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.
A) How many gallons of water leak out of the tank from time t=0 to t=3 minutes?
B) How many gallons of water are in the tank at time t=3 minutes?
C) write an expression for A(t) the total number of gallons of water in the tank at time t?
D)at what time t, for o<t<120, is the amount of water in the tank a maximum? justify the answer
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?
A) 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.
The leakage rate is given by (t+1)^(1/2) gallons per minute. To find the total amount of leakage, we integrate this function over the interval [0, 3]:
∫[(t+1)^(1/2)] dt
To evaluate this integral, we can use the substitution method. Let u = t + 1, then du = dt. Substituting these values, the integral becomes:
∫u^(1/2) du
Integrating, we get:
(2/3) u^(3/2) + C
Now we substitute back u = t + 1:
(2/3) (t + 1)^(3/2) + C
Evaluating this expression over the interval [0, 3], we get:
[(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)
= 16/3 - 2/3
= 14/3
Therefore, the number of gallons of water that leak out of the tank from time t=0 to t=3 minutes is 14/3 gallons.
B) To find the number of gallons of water in the tank at time t=3 minutes, we need to subtract the total amount of water that leaked out from the initial amount of water.
The initial amount of water is given as 30 gallons. From part A, we know that 14/3 gallons leaked out from time t=0 to t=3 minutes. Subtracting these values:
30 - 14/3 = (90/3) - (14/3) = 76/3
Therefore, there are 76/3 gallons of water in the tank at time t=3 minutes.
C) To write an expression for A(t), the total number of gallons of water in the tank at time t, we need to consider the amount of water that was pumped into the tank minus the amount of water that leaked out.
The pumping rate is given as 8 gallons per minute. Integrating this constant rate over the interval [0, t], we get:
∫8 dt = 8t + C
The leakage rate is given as (t+1)^(1/2) gallons per minute. Integrating this function over the interval [0, t], we get:
∫[(t+1)^(1/2)] dt = (2/3) (t + 1)^(3/2) + D
Combining these two expressions, we get:
A(t) = [8t + C] - [(2/3) (t + 1)^(3/2) + D]
Where C and D are constants of integration.
D) To find the time t at which the amount of water in the tank is at a maximum, we need to find the critical point of the function A(t).
First, we take the derivative of A(t) with respect to t:
A'(t) = 8 - (2/3) * (3/2) * (t+1)^(1/2)
= 8 - (2/3) * (3/2) * (t+1)^(1/2)
= 8 - (t+1)^(1/2)
To find the critical point, we set A'(t) equal to zero and solve for t:
8 - (t+1)^(1/2) = 0
(t+1)^(1/2) = 8
Squaring both sides, we get:
t + 1 = 64
t = 63
Therefore, at t=63 minutes, for 0 < t < 120, the amount of water in the tank is at a maximum.
Note: To justify this answer, you can also check the second derivative of A(t) at t=63 to confirm that it is negative, indicating a maximum point.
To solve the given problem step-by-step, let's break it down:
A) How many gallons of water leak out of the tank from time t=0 to t=3 minutes?
To find the number of gallons leaked out from time t=0 to t=3 minutes, we need to calculate the integral of the leak rate from t=0 to t=3.
Integration of (t+1)^(1/2) dt from 0 to 3:
∫(t+1)^(1/2) dt from 0 to 3
= [2/3*(t+1)^(3/2)] from 0 to 3
= 2/3*(3+1)^(3/2) - 2/3*(0+1)^(3/2)
= 2/3*(4^(3/2)) - 2/3
≈ 3.079 gallons (rounded to three decimal places)
So, approximately 3.079 gallons of water leak out of the tank from time t=0 to t=3 minutes.
B) How many gallons of water are in the tank at time t=3 minutes?
The number of gallons of water in the tank at a given time t is the integral of the pump rate from t=0 to t.
Integration of 8 dt from 0 to 3:
∫8 dt from 0 to 3
= 8t from 0 to 3
= 8*3 - 8*0
= 24 gallons
So, there are 24 gallons of water in the tank at time t=3 minutes.
C) Writing an expression for A(t), the total number of gallons of water in the tank at time t:
A(t) = Initial amount of water + Integral of pump rate from t=0 to t - Integral of leak rate from t=0 to t
Given:
Initial amount of water = 30 gallons
Pump rate = 8 gallons per minute
Leak rate = (t+1)^(1/2) gallons per minute (for 0<t<120)
So, the expression for A(t) becomes:
A(t) = 30 + ∫8 dt from 0 to t - ∫(t+1)^(1/2) dt from 0 to t
D) At what time t, for 0<t<120, is the amount of water in the tank a maximum? Justify the answer.
To find the time t at which the amount of water in the tank is maximum, we need to find the critical points of the expression A(t). We can do this by differentiating A(t) with respect to t and finding where the derivative equals zero.
dA(t)/dt = 0 + 8 - (t+1)^(1/2) = 0
Simplifying, we have:
8 = (t+1)^(1/2)
Squaring both sides:
64 = t + 1
t = 63
Thus, the amount of water in the tank is maximum at t=63 minutes. This can be justified by testing the derivative on either side of t=63 to confirm that the derivative changes sign, implying a maximum point.