Problem 101 - University Calculus: Early Transcendentals

Question

The velocity of a heavy meteorite entering Earth's atmosphere is inversely proportional to \(\sqrt{s}\) when it is \(s \mathrm{km}\) from Earth's center. Show that the meteorite's acceleration is inversely proportional to \(s^{2}\).

Step-by-Step Solution

Answer

The acceleration is inversely proportional to \( s^2 \).

Step 1: Express the relationship between velocity and distance

According to the problem, the velocity \( v \) of the meteorite is inversely proportional to \( \sqrt{s} \). This relationship can be expressed as \( v = \frac{k}{\sqrt{s}} \) where \( k \) is a constant of proportionality.

Step 2: Differentiate to find acceleration

The acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \). First, express \( s \) in terms of time, then differentiate \( v = \frac{k}{\sqrt{s}} \) with respect to \( t \) using the chain rule: \( a = \frac{d}{dt}\left( \frac{k}{\sqrt{s}} \right) = -\frac{k}{2s^{3/2}} \cdot \frac{ds}{dt} \).

Step 3: Establish relationship between acceleration and distance

Since velocity \( v = \frac{ds}{dt} = \frac{k}{\sqrt{s}} \), substitute this information into the expression for acceleration: \( a = -\frac{k}{2s^{3/2}} \cdot \frac{k}{\sqrt{s}} = -\frac{k^2}{2s^2} \). Hence, the acceleration \( a \) is inversely proportional to \( s^2 \).

Key Concepts

Velocity-Distance Relationship

The velocity-distance relationship describes how the speed of an object changes depending on its distance from a reference point. In this scenario, the velocity of a meteorite is inversely proportional to the square root of the distance from Earth's center. This specific relationship can be mathematically expressed as \( v = \frac{k}{\sqrt{s}} \), where:

• \( v \) represents the velocity of the meteorite.
• \( s \) represents the distance from the Earth's center.
• \( k \) is a constant of proportionality.

This means that as the distance \( s \) decreases, the speed of the meteorite increases, and vice versa. This kind of relationship is often observed in physical phenomena involving forces like gravity, where an object's speed or force changes based on how far it is from another object. Understanding this relationship aids in predicting how the meteorite will behave as it gets closer to Earth's surface.

Differentiation in Calculus

Differentiation is a fundamental concept in calculus, used to find the rate at which a quantity changes. In this problem, we've used differentiation to find the meteorite's acceleration. Acceleration is the rate of change of velocity with respect to time. To determine acceleration from velocity, we took the derivative of the velocity function with respect to time. This involves the chain rule, a handy tool in calculus when dealing with functions of functions.For the meteorite, the velocity is expressed as \( v = \frac{k}{\sqrt{s}} \). To find acceleration \( a \), differentiate \( v \) with respect to time \( t \):

• Apply the chain rule: First, differentiate with respect to \( s \), then multiply by \( \frac{ds}{dt} \).
• This gives \( a = \frac{d}{dt}\left( \frac{k}{\sqrt{s}} \right) = -\frac{k}{2s^{3/2}} \cdot \frac{ds}{dt} \).

This formula shows how calculus allows us to describe the changing conditions of velocity and distance, key for analyzing real-world dynamics such as a meteorite's descend.

Proportionality Constant

The proportionality constant \( k \) is crucial in understanding how one variable changes in relation to another. In the given problem, \( k \) helps us express and quantify the inverse proportionality between velocity and the square root of distance. It is a specific value that remains the same even as the values of velocity and distance change, making it a useful reference for calculations.Understanding \( k \) is important because:

• It determines the scale of the relationship; different values of \( k \) can describe different scenarios of how fast or slow changes happen.
• It provides a connection between different equations. For instance, \( k \) is used both in describing velocity and in calculating acceleration.

In our exercise, knowing \( k \) allows us to link the movement and force acting on the meteorite, ultimately simplifying complex natural motions into comprehensible mathematical expressions.