Question

A particle moves in a straight line with the velocity function \(v(t)=\sin (\omega t) \cos ^{2}(\omega t) .\) Find its position function \(x=f(t)\) if \(f(0)=0 .\)

Short answer

Integrate the velocity function using trigonometric identities and initial conditions to find the position function.

Step-by-step solution

Recognize the Integration Requirement

Since velocity is the derivative of position with respect to time, the position function can be found by integrating the velocity function. Thus, we need to integrate the given velocity function \(v(t) = \sin(\omega t) \cos^2(\omega t)\) to find \(x=f(t)\).

Simplify the Integrand Using Trigonometric Identities

To make the integration easier, apply trigonometric identities. Note that \(\cos^2(\omega t) = \frac{1 + \cos(2\omega t)}{2}\). Rewrite the velocity function:\[v(t) = \sin(\omega t) \cdot \frac{1 + \cos(2\omega t)}{2} = \frac{1}{2}(\sin(\omega t) + \sin(\omega t) \cos(2\omega t)).\]

Integration of the Simplified Form

The function can now be split into two terms:\[v(t) = \frac{1}{2}\sin(\omega t) + \frac{1}{2}\sin(\omega t) \cos(2\omega t).\]Integrate each term separately:1. The integral of \(\sin(\omega t)\) is \(-\frac{1}{\omega} \cos(\omega t)\).2. Apply the substitution \(u = \omega t\) and then use by-parts or other suitable trigonometric identity for \(\sin(u) \cos(2u)\).

Apply the Initial Condition

After obtaining the indefinite integral \(f(t)\), use the initial condition \(f(0) = 0\) to solve for any constant of integration. Since this is a definite integral from 0 to \(t\), ensure that the solution satisfies this boundary.

Combine the Resulting Expressions for the Position Function

Combine the integrals from each term. Solve using the given initial condition to find the full expression for \(f(t)\). This will provide \(f(t) = -\frac{1}{2\omega}\cos(\omega t) + C\), where \(C\) is adjusted to meet the initial condition.

Key concepts

Velocity and Position

The concepts of velocity and position are closely linked in calculus, especially in the context of motion along a straight line. Velocity describes the rate at which an object changes its position over time. In mathematical terms, it is the derivative of the position function with respect to time. This means if you have a velocity function, such as \( v(t) = \sin(\omega t) \cos^2(\omega t) \), you can find the position function by integrating the velocity function.

The relationship between velocity and position is pivotal in calculus problem-solving because it allows you to move from understanding how fast an object is moving to where exactly it is located over time. Integration of the velocity function essentially "accumulates" the area under the velocity curve, representing how far the object has traveled from its initial position.

In the given exercise, using the integral to find position involves recognizing that the definite integral of velocity over a time interval gives the net change in position. Since the particle starts from rest at \( f(0) = 0 \), the integration will solve directly for \( f(t) \). This direct computation highlights the powerful link between derivatives (velocity) and integrals (position), offering essential insights into the world of dynamic systems.

Trigonometric Identities in Integration

Trigonometric identities are useful tools when integrating complex functions, as they allow for simplification of integrands. For example, in the given problem, we have the trigonometric expression \( \sin(\omega t) \cos^2(\omega t) \). Simplifying this using the trigonometric identity \( \cos^2(\omega t) = \frac{1 + \cos(2\omega t)}{2} \) enables easier integration by splitting the expression into more manageable parts.

By rewriting the velocity function as \[ v(t) = \frac{1}{2}( \sin(\omega t) + \sin(\omega t) \cos(2\omega t) ) \], the integration process becomes more straightforward. Each component can then be handled separately, simplifying the overall task and reducing potential errors.

Simplification using identities is a frequent step in solving calculus problems because it transforms expressions in ways that reveal simpler, more familiar forms, particularly those integral results we readily know. This transformation effectively reduces what might be a challenging integral into a sum of integrals of basic functions, making it easier to apply standard integration techniques.

Initial Conditions in Differential Equations

Initial conditions are a crucial part of solving differential equations because they allow you to determine specific solutions to otherwise general solutions obtained through integration. In the context of finding a position function from a velocity function, an initial condition such as \( f(0) = 0 \) helps determine the constant of integration.

When you integrate a function, the result includes a constant term, \( C \), because indefinite integration reverses differentiation and the derivative of a constant is zero. This means there could be infinitely many functions differing only by constant values. Initial conditions eliminate this ambiguity by providing a specific starting point to fix this constant.

In this exercise, applying the initial condition \( f(0) = 0 \) ensures that the position function accurately reflects the starting point of the particle at time zero. By plugging this value back into the integrated function, you can solve for the constant \( C \), fully determining the particle's position function. Correctly using initial conditions thus bridges the gap between mathematical expressions and real-world scenarios, leading to practical, applicable solutions in physics and engineering.