A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per gallon is added to the tank at 6 gal/min, and the resulting solution leaves at the same rate. Find the quantity Q(t) of salt in the tank at time t > 0.

Salt flows into the tank at a rate of

(1/2 lb/gal) * (6 gal/min) = 3 lb/min

and flows out at a rate of

(Q(t)/60 lb/gal) * (6 gal/min) = 6Q(t) lb/min

The net rate of change of the amount of salt in the tank at time is then governed by

Solve for :

The tank starts with 10 lb of salt, so that Q(0) = 10. This gives us

so that the amount of salt in the tank at time is given by

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt

you can not solvr this without numbers