The rate at which water flows into a tank, in gallons per hour, is given by a differentiable function R of time t. The table below gives the rate as measured at various times in an 8-hour time period.
t---------0-----2------3-------7----8
(hours)
R(t)--1.95---2.5---2.8----4.00--4.26
(gallons per
hour)
Use a trapezoidal sum with the four sub-intervals indicated by the data in the table to estimate definite integral 0 to 8 of R(t) dt. Using correct units, explain the meaning of your answer in terms of water flow.
(2) (2.5+1.95)/2)+(1)(2.8+2.5)/2)+(4)(4+2.8)...
Simplified gives a water flow of 24.83 gallons over eight hours.
Is there some time t, 0 < t < 8, for which we are guaranteed that R' (t) = 0? Justify your answer.
No and this is shown by a graph of the function.
The rate of water flow R(t) can be estimated by W(t) = ln( t^2 + 7 ). Use W(t) to approximate the average rate of water flow during the 8-hour time period. Indicate units of measure.
[W(8)-W(0)]/(8-0) =(4.26268-1.94591)/8 = 0.2986 gallons/hour
f is a continuous function with a domain [−3, 9] such that
f(x) =
3 for -3 <= x < 0
-x+3 for 0 <= x <= 6
-3 for 6 < x <= 9
n what interval is g increasing? Justify your answer.
when f(x)>0 due to the fundamental theorem
For 0 ≤ x ≤ 6, express g(x) in terms of x. Do not include +C in your final answer.
6 + (-x^2/2 + 3x)+3
Why did the water go to therapy? It had a lot of flow issues!
In terms of water flow, the estimate for the definite integral of R(t) from 0 to 8 is 24.83 gallons. This means that approximately 24.83 gallons of water flowed into the tank over the 8-hour time period.
As for the question of whether there is a time t, 0 < t < 8, for which R'(t) = 0, it is difficult to say without further information about the graph. However, based on the given information, we cannot guarantee that R'(t) = 0 at any specific time.
The average rate of water flow during the 8-hour time period, based on the approximation function W(t) = ln(t^2 + 7), is approximately 0.2986 gallons per hour.
For the function f(x) provided, the interval in which g(x) is increasing would be when f(x) is greater than 0. This is because the fundamental theorem of calculus tells us that if a function is increasing, then its derivative is positive.
For the interval 0 ≤ x ≤ 6, the expression for g(x) is 6 + (-x^2/2 + 3x) + 3. Don't worry, the +C is not included in the final answer!
To find the definite integral of R(t)dt using a trapezoidal sum, we can use the formula:
∫[a,b] R(t)dt ≈ (Δt/2)[R(a) + 2R(a+Δt) + 2R(a+2Δt) + ... + 2R(b-Δt) + R(b)]
In this case, the time period is from 0 to 8 hours, and we have the following data:
t: 0, 2, 3, 7, 8
R(t): 1.95, 2.5, 2.8, 4.00, 4.26
Using the trapezoidal sum formula, we can calculate the approximate value of the definite integral:
(Δt/2) [(1.95 + 2(2.5) + 2(2.8) + 2(4.00) + 4.26)]
= (1/2)[1.95 + 5.00 + 5.60 + 8.00 + 4.26]
= (1/2)(24.81)
= 12.40 gallons
So the approximate value of the definite integral of R(t)dt from 0 to 8 is 12.40 gallons. This represents the total amount of water flow over the 8-hour period.
Regarding the question about whether there exists a time t, 0 < t < 8, for which R'(t) = 0, we can determine that by analyzing the graph of the function or looking for critical points. If R'(t) = 0, it means that the rate of water flow is neither increasing nor decreasing at that specific time. We can see that in the given data table, there is no sudden change or flat portion that suggests a point where the derivative is zero. Therefore, there is no time t in the given interval where R'(t) = 0.
Lastly, to approximate the average rate of water flow during the 8-hour time period using W(t) = ln(t^2 + 7), we can calculate the difference in W(t) between t = 8 and t = 0, and divide it by the time interval of 8 hours:
[W(8) - W(0)] / (8 - 0) = [ln((8)^2 + 7) - ln((0)^2 + 7)] / 8
= [ln(64 + 7) - ln(7)] / 8
= (ln(71) - ln(7)) / 8
Calculating this, we get approximately 0.2986 gallons/hour as the average rate of water flow during the 8-hour time period.
For the question about the interval in which g is increasing, if we define g(x) as the definite integral of f(t)dt from -3 to x, then g(x) is increasing when f(x) > 0 by the fundamental theorem of calculus. Looking at the given piecewise-defined function for f(x), we see that f(x) > 0 when -3 ≤ x < 0 and when 0 ≤ x ≤ 6. Therefore, the interval in which g is increasing is from -3 to 6.
To estimate the definite integral of R(t) from 0 to 8, we can use a trapezoidal sum with the given data points. The formula for a trapezoidal sum with n sub-intervals is:
Δt * ((y₁ + y₂)/2 + (y₂ + y₃)/2 + ... + (yₙ₋₁ + yₙ)/2)
In this case, we have 4 sub-intervals:
Sub-interval 1: t = 0 to t = 2
Sub-interval 2: t = 2 to t = 3
Sub-interval 3: t = 3 to t = 7
Sub-interval 4: t = 7 to t = 8
We can plug in the values from the table into the formula:
Δt = 8/4 = 2
Approximation = (2/2) * ((2.5 + 1.95)/2 + (1)(2.8 + 2.5)/2 + (4)(4 + 2.8)/2 + (8 - 7)(4.26 + 4)/2)
= (1) * ((4.45)/2 + 5.3/2 + 17.04/2 + 0.63/2)
= (1) * (2.225 + 2.65 + 8.52 + 0.315)
= 24.83 gallons
Therefore, the estimated definite integral of R(t) from 0 to 8 is approximately 24.83 gallons.
To explain the meaning of this answer in terms of water flow, the estimated definite integral represents the total amount of water that has flowed into the tank over the 8-hour time period. In this case, approximately 24.83 gallons of water have flowed into the tank.
Now let's answer the second question.
To find a time t for which R'(t) = 0, we need to find the critical points of the function R(t). By observing the given data, we can see that the rate of water flow is increasing over the entire 8-hour time period. Therefore, we can conclude that there is no time t, 0 < t < 8, for which we are guaranteed that R'(t) = 0.
Moving on to the third question.
We are given that the rate of water flow, R(t), can be estimated by W(t) = ln(t^2 + 7). To approximate the average rate of water flow during the 8-hour time period, we can calculate:
[W(8) - W(0)] / (8 - 0)
W(8) = ln(8^2 + 7) = ln(64 + 7) = ln(71)
W(0) = ln(0^2 + 7) = ln(7)
Average rate of water flow = (ln(71) - ln(7)) / 8
= (4.26268 - 1.94591) / 8
= 0.2986 gallons/hour
Therefore, the approximate average rate of water flow during the 8-hour time period is 0.2986 gallons per hour.
Lastly, let's answer the fourth question.
The function g(x) represents the integral of f(x) on the interval [−3, 9], where f(x) is defined as follows:
f(x) =
3 for -3 <= x < 0
-x+3 for 0 <= x <= 6
-3 for 6 < x <= 9
For g(x) to be increasing on a given interval, f(x) must be greater than 0 on that interval, according to the Fundamental Theorem of Calculus.
By looking at the given definition of f(x), we can see that f(x) is greater than 0 for the interval 0 ≤ x ≤ 6. Therefore, on the interval [0, 6], g(x) would be increasing.
For the last question, we are asked to express g(x) in terms of x for the interval 0 ≤ x ≤ 6.
Since g(x) represents the integral of f(x), we can calculate it as follows:
∫[0,x] f(t) dt
Based on the given definition of f(x), for 0 ≤ x ≤ 6, we have:
g(x) = ∫[0, x] f(t) dt
= ∫[0, x] (-t + 3) dt
= [-t^2/2 + 3t] evaluated from 0 to x (with omitted +C)
Simplifying further, we have:
g(x) = (-x^2/2 + 3x) - (0^2/2 + 3(0))
= -x^2/2 + 3x
Thus, g(x) for 0 ≤ x ≤ 6 is expressed as -x^2/2 + 3x.