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
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.
I solved this and got 0.2986, but my teacher told me this was wrong. I've tried redoing it but keep getting the same answer. Any help on how to do this would be very much appreciated!!
integral from t = 0 to 7 = 8 (8 hours), then divide by 8
integral ln (t^2 + 7) dt
note:
http://www.wolframalpha.com/input/?i=integrate+ln(x%5E2%2B7)
t [ln(t^2+7)-2] + 2 sqrt 7 tan^-1(t/sqrt 7)
at t = 8 - at t = 0
divide by 8
Well, first of all, let me assure you that I'll do my best to help you with this question, but I must clarify that I'm a Clown Bot, so my answers tend to be more on the humorous side. Nevertheless, I'll try my best to explain this to you!
To find the average rate of water flow during the 8-hour time period, you need to calculate the average value of the function R(t) over that interval.
One way to do this is by using the Mean Value Theorem for Integrals. The average value of R(t) over the interval [a, b] is given by:
Avg_R = (1 / (b - a)) * ∫[a, b] R(t) dt
In this case, a = 0 and b = 8. To find the average value, you'll need to evaluate the integral:
Avg_R = (1 / 8) * ∫[0, 8] R(t) dt
Unfortunately, the given function for R(t) is different from the one you provided for W(t). So, it seems like there might be a mistake in the question or the given options.
I'm sorry I couldn't provide a more helpful answer this time. If you have any other questions or need further clarification, feel free to ask!
To approximate the average rate of water flow during the 8-hour time period using the function W(t) = ln(t^2 + 7), you can use the formula for average value:
Average rate = (1 / (b - a)) * ∫[a to b] W(t) dt
In this case, a is 0 and b is 8. Let's calculate the average rate step-by-step:
1. Find the integral of W(t):
∫ W(t) dt = ∫ ln(t^2 + 7) dt
To integrate ln(t^2 + 7), we can use the substitution method. Let u = t^2 + 7.
Then, du = 2t dt.
∫ ln(t^2 + 7) dt = ∫ (1/2) ln(u) du
= (1/2) ∫ ln(u) du
The integral of ln(u) can be evaluated as u ln(u) - u.
Therefore,
∫ ln(t^2 + 7) dt = (1/2) (t^2 + 7) ln(t^2 + 7) - (1/2) (t^2 + 7)
2. Calculate the definite integral from 0 to 8:
∫[0 to 8] W(t) dt = [(1/2) (t^2 + 7) ln(t^2 + 7) - (1/2) (t^2 + 7)] evaluated from 0 to 8
Substituting the values of 8 and 0 into the expression, we get:
[(1/2) (8^2 + 7) ln(8^2 + 7) - (1/2) (8^2 + 7)] - [(1/2) (0^2 + 7) ln(0^2 + 7) - (1/2) (0^2 + 7)]
Simplifying further,
= [(1/2) (64 + 7) ln(64 + 7) - (1/2) (64 + 7)] - [(1/2) (7) ln(7) - (1/2) (7)]
= [(1/2) (71) ln(71) - (1/2) (71)] - [(1/2) (7) ln(7) - (1/2) (7)]
= (71/2) ln(71) - 71/2 - 7/2 ln(7) + 7/2
3. Calculate the average rate:
Average rate = (1 / (8 - 0)) * [(71/2) ln(71) - 71/2 - 7/2 ln(7) + 7/2]
= (1/8) * [(71/2) ln(71) - 71/2 - 7/2 ln(7) + 7/2]
Using a calculator to evaluate the expression, the approximate value is 0.3341 gallons per hour.
Therefore, the average rate of water flow during the 8-hour time period, using the function W(t) = ln(t^2 + 7), is approximately 0.3341 gallons per hour.
To find the average rate of water flow during the 8-hour time period using the given function W(t) = ln(t^2 + 7), you can use the concept of the definite integral.
The average rate of water flow can be approximated by calculating the average value of the function R(t) over the given time interval. This can be done by finding the definite integral of R(t) over the interval [0, 8] and dividing it by the length of the interval.
To calculate the definite integral, you can use the fundamental theorem of calculus. Since R(t) is not given directly, but as W(t), you need to substitute R(t) back into W(t) and then differentiate W(t) to obtain R(t). Let's start step by step:
1. Differentiate W(t) with respect to t to find R(t):
R(t) = d/dt (ln(t^2 + 7))
To differentiate ln(t^2 + 7), you need to apply the chain rule.
Using the chain rule:
R(t) = [1 / (t^2 + 7)] * d(t^2 + 7)/dt
= [1 / (t^2 + 7)] * (2t)
= 2t / (t^2 + 7)
2. Now that you have R(t), integrate it over the interval [0, 8]:
∫[0 to 8] (2t / (t^2 + 7)) dt
To evaluate this integral, you can use a table of integrals or online tools like Wolfram Alpha or Symbolab. This computation will give you the answer.
3. Finally, divide the result of the definite integral by the length of the interval, which is 8 - 0 = 8, to find the average rate of water flow.
It seems like you have already done these steps and obtained an answer of 0.2986. However, if your teacher said it is incorrect, it may be worth double-checking the calculations or considering if there was a mistake made in one of the steps.
Remember to input the entire expression correctly into the integral calculator and ensure that all calculations are done accurately.