Problem 41 - Calculus Volume 1

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 are walking to a bus stop at a right-angle corner. You move north at a rate of \(2 \mathrm{~m} / \mathrm{sec}\) and are \(20 \mathrm{~m}\) south of the intersection. The bus travels west at a rate of \(10 \mathrm{~m} /\) sec away from the intersection - you have missed the bus! What is the rate at which the angle between you and the bus is changing when you are \(20 \mathrm{~m}\) south of the intersection and the bus is \(10 \mathrm{~m}\) west of the intersection?

Step-by-Step Solution

Answer

The angle is changing at approximately 0.036 rad/s.

Step 1: Understand the Problem

The problem involves a swimming pool shaped like a hemisphere being filled at a constant rate and calculating a changing angle formed between two moving points, one moving north and one moving west, from a corner intersection.

Step 2: Identify Known Values

For the pool part, you know the volume rate of change is \( \frac{dV}{dt} = 25 \, \text{ft}^3/\text{min} \) and the radius \( r = 10 \text{ ft} \). For the angle problem, your distance south is 20 m, the bus's distance west is 10 m, and their respective rates are \( 2 \, \text{m/s} \) north for you and \( 10 \, \text{m/s} \) west for the bus.

Step 3: Calculate the Volume of the Hemisphere

The volume of a hemisphere is given by \( V = \frac{2}{3} \pi r^3 \). Using \( r = 10 \), we find \( V = \frac{2}{3} \pi (10)^3 = \frac{2000}{3}\pi \).

Step 4: Relate Changes in Volume to Rate of Filling

Differentiate the volume formula with respect to time: \( \frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt} \). Given \( \frac{dV}{dt} = 25 \, \text{ft}^3/\text{min} \), solve for \( \frac{dr}{dt} \) when \( r = 10 \).

Step 5: Set Up the Angle Problem

You're dealing with the rate of change of an angle between you and the bus. Use the coordinates: your position is (0, y) where y changes with \( \dot{y} = -2 \), and the bus position is (x, 0) where x changes with \( \dot{x} = 10 \).

Step 6: Use Trigonometry to Relate Positions

The positions form a triangle with the intersection, where \( \theta \) is the angle opposite the right angle at the intersection. Use \( \tan(\theta) = \frac{y}{x} \) which implies \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \).

Step 7: Differentiate to Find Rate of Change of \( \theta \)

Differentiate \( \theta = \tan^{-1}(\frac{y}{x}) \) with respect to time to find \( \frac{d\theta}{dt} \). Remember: for \( \frac{d\tan^{-1}(u)}{dt} = \frac{1}{1+u^2}\cdot\frac{du}{dt} \), use \( u=\frac{y}{x} \).

Step 8: Plug in Known Values to Differentiate

Plug \( x = 10 \), \( y = -20 \), \( \dot{x} = 10 \), \( \dot{y} = -2 \) into \( \frac{d\theta}{dt} = \frac{1}{1+\left(\frac{y}{x}\right)^2}\left(\frac{x\dot{y} - y\dot{x}}{x^2}\right) \) to calculate the value.

Step 9: Calculate Final Answer

Simplify and substitute the given values: \( \frac{d\theta}{dt} = \frac{1}{1+(\frac{-20}{10})^2} \cdot \left( \frac{10(-2) - (-20)(10)}{10^2} \right) = \frac{1}{5} \left( \frac{-20 + 200}{100} \right) = \frac{18}{500} = \frac{9}{250} \approx 0.036 \, \text{rad/s} \).

Key Concepts

hemisphere volume

A hemisphere is essentially half of a sphere, and calculating its volume is a slight variation of calculating the volume of a full sphere. The formula for the volume of a full sphere is given by \( V = \frac{4}{3} \pi r^3 \). For a hemisphere, you simply take half of this value:

• Volume of hemisphere: \( V = \frac{2}{3} \pi r^3 \)

Where \( r \) is the radius of the hemisphere. In our exercise, the radius of the pool is given as \( 10 \text{ ft} \), so the volume of the pool is \( \frac{2000}{3}\pi \ \text{ft}^3 \). Understanding this concept is crucial in problems involving the rate of change of volume, especially when the object is being filled or emptied over time.

trigonometry

Trigonometry often comes into play when dealing with angles and their relationships in triangles. In this exercise, we are looking at a scenario where two people are moving in such a way that their paths form a right triangle with a corner intersection. Here,

• Using trigonometric functions like tangent and inverse tangent helps to relate the positions of the two moving entities.
• Tangent is commonly used because it deals with the ratio of the opposite side to the adjacent side in a right triangle.

In the exercise, the tangent of angle \( \theta \) is given by the ratio \( \frac{y}{x} \), where \( y \) is the north-south distance (your position relative to the corner) and \( x \) is the east-west distance (the bus's position relative to the corner). We use \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \) to determine the actual angle between you and the bus.

angle rate of change

The rate of change of an angle is an interesting concept, often requiring the application of related rates in calculus. In our context, we are interested in how quickly the angle \( \theta \) is changing over time as one moves north and the other moves west.

• To find this, we represent the angle \( \theta \) in terms of a trigonometric function: \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \).
• The change in \( \theta \) over time involves differentiating this expression with respect to time.

This leads to the derivative \( \frac{d\theta}{dt} \). Using various chain rules and the given values, we can solve and find \( \frac{d\theta}{dt} = 0.036 \text{ rad/s} \), which tells us how fast the angle is changing as the position of both entities vary.

differentiation with respect to time

Differentiation is a technique used in calculus to find rates of change. When we differentiate with respect to time, we're essentially seeing how one quantity changes as time progresses. In the problem, this is utilized to determine both the rate at which the volume of the hemisphere increases and the rate at which the angle between two moving objects changes.

• For the hemisphere, differentiation helps relate changes in the pool's water volume with respect to the filling time.
• For the angle between you and the bus, we differentiate a trigonometric expression with respect to time.

By setting up equations that describe these situations (such as \( \frac{dV}{dt} \) for volume change and \( \frac{d\theta}{dt} \) for angular change) and differentiating them, we solve for the unknown rates, giving us insight into the dynamics of the scenario.