Problem 152 - Calculus Volume 1

Question

For the following exercises, the given functions represent the position of a particle traveling along a horizontal line. a. Find the velocity and acceleration functions. b. Determine the time intervals when the object is slowing down or speeding up. $$s(t)=\frac{t}{1+t^{2}}$$

Step-by-Step Solution

Answer

Velocity: \(v(t) = \frac{1 - t^2}{(1+t^2)^2}\); Acceleration: \(a(t) = \frac{4t^3 - 6t}{(1+t^2)^3}\). Speeding up: \((-\infty, -1) \cup (0, \frac{\sqrt{3}}{2})\); Slowing down: \((-1, 0) \cup (\frac{\sqrt{3}}{2}, +\infty)\).

Step 1: Find the Velocity Function

To find the velocity function, differentiate the position function \(s(t)\) with respect to time \(t\). Given \(s(t) = \frac{t}{1+t^2}\), use the quotient rule for derivatives, which states if \(f(t) = \frac{g(t)}{h(t)}\), then \[ f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{[h(t)]^2}. \]Here, \(g(t) = t\) and \(h(t) = 1 + t^2\), with \(g'(t) = 1\) and \(h'(t) = 2t\). Now, apply the quotient rule:\[ v(t) = \frac{(1)(1+t^2) - (t)(2t)}{(1+t^2)^2} = \frac{1 + t^2 - 2t^2}{(1+t^2)^2} = \frac{1 - t^2}{(1+t^2)^2}. \]So, the velocity function is \(v(t) = \frac{1 - t^2}{(1+t^2)^2}\).

Step 2: Find the Acceleration Function

To find the acceleration function, differentiate the velocity function \(v(t) = \frac{1-t^2}{(1+t^2)^2}\) with respect to time \(t\). Again, use the quotient rule:Let \(g(t) = 1 - t^2\) and \(h(t) = (1+t^2)^2\), with \(g'(t) = -2t\) and \(h'(t) = 2(1+t^2)(2t) = 4t(1+t^2)\). Now, apply the quotient rule:\[ a(t) = \frac{(-2t)(1+t^2)^2 - (1-t^2)(4t(1+t^2))}{(1+t^2)^4}. \]Simplifying gives:\[ a(t) = \frac{-2t(1+t^2)^2 - 4t(1-t^2)(1+t^2)}{(1+t^2)^4} = \frac{-2t(1+t^2) - 4t(1-t^2)}{(1+t^2)^3} = \frac{-2t - 4t + 4t^3}{(1+t^2)^3}. \]So the acceleration function is \(a(t) = \frac{4t^3 - 6t}{(1+t^2)^3}\).

Step 3: Determine Intervals of Speeding Up or Slowing Down

A particle speeds up when both velocity and acceleration have the same sign, and slows down when they have opposite signs. To assess this, examine the signs of \(v(t) = \frac{1-t^2}{(1+t^2)^2}\) and \(a(t) = \frac{4t^3 - 6t}{(1+t^2)^3}\).1. **Sign of \(v(t)\):** - \(v(t) > 0\) when \(1 - t^2 > 0\), i.e., \(-1 < t < 1\). - \(v(t) < 0\) when \(1 - t^2 < 0\), i.e., \(t < -1\) or \(t > 1\).2. **Sign of \(a(t)\):** - Factoring: \(a(t) = \frac{2t(2t^2 - 3)}{(1+t^2)^3}\). - \(a(t) = 0\) at \(t = 0, \pm\frac{\sqrt{3}}{2}\). - \(a(t) > 0\) for \(t < -1\) or \(0 < t < \frac{\sqrt{3}}{2}\). - \(a(t) < 0\) otherwise.By analyzing these intervals:- **Speeding Up:** When \(t < -1\) and when \(0 < t < \frac{\sqrt{3}}{2}\).- **Slowing Down:** When \(-1 < t < 0\) and when \(\frac{\sqrt{3}}{2} < t\).

Step 4: Conclude the Time Intervals

From the analysis in Step 3, conclude that the particle is **speeding up** on \((-\infty, -1) \cup (0, \frac{\sqrt{3}}{2})\) and is **slowing down** on \((-1, 0) \cup (\frac{\sqrt{3}}{2}, +\infty)\).

Key Concepts

Velocity Function

Understanding velocity is crucial when it comes to particle motion analysis in calculus. The velocity function provides the rate at which a particle's position changes over time. To find this function from the position function, we need to differentiate the given position function with respect to time.

In our exercise, the position function is given as \(s(t) = \frac{t}{1+t^2}\). To find the velocity \(v(t)\), we apply the quotient rule for derivatives, which is useful when we are dealing with division in our function.

• The quotient rule formula: If \(f(t) = \frac{g(t)}{h(t)}\), then \(f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{[h(t)]^2}\).
• In our case: \(g(t) = t\) and \(h(t) = 1 + t^2\), meaning \(g'(t) = 1\) and \(h'(t) = 2t\).

Using the quotient rule, we find the velocity function:\[v(t) = \frac{1 - t^2}{(1+t^2)^2}.\] The velocity function tells us how quickly and in which direction the particle is moving at any given time \(t\).

Acceleration Function

Acceleration is the derivative of the velocity function, providing the rate at which the velocity of a particle changes over time. Simply put, it tells us if the particle is speeding up or slowing down.

For our task, we must differentiate the previously found velocity function, \(v(t) = \frac{1-t^2}{(1+t^2)^2}\), again employing the quotient rule. This step gives us valuable insight into the dynamics of the particle.

• Let \(g(t) = 1 - t^2\) and \(h(t) = (1+t^2)^2\), with \(g'(t) = -2t\) and \(h'(t) = 4t(1+t^2)\).
• The acceleration function applies the same quotient rule, yielding: \[a(t) = \frac{4t^3 - 6t}{(1+t^2)^3}.\]

This function assists in further understanding how the particle's speed is changing over time and in identifying when the particle experiences changes in its motion.

Quotient Rule

The quotient rule is a fundamental tool in calculus for derivatives. It is particularly helpful when working with functions that are fractions. This rule assists us in finding the derivative of a function that is the division of two other functions.

• Mathematically, it is expressed as: If \(f(t) = \frac{g(t)}{h(t)}\), the derivative is \(f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{[h(t)]^2}\).
• This rule needs identifying the parts of the function: the numerator \(g(t)\) and the denominator \(h(t)\), followed by their respective derivatives \(g'(t)\) and \(h'(t)\).

Using the quotient rule efficiently allows computation of velocity and acceleration functions in particle motion exercises such as this one, where repeated differentiation is often crucial.

Particle Motion Analysis

Particle motion analysis involves understanding how a particle moves over time based on its position, velocity, and acceleration functions. It's an essential concept in study of motion across various fields.

The analysis begins with:

• Computing the velocity to gauge speed and direction. A positive velocity indicates movement in one direction, while a negative velocity indicates the opposite.
• The acceleration is calculated next to observe changes in the particle's speed over time. It's positive if the particle is speeding up and negative if it's slowing down.

To determine when the particle speeds up or slows down, compare the signs of velocity and acceleration:

• Particles speed up when both velocity and acceleration are of the same sign.
• Particles slow down when their signs differ.

In our example, we found that the particle speeds up for \(t < -1\) and \(0 < t < \frac{\sqrt{3}}{2}\). Conversely, it slows down for \(-1 < t < 0\) and \(\frac{\sqrt{3}}{2} < t\). This comprehensive understanding supports the overarching analysis of the particle's motion.