Problem 30 - Calculus Single Variable

Question

A rocket that is rising vertically is being tracked by a ground-level camera located 3 mi from the point of blastoff. When the rocket is 2 mi high, its speed is 400 mph. At what rate is the (acute) angle between the horizontal and the camera's line of sight changing?

Step-by-Step Solution

Answer

The rate of change of the angle is approximately \(92.31\) radians per hour.

Step 1: Draw a diagram and assign variables

Visualize the problem by drawing a right triangle where the rocket is the vertical side, the distance from the launch point to the camera is the horizontal side, and the hypotenuse is the camera's line of sight. Let \(y\) be the height of the rocket, \(x = 3\) miles (constant), and \(\theta\) be the angle between the horizontal side and the hypotenuse.

Step 2: Write the relationship using trigonometry

Using trigonometric ratios, we know that \(\tan \theta = \frac{y}{x}\). Here \(x = 3\) miles, so \(\tan \theta = \frac{y}{3}\).

Step 3: Differentiate with respect to time

Differentiate the equation \(\tan \theta = \frac{y}{3}\) with respect to \(t\) (time): \( \sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{3} \cdot \frac{dy}{dt} \).

Step 4: Evaluate derivatives

Given \(\frac{dy}{dt} = 400\) mph and \(y = 2\) miles when finding the rate of change of \(\theta\): calculate \( \sec^2 \theta \).

Step 5: Find \(\sec^2 \theta\) using triangle ratios

Using \(y = 2\) miles and \(x = 3\) miles, the camera sees a hypotenuse (distance from the camera to the rocket) of \( \sqrt{2^2 + 3^2} = \sqrt{13}\) miles. Thus, \( \sec \theta = \frac{\sqrt{13}}{3} \). Therefore, \( \sec^2 \theta = \frac{13}{9} \).

Step 6: Solve for \(\frac{d\theta}{dt}\)

Substitute into the differentiated equation: \( \frac{13}{9} \cdot \frac{d\theta}{dt} = \frac{1}{3} \cdot 400 \). Solve for \(\frac{d\theta}{dt} \):\[ \frac{d\theta}{dt} = \frac{400}{3} \cdot \frac{9}{13} = \frac{1200}{13} \text{ radians per hour}. \]

Key Concepts

Trigonometry in Rocket Motion

Trigonometry plays a crucial role in solving problems involving angles and lengths, especially when dealing with objects like rockets and cameras. In this scenario, we have a right triangle formed by the rocket's height, the distance from the camera to the launch site, and the camera's line of sight. By using the tangent ratio, we establish a relationship between the rocket’s height and the angle of elevation observed by the camera:

• The tangent of an angle in a right triangle (b8) is the ratio of the opposite side to the adjacent side.
• Here, b8 is the angle between the horizontal ground (adjacent) and the camera's line of sight (hypotenuse).
• The relationship is tan b8 = \frac{y}{x}\, where \(y\) is the height of the rocket, and \(x\) is the fixed distance from the camera to the launch point.

Understanding this relationship helps us predict how angle changes as the rocket moves.

Differentiation in Motion Analysis

Differentiation is a powerful mathematical tool used to find how quantities change over time. In our problem, we want to know how rapidly the angle between the camera's line of sight and the ground is changing as the rocket ascends.

• We differentiate the equation \(\tan \theta = \frac{y}{3}\) with respect to time \(t\), leading to \( \sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{3} \cdot \frac{dy}{dt} \).
• The derivative \( \frac{d\theta}{dt} \) represents the rate of change of the angle \(\theta\) over time. It tells us how quickly the camera needs to adjust its angle to keep tracking the rocket efficiently.

This process allows us to convert information about the rocket's speed directly into information about how fast the viewing angle changes, which is vital in real-time tracking situations.

Rocket Motion and Dynamics

The dynamic motion of rockets provides fascinating scenarios for application of mathematical concepts like trigonometry and differentiation. As the rocket ascends, its changing position affects several factors:

• The height \(y\) of the rocket increases, affecting the angle \(\theta\) observed by the camera.
• Rocket speed, in this case 400 mph, contributes directly to the rate at which height changes over time (\(\frac{dy}{dt}\)).

Tracking changes accurately requires understanding how rapid movements translate into positional data that external observers like cameras must account for to effectively monitor ascent.

Adjusting Camera Angles to Track Rockets

In this exercise, a critical aspect is how a ground-level camera adjusts its angle to maintain a clear line of sight to a moving rocket.

• The angle of sight, \(\theta\), changes as the rocket changes altitude, requiring the camera to pivot continuously.
• To determine the rate of this change, solving \( \frac{d\theta}{dt} = \frac{1200}{13} \text{ radians per hour} \) gives the velocity of the camera's pivot.
• This ensures that the camera's focus remains on the rocket, enabling effective tracking and data recording for analysis.

This adjustment is crucial for applications like space missions or test flights, where seamless tracking leads to valuable data collection and mission success.