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Chapter 4: Problem 14

A hot air balloon lifts off from ground rising vertically. From 100 feet away, a \(5^{\prime}\) woman tracks the path of the balloon. When her sightline with the balloon makes a \(45^{\circ}\) angle with the horizontal, she notes the angle is increasing at about \(5^{\circ} / \mathrm{min} .\) (a) What is the elevation of the balloon? (b) How fast is it rising?

Short Answer

Expert verified
(a) 100 feet; (b) \(\frac{25\pi}{18}\) feet/min.

Step by step solution

01

Understand the Problem

The problem involves a right triangle formed by the woman, the ground, and the vertical height of the balloon. The woman stands 100 feet away from the base of the balloon's path. We are given an angle of elevation of \(45^\circ\) and an angle increase rate of \(5^\circ/\text{min}\).
02

Use Right Triangle Trigonometry

At an angle of \(45^\circ\), the balloon's height \(h\) can be found using the tangent function: \(\tan(\theta) = \frac{h}{100}\). Since \(\tan(45^\circ) = 1\), we have \(h = 100\) feet.
03

Establish a Relationship for Related Rates

Use the tangent function for a general angle \(\theta(t)\): \(\tan(\theta) = \frac{h(t)}{100}\). Differentiate both sides with respect to time \(t\) to establish a relation between the rate of change of the height \(\frac{dh}{dt}\) and \(\frac{d\theta}{dt}\).
04

Differentiate the Equation

By differentiating \(\tan(\theta) = \frac{h}{100}\), we get \(\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dh}{dt}\).
05

Plug in the Values

At \(\theta = 45^\circ\), \(\sec^2(45^\circ) = 2\) and \(\frac{d\theta}{dt} = 5^\circ/\text{min}\). Converting units, \(\frac{d\theta}{dt} = \frac{5 \pi}{180}\, \text{rad/min}\). Now substitute into the differentiated equation: \(2 \cdot \frac{5\pi}{180} = \frac{1}{100} \cdot \frac{dh}{dt}\).
06

Solve for \(\frac{dh}{dt}\)

Solving the equation, \(\frac{dh}{dt} = 100 \cdot 2 \cdot \frac{5\pi}{180} = \frac{500\pi}{180}\). This simplifies to \(\frac{25\pi}{18}\, \text{feet/min}\).
07

Final Answer

The balloon is 100 feet elevated at the given angle, and its rising speed is approximately \(\frac{25\pi}{18}\, \text{feet/min}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right Triangle Trigonometry
Right triangle trigonometry involves the study of relationships between the angles and sides of right triangles. In the context of our hot air balloon problem:
  • The woman, standing 100 feet away, and the vertical rise of the balloon form a right triangle.
  • The horizontal distance from the woman to the balloon's path acts as one leg of the triangle.
  • The vertical distance, or the height of the balloon at any point, is the other leg.
To find the height of the balloon when the angle of elevation is 45 degrees:
  • Use the tangent function, which relates the opposite side (height of the balloon) to the adjacent side (distance from the woman).
  • At an angle of 45 degrees, the tangent is equal to 1, indicating that both legs of the triangle (the distance and the height) are equal.
  • Thus, the balloon is 100 feet above the ground.
Angle of Elevation
The angle of elevation is the angle formed between the horizontal line from the observer and their line of sight to an object above the horizontal plane. In our problem:
  • The angle of elevation is initially given as 45 degrees when the woman observes the balloon.
  • This angle is crucial because it helps relate the triangle's sides through trigonometric functions, like tangent.
  • As the balloon rises, this angle increases, which is a dynamic aspect of the problem.
The angle is increasing at a rate of 5 degrees per minute.
  • This rate of change of the angle is important for calculating how fast the balloon is ascending.
  • Knowing how the angle changes over time can allow us to apply differentiation to find the related rates of change for other variables, like height.
Differentiation
Differentiation is a fundamental concept in calculus used to determine how a function changes as its input changes. In the context of related rates, we use differentiation to relate the rates of change of different quantities. For our scenario:
  • The relationship between the balloon's height and the angle of elevation is governed by the tangent function.
  • We differentiate the function that describes this relationship (\( \tan(\theta) = \frac{h}{100} \)) with respect to time to find the rate at which the height changes as the angle changes.
After differentiating, we obtain:
  • \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dh}{dt} \)
  • This equation connects the rate of change of the angle, \( \frac{d\theta}{dt} \), and the rate of change of height, \( \frac{dh}{dt} \).
Substituting known values and solving this equation yields the rising speed of the balloon.
Tangent Function
The tangent function is one of the primary trigonometric functions that express relationships between the angles and sides of right triangles. For our problem:
  • The tangent of an angle is defined as the ratio of the opposite side to the adjacent side in a right triangle.
  • Here, it links the height of the balloon to the horizontal distance (100 feet) from the observer.
When the angle of elevation is 45 degrees:
  • The tangent of 45 degrees is 1, indicating that the length of the opposite side (balloon's height) equals the length of the adjacent side (100 feet).
  • This fact directly gives the height of the balloon, showing the power of the tangent function in solving such problems.
The tangent function is also central when establishing the related rates equations, allowing us to connect changes in angle to changes in height.

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