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.
To find out how much water is in the tank at time t = 10 minutes, we need to calculate the net change in the amount of water in the tank over the interval of time from t = 0 to t = 10 minutes.
The given rate at which water flows into the tank is given by F(t) = (6 + t) / (1 + t), measured in gallons per minute.
The given rate at which water empties out of the tank is given by E(t) = ln(t + 2) / (t + 1), also measured in gallons per minute.
To calculate the net change in water, we need to find the integral of the difference between the inflow rate (F(t)) and the outflow rate (E(t)) over the time interval from t = 0 to t = 10. Let's call this integral N(t), representing the net change in the amount of water in the tank at time t.
N(t) = ∫[0 to 10] (F(t) - E(t)) dt
Now, let's substitute the given expressions for F(t) and E(t) into the above integral:
N(t) = ∫[0 to 10] [(6 + t) / (1 + t) - ln(t + 2) / (t + 1)] dt
To evaluate the integral, we can use calculus techniques such as integration by substitution or integration by parts. However, since it involves a complicated expression, a numerical approximation method like numerical integration may be more suitable.
Using numerical integration techniques like Simpson's rule or the trapezoidal rule, we can approximate the value of the integral N(t) for the interval [0, 10]. This approximation will give us the net change in the amount of water in the tank over that time interval.
Finally, to find the amount of water in the tank at time t = 10 minutes, we need to determine the initial amount of water in the tank at t = 0 and then add the net change in water over the interval [0, 10] to it.
However, without additional information about the initial amount of water in the tank at t = 0, we cannot determine the exact amount of water in the tank at t = 10 minutes.
To find the amount of water in the tank at t = 10 minutes, we need to calculate the net flow rate of water into the tank minus the rate at which it is emptied.
Let's start by finding the net flow rate F(t) - E(t) at t = 10 minutes.
Given:
F(t) = (6 + t) / (1 + t)
E(t) = ln(t + 2) / (t + 1)
To find F(10):
F(10) = (6 + 10) / (1 + 10)
F(10) = 16 / 11
F(10) = 1.4545 (approximately)
To find E(10):
E(10) = ln(10 + 2) / (10 + 1)
E(10) = ln(12) / 11
E(10) ≈ 0.1462 (approximately)
Now, let's find the net flow rate at t = 10:
Net flow rate = F(10) - E(10)
Net flow rate ≈ 1.4545 - 0.1462
Net flow rate ≈ 1.3083 (approximately) gallons per minute
To find the amount of water in the tank at t = 10 minutes, we multiply the net flow rate by the time:
Amount of water = Net flow rate * t
Amount of water ≈ 1.3083 * 10
Amount of water ≈ 13.083 (approximately) gallons
Therefore, there are approximately 13 gallons of water in the tank at t = 10 minutes.
F of t equals the quotient of 6 plus t and the quantity 1 plus t??
How about
F(t)=(6+t)/(1+t)
E(t)=ln(t+2)/(t+1)
Since these are the rates of change, the amount of water will be
V(t)=∫[0,10] F(t)-E(t) dt
=∫[0,10] (6+t)/(1+t)-ln(t+2)/(t+1) dt
Now, (6+t)/(1+t) = 1 + 5/(1+t)
So, ∫(6+t)/(1+t) dt = t + 5ln(1+t)
Now, ln(t+2)/(t+1) does not integrate using elementary functions.
If I parsed your words wrong, then fix things, and you should be able to finish up where I have left off.