Solve the following questions, please. A 50 gal tank contains 20 gal of pure wat

Question

Solve the following questions, please. A 50 gal tank contains 20 gal of pure water. Brine containing 1/2 lb/gal of salt enters the tank at the rate of 4 gal/min. The well stirred solution leaves the tank at the rate of 2 gal/min. What is the amount of salt in the tank at the precise moment that it is full? A swimming pod initially contains 10,000 gal of water which is 1% chlorine. Pure water flows into the pool at the rate of 20 gal/min and leaves at the rate of 30 gal/min. What is the concentration of chlorine in the pool after 1 hour. When theoretically will the tank become empty? Solve the IVP y" - 2y' + 5y = 0, y(pi) = e^pi, y'(pi) = 2e^pi. Solve the IVP (3x^2y - sin (x + y) - ye^xy) dx + (x^3 - sin (x + y) - xe^xy) dy = 0, y (0) = 0. Suppose that the population of an animal p(t) increases at a rate which is proportional to the square of the population. If the population after a minute is p(1) = 10 animals and after 2 minutes it is 20 animals, find the population of the animals at time t. Explain what happens to the population as t increases.

Explanation / Answer

1)let amount of salt in tank at time t is y(t)

rate of salt input =4*(1/2) =2 lb/min

rate of salt output =2*((y(t))/(20+(4-2)t)) =2*((y(t))/(20+2t))=((y(t))/(10+t)) lb/min

net rate y'(t)=rate of salt input -rate of salt output

y'(t) =2 -((y(t))/(10+t)), initial amount of salt in tank y(0) =0 (since it is pure water)

y'(t) +((y(t))/(10+t)) =2

dy/dt +(y/(10+t)) =2

dy(10+t) + ydt =2(10+t) dt

(y(10+t))'=(20+2t) dt

integrate on both sides

(y(10+t))'=(20+2t) dt

y(10+t)=(20t+t2) +C

y=[(20t+t2) +C]/(10+t)

amount of salt in tank at time t is y(t)=[(20t+t2) +C]/(10+t)

y(0)=0

[(20*0 +02) +C]/(10+0) =0

C/10=0

C=0

amount of salt in tank at time t is y(t)=(20t+t2)/(10+t)

time taken for the tank to get full ,t=(50-20)/(4-2)

time taken for the tank to get full ,t=15 min

amount of salt in tank when tank is full =y(15)

amount of salt in tank when tank is full =((20*15)+152)/(10+15)

amount of salt in tank when tank is full =(300+225)/(25)

amount of salt in tank when tank is full =21 lb

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