Problem 12 - The Calculus with Analytic Geometry

Question

The motion of a particle is along a horizontal line according to the given equation of motion, where \(s \mathrm{ft}\) is the directed distance of the particle from a point \(O\) at \(t \mathrm{sec}\). The positive direction is to the right. Determine the intervals of time when the particle is moving to the right and when it is moving to the left. Also determine when the particle reverses its direction. Show the behavior of the motion by a figure similar to Fig. 3.2.2, choosing values of \(t\) at random but including the values of \(t\) when the particle reverses its direction. $$ s=\frac{t}{1+t^{2}} $$

Step-by-Step Solution

Answer

The particle moves right for \( -1 < t < 1 \) and left for either \( t < -1 \) or \( t > 1 \). Direction reverses at \( t = \begin{array}{cc} -1 \) and \( 1 \).

Step 1: Find the Velocity

To determine the intervals when the particle is moving to the right or left, first find the velocity by taking the derivative of the position function: $$s = \frac{t}{1+t^2}$$. Use the quotient rule: \( \frac{d}{dt} \bigg(\frac{u}{v}\bigg) = \frac{u'v - uv'}{v^2} \). Here, \( u = t \) and \( v = 1 + t^2 \).

Step 2: Apply the Quotient Rule

Differentiate: \( u' = 1 \) and \( v' = 2t \). Substitute into the quotient rule: \( \frac{d}{dt}\bigg(\frac{t}{1+t^2}\bigg) = \frac{(1)(1+t^2) - (t)(2t)}{(1+t^2)^2} = \frac{1 + t^2 - 2t^2}{(1+t^2)^2} = \frac{1 - t^2}{(1+t^2)^2} \).

Step 3: Set the Velocity to Zero

To find where the particle changes direction, set the velocity function to zero: \( \frac{1 - t^2}{(1+t^2)^2} = 0 \). This gives the equation \( 1 - t^2 = 0 \), solving for \( t \) yields \( t = \begin{array}{cc} 1 \) and \( t = -1 \).

Step 4: Determine the Behavior of Velocity

Analyze the sign of \( \frac{1 - t^2}{(1+t^2)^2} \) in the intervals: \( t < -1 \), \( -1 < t < 1 \), and \( t > 1 \). If \( t < -1 \), \( 1 - t^2 \) is negative; if \( -1 < t < 1 \), \( 1 - t^2 \) is positive; if \( t > 1 \), \( 1 - t^2 \) is negative.

Step 5: Identify Intervals of Motion

The particle moves to the left (velocity negative) for \( t < -1 \) and \( t > 1 \), and to the right (velocity positive) for \( -1 < t < 1 \).

Step 6: Determine Direction Changes

The particle reverses its direction at \( t = -1 \) and \( t = 1 \).

Step 7: Plot the Motion

On a graph, plot the position function \( s(t) = \frac{t}{1+t^2} \). Ensure to label and include the critical points at \( t = -1 \) and \( t = 1 \). Demonstrate the behavior of the motion showing the intervals of leftward and rightward motion.

Key Concepts

quotient rule

The quotient rule is instrumental in finding derivatives of functions that are ratios of two differentiable functions. In this exercise, the position function is given by $$ s = \frac{t}{1+t^2} $$ To find the velocity, we need to differentiate this function using the quotient rule. The quotient rule formula is: $$ \frac{d}{dt} \bigg(\frac{u}{v}\bigg) = \frac{u'v - uv'}{v^2} $$ Here, in our context:

• Let $$ u = t $$
• and $$ v = 1 + t^2 $$

Knowing these, we can apply the quotient rule easily.

velocity derivative

To analyze the motion of the particle, we first need to compute its velocity by differentiating the position function. Using our earlier components, we differentiate:

• The derivative of $$ u = t $$ is $$ u' = 1 $$ .
• The derivative of $$ v = 1 + t^2 $$ is $$ v' = 2t $$ .

Substituting into the quotient rule formula, we get: $$ \frac{d}{dt}\bigg(\frac{t}{1+t^2}\bigg) = \frac{(1)(1+t^2) - (t)(2t)}{(1+t^2)^2} $$. Simplifying this expression further, we obtain: $$ \frac{1 + t^2 - 2t^2}{(1+t^2)^2} = \frac{1 - t^2}{(1+t^2)^2} $$ This is the velocity function, which helps us understand the direction and speed of the particle's motion.

direction reversal

To find when the particle changes direction, we need to determine where the velocity function is equal to zero. By setting: $$ \frac{1 - t^2}{(1+t^2)^2} = 0 $$ we solve for the critical points. The equation simplifies to: $$ 1 - t^2 = 0 $$ which gives: $$ t = 1 $$ and $$ t = -1 $$ . Therefore, the particle reverses its direction at these time values.
At these points, the velocity function transitions between positive and negative, indicating a change in direction.

intervals of motion

To find when the particle moves to the right or left, we analyze the sign of the velocity function $$ \frac{1 - t^2}{(1+t^2)^2} $$ on different intervals. The sign of the numerator (1 - t^2) determines the direction:

• For $$ t < -1 $$ : $$ 1 - t^2 $$ is negative.
• For $$ -1 < t < 1 $$ : $$ 1 - t^2 $$ is positive.
• For $$ t > 1 $$ : $$ 1 - t^2 $$ is negative.

Thus, in these intervals:

• The particle moves to the left (negative velocity) for: $$ t < -1 $$ and $$ t > 1 $$ .
• It moves to the right (positive velocity) for: $$ -1 < t < 1 $$ .

These intervals illustrate the consistent movement direction on a graph, with direction changes at $$ t = -1 $$ and $$ t = 1 $$.