Problem 38
Question
Consider a pool shaped like the bottom half of a sphere, that is being filled at a rate of 25 \(\mathrm{ft}^{3} / \mathrm{min}\). The radius of the pool is \(10 \mathrm{ft}\). You stand \(40 \mathrm{ft}\) from a bottle rocket on the ground and watch as it takes off vertically into the air at a rate of \(20 \mathrm{ft} /\) sec. Find the rate at which the angle of elevation changes when the rocket is \(30 \mathrm{ft}\) in the air.
Step-by-Step Solution
Verified Answer
The angle of elevation changes at \( \frac{4}{25} \) radians per second.
1Step 1: Set Up Problem and Understand Variables
We have a right triangle formed by the observer, the rocket, and the ground. The observer is 40 ft from the launch point. As the rocket rises, we maintain this fixed base. Let \( \theta \) be the angle of elevation from the observer to the rocket and \( h(t) \) the height of the rocket at time \( t \). So, \( h(t) = 30 \) ft when the angle of elevation is sought.
2Step 2: Apply Trigonometric Relationship
Use the right triangle relationship: \( \tan(\theta) = \frac{h(t)}{40} \), where \( h(t) \) is the rocket's height. Since \( h(t) = 30 \) ft, substitute this value: \( \tan(\theta) = \frac{30}{40} = \frac{3}{4} \). From this, find \( \theta \).
3Step 3: Differentiate with Respect to Time
Differentiate the equation \( \tan(\theta) = \frac{h(t)}{40} \) with respect to time \( t \): \[ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{40} \cdot \frac{dh(t)}{dt} \]Given \( \frac{dh(t)}{dt} = 20 \text{ ft/sec} \), plug this into the equation and solve for \( \frac{d\theta}{dt} \).
4Step 4: Solve for \( \frac{d\theta}{dt} \)
We know \( \sec(\theta) = \frac{5}{4} \) from \( \cos(\theta) = \frac{4}{5} \) (by drawing the triangle and using Pythagorean identity). Now substitute these into the differentiated equation:\[ (\frac{5}{4})^2 \cdot \frac{d\theta}{dt} = \frac{1}{40} \times 20 \]This gives:\[ \frac{d\theta}{dt} = \frac{4}{25} \text{ radians/sec} \]
5Step 5: Interpret Solution
Give a clear conclusion: When the rocket is 30 ft high, the angle of elevation \( \theta \) increases at a rate of \( \frac{4}{25} \) radians per second. This specifies how fast the observer's viewing angle changes.
Key Concepts
Rate of ChangeTrigonometric RelationshipsDifferentiation
Rate of Change
The concept of "rate of change" is central to understanding how one quantity varies with respect to another, often with time. Imagine observing how fast ice melts on a hot day or how quickly a car speeds up on the highway. These are practical examples of a rate of change.
In our problem, we're dealing with a changing angle due to a rocket's vertical ascent. The key task is to determine how the angle of elevation, seen by an observer, changes over time. This is represented mathematically as \( \frac{d\theta}{dt} \), where \( \theta \) is the angle of elevation.
Whenever something's rate is given, like the rocket's ascent at 20 ft/sec, it hints that calculus will come into play. You'll principally observe how this influences another variable — here, the angle viewed from a fixed point. Understanding rate of change gives insight into dynamic systems, showing how quickly or slowly events transpire.
In our problem, we're dealing with a changing angle due to a rocket's vertical ascent. The key task is to determine how the angle of elevation, seen by an observer, changes over time. This is represented mathematically as \( \frac{d\theta}{dt} \), where \( \theta \) is the angle of elevation.
Whenever something's rate is given, like the rocket's ascent at 20 ft/sec, it hints that calculus will come into play. You'll principally observe how this influences another variable — here, the angle viewed from a fixed point. Understanding rate of change gives insight into dynamic systems, showing how quickly or slowly events transpire.
Trigonometric Relationships
Trigonometry provides a powerful toolkit for analyzing angles and distances. It helps us relate these quantities through right triangles — shapes that recur in real-world scenarios.
For the rocket problem, the observer, the rocket, and the ground form a right triangle. The angle of elevation \( \theta \) changes as the rocket climbs. We utilize the tangent function here because it connects the opposite side of our triangular view (rocket's height) with the adjacent side (the fixed distance from the observer to the launch point).
The specific relationship we've used is \( \tan(\theta) = \frac{h(t)}{40} \), which links the angle \( \theta \) to the rocket's height \( h(t) \). By plugging in known values, like when the rocket is 30 ft high, we simplify the problem to find \( \theta \).
Understanding trigonometric relationships allows us to efficiently connect physical situations involving angles and distances, easing the process of solving numerous real-world problems.
For the rocket problem, the observer, the rocket, and the ground form a right triangle. The angle of elevation \( \theta \) changes as the rocket climbs. We utilize the tangent function here because it connects the opposite side of our triangular view (rocket's height) with the adjacent side (the fixed distance from the observer to the launch point).
The specific relationship we've used is \( \tan(\theta) = \frac{h(t)}{40} \), which links the angle \( \theta \) to the rocket's height \( h(t) \). By plugging in known values, like when the rocket is 30 ft high, we simplify the problem to find \( \theta \).
Understanding trigonometric relationships allows us to efficiently connect physical situations involving angles and distances, easing the process of solving numerous real-world problems.
Differentiation
Differentiation is a cornerstone of calculus, enabling us to determine how a function changes. In practical terms, it's about finding rates of change — like speed, acceleration, or, in our case, how fast an angle increases.
When dealing with the rocket's changing angle of elevation, we differentiated the trigonometric function \( \tan(\theta) = \frac{h(t)}{40} \) with time \( t \). This process reveals \( \frac{d\theta}{dt} \), the rate at which \( \theta \) changes over time.
The formula \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{40} \cdot \frac{dh(t)}{dt} \) emerges from this differentiation. Plugging in the known rate of the rocket's ascent (20 ft/sec) allows us to solve for \( \frac{d\theta}{dt} \).
This lets us calculate how rapidly the observer's angle of elevation changes, demonstrating differentiation's power in translating dynamic situations into comprehensible rates.
When dealing with the rocket's changing angle of elevation, we differentiated the trigonometric function \( \tan(\theta) = \frac{h(t)}{40} \) with time \( t \). This process reveals \( \frac{d\theta}{dt} \), the rate at which \( \theta \) changes over time.
The formula \( \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{40} \cdot \frac{dh(t)}{dt} \) emerges from this differentiation. Plugging in the known rate of the rocket's ascent (20 ft/sec) allows us to solve for \( \frac{d\theta}{dt} \).
This lets us calculate how rapidly the observer's angle of elevation changes, demonstrating differentiation's power in translating dynamic situations into comprehensible rates.
Other exercises in this chapter
Problem 37
For the following exercises, draw the situations and solve the related-rate problems. You are stationary on the ground and are watching a bird fly horizontally
View solution Problem 38
For the following exercises, draw the situations and solve the related-rate problems. You stand 40 ft from a bottle rocket on the ground and watch as it takes o
View solution Problem 41
For the following exercises, draw the situations and solve the related-rate problems. You are walking to a bus stop at a right-angle corner. You move north at a
View solution Problem 41
Consider a pool shaped like the bottom half of a sphere, that is being filled at a rate of 25 \(\mathrm{ft}^{3} / \mathrm{min}\). The radius of the pool is \(10
View solution