The following table shows the rate of water flow (in L/min) through a dam.
t (min) 1 5 9 13 17 21 25
V'(t) (L/min) 6 6 2 2 3 6 2
Approximate the total volume of water that passed through the dam from t=1 to t=25 using Simpson's rule, with n=6.
To approximate the total volume of water that passed through the dam from t=1 to t=25 using Simpson's rule, we first need to find the values of the function at equally spaced points within the interval [1,25].
Given the table, we have the following values for t and V'(t):
t (min) 1 5 9 13 17 21 25
V'(t) 6 6 2 2 3 6 2
Since we are asked to use Simpson's rule with n=6, we will divide the interval [1, 25] into 6 subintervals of equal width.
The width of each subinterval, Δx, can be calculated by dividing the total width of the interval by the number of subintervals:
Δx = (25 - 1) / 6 = 4
Next, we can calculate the values of the function at the equally spaced points within the interval. Since we are given the values of the derivative V'(t), we need to integrate it to get the original function V(t).
To do this, we can use the following formula:
V(t) = ∫ V'(t) dt
Considering V'(t) as a discrete data set and using Simpson's rule to approximate the integral, we can calculate V(t) at each equally spaced point.
Applying Simpson's rule to calculate V(t) at each point, we get:
V(t_0) = (Δx/3) * [V'(t_0) + 4V'(t_1) + V'(t_2)]
V(t_1) = (Δx/3) * [V'(t_1) + 4V'(t_2) + V'(t_3)]
V(t_2) = (Δx/3) * [V'(t_2) + 4V'(t_3) + V'(t_4)]
V(t_3) = (Δx/3) * [V'(t_3) + 4V'(t_4) + V'(t_5)]
V(t_4) = (Δx/3) * [V'(t_4) + 4V'(t_5) + V'(t_6)]
V(t_5) = (Δx/3) * [V'(t_5) + 4V'(t_6) + V'(t_7)]
Substituting the values of V'(t) from the table:
V(t_0) = (4/3) * [6 + 4*6 + 2] = 40/3
V(t_1) = (4/3) * [6 + 4*2 + 2] = 16/3
V(t_2) = (4/3) * [2 + 4*2 + 3] = 22/3
V(t_3) = (4/3) * [2 + 4*3 + 6] = 28/3
V(t_4) = (4/3) * [3 + 4*6 + 2] = 34/3
V(t_5) = (4/3) * [6 + 4*2 + 0] = 20/3
Finally, to approximate the total volume of water that passed through the dam from t=1 to t=25, we can sum up the calculated volumes V(t_i) for each subinterval:
Approximate Total Volume = V(t_0) + V(t_1) + V(t_2) + V(t_3) + V(t_4) + V(t_5)
= (40/3) + (16/3) + (22/3) + (28/3) + (34/3) + (20/3)
= 160/3 ≈ 53.33 L
Therefore, the approximate total volume of water that passed through the dam from t=1 to t=25 is approximately 53.33 liters.