Research on Surveillance Cameras

Question

Research reports indicate that surveillance cameras at major intersections dramatically reduce the number of drivers who barrel through red lights. The cameras automatically photograph vehicles that drive into intersections after the light turns red. Vehicle owners are then mailed citations instructing them to pay a fine or sign an affidavit that they weren't driving at the time. The function

N(t) = 4.50t³ − 21.03t² + 42.06t + 69.5    (0 ≤ t ≤ 4)

gives the number, N(t),

 of communities in an undetermined first-world country using surveillance cameras at intersections in year t, with t = 0

 corresponding to 2003.

(b). When was the number of communities using surveillance cameras at intersections changing least rapidly? (Round your answers to two decimal places.)

t = ?



What is the rate of increase?

?  communities using a security camera/year

Answer 1

The question here is asking for the time at which the rate of change is changing at the slowest rate, meaning that we are looking for the point at which the first derivative of the function (which represents the rate of change of the number of surveillance cameras with respect to time) reaches a minimum. First, we find the first derivative function, which can easily be done with power rule:

dN/dt = 4.50*3*t² - 21.03*2*t + 42.06

Next, we find the second derivative by applying power rule a second time on the first derivative function we found above:

d²N/dt² = 4.50*3*2*t - 21.03*2 = 27t - 42.06

Since we are trying to find the point at which dN/dt reaches a minimum, we set the second derivative equal to zero and solve for t.

27t - 42.06 = 0

27t = 42.06

t = 42.06/27 = 1.557777777...

In this case, we are looking at the entire interval 0 ≤ t ≤ 4 to determine when dN/dt is at a minimum, meaning we are looking for an absolute minimum. What this means is that in addition to testing the time value we found above of t=1.557777..., we also need to test t=0 and t=4 to see which value leads to the smallest value of dN/dt. We do so by plugging in the values of t=0, t=1.5577777..., and t=4 to the equation for dN/dt that we found above.

dN/dt = 4.50*3*t² - 21.03*2*t + 42.06

dN/dt = 13.5*t² - 42.06*t + 42.06

plugging in the values, we get:

N'(0) = 42.06

N'(1.55777777) = 9.3

N'(4) = 89.82

We see that N'(1.5577777) yields the smallest value, telling us that the time at which the number of communities using surveillance cameras at intersections was changing least rapidly was indeed at t=1.5577777..., or rounded t=1.56.

We also already found that the rate of change that corresponded to that time value was N'(1.55777777) = 9.3.

Hope that helps!