Question

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

Answer 1

Answer:

Step-by-step explanation:

The complete question is

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.

Let C(t) be the amount of water in the tank at time t. We now that the rate of change of the tank is given by

$\frac{dC}{dt}=[\tex]rate at which water flows in- rate at which water flows out. Then [tex]\frac{dC}{dt}=\frac{t+6}{t+1}-\frac{\ln(t+2)}{(t+1)}[\tex]so, the desired expression is obtained by integrating with respect to t. This leads us to [tex]C(t) = \int \frac{t+1}{t+1}+ \frac{5}{t+1} - \frac{\ln(t+2)}{(t+1)} dt=t+ 5 \ln (|t+1|)-\int \frac{\ln(t+2)}{(t+1)} dt +C$

Unfortunately, the integral $\int \frac{\ln(t+2)}{(t+1)} dt$ cannot be expressed using fundamental functions. So, the problem cannot have an specific function (if you are willing to know the complete answer, the integral of this function uses the  polylogarithm function with n=2).

Since you want the exact amount of water at time, you need to give C a value, that is, you need to know an initial condition for the problem. This means, you need to know the amount of water in the tank at time 0