Question

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Sec 1.4: Linear equations and the integrating factor 8 Previous A fish tank initially contains 10 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 5 liters per minute. The solution is mixed well and drained at 5 liters per minute. Let r be the amount of salt, in grams, in the fish tank after t minutes have elapsed. Find a formula for the of change in the amount of salt, dr/dt, in terms of the amount of salt in the solution a and the unknown concentration of incoming brine c. rate da grams/minute help (formulas) dt Find a formula for the amount of salt, in grams, after t minutes have elapsed. Your answer should be in terms of e and t. grams help (formulas) r(t) In 35 minutes there are 25 grams of salt in the fish tank. What is the concentration of salt in the incqming brine? g/L help (numbers) Submit answer

Answer 1

x= amount of salt in solution rias ll minutes ro = 5L (minutes (i= c = incoming concentration. vet) = 10 + (0; -910) t = 10 + CS-5) + =10 tot luct) = 10 co= out going Cost concentration. = 0.13 DE is given by. di = Mici - Moco at = 5(C) - 5 (0.12) The house - 56-0,5x) * (a) da = 50-0.52 at

habe -0.5 (+ r) har en 2-0.5 (*- eo) olz = -0.9 (OC-10c) da =-0.5 at X-100 Samor dx = fro.s at en COC-100) = -0.5t to . -0,5t to x-106 = 0.st.ee x= 100 rsieost -♡ here so pure water initially x100=o.

A . 010c tc, e o= 100 trie 0=10C +C, C13-100 put in ean 0 x= 106 = 100 e oste (6) in 35 minutes, 25 grams of x (35)=25 there salt are . -0.5(35) 25 = 10c. loc éº. 25 = 10c-loce 17.5 25 = 100 - 100 (6) 25 = 10(-0 100 - 25 (= 25 (c =2.5 g h a co