To find the total number of cars, we need to integrate. Part (a) To find the total number of cars, we need to integrate. # of cars = (82 + 4 sin(t/2)) dt 30 2474.077 Keep in mind that you must be in RADIANS. 2474 cars

Since F’(7) is negative, the traffic flow is decreasing at t=7. Part (b) Since F(t) is the traffic flow, then F’(t) would be the RATE OF CHANGE of the traffic flow. We were asked if the traffic flow was increasing or decreasing at t=7. For this, we’ll need the derivative. F’(t) = 2 cos (t/2) -1.872 or -1.873 Since F’(7) is negative, the traffic flow is decreasing at t=7.

Average value = (82 + 4 sin(t/2)) dt 1 Part (c) NOTE: Part (c) uses a formula from Calculus, while Part (d) is nothing more than an Algebra 1 problem. The AP Test graders don’t expect you to show the work for this integration (u-subs, etc). That just wastes your valuable testing time. Instead, simply show them your integral and then evaluate it using the graphing calculator. Average value = (82 + 4 sin(t/2)) dt 10 15 1 15-10 Average value = 81.899 cars/minute

Keep in mind that you must be in RADIANS. Part (d) The AVERAGE RATE OF CHANGE is not Calculus. It’s the slope formula from Algebra 1. Keep in mind that you must be in RADIANS. F(15) – F(10) 15 - 10 85.752 – 78.164 15 - 10 AVG. RATE OF CHANGE = 1.517 or 1.518 cars/min2