Question

Chapter 2, Section 2.3, Additional Question 02 A tank originally contains 100 gal of fresh water. Then water containing = lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 10 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min. Round your answer to two decimal places. Ibs the absolute tolerance is +/-0.01 Click if you would like to Show Work for this question: Open Show Work

Answer 1

Let a be the amount of salt in tank. Rate of change of amt of salt = date in- rate out d@ = 4X2 - 90 - - de 2 -t/50 lat = 1 salt/gal x 2 gallmin - 2 gallmin x a salt 10o gal de +Q=1 0 If ze S et soth of is alt) e 150 - let/so idt a(t) e 150 - so ettrotic alt) - 50 + cen at t=0 amount of salt in tank iso 810) = 0 = 50+ Le = 50+ [c2-50 84) = 50-soe elso OLIO) = 50 - 50 e sº = 9.06 (approx) - This is the new initial value for the next equation after 10 minutes input of salt is zero) e oso

New is differential equation do = 0 - 2 x 10 dt 100 (rate in is zero) da = -a -tldt 10 1o 12) = - Itta, 10/= -4/1ota at a. Be tota (= exponent is always the) de cetlio ( where Cage) t²0 810) 29.06( from ) 8, 9.06e-thio 03 (10) = 9.06€ 9.066 = 3.332 lb (approx) after additional 10 minutes ant of salt is 10 minute Thus