Problem 105 - Precalculus

Question

A video camera is focused on a rocket on a launching pad 2 miles from the camera. The angle of elevation from the ground to the rocket atter \(x\) secondsis \(\frac{\pi}{120} x\) a. Write a function expressing the altitude \(h(x),\) in miles, of the rocket above the ground after \(x\) seconds. Ignore the curvature of the Earth. b. Graph \(h(x)\) on the interval \((0,60)\) . c. Evaluate and interpret the values \(h(0)\) and \(h(30)\) . d. What happens to the values of \(h(x)\) as \(x\) approaches 60 seconds? Interpret the meaning of this in terms of the problem.

Step-by-Step Solution

Answer

a. \( h(x) = 2 \tan\left(\frac{\pi}{120}x\right) \). c. \( h(0) = 0 \), rocket is on the ground; \( h(30) = 2 \), rocket is 2 miles high. d. As \( x \to 60 \), \( h(x) \to \infty \), indicating rapid ascent.

Step 1: Understand the Scenario

We have a camera 2 miles from a launch pad and an angle of elevation that follows the formula \( \theta = \frac{\pi}{120}x \), where \(x\) is the time in seconds. We need to express the altitude \(h(x)\) in terms of \(x\).

Step 2: Model the Scenario Using Trigonometry

In this scenario, the distance from the camera to the launch pad represents the adjacent side of a right triangle, and the altitude of the rocket is the opposite side. Using the tangent function, we have \( \tan(\theta) = \frac{h}{2} \). Here, \( \theta = \frac{\pi}{120}x \).

Step 3: Derive the Function for Altitude

By substituting \( \theta \) into the tangent equation, we get \( \tan\left(\frac{\pi}{120}x\right) = \frac{h(x)}{2} \). Solving for \(h(x)\), we find \( h(x) = 2 \tan\left(\frac{\pi}{120}x\right) \).

Step 4: Graph the Function on Given Interval

We need to plot the function \( h(x) = 2 \tan\left(\frac{\pi}{120}x\right) \) on the interval \((0, 60)\). This involves calculating values at selected points within this interval to visualize how the altitude changes over time.

Step 5: Evaluate h(0)

Substitute \( x = 0 \) into the function: \( h(0) = 2 \tan\left(\frac{\pi}{120} \cdot 0\right) = 2 \cdot 0 = 0 \). The initial height is 0 miles, indicating the rocket is on the ground.

Step 6: Evaluate h(30)

Substitute \( x = 30 \) into the function: \( h(30) = 2 \tan\left(\frac{\pi}{120} \cdot 30\right) = 2 \tan\left(\frac{\pi}{4}\right) = 2 \cdot 1 = 2 \). The rocket's altitude at 30 seconds is 2 miles.

Step 7: Consider Behavior as x Approaches 60

As \( x \rightarrow 60 \), \( \theta \rightarrow \frac{\pi}{2} \), making \( \tan(\theta) \) approach infinity. So, \( h(x) \rightarrow \infty \). This suggests that the rocket continues to ascend rapidly, approaching a vertical position as x nears 60.

Key Concepts

Angle of Elevation

When you hear about the angle of elevation, think about tilting your head upwards to focus on an object above the ground. Imagine standing at the base of a rocket launch pad and looking upwards as the rocket flies into the sky. The angle between your line of sight and the flat ground is known as the angle of elevation.
In our problem, the angle of elevation is crucial because it changes as the rocket gains height over time. The exercise specifically provides the formula for this angle as a function of time: \[ \theta = \frac{\pi}{120} \times x \]where \(x\) is the time in seconds.
This means every second, the angle grows slightly larger, reflecting how the rocket rises higher.

• The formula links time directly to the angle, making it possible to predict how high the rocket will be as time goes on.
• Notice that this angle originates from a horizontal line up to the sight line reaching the rocket. It starts at \(0\) when the rocket is on the ground.

The angle of elevation plays a pivotal role in calculating the rocket's altitude, represented by height \(h(x)\) in this problem, using trigonometric functions.

Tangent Function

The tangent function is an essential tool in trigonometry, especially when dealing with right triangles. In our case, it links the angle of elevation with the rocket's height above the launch pad. Imagine a right triangle formed by the line of sight from the camera to the rocket. The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side.
Here, the tangent function situates the angle formula \[ \tan(\theta) = \frac{h(x)}{2} \]in support of finding the rocket’s altitude, \(h(x)\).
This is derived because of how the rocket, camera, and ground form a right triangle:

• Opposite side: The rocket’s height \(h(x)\).
• Adjacent side: The fixed distance, 2 miles, from the camera to the pad.

This configuration allows us to solve for \(h(x)\), indicating how the tangent function makes use of the angle to calculate variable heights overtime.By rearranging the formula, we derive the altitude function: \[ h(x) = 2 \tan\left(\frac{\pi}{120} x \right) \] This expression clearly shows how the rocket's height is directly related to time through the tangent of the angle of elevation, providing insights into the rocket’s aviation path.

Graphing Functions

Graphing functions enables us to visually interpret mathematical relationships, making it easier to understand how certain variables interact over time. In this exercise, we are given the function \[ h(x) = 2 \tan\left(\frac{\pi}{120}x\right) \] which models the rocket's height based on time \(x\).
Graphing this function on the interval \( (0, 60) \) seconds offers a clearer perspective of the rocket’s ascent.

• When you graph trigonometric functions like tangent, it’s essential to identify key periods and behavior such as increasing to infinity where the tangent approaches asymptotes.
• By plotting points at specific seconds, like \(x = 0\), \(x = 30\), and observing what happens as \(x\) nears \(60\), the changes in altitude become evident.

This graphical approach reveals the height \(h(x)\), starting from 0 miles, reaching 2 miles at 30 seconds, and increasing rapidly as it approaches 60 seconds. This can be visualized as the curve steeply climbs upward. Such a representation indicates how quickly the rocket ascends, aligning with real-world rocket accelerations. Therefore, graphing is a crucial step in comprehending these dynamics and the role of trigonometric functions in translating angles and distances into visible data.