Problem 46 - University Calculus: Early Transcendentals

Question

Give the velocity \(v=d s / d t\) and initial position of an object moving along a coordinate line. Find the object's position at time \(t\). $$v=\frac{2}{\pi} \cos \frac{2 t}{\pi}, \quad s\left(\pi^{2}\right)=1$$

Step-by-Step Solution

Answer

The object's position is \( s(t) = \sin \left( \frac{2t}{\pi} \right) + 1 \).

Step 1: Understand the Problem

We're given the velocity of an object as a function of time, \( v = \frac{2}{\pi} \cos \frac{2t}{\pi} \), and an initial condition \( s(\pi^2) = 1 \). Our task is to find the object's position function \( s(t) \).

Step 2: Integrate Velocity to Find Position

Since velocity \( v = \frac{ds}{dt} \), we can find the position \( s(t) \) by integrating the velocity with respect to time: \[ s(t) = \int \frac{2}{\pi} \cos \frac{2t}{\pi} \, dt. \]

Step 3: Solve the Integral

The integral can be solved using a substitution. Let \( u = \frac{2t}{\pi} \), then \( du = \frac{2}{\pi} dt \). Rewriting the integral, we have:\[ s(t) = \int \cos u \, du \]. The integral of \( \cos u \) with respect to \( u \) is \( \sin u \). Thus:\[ s(t) = \sin \left( \frac{2t}{\pi} \right) + C \] where \( C \) is an integration constant.

Step 4: Apply Initial Condition

We use the initial condition \( s(\pi^2) = 1 \) to solve for \( C \). Substituting \( t = \pi^2 \):\[ 1 = \sin \left( \frac{2 \pi^2}{\pi} \right) + C \].\[ 1 = \sin(2\pi) + C \].Since \( \sin(2\pi) = 0 \), we have \( 1 = 0 + C \) which means \( C = 1 \).

Step 5: Write the Final Position Equation

Substituting the value of \( C \) into the position function, we have:\[ s(t) = \sin \left( \frac{2t}{\pi} \right) + 1 \].

Key Concepts

Integration

Integration is a key concept in calculus, which allows us to find a function's antiderivative or the area under its curve. Think of it as the reverse operation of differentiation. For this exercise, since we're dealing with velocity and want to derive the position function, integration helps us determine how an object's position changes over time based on its velocity.

• Given the velocity function expressed as a derivative of the position function, integrating it will give back the original position function. In this problem, velocity is given by the equation: \[ v(t) = \frac{2}{\pi} \cos \frac{2t}{\pi} \]
• To find the position, we integrate the velocity function as follows:\[ s(t) = \int \frac{2}{\pi} \cos \frac{2t}{\pi} \, dt \]

This integration process involves a substitution method which simplifies the integral and allows us to find a function that describes the position.

Velocity

Velocity is defined as the rate of change of position with respect to time. In simple terms, it tells us how fast an object moves and in which direction. The velocity function provided \( v = \frac{2}{\pi} \cos \frac{2t}{\pi} \) expresses this rate of change as a trigonometric function.
Understanding the velocity function:

• Velocity indicates both speed and direction, where positive values can indicate movement forward, and negative values indicate movement backward.
• In this case, the velocity function is periodic because it involves a cosine function, meaning that the velocity changes in a cyclic manner over time.

By analyzing the velocity function, we can predict how the position of the object will change as time progresses.

Position function

The position function, often denoted as \(s(t)\), describes the location of an object at any given time \(t\). By integrating the velocity, we obtain the position function, which tells us exactly where the object is along a coordinate line.
For this example, integrating the velocity function results in the position function:\[ s(t) = \sin \left( \frac{2t}{\pi} \right) + C \]

• This tells us the object's position in terms of time, incorporating a constant \(C\), which is determined using initial conditions.
• The periodic nature of this function, due to the sine term, suggests the object moves back and forth over time.

The position function is foundational for understanding an object's motion because it shows both the object's path and its location at any specified time.

Initial condition

Initial conditions are vital when solving differential equations because they allow us to determine the specific solution to the equations. These are data points that tell us where the object was at a specific time. In our example, the initial condition provided is \(s(\pi^2) = 1\).
How initial conditions refine our solution:

• They help in determining the constant \(C\) that appears after integrating the velocity function.
• By applying the initial condition, we substitute \(t = \pi^2\) and \(s = 1\) into our position function so we can solve for \(C\): \[ 1 = \sin(2\pi) + C \]
• This particular initial condition led to finding \(C = 1\).

Using the initial condition is the key step that finalizes our position function, allowing us to have a complete and accurate model of the object's motion over time.