Question

A security camera scans a long straight fence that encloses a section of a military base. The camera is mounted on a post that is located \(5 \mathrm{m}\) from the midpoint of the fence. The camera makes one complete rotation in 60 s. a) Determine the tangent function that represents the distance, \(d\), in metres, along the fence from its midpoint as a function of time, \(t,\) in seconds, if the camera is aimed at the midpoint of the fence at \(t=0\) b) Graph the function in the interval \(-15 \leq t \leq 15\) c) What is the distance from the midpoint of the fence at \(t=10 \mathrm{s},\) to the nearest tenth of a metre? d) Describe what happens when \(t=15 \mathrm{s}\)

Short answer

The distance function is \(d(t) = 5 \tan(\frac{\pi}{30}t)\). At \(t=10s,\) \(d \approx 8.7m\). At \(t=15s,\rightarrow\) distance is undefined as function approaches a vertical asymptote.

Step-by-step solution

- Determine the angular velocity of the camera

Since the camera completes one rotation in 60 seconds, calculate the angular velocity, \(\theta(t)\), using \(\theta(t) = 2\frac{\taan}{60}*t\).

- Find the tangent function for distance

Use the relationship between the distance and the tangent of the angle, with the camera at 5 meters from the midpoint. The distance function is \(d(t) = 5 \tan(2\frac{\taan}{60}t)\).

- Graph the function within the specified interval

Plot the function within the interval \(-15 \leq t \leq 15 \) using a graphing tool. Mark points and asymptotes where necessary.

- Calculate the distance at t=10 seconds

Substitute \(t = 10\) into the distance function and compute: \(d(10) = 5 \tan(2\frac{\taan}{60}*10) = 5 \tan(\frac{\taan}{3}t)\),

- Determine the behavior at t=15 seconds

Analyze the tangent function near \(t=15\) by noting where the tangent function approaches infinity, thus concluding the distance increases dramatically.

Key concepts

tangent function

Let's start with understanding the **tangent function** and its role in this problem. The tangent function is one of the primary trigonometric functions, which relates an angle of a right triangle to the ratio of the opposite side to the adjacent side. The formula is given by: \[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]. When the camera rotates to scan the fence, it forms an angle with the base position. This angle affects the distance the camera covers along the fence. By expressing the distance as a function of the tangent of this angle, we can accurately track the position of the camera along the fence over time.

angular velocity

Now, let’s breakdown **angular velocity** and its calculation in this context. Angular velocity measures how fast an object rotates or revolves relative to another point. For the security camera, the angular velocity determines how quickly it scans the fence as it rotates. Given that the camera completes one full rotation (360 degrees or \(2\theta\to radians\)) in 60 seconds, we find the angular velocity by the formula: \(\theta(t) = \frac{2\theta\to}{60} \times t \). This simplifies to: \(\theta(t) = \frac{\theta\to}{30} \times t \). This equation quantifies the angle as a function of time.

graphing trigonometric functions

Next, we focus on **graphing trigonometric functions**. Graphing the function \(d(t) = 5 \times \tan\big(\frac{\theta\to}{30} \times t\big)\) between the interval \(-15 \text{to} 15\) seconds will help us visualize the distance covered by the camera over time. The tangent function produces a periodic graph with asymptotes where the function approaches infinity. Here’s how to do it:

• Create a coordinate plane with \(t\) (in seconds) on the x-axis and \(d(t)\) (distance in meters) on the y-axis.
• For each value of \(t\) within the interval, compute \(\frac{\theta\to}{30} \times t\).
• Apply the tangent function to each value to get \(5 \times \tan\big(\frac{\theta\to}{30} \times t\big)\).

Mark the periodic nature and vertical asymptotes to complete the graph. This visual representation shows how the distance changes and where it becomes undefined (approaching infinity).

distance calculation

Finally, let’s tackle the **distance calculation**. To find the distance from the midpoint of the fence at a specific time, substitute that time into the distance function. For \(t=10\) seconds:

• First, calculate the angle: \(\theta(10) = \frac{\theta\to}{30} \times 10 = \frac{\theta\to}{3}\).
• Compute the tangent: \(\tan\big(\frac{\theta\to}{3}\big)\).
• Multiply by 5 meters (distance from the midpoint): \(d(10) = 5 \times \tan\big(\frac{\theta\to}{3}\big)\).

This gives us the distance along the fence from the midpoint when the camera has been rotating for 10 seconds. As for \(t=15\) seconds, since \(\tan\big(\frac{\theta\to}{2}\big)\) approaches infinity (an asymptote), the distance likewise approaches infinity, indicating a dramatic increase.