A tank initially contains 15 liters of water with 17 grams of salt in solution.

Question

A tank initially contains 15 liters of water with 17 grams of salt in solution. Water containing salt with a concentration of 1 grams per liter is poured in the tank at a rate of 3 liters per minute. The well-stirred water is allowed to pour out the tank at a rate of 1 liters per minute.

(2/10) Find the function V volume of water in the tank, in liters.

(2/10) Find the differential equation satisfied by the function Q, the total amount of salt in the tank measured in grams.

(4/10) Integrate the separable equation in part (b) to find the function Q, the amount of salt grams in the tank as function of time,

Explanation / Answer

1)v(t)=15 +(3-1)*t

v=15+2t

1)Q'(t)=rate at which salt enters tank -rate at which salt exits the tank

Q'(t)=(flow rate in *concentration)-(flow rate out *concentration)

Q'(t)=1*3 - 1*Q(t)/(15+2t)

Q'(t)=3 - Q(t)/(15+2t)

dQ/dt =3 -Q/(15+2t)

dQ +Q/(15+2t) dt=3dt

integrating factor =e1/(15+2t) dt

integrating factor =e(1/2)ln(15+2t)

integrating factor= (15+2t)1/2

multiply on both sides by integrating factor = (15+2t)

(15+2t)1/2dQ +(15+2t)1/2Q/(15+2t) dt=3(15+2t)1/2dt

(15+2t)1/2dQ +Q/(15+2t)1/2 dt=3(15+2t)1/2dt

((15+2t)1/2Q)'=3(15+2t)1/2dt

integrate on both sides

((15+2t)1/2Q)'=3(15+2t)1/2dt

((15+2t)1/2Q)=3(15+2t)1/2dt

((15+2t)1/2Q)=3(1/3)(15+2t)3/2+c

((15+2t)1/2Q)=(15+2t)3/2+c

Q=(15+2t)+[c/(15+2t)1/2]

initially contains 15 liters of water with 17 grams of salt in solution.

=>>Q(0)=17

17=(15+0)+[c/(15+0)1/2]

c/(15)1/2=2

c=2(15)1/2

Q=(15+2t)+[2(15)1/2/(15+2t)1/2]

Q(t)=(15+2t)+[2(15/(15+2t))1/2] is the solution

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