Problem 43 - Thomas Calculus in SI Units

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 \(t\). $$\boldsymbol{v}=\sin \pi t, \quad s(0)=0$$

Step-by-Step Solution

Answer

The position function is \( s(t) = -\frac{1}{\pi} \cos(\pi t) + \frac{1}{\pi} \).

Step 1: Understand the Problem

We are given the velocity function \( v = \sin(\pi t) \) and the initial position \( s(0) = 0 \). Our task is to find the object's position function \( s(t) \) at any time \( t \). The velocity is the derivative of the position with respect to time, so we will integrate \( v \) to find \( s(t) \).

Step 2: Set Up the Integral

To find the position function \( s(t) \), we need to integrate the velocity function \( v = \sin(\pi t) \). Thus, we need to find \( s(t) = \int \sin(\pi t) \, dt \).

Step 3: Solve the Integral

To integrate \( \sin(\pi t) \), we use the substitution method: let \( u = \pi t \), hence \( du=\pi \, dt \) or \( dt = \frac{du}{\pi} \). The integral becomes \( \int \sin(u) \frac{du}{\pi} = \frac{1}{\pi} \int \sin(u) \, du \). The integral of \( \sin(u) \) is \(-\cos(u) \), so \[ \frac{1}{\pi} \int \sin(u) \, du = -\frac{1}{\pi} \cos(u) + C \]. Replace \( u \) back with \( \pi t \), we get \[ s(t) = -\frac{1}{\pi} \cos(\pi t) + C \].

Step 4: Apply Initial Condition

Use the initial condition \( s(0) = 0 \) to solve for \( C \). Substitute \( t = 0 \) into the position function: \[ s(0) = -\frac{1}{\pi} \cos(0) + C = 0 \]. Since \( \cos(0) = 1 \), this becomes \(-\frac{1}{\pi} \times 1 + C = 0 \). Thus, \( C = \frac{1}{\pi} \).

Step 5: Write the Position Function

With the constant \( C \) found, the position function is \( s(t) = -\frac{1}{\pi} \cos(\pi t) + \frac{1}{\pi} \).

Key Concepts

Integration Techniques

When dealing with calculus problems involving motion, integration techniques are crucial for finding position from velocity. In the original exercise, we are given the velocity function \( v = \sin(\pi t) \) and need to find the position function \( s(t) \).
To solve this, the integration of the velocity function is required. Integration effectively reverses differentiation, allowing us to 'accumulate' the effect of a function over time.

• **Integration by substitution**: This technique was applied in the solution by substituting \( u = \pi t \). This simplified the integral, making the calculation more straightforward.
• **The integral of \( \sin(u) \) is known to be \(-\cos(u)\)**: After substitution, the integration becomes manageable due to this standard integral form.

Use integration whenever you see a problem transitioning from rates of change (like velocity) to cumulative effect (like position). It's all about finding the area under the curve, which gives a complete movement story over time.

Initial Value Problem

An initial value problem adds a specific condition to the solution of a differential equation, giving it 'initial' context.
In this case, we have the initial condition \( s(0) = 0 \), meaning that at time \( t = 0 \), the object's position is zero. After integrating, you get a general position function \( s(t) \) that includes a constant \( C \). This constant represents unknown factors from the indefinite integral.

• **Apply the initial condition**: Plugging in the initial values into the position function helps you solve for any unknown constants. In our exercise, solving for \( C \) was achieved by substituting \( t = 0 \) and setting \( s(0) = 0 \). This step guarantees us a unique solution fitting the initial conditions.

Initial value problems are about customizing the general solution to fit specific scenarios. Here, it ensures the function behaves as required right from the starting point.

Velocity and Position Relationship

The relationship between velocity and position is foundational in kinematics, the branch of physics and calculus that studies motion. In calculus, the velocity of an object is the derivative of its position function with respect to time. Conversely, to find the position from velocity, one must integrate the velocity function.

• **Velocity as a derivative**: It signifies how fast and in which direction the position is changing over time. Given by \( v = \frac{ds}{dt} \), it tells us how incremental changes lead to overall displacement, which is the position.
• **From velocity to position**: Integration is needed to reverse this derivative relationship. In our example, integrating the velocity function \( v = \sin(\pi t) \) gives the function of position \( s(t) \).

Understanding this relationship helps decipher complex motion problems by dissecting how change in speed and duration affects the distance covered. Thus, knowing how to derive position from velocity using integration is key in many physics and engineering applications.