(Solved): Consider two interconnected tanks as shown in the figure above. Tank 1 initial contains \( 50 \math ...

Consider two interconnected tanks as shown in the figure above. Tank 1 initial contains \( 50 \mathrm{~L} \) (liters) of water and \( 180 \mathrm{~g} \) of salt, while tank 2 initally contains \( 40 \mathrm{~L} \) of water and \( 465 \mathrm{~g} \) of salt. Water containing \( 15 \mathrm{~g} \mathrm{~L} \) of salt is poured into tank1 at a rate of \( 1.5 \mathrm{~L} \) min while the mixture flowing into tank 2 contains a salt concentration of \( 45 \mathrm{~g} \mathrm{~L} \) of salt and is flowing at the rate of \( 3.5 \mathrm{Lmin} \). The two connecting fubes have a flow rate of \( 4 \mathrm{~L} / \mathrm{min} \) from tank 1 to tank 2 ; and of \( 2.5 \mathrm{Lmin} \) from tank 2 back to tank 1 . Tank 2 is drained at the rate of \( 5 \mathrm{~L} / \mathrm{min} \). 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 number of grams \( (g) \) of salt 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 \( q(t) \).) \( q^{\prime}= \) Give the initial values \[ \left[\begin{array}{l} p(0) \\ q(0) \end{array}\right]=\left[\begin{array}{l} ] \end{array}\right] \] Show the equation that needs to be solved to find a constant solution to the differential equation: A constant solution is obtained if \( p(t)= \) for all time 1 and \( q(t)= \) for all time \( t \).