Question

water leaks out of a tank at a rate of r(t)=13−t23 for t≥0, measured in gallons per minute. initially the tank has 60 gallons of water in the tank. how much water is left in the tank after 3 minutes?

Answer 1

After 3 minutes, there is -1 gallon of water left in the tank, which means the tank is empty.

Step-by-step explanation:

The rate of water leaking out of the tank is given by the function r(t) = 13 - t/2/3 for t ≥ 0. To find how much water is left in the tank after 3 minutes, we need to calculate the integral of r(t) from 0 to 3 and subtract that amount from the initial amount in the tank.

First, let's find the integral of r(t). The integral of 13 is 13t. To find the integral of -t/2/3, we can use the power rule which states that the integral of x^n is (x^(n+1))/(n+1). So, the integral of -t/2/3 is (-t^(1+1))/(2/3+1) = (-t^2)/(2/3+1) = (-t^2)*3/5 = -3/5*t^2.

Now, let's calculate the definite integral of r(t) from 0 to 3:

1. Substitute 3 into r(t): 13 - 3/2/3 = 13 - 1 = 12.
2. Substitute 0 into r(t): 13 - 0 = 13.
3. Subtract the result from step 2 from the result from step 1: 12 - 13 = -1.

Therefore, after 3 minutes, there is -1 gallon of water left in the tank. Since we can't have a negative amount of water in the tank, we can conclude that the tank is empty after 3 minutes.

Answer 2

After 3 minutes, there would be 24 gallons of water left in the tank, given the rate of leakage and initial water volume.

Step-by-step explanation:

The student's question involves the rate of water leaking out of a tank, where the rate is given as a function of time r(t) = 13 - t2/3 for t ≥ 0, measured in gallons per minute. Initially, there are 60 gallons of water in the tank. The task is to calculate how much water is left in the tank after 3 minutes. To find the amount of leaked water, we would integrate the rate function r(t) from 0 to 3 minutes.

Step-by-step solution:

1. Set up the integral for the volume of water leaked: ∫ r(t) dt from t = 0 to t = 3.
2. Perform the integration: Integral of (13 - t2/3) dt = 13t - (t3/9) + C, where C is the constant of integration.
3. Calculate the definite integral by substituting the limits of integration: (13 * 3 - (33/9)) - (13 * 0 - (03/9)) = 39 - 3 = 36 gallons.
4. Subtract the volume of leaked water from the initial volume to find the remaining water: 60 - 36 = 24 gallons.

Therefore, after 3 minutes, there would be 24 gallons of water left in the tank.