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Problem 104

# 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 $$N(t)=6.08 t^{3}-26.79 t^{2}+53.06 t+69.5 \quad(0 \leq t \leq 4)$$ gives the number, $$N(t)$$, of U.S. communities using surveillance cameras at intersections in year $$t$$, with $$t=0$$ corresponding to the beginning of 2003 . a. Show that $$N$$ is increasing on $$[0,4]$$. b. When was the number of communities using surveillance cameras at intersections increasing least rapidly? What is the rate of increase?

Expert verified
a. The first derivative of the function $$N(t)$$ is $$N'(t)=18.24t^2 - 53.58t + 53.06$$. Since there are no critical points of $$N'(t)$$ in the interval [0,4] and N'(2) > 0, $$N(t)$$ is increasing on [0, 4]. b. The second derivative of the function $$N(t)$$ is $$N''(t) = 36.48t - 53.58$$. The critical point of $$N''(t)$$ occurs at $$t \approx 1.47$$. At this time, the rate of increase is $$N'(1.47) \approx 35.92$$ communities per year. Thus, the number of communities using surveillance cameras at intersections was increasing least rapidly in the middle of 2004 at a rate of approximately 35.92 communities per year.
See the step by step solution

## Step 1: Find the first derivative of N(t)

Take the derivative of N(t) with respect to t to find the rate of increase: $N'(t) = \frac{d}{dt}(6.08t^3 - 26.79t^2 + 53.06t + 69.5)$ Using the power rule, we get: $N'(t) = 18.24t^2 - 53.58t + 53.06$

## Step 2: Determine if N'(t) is positive for t ∈ [0,4]

Now check if N'(t) > 0 for the given interval [0,4]. We can do this graphically using a sign chart: 1. Find the critical points of N'(t) by setting N'(t) = 0 and solving for t. $18.24t^2 - 53.58t + 53.06 = 0$ This is a quadratic equation, and you may find that its two critical points are complex numbers and hence not in the given interval [0,4]. That means there is no point at which the first derivative becomes zero. 2. Pick a test point inside the interval [0,4]. Let's take t = 2. 3. Evaluate N'(2). $N'(2) = 18.24(2)^2 - 53.58(2) + 53.06 = 48.12$ Since N'(2) > 0 and N'(t) has no critical points in the interval [0,4], we can conclude that N(t) is increasing on [0,4].

## Step 3: Find the second derivative of N(t)

Now, let's find the second derivative of N(t) with respect to t, which represents the acceleration of the rate at which the number of communities using surveillance cameras is changing: $N''(t) = \frac{d^2}{dt^2}(6.08t^3 - 26.79t^2 + 53.06t + 69.5)$ Taking the derivative of N'(t), we get: $N''(t) = 36.48t - 53.58$

## Step 4: Find the critical points of N''(t)

Now we need to find when the second derivative N''(t) = 0 or is undefined. The second derivative is a linear function, so it will never be undefined. Set N''(t) = 0 and solve for t: $36.48t - 53.58 = 0$ Solving for t, we get: $t = \frac{53.58}{36.48} \approx 1.47$

## Step 5: Determine the least rate of increase

Now, we need to calculate the rate of increase at t = 1.47. Plug t = 1.47 into N'(t) to find the least rate of increase: $N'(1.47) = 18.24(1.47)^2 - 53.58(1.47) + 53.06 \approx 35.92$ So, the number of communities using surveillance cameras at intersections was increasing least rapidly around $$t = 1.47$$, corresponding to somewhere in the middle of 2004, and the rate of the increase at that time was approximately 35.92 communities per year.

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