AP Exam 2006AP Exam 2006

Open Ended Problem #2Open Ended Problem #2

Graphing Calculator PermittedGraphing Calculator Permitted

At an intersection in Thomasville, Oregon, cars turn left at the rate

cars per hour over the time interval from 0 to 18 hours. The graph of y = L(t) is shown below.

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200

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100

80

60

40

20

5 10 15 20 25

3sin60)( 2 t

ttLN

um

ber

of C

ars

turn

ing

left

Time in hours

L(t)

t

1. To the nearest whole number, find the total number of cars turning left at the intersection over the time interval from 0 to 18 hours. (Click the mouse to see solution.)

Since L(t) is the rate of cars turning left in cars per hour and t is the time in hours, to determine the total number of cars that turned left during a given time, you must multiply cars per hour times hours. We need to find the area under the curve from 0 to 18 hours. We can determine this value by integrating.

Use your graphing calculator to determine the integral.

8237.16573

sin6018

0

2

dt

tt

We’ll round this answer to 1657 cars.

18

0

2

3sin60 dt

tt

2. Traffic engineers will consider turn restrictions when the number of cars turning left exceeds 150 cars per hour. Find all values of t for which the number of cars turning left exceeds 150 cars per hour and compute the average of the function L(x) over this time interval. Indicate units of measure.We want to determine when the rate of left hand turns is greater than or equal to 150. So, set up an inequality like this:

To determine the times t when the inequality is true, examine the graph of this function

1503

sin60)( 2

tttf

and use your graphing calculator to determine the t-intercepts. The graph crosses the t-axis when the function equals 150 and the intervals of the graph above the t-axis represent time intervals when the function exceeds 150.

1503

sin60 2

t

t

The graph crosses the t-axis at 12.42831 and 16.121657. The values of t for which L(t) is greater than 150 would include:

.121657.1642831.12 tThe average value of L(t) over this time period is found by computing the average value of the function using the formula:

b

a

dttLab

L(t) )(1

of Value Average

121657.16

42831.12

2

3sin60

42831.12121657.16

1dt

tt

)54986.736)(270757(.

cars/hour 199.426

3. Traffic engineers will install a signal if there is any two-hour time interval during which the product of the total number of cars turning left and the total number of oncoming cars traveling straight through the intersection is greater than 200,000. In every two-hour time interval, 500 oncoming cars travel straight through the intersection. Does this intersection require a traffic signal? Explain the reasoning that leads to your conclusion.

Begin with an inequality:

(Total cars Turning left) (Total cars proceeding straight) > 200,000

OR

4003

sin60

000,2005003

sin60

2

2

dtt

t

dtt

t

We need to find a two hour interval for which the integral is greater than 400. It seems logical to assume that the integral would have its greatest value near the maximum value of the function, so let’s find the x-value corresponding to the maximum value of the function. Plot the function with your graphing calculator and determine the maximum value of the function and its corresponding x-coordinate.

Since the maximum traffic flow occurs at 14.29 hours, find the total number of cars turning left 1 hour before and after the peak time and see if that value is greater than 400.

Integrating from 13.29 to 15.25 hours gives a value of 435.8448 cars turning left. Therefore, the intersection does require a traffic signal.

The EndThe End