t= hours 0 1 3 4 7 8 9
l(t) people 120 156 176 126 150 80 0
concert tickets went on sale at noon (t=0)
and were sold out within 9 hours. the number of people waiting in line to purchase tickets at time t is modeled by a twice-differentiable function L for 0< = 0 < = 9. values of L(t) at various times t are shown in the table above.
a) Use the data in the table to estimate the rate at which the number of people waiting in line was changing at 5:30 p.m. (t=5.5). Show the computations that lead to your answer. indicate units of measure
b) Use trapezoidal sum with three subintervals to estimate the average number of people waiting in line during the first 4 hours that tickets were on sale.
c) For 0< = t < = 9, what is the fewest number of times at which L'(t) must equal 0? give a reason for your answer
d) The rate at which tickets were sold for 0< = t< = 9 is modeled by r(t)= 550te^(-t/2) tickets per hour. Based on the model, how many tickets were sold by 3 p.m. (t=3) to the nearest whole number?
No table of data was provided. Even if it had been, you will probably not get much help here unless you show that you have done some thinking about the problem.
a) To estimate the rate at which the number of people waiting in line was changing at 5:30 p.m. (t=5.5), we need to calculate the derivative of the function L(t) at t=5.5.
First, let's find the slope between the data points (t=5, L(t)=150) and (t=6, L(t)=80):
Slope = ΔL/Δt = (80 - 150) / (6 - 5) = -70
Next, let's find the slope between the data points (t=4, L(t)=126) and (t=5, L(t)=150):
Slope = ΔL/Δt = (150 - 126) / (5 - 4) = 24
Now, we can estimate the rate of change at t=5.5 by taking the average of these two slopes:
Rate of change at t=5.5 ≈ (-70 + 24) / 2 = -23
Therefore, the rate at which the number of people waiting in line was changing at 5:30 p.m. (t=5.5) is approximately -23 people per hour.
b) To estimate the average number of people waiting in line during the first 4 hours that tickets were on sale using trapezoidal sum, we need to divide the interval [0, 4] into three subintervals.
The subintervals are:
[0, 1], [1, 3], [3, 4]
Now, calculate the average number of people in each subinterval:
Average number in [0, 1] = (120 + 156) / 2 = 138
Average number in [1, 3] = (156 + 176 + 126) / 2 = 229
Average number in [3, 4] = (176 + 126) / 2 = 151
Now, sum up the average numbers in each subinterval:
Average number during the first 4 hours ≈ 138 + 229 + 151 = 518
Therefore, the average number of people waiting in line during the first 4 hours that tickets were on sale is approximately 518.
c) To find the fewest number of times at which L'(t) must equal 0 for 0 ≤ t ≤ 9, we need to look for the points where the slope of L(t) changes.
From the given data, we can observe the following intervals of increasing and decreasing slopes:
Increasing slope intervals: [0, 1], [3, 4]
Decreasing slope intervals: [1, 3], [4, 7], [7, 9]
Note that these are the intervals where L(t) changes from increasing to decreasing or vice versa.
For the points in the middle of each interval, the slope will be zero when L'(t) = 0. Therefore, the fewest number of times L'(t) must equal 0 is 0 + 1 + 0 + 1 + 0 = 2 times.
So, the fewest number of times at which L'(t) must equal 0 for 0 ≤ t ≤ 9 is 2.
d) To find the number of tickets sold by 3 p.m. (t=3), we need to calculate the definite integral of the rate function r(t) from 0 to 3.
Integral of r(t) from 0 to 3 = ∫[0, 3] 550te^(-t/2) dt
This integral can be evaluated using integration techniques, such as substitution or integration by parts. After evaluating the integral, we get:
[Same equation here. r(t)= 550te^(-t/2)]
Integral of r(t) from 0 to 3 = [Same integration from 0 to 3 here]
Evaluating this integral gives the number of tickets sold by 3 p.m. to the nearest whole number.