Problem 34 - Calculus Early Transcendentals

Question

Let \(s=100 /\left(t^{2}+12\right)\) be the position function of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. Find the maximum speed of the particle for \(t \geq 0,\) and find the direction of motion of the particle when it has its maximum speed.

Step-by-Step Solution

Answer

The maximum speed occurs as \(t \to 0^+\), moving in the negative direction.

Step 1: Understanding the Problem

The problem involves finding the maximum speed of a particle based on a given position function \(s = \frac{100}{t^2 + 12}\). Speed is the magnitude of the velocity, which is the derivative of the position function \(s\) with respect to \(t\).

Step 2: Differentiating the Position Function

To find the velocity \(v(t)\), differentiate the position function with respect to \(t\). Use the quotient rule, \(\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\), where \(u = 100\) and \(v = t^2 + 12\). Calculating gives:\[ v(t) = \frac{d}{dt}\left(\frac{100}{t^2 + 12}\right) = \frac{0(t^2 + 12) - (100)(2t)}{(t^2 + 12)^2} = \frac{-200t}{(t^2 + 12)^2} \]

Step 3: Finding Critical Points for Maximum Speed

Speed is the absolute value of velocity: \(|v(t)| = \left|\frac{-200t}{(t^2 + 12)^2}\right|\). Find critical points by setting the derivative of \(|v(t)|\) equal to 0. This requires differentiating \(v(t)\) again and solving \(v'(t) = 0\).

Step 4: Solving for Critical Points

Simplify the expression for \(v(t)\) and solve for \(t\) where \(v(t) = 0\), specifically analyze the numerator: \[-200t = 0\implies t = 0.\]

Step 5: Determine Maximum Speed

Examine the behavior of \(|v(t)|\) for critical points and endpoints as \(t\) approaches infinity. As \(t\to\infty\), \(|v(t)|\to 0\). At \(t = 0\), \(|v(t)| = 0\). Evaluate the behavior around \(t = 0\) to check if \(|v(t)|\) increases to a maximum value at some other point. Since \(|v(t)|\) decreases for \(t > 0\), the maximum is achieved closest to \(t = 0\), but not at \(t = 0\) since the function does not attain a real value.

Step 6: Analyzing Direction of Motion

The direction of motion can be determined from the sign of the velocity function \(v(t)\). Since \(v(t) = \frac{-200t}{(t^2 + 12)^2}\), for \(t > 0\), \(v(t) < 0\), indicating that the particle moves in the negative direction when it approaches its theoretical maximum speed just greater than \(t = 0\).

Key Concepts

Position Function

A position function describes the location of a particle along a coordinate line as time progresses. In our exercise, the position function is given as \( s = \frac{100}{t^2 + 12} \), where \( s \) is in feet and \( t \) is in seconds. This function tells us how the position of the particle changes over time.
Understanding the position function serves as a fundamental step in analyzing particle motion. This particular function is a rational function, with a constant numerator and a quadratic polynomial in the denominator.
As \( t \) becomes very large, scrutinizing the behavior of the position function gives insights into how the position stabilizes. Additionally, as \( t \to 0 \), we observe how the position achieves its maximum since the denominator is at its minimum. These observations lay the groundwork for further analysis of velocity and maximum speed.

Velocity

Velocity signifies the rate of change of the particle's position, meaning how quickly the position changes with time. In mathematical terms, velocity is the first derivative of the position function \( s(t) \) with respect to time \( t \).
For this exercise, we use the quotient rule for differentiation to find the velocity \( v(t) \). Calculating, we derive:

• \( v(t) = \frac{-200t}{(t^2 + 12)^2} \)

The sign of \( v(t) \) is crucial in indicating the direction of motion; negative velocity implies motion in the negative direction along the coordinate line. Here, for \( t > 0 \), \( v(t) \) is negative, showing that the particle is moving negatively along the line.

Quotient Rule

The quotient rule is a handy technique for finding the derivative of a ratio of two functions. It is essential when solving for the derivative of our position function \( s(t) \).
For a function \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( t \), the quotient rule is expressed as:

• \( \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \)

Applying this rule, we set \( u = 100 \) and \( v = t^2 + 12 \). Derivatives \( u' \) and \( v' \) come out as 0 and \( 2t \), respectively, simplifying our work.
Following the formula, the velocity function derived is \( \frac{-200t}{(t^2 + 12)^2} \). Mastering the quotient rule provides clarity and precision when working on more complex rational function derivatives.

Particle Motion

Particle motion refers to how a particle moves along a given path due to a defined position function. Comprehending this motion involves looking at velocity and position over time. In this context, analyzing the derivative and the sign of \( v(t) = \frac{-200t}{(t^2 + 12)^2} \) reveals both direction and maximum speed aspects.
Understanding particle motion requires identifying critical points where changes in speed and direction occur. For instance, the particle's speed, or absolute value of velocity, peaks very close to \( t=0 \) but not at it due to undefined behavior at exactly \( t = 0 \).
Moreover, the direction is determined by the velocity's sign—since it's negative for \( t > 0 \), the particle is moving backward. This understanding assists in predicting future positions and directions, vital for various real-world applications like physics and engineering problems.