Problem 29 - Precalculus Mathematics for Calculus

Question

The displacement \(s\) (in meters) of a particle moving in a straight line is given by the equation of motion \(s=4 t^{3}+6 t+2,\) where \(t\) is measured in seconds. Find the instantaneous velocity of the particle \(s\) at times \(t=a\) \(t=1, t=2, t=3\)

Step-by-Step Solution

Answer

Instantaneous velocities are: \( v(a) = 12a^2 + 6; v(1) = 18; v(2) = 54; v(3) = 114.\)

Step 1: Understanding Instantaneous Velocity

Instantaneous velocity is the derivative of the displacement with respect to time. This means we need to find the derivative of the given function, which describes displacement as a function of time.

Step 2: Differentiate the Displacement Function

We have the displacement function: \[ s(t) = 4t^3 + 6t + 2. \]The instantaneous velocity \( v(t) \) is the derivative \( s'(t) \). We differentiate term by term:- The derivative of \( 4t^3 \) is \( 12t^2 \),- The derivative of \( 6t \) is \( 6 \),- The derivative of \( 2 \) is \( 0 \).Thus, the instantaneous velocity is:\[ v(t) = 12t^2 + 6. \]

Step 3: Calculate the Instantaneous Velocity at Different Times

We need to find \( v(t) \) at \( t = a, t = 1, t = 2, \) and \( t = 3 \):- When \( t = a \): \[ v(a) = 12a^2 + 6. \]- When \( t = 1 \): \[ v(1) = 12(1)^2 + 6 = 18. \]- When \( t = 2 \): \[ v(2) = 12(2)^2 + 6 = 54. \]- When \( t = 3 \): \[ v(3) = 12(3)^2 + 6 = 114. \]

Key Concepts

Displacement Function

The displacement function is a mathematical expression that describes how an object moves over time along a straight path. In this context, the displacement function is given by \[s(t) = 4t^3 + 6t + 2.\] This equation allows us to determine the position of a particle at any given time, \( t \).

• The term \( 4t^3 \) relates to the changing rate of displacement over time, showing more complex motion.
• \( 6t \) indicates a uniform rate of change, like consistent speed.
• \( 2 \) is a constant, defining an initial position offset.

Understanding this function is crucial for figuring out quantities like instantaneous velocity, which gives us insight into the speed of the particle at specific moments.

Derivative

A derivative is a fundamental concept in calculus that measures how a function changes as its input changes. When we speak of derivatives in relation to the displacement function, we mean the derivative of the function with respect to time, \( t \).For our displacement function, the derivative is found by differentiating each term:

• The derivative of \( 4t^3 \) is \( 12t^2 \), showing how this part of the function changes.
• The derivative of \( 6t \) is \( 6 \), indicating the rate of change for that linear part.
• The derivative of \( 2 \) is \( 0 \), as constants do not change, thus have no rate of change.

The derivative function, which is \( v(t) = 12t^2 + 6 \), gives us the instantaneous rate of change of the displacement, synonymous with what we calculate as instantaneous velocity.

Differentiation

Differentiation is the process of finding the derivative of a function. It is a technique used to calculate how a function changes at any given point. By applying differentiation to the displacement function \[s(t) = 4t^3 + 6t + 2,\] we can compute how the particle's position alters over time. This process involves tackling each term separately: raising each power down and multiplying it by the coefficient. Once differentiated, the resulting function \[v(t) = 12t^2 + 6\] steps away from position-based analysis and gives us velocity, which tells us how fast and in what manner the position changes.

Calculus Concepts

Understanding calculus concepts is essential for analyzing continuous change. In physics and mathematics, it serves as a vital tool to interpret and solve problems involving rates of change, such as deriving velocity from a displacement function.

• Functions: Descriptions of how one quantity changes with respect to another, such as displacement over time.
• Derivatives: These tell us about the rate at which things change, such as determining velocity, which is essential to measure how fast an object moves.
• Instantaneous Rates: Differential calculus provides the mechanism to find instantaneous velocity, a key calculus concept allowing for precise measurement of change at any specific time point.

By grasping these calculus concepts, students gain the skills to dive deeper into advanced mathematics and practical applications in fields such as physics, engineering, and beyond.