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?
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?
nice...this is from the ap test.. i didn't know how to do it either :/
HAHAH!
i need help on this one too :(
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 can use the concept of average rate of change between two given points.
We can see from the table that at t=5 and t=6, the number of people waiting in line was 150 and 80, respectively. So, the time interval is 6 - 5 = 1 hour, and the change in the number of people waiting in line is 80 - 150 = -70 (negative because the number of people decreased).
To find the rate of change, we divide the change in the number of people waiting in line by the time interval:
Rate of change = Change in people / Time interval
= -70 / 1
= -70
Therefore, the rate at which the number of people waiting in line was changing at 5:30 p.m. (t=5.5) is -70 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 with three subintervals, we can divide the time interval into three equal parts: 0 to 1 hour, 1 to 3 hours, and 3 to 4 hours.
For the first subinterval (0 to 1 hour), the number of people waiting in line is 120.
For the second subinterval (1 to 3 hours), the number of people waiting in line is the average of the values at time 1 and time 3, which is (156 + 176) / 2 = 166.
For the third subinterval (3 to 4 hours), the number of people waiting in line is 126.
Now, we can calculate the total average number of people waiting in line by summing up the results from the three subintervals and dividing by the total number of subintervals (3):
Average number of people waiting in line = (120 + 166 + 126) / 3
= 412 / 3
= 137.33 (approximately)
Therefore, the average number of people waiting in line during the first 4 hours that tickets were on sale is approximately 137.33.
c) To find the fewest number of times at which L'(t) must equal 0, we need to look for the points where the rate at which the number of people waiting in line is changing is zero, which corresponds to the points where the derivative of the function L(t) is zero.
From the given data, we can see that the rate of change switches from positive to negative around t = 3 and from negative to positive around t = 8. Therefore, L'(t) equals 0 at these two points.
Hence, there are at least two points, t = 3 and t = 8, where L'(t) must equal 0.
d) To find the number of tickets sold by 3 p.m. (t=3), we can integrate the rate function r(t) from 0 to 3. The integral of r(t) gives us the cumulative number of tickets sold up to that point.
Integrating r(t):
∫(0 to 3) 550te^(-t/2) dt
This integral can be solved using calculus techniques. The result is:
-2200e^(-3/2) + 1760
Now, rounding the result to the nearest whole number, we get:
-2200e^(-3/2) + 1760 ≈ 1348
Therefore, approximately 1348 tickets were sold by 3 p.m.