Water is Pumped into an underground tank at a constant rate of 8 gallons per minute.Water leaks out of the tank at the rate of (t+1)^½ gallons per minute, for 0<t<120 minutes. At time t=0, the tank contains 30 gallons of water.
How many gallons of water leak out of the tank from time t=0 to t=3 minutes?
the effective rate of change of the water is dV/dt = 8 - (t+1)^½ + c
so integrating we get
V = 8t - (2/3)(t+1)^(3/2) + c , where c is a constant
but when t=0 , V = 30
30 = 0 - 2/3(1)^(3/2) + c
c = 92/3
so V = 8t - (2/3)(t+1)^(3/2) + 92/3
when t = 3 minutes
V = 24 - (2/3)(4)^(3/2) + 92/3
= 24 - 16/3 + 92/3
= 148/3
so the increase is 19.3333333
but we poured in 8 gallons/min for 3 minutes which is 24 gallons.
so the leakage must be 24 - 19.3333
or 4.66666 gallons
You work is extremely confusing. You started off with dv/dt and then the next line you have V = 8t - (2/3)(t+1)^(3/2) + c? wouldn't the derivative of sqrt(t+1) be (1/2)(t+1)^(-3/2)?
To find the number of gallons of water that leak out of the tank from time t=0 to t=3 minutes, we need to calculate the integral of the leakage rate function from 0 to 3 minutes.
The leakage rate at time t is given by (t+1)^(1/2) gallons per minute.
To find the integral from 0 to 3, we write the integral as follows:
∫[(t+1)^(1/2)] dt from 0 to 3
Now, let's calculate the integral.
First, let's find the antiderivative of (t+1)^(1/2).
Let u = t+1, then du = dt.
So, the integral becomes:
∫[(t+1)^(1/2)] dt from 0 to 3
= ∫u^(1/2) du from 0 to 3
= (2/3) * u^(3/2) evaluated from 0 to 3
= (2/3) * (3+1)^(3/2) - (2/3) * (0+1)^(3/2)
= (2/3) * 4^(3/2) - (2/3) * 1^(3/2)
= (2/3) * 8 - (2/3) * 1
= (16/3) - (2/3)
= 14/3
Therefore, the total amount of water that leaks out of the tank from time t=0 to t=3 minutes is 14/3 gallons.
To determine the number of gallons of water that leak out of the tank from time t=0 to t=3 minutes, we need to find the integral of the leakage rate function over the given time interval.
The leakage rate at any given time t is given by (t+1)^½ gallons per minute.
To find the number of gallons leaked out, we need to find the definite integral of the leakage rate function from t=0 to t=3. The integral represents the accumulated leakage over this time interval.
∫[(t+1)^½] dt from 0 to 3
To solve this definite integral, we can use the power rule for integration, which states that the integral of (x^n) dx is equal to (1/(n+1)) * (x^(n+1)) + C, where C is the constant of integration.
Applying the power rule, we have:
∫[(t+1)^½] dt = (2/3) * (t+1)^(3/2)
Now we can evaluate the definite integral from 0 to 3:
[(2/3) * (t+1)^(3/2)] evaluated from 0 to 3
= [(2/3) * (4^(3/2) - 1^(3/2))]
= [(2/3) * (8 - 1)]
= [(2/3) * 7]
= 14/3
Therefore, the number of gallons of water that leak out of the tank from time t=0 to t=3 minutes is approximately 4.67 gallons.