Question

Water flows into a tank according to the rate F(t)= (t+6)/(1+t), and at the same time empties out at the rate E(t)= (ln(t+2))/(t+1), with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest galllon, is in the tank at time t=10 minutes.

PLEASE HELP; I AM CONFUSED!

Answer 1

To find the amount of water in the tank at time t = 10 minutes, we need to calculate the net rate of change of water in the tank at that time. Given the inflow rate F(t) = (t+6)/(1+t) and outflow rate E(t) = (ln(t+2))/(t+1), we can substitute t = 10 to find the respective rates, and then integrate the net rate of change from t = 0 to t = 10 to find the amount of water in the tank.

Step-by-step explanation:

To find the amount of water in the tank at time t = 10 minutes, we need to calculate the net rate of change of water in the tank at that time. The net rate of change is the difference between the inflow and outflow rates.

Given inflow rate F(t) = (t+6)/(1+t) and outflow rate E(t) = (ln(t+2))/(t+1), we can substitute t = 10 into both equations to find the respective rates. F(10) = (10+6)/(1+10) = 16/11 gallons per minute and E(10) = (ln(12))/(11) ≈ 0.188 gallons per minute.

Therefore, the net rate of change is 16/11 - 0.188 ≈ 1.276 gallons per minute. To find the amount of water in the tank at time t = 10 minutes, we integrate the net rate of change from t = 0 to t = 10: ∫[0 to 10] (16/11 - 0.188) dt ≈ 11.618 gallons. Therefore, there are approximately 11.618 gallons of water in the tank at t = 10 minutes.

Answer 2

Water flows into a tank according to the rate F of t equals the quotient of 6 plus t and the quantity 1 plus t , and at the same time empties out at the rate E of t equals the quotient of the natural log of the quantity t plus 2 and the quantity t plus 1 , with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest gallon, is in the tank at time t = 10 minutes