Problem 99 - Thomas Calculus

Question

Falling meteorite 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 meteorite's acceleration is inversely proportional to \( s^2 \).

Step 1: Understanding Velocity Proportionality

The problem states that the velocity \( v \) of the meteorite is inversely proportional to \( \sqrt{s} \), where \( s \) is the distance from Earth's center. This can be expressed as \( v = \frac{k}{\sqrt{s}} \) for some constant \( k \).

Step 2: Express Velocity in Terms of s

Given \( v = \frac{k}{\sqrt{s}} \), we can rewrite this as \( v = k s^{-0.5} \). This makes it easier to differentiate with respect to \( s \).

Step 3: Differentiate Velocity to Find Acceleration

The acceleration \( a \) is the derivative of velocity with respect to time. Using the chain rule, \( a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} \). However, since \( \frac{ds}{dt} = v \), we have \( a = v \frac{dv}{ds} \).

Step 4: Differentiate v with respect to s

Differentiate \( v = k s^{-0.5} \) with respect to \( s \): \( \frac{dv}{ds} = k(-0.5)s^{-1.5} = -0.5k s^{-1.5} \).

Step 5: Substitute and Simplify

Substitute \( v = k s^{-0.5} \) and \( \frac{dv}{ds} = -0.5k s^{-1.5} \) into the expression for acceleration: \( a = k s^{-0.5} (-0.5k s^{-1.5}) = -0.5k^2 s^{-2} \).

Step 6: Conclude Acceleration Proportionality

We find \( a = -0.5k^2 s^{-2} \), which shows the acceleration is inversely proportional to \( s^{2} \), as \( a \propto \frac{1}{s^2} \).

Key Concepts

Inverse Proportionality

In mathematics, inverse proportionality describes a relationship between two variables where one variable increases as the other decreases. If one doubles, the other halves, and vice versa. In the example of the falling meteorite, the velocity of the meteorite is inversely proportional to the square root of the distance from Earth's center. This means that as the meteorite gets closer to Earth, and the distance decreases, its velocity increases.

This relationship can be expressed with the equation:

• \( v = \frac{k}{\sqrt{s}} \)

where \( v \) is velocity, \( s \) is distance, and \( k \) is a constant. Understanding this concept is essential to solving the problem involving the meteorite's acceleration.

Effectively, inverse proportionality helps reveal how variables impact each other's magnitude within dynamic systems.

Differentiation

Differentiation is a core calculus concept that helps us understand how a function changes. It allows us to determine the rate at which one quantity changes with respect to another. In the context of the given problem, we use differentiation to find the acceleration of the meteorite based on its velocity.

By differentiating the expression for velocity \( v = k s^{-0.5} \) with respect to \( s \), we are able to find how the velocity changes as the distance changes. The process is straightforward:

• Differentiate \( k s^{-0.5} \) to get \( \frac{dv}{ds} = -0.5k s^{-1.5} \)

Differentiation, therefore, provides a powerful tool for analyzing and solving dynamic problems where rates of change are critical.

Velocity and Acceleration

Velocity and acceleration are key concepts in physics describing motion. Velocity tells us how fast something moves in a specific direction, while acceleration describes how velocity changes with time.

In the problem of the meteorite, velocity is initially given as inversely proportional to the square root of \( s \). To find acceleration, we need to calculate how the velocity changes as the meteor approaches Earth.

Using the derivative of velocity, we express acceleration \( a \) in terms of velocity and its rate of change with respect to \( s \):

• \( a = v \frac{dv}{ds} \)

This formula helps us understand that acceleration is not just about the speed of the meteorite, but how rapidly that speed changes over time as the distance changes.

Chain Rule

The chain rule is a fundamental differentiation technique in calculus used to differentiate composite functions. It allows you to find the rate of change of one variable with respect to another, keeping in mind the dependencies between several variables.

In the meteorite problem, we use the chain rule to express acceleration, which involves differentiating velocity with respect to time \( t \) but doing so indirectly via \( s \). This involves the equation:

• \( a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} \)

Using the given that \( \frac{ds}{dt} = v \), we find:

• \( a = v \frac{dv}{ds} \)

The chain rule, therefore, is a powerful framework for tackling complex problems that require understanding interconnected variable changes.