1. The problem dA/dt = -.02A dB/dt = .01A - .04B dC/dt = .01A + .04B A(0)=1000 B(0)=0 C(0)=0 Come up with a sketch and a word problem involving three tanks for which the system of differential equations with initial conditions above applies. Have fresh water flowing into tank A, with the volumes of brine in B and C remaining constant. Assume that the third tank has such capacity that it would not overflow during the time period under consideration. Use different values for the amounts of brine in tanks B and C (for example 1,000 gallons in tank B and 2,000 gallons in tank C [PS : don't use 1000 and 2000]) 2. equations n/a 3. Attempt at solution Can someone help me interpret this problem. there are multiple aspects of this system that don't make sense to me. first is that if A flows into B then the rate out of A should be equal to the rate in for B. but instead it is half (.02A in dA/dt and .01A in dB/dt). and I also don't understand why dC/dt is a sum instead of a difference. If rate of change is rate in minus rate out then why is dC/dt a sum?