in Tank 1 as function of time per min and q(t) = amount of salt, in grams, in Tank 2 as function of time per min The initial conditions are given by, p(0) 215 grams of salt g(0) 355 grams of salt Thus the derivative is p'(t) = The rate of change with respect to time of the amount of salt in Tank 1 g(t) = The rate of change with respect to time of the amount of salt in Tank 2

The differential equations are given by, p (t) (Incoming flow in Tank 1) (Concentration of salt in flow) +(Incoming flow from Tank 2) (Concentration of salt in Tank 2) -(Outgoing flow from Tank 1)(Concentration in Tank 1) (t)-(Incoming flow) (concentration of salt in flow p(t) (Volume of water in Tank 2) +(Incoming flow from Tank 2) q (t) -(Outgoing flow from Tank 1) Volume of water in Tank1 「g(t) p (t) 4 L/min)+ 90 L 10 L p'(t)-200 + PO-40 g/mi 10

f' (t)-(Incoming fow)(Concentration of salt in flow) p(t) Volume of water in Tank 1 Incoming flow from Tank 1) g(t) -( Outgoing flow from Tank 2 ) Volume of water in Tank 2 q(t) d (t)-(45L) (1L/min) (7 Li-(8 L/min) 90 L 10 L 7pt) 4q(t) t)=45+ 10 Therefore p'(t) and q'(t) in the matrix form are given by, 「p(01 p'(t)1-13-101-p( )1,「2001 215 wi L355 g(01 10 45