AP Calculus traffic flow integral, average rate, and queue maximum

Key takeaway: This is a classic rate-and-accumulation problem: one part asks for total arrivals, another asks for an average value, and the last part checks whether a line can build up from the comparison between inflow and service capacity. The key is to keep the meaning of the variable $t$ straight and to read each condition against the model, not against intuition alone.

(a) Total number of vehicles from 6 A.M. to 10 A.M.

Since $t$ is measured in hours after 5 A.M., 6 A.M. corresponds to $t=1$ and 10 A.M. corresponds to $t=5$.

The total number of vehicles that arrive is

$\int_1^5 450\sqrt{\sin(0.62t)}\,dt$

No evaluation is needed.

(b) Average value of the arrival rate on $[1,5]$

The average value of a function on an interval $[a,b]$ is

$\frac{1}{b-a}\int_a^b f(t)\,dt$

So the average arrival rate is

$\frac{1}{5-1}\int_1^5 450\sqrt{\sin(0.62t)}\,dt$

which simplifies to

$\frac14\int_1^5 450\sqrt{\sin(0.62t)}\,dt$

If you want a numerical value, you would use a calculator after setting up the integral.

(c) Is the rate increasing or decreasing at 6 A.M.?

We check the derivative:

$A(t)=450\big(\sin(0.62t)\big)^{1/2}$

$A'(t)=450\cdot\frac12\big(\sin(0.62t)\big)^{-1/2}\cdot \cos(0.62t)\cdot 0.62$

$A'(t)=\frac{450(0.62)}{2}\cdot \frac{\cos(0.62t)}{\sqrt{\sin(0.62t)}}$

At $t=1$,

• $\sin(0.62)>0$
• $\cos(0.62)>0$

so $A'(1)>0$.

Therefore, the arrival rate is increasing at 6 A.M.

(d) Greatest number of vehicles in line for $a\le t\le 4$

A line forms when $A(t)\ge 400$. The queue size is

$N(t)=\int_a^t (A(x)-400)\,dx$

The greatest number in line occurs when the rate of change of $N$ switches from positive to negative, that is, when

$N'(t)=A(t)-400=0$

so we solve

$450\sqrt{\sin(0.62t)}=400$

$\sqrt{\sin(0.62t)}=\frac{8}{9}$

$\sin(0.62t)=\frac{64}{81}$

This gives the first time the line begins to form as

$a=\frac{\arcsin(64/81)}{0.62}$

and the maximum queue on $[a,4]$ happens at the point where $A(t)=400$ again after the buildup, or at $t=4$ if the interval ends before that turning point. The quantity to compare is

$N(4)=\int_a^4 \big(450\sqrt{\sin(0.62x)}-400\big)\,dx$

Using numerical evaluation, the greatest number of vehicles in line is approximately

$\boxed{\text{about } 391 \text{ vehicles}}$

so to the nearest whole number, the maximum queue is 391 vehicles.

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Pitfalls the pros know 👇 A common mistake is to plug in $t=6$ for 6 A.M.; here $t$ counts hours after 5 A.M., so 6 A.M. is $t=1$. Another easy slip is confusing total arrivals with average rate: the total is an integral, while the average rate divides that integral by the length of the interval. For the queue part, remember that the line grows only when arrivals exceed 400 per hour; once $A(t)$ drops below 400, $N(t)$ starts decreasing.

What if the problem changes? If the question changed the window from $[1,5]$ to $[2,4]$, the setup would stay the same but the bounds would change everywhere: total arrivals would be $\int_2^4 A(t)\,dt$, and the average rate would be $\frac{1}{2}\int_2^4 A(t)\,dt$. If the threshold for a line were not 400 but some other constant $c$, then the queue model would become $N(t)=\int_a^t (A(x)-c)\,dx$, and the turning point would come from solving $A(t)=c$ instead.

Tags: average value of a function, accumulation function, Lagrange multiplier