Problem 42 - Calculus Single Variable

Question

A position \(p(t)\) is given. Calculate the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ). \(p(t)=3 \sin (2 t) \mathrm{m}\)

Step-by-Step Solution

Answer

The acceleration is given by \( a(t) = -12 \sin(2t) \) m/s².

Step 1: Understand the Problem

Given the position function of an object, \( p(t) = 3 \sin (2t) \) meters, we need to find the acceleration function. To do this, we first find the velocity function, which is the derivative of the position function with respect to time \( t \), and then find the acceleration function, which is the derivative of the velocity function.

Step 2: Calculate the Velocity

The velocity function \( v(t) \) is the first derivative of the position function \( p(t) \). Compute the derivative:\[v(t) = \frac{d}{dt}(3 \sin(2t))\]Using the chain rule, the derivative of \( \sin(2t) \) is \( 2 \cos(2t) \), so:\[v(t) = 3 \times 2 \cos(2t) = 6 \cos(2t)\]

Step 3: Calculate the Acceleration

The acceleration function \( a(t) \) is the derivative of the velocity function \( v(t) \). Compute the derivative:\[a(t) = \frac{d}{dt}(6 \cos(2t))\]Using the chain rule again, the derivative of \( \cos(2t) \) is \(-2 \sin(2t)\), so:\[a(t) = 6 \times -2 \sin(2t) = -12 \sin(2t)\]Thus, the acceleration is \( a(t) = -12 \sin(2t) \) m/s².

Key Concepts

Understanding the Position Function

A position function describes the location of a point or object on a straight line or in space at any given time. In this exercise, the position is given by the function \( p(t) = 3 \sin(2t) \), measured in meters. This type of function tells us where an object is located at each time \( t \).
Understanding the structure of the position function is crucial as it serves as the base for calculating velocity and acceleration. Here, it’s important to recognize that the function is defined in terms of a sine wave, which implies the motion is oscillatory.

• The amplitude is 3, indicating how far the motion occurs from the center or equilibrium position.
• The coefficient of \( t \) inside the sine function shows the frequency of oscillations.

The position function provides a starting point to derive other important kinematic quantities like velocity and acceleration.

Calculating the Velocity Function

Velocity is the derivative of the position function with respect to time \( t \). In simple terms, it tells us how fast the position is changing at any given moment. In this problem, the velocity \( v(t) \) is found by differentiating the position function \( p(t) = 3 \sin(2t) \):
We apply the derivative rule and use the chain rule (which we'll touch on later) to find:
\[ v(t) = \frac{d}{dt}(3 \sin(2t)) = 3 \times 2 \cos(2t) = 6 \cos(2t) \]

• The velocity function \( v(t) = 6 \cos(2t) \) signifies the rate of change of position with respect to time.
• It describes how fast the oscillation occurs and in which direction, positive or negative.

This is critical since velocity not only tells us how fast something is moving but also its direction of movement.

Finding the Acceleration Function

Acceleration gives us the rate at which the velocity changes over time. Just like velocity, it can be found by taking the derivative, this time from the velocity function.
So we differentiate \( v(t) = 6 \cos(2t) \) using the chain rule once more:
\[ a(t) = \frac{d}{dt}(6 \cos(2t)) = 6 \times -2 \sin(2t) = -12 \sin(2t) \]

• The acceleration function \( a(t) = -12 \sin(2t) \) provides information on how the speed of the object is increasing or decreasing.
• A negative value indicates acceleration in the opposite direction to the motion, effectively a deceleration.

Acceleration is essential in understanding how an object’s motion changes, helping us to predict future positions and velocities.

Understanding Derivatives in Calculus

Derivatives are fundamental in calculus, serving as a tool to measure how a function changes as its input changes. In this context, they allow us to move from position to velocity, and from velocity to acceleration.

• The first derivative of a position function yields the velocity function.
• The second derivative provides acceleration.

Intuitively, derivatives give us an object's instantaneous rate of change at any point in time. They are foundational in analyses related to motion and change, presenting a way to precisely predict future states or changes in a system based on current behavior.

Introducing the Chain Rule

The chain rule is a critical differentiation technique in calculus, especially useful for composite functions, where one function is "inside" another. In these solutions, it allows us to differentiate \( \sin(2t) \) and \( \cos(2t) \).
To find the derivative of composite functions like these, the chain rule dictates:
1. Differentiate the outer function.
2. Multiply by the derivative of the inner function.
In our problem, we apply this rule twice:

• First, when differentiating \( \sin(2t) \) to find velocity, we multiply \( \cos(2t) \) by the derivative of \( 2t \), which is \( 2 \).
• Then, while finding acceleration, differentiating \( \cos(2t) \), the derivative is \(-2 \sin(2t)\).

This rule is essential for solving many calculus problems, especially those involving nested functions.