The rate at which water flows into a tank, in gallons per hour, is given by a differentiable, increasing function R of time t. The table below gives the rate as measured at various times in an 8-hour time period.
t (hours)
0 2 3 7 8
R(t) (gallons per hour)
1.95 2.5 2.8 4 4.26
1. Use a trapezoidal sum with the four sub-intervals indicated by the data in the table to estimate integral[0 to 8](R(t)dt). Using correct units, explain the meaning of your answer in terms of water flow.
2. Is there some time t, 0 < t < 8, such that R′(t) = 0? Justify your answer.
3. The rate of water flow R(t) can be estimated by W(t) = ln(t2 + 7). Use W(t) to approximate the average rate of water flow during the 8-hour time period. Indicate units of measure.
1. To estimate the integral [0 to 8] (R(t) dt) using a trapezoidal sum with the given data, we can divide the interval [0, 8] into four sub-intervals: [0, 2], [2, 3], [3, 7], and [7, 8].
The formula for the trapezoidal sum is:
Δt * [(f(a) + f(b)) / 2 + ∑(f(xi))], where Δt is the width of each sub-interval, f(a) and f(b) are the function values at the endpoints, and ∑(f(xi)) represents the sum of the function values for the intermediate points.
For the given data, the trapezoidal sum can be calculated as follows:
Δt1 = 2 - 0 = 2
Δt2 = 3 - 2 = 1
Δt3 = 7 - 3 = 4
Δt4 = 8 - 7 = 1
Trapezoidal sum = Δt1 * [(R(0) + R(2)) / 2] + Δt2 * [(R(2) + R(3)) / 2] + Δt3 * [(R(3) + R(7)) / 2] + Δt4 * [(R(7) + R(8)) / 2]
= 2 * [(1.95 + 2.5) / 2] + 1 * [(2.5 + 2.8) / 2] + 4 * [(2.8 + 4) / 2] + 1 * [(4 + 4.26) / 2]
≈ 7.405 gallons
Therefore, the estimated value of the integral [0 to 8] (R(t) dt) is approximately 7.405 gallons. This represents the total amount of water that flowed into the tank during the 8-hour time period.
2. To determine if there is some time t, 0 < t < 8, such that R'(t) = 0, we need to find the critical points of the function R(t).
The critical points occur when the derivative R'(t) is equal to zero or does not exist. Looking at the given data, we can see that the rate of water flow R(t) is increasing for all the recorded times. This implies that R'(t) is always positive.
Since R'(t) is always positive, there is no time t, 0 < t < 8, where R'(t) equals zero. Therefore, there is no such time where the rate of water flow is not changing.
3. To approximate the average rate of water flow during the 8-hour time period using the function W(t) = ln(t^2 + 7), we need to evaluate the integral [0 to 8] (W(t) dt) and divide it by the total time.
The integral [0 to 8] (W(t) dt) can be calculated as follows:
∫[0 to 8] (W(t) dt) = ∫[0 to 8] (ln(t^2 + 7) dt)
Using the fundamental theorem of calculus, we can find the antiderivative F(t) of ln(t^2 + 7):
F(t) = ∫ ln(t^2 + 7) dt
= t * ln(t^2 + 7) - 2t + C
Evaluating the integral from 0 to 8 gives:
∫[0 to 8] (W(t) dt) = F(8) - F(0)
= (8 * ln((8^2) + 7) - 2(8)) - (0 * ln((0^2) + 7) - 2(0))
≈ 13.291
The average rate of water flow during the 8-hour time period can be approximated by dividing the integral by the total time:
Average rate = (1 / 8) * ∫[0 to 8] (W(t) dt)
= (1 / 8) * 13.291
≈ 1.661 gallons per hour
Therefore, the average rate of water flow during the 8-hour time period, using the estimation from the function W(t) = ln(t^2 + 7), is approximately 1.661 gallons per hour.
1. To estimate the integral of R(t) from 0 to 8 using a trapezoidal sum, we can use the formula:
∫[0 to 8](R(t)dt) ≈ (h/2) * [R(a) + 2R(a+h) + 2R(a+2h) + ... + 2R(a+(n-1)h) + R(b)]
where h is the width of each subinterval, a and b are the initial and final values of t, and n is the total number of subintervals.
In this case, we have four subintervals (n = 4), so h = (8-0)/4 = 2.
Using the given data, we have:
a = 0, b = 8
R(a) = R(0) = 1.95
R(a+h) = R(2) = 2.5
R(a+2h) = R(4) = 2.8
R(a+3h) = R(6) = 4
R(b) = R(8) = 4.26
Plugging these values into the formula, we get:
∫[0 to 8](R(t)dt) ≈ (2/2) * [1.95 + 2(2.5) + 2(2.8) + 2(4) + 4.26]
≈ 1 * [1.95 + 5 + 5.6 + 8 + 4.26]
≈ 1 * 24.81
≈ 24.81 gallons
The meaning of this answer in terms of water flow is that the estimated total amount of water that flowed into the tank over the 8-hour time period is approximately 24.81 gallons.
2. To find if there exists a time t, 0 < t < 8, such that R'(t) = 0, we need to look for any values of t between 0 and 8 that correspond to a local maximum or minimum in the rate of water flow.
To do this, we can examine the given data and determine the pattern of the rate of water flow. By observing the table, we can see that the rate is increasing as time progresses. Since R(t) is an increasing function and there are no decreasing intervals between the given data points, we can conclude that R'(t) is always positive for any time t, 0 < t < 8.
Therefore, there is no time t between 0 and 8 such that R'(t) = 0.
3. To approximate the average rate of water flow during the 8-hour time period using the function W(t) = ln(t^2 + 7), we need to evaluate the integral of W(t) from 0 to 8 and then divide it by the total time.
The average rate of water flow can be calculated as:
Average rate = (1/8) * ∫[0 to 8](W(t)dt)
To find ∫[0 to 8](W(t)dt), we can integrate the function W(t) using the rules of integration:
∫[0 to 8](W(t)dt) = ∫[0 to 8](ln(t^2 + 7)dt)
After performing the integration and evaluating it from 0 to 8, we get:
∫[0 to 8](ln(t^2 + 7)dt) ≈ 3.846
Therefore, the average rate of water flow during the 8-hour time period, estimated using W(t) = ln(t^2 + 7), is approximately (1/8) * 3.846 = 0.481 gallons per hour.
#1. There are many handy online calculators you can use to verify your results.
#2. I doubt it, since the values are increasing
#3. The average rate of flow over the interval is
(∫[0,8] W(t) dt)/(8-0)
= 1/8 ∫[0,8] ln(t^2+7) dt = 3.09 gal/hr
Doing the integral takes some maneuvering.
Using integration by parts,
u = ln(t^2+7)
du = 2t/(t^2+7) dt
dv = dt
v = t
∫ln(t^2+7) dt = t ln(t^2+7) - 2∫t^2/(t^2+7) dt
Now, t^2/(t^2+7) = 1 - 1/(t^2+7) so we have
∫ln(t^2+7) dt = t ln(t^2+7) - 2(∫1 - 1/(t^2+7)) dt
= t ln(t^2+7) - 2t + 2√7 arctan(t/√7)