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Chapter 4: Problem 43

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 \(t.\) \begin{equation}v=\sin \pi t, \quad s(0)=0\end{equation}

Short Answer

Expert verified
The object's position at time \( t \) is \( s(t) = -\frac{1}{\pi} \cos(\pi t) + \frac{1}{\pi} \).

Step by step solution

01

Understanding the Problem

In this exercise, we are given a velocity function \( v = \sin \pi t \) and an initial position \( s(0) = 0 \). Our task is to find the position function \( s(t) \) describing the object's position at time \( t \).
02

Set Up the Integration

To find the position function \( s(t) \), we need to integrate the velocity function with respect to time. This will give us an expression for \( s(t) \). Recall that the position function is the integral of the velocity function: \[s(t) = \int v \, dt = \int \sin \pi t \, dt\]
03

Perform the Integration

Integrate the velocity function \( \sin \pi t \). Use the substitution \( u = \pi t \), so \( du = \pi dt \) or \( dt = \frac{du}{\pi} \): \[\int \sin \pi t \, dt = \int \sin u \, \frac{du}{\pi} = \frac{1}{\pi} \int \sin u \, du \] The integral of \( \sin u \) is \( -\cos u + C \). Therefore, \[\frac{1}{\pi} \int \sin u \, du = -\frac{1}{\pi} \cos u + C \]Substituting back \( u = \pi t \), we get: \[s(t) = -\frac{1}{\pi} \cos(\pi t) + C\]
04

Apply Initial Condition

We need to find the constant \( C \) using the initial condition \( s(0) = 0 \). Substitute \( t = 0 \) and \( s(0) = 0 \) into the position function: \[0 = -\frac{1}{\pi} \cos(\pi \times 0) + C\] Since \( \cos(0) = 1 \), this simplifies to \[0 = -\frac{1}{\pi} \times 1 + C \] Solving for \( C \), we find \( C = \frac{1}{\pi} \).
05

Write the Final Position Function

Substitute \( C = \frac{1}{\pi} \) back into the position function:\[s(t) = -\frac{1}{\pi} \cos(\pi t) + \frac{1}{\pi}\]This is the position function of the object at time \( t \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity Function
In calculus, the velocity function is a fundamental concept used to describe how fast an object moves along a path over time. It is a derivative. This means, it is the rate of change of the position function with respect to time.
  • For example, if the velocity function is given as \( v = \sin \pi t \), it tells us how the speed of the object varies depending on the time \( t \).
  • This specific function, \( \sin \pi t \), is a trigonometric function, which has a periodic nature. It oscillates between -1 and 1, providing a smooth wave-like pattern for speed.
Knowing the velocity lets us understand the dynamics of an object's movement along a line or any other path.
Position Function
The position function, denoted as \( s(t) \), describes the exact location of an object at any time \( t \). To find it, we integrate the velocity function. This is because integration is the inverse operation of differentiation.
  • In our example, by integrating \( \sin \pi t \), we aim to find the position function \( s(t) \).
  • Therefore, \( s(t) = \int \sin \pi t \, dt \).
After performing the integration, our solution gives us a formula to calculate an object's exact position as time progresses.
Initial Condition
The initial condition is crucial in solving differential equations, like the ones involving velocity and position functions. It provides a specific value of the position function at a certain point, often at the beginning of the timeline.
  • In the problem, the initial condition is given as \( s(0) = 0 \). This means that at time \( t = 0 \), the object is at the origin or starting point on the coordinate line.
  • Such conditions allow us to solve for constants of integration that appear after integrating a function.
Applying this ensures that our solution perfectly aligns with the given conditions at the start.
Trigonometric Integration
Trigonometric integration involves integrating functions that contain trigonometric expressions like \( \sin(x) \) or \( \cos(x) \). It requires certain substitutions to simplify the process.
  • For instance, in our exercise, transitioning to \( u = \pi t \) helps in performing the integration of \( \sin \pi t \).
  • Through substitution, the integration simplifies to \( \int \sin u \, \frac{du}{\pi} \), which becomes \( -\frac{1}{\pi} \cos u + C \) after integrating.
Understanding trigonometric integration is key in tackling exercises involving periodic functions, common in physics and engineering applications.

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Most popular questions from this chapter

Production level Suppose that \(c(x)=x^{3}-20 x^{2}+20,000 x\) is the cost of manufacturing \(x\) items. Find a production level that will minimize the average cost of making \(x\) items.

You are to construct an open rectangular box with a square base and a volume of 48 \(\mathrm{ft}^{3} .\) If material for the bottom costs \(\ 6 / \mathrm{ft}^{2}\) and material for the sides costs \(\ 4 / \mathrm{ft}^{2},\) what dimensions will result in the least expensive box? What is the minimum cost?

Finding displacement from an antiderivative of velocity a. Suppose that the velocity of a body moving along the s-axis is $$ \frac{d s}{d t}=v=9.8 t-3 $$ \begin{equation} \begin{array}{l}{\text { i) Find the body's displacement over the time interval from }} \\ {t=1 \text { to } t=3 \text { given that } s=5 \text { when } t=0 \text { . }} \\ {\text { ii) Find the body's displacement from } t=1 \text { to } t=3 \text { given }} \\ {\text { that } s=-2 \text { when } t=0 \text { . }} \\ {\text { iii) Now find the body's displacement from } t=1 \text { to } t=3} \\ {\text { given that } s=s_{0} \text { when } t=0 .}\end{array} \end{equation} b. Suppose that the position \(s\) of a body moving along a coordinate line is a differentiable function of time \(t .\) Is it true that once you know an antiderivative of the velocity function \(d s / d t\) you can find the body's displacement from \(t=a\) to \(t=b\) even if you do not know the body's exact position at either of those times? Give reasons for your answer.

Frictionless cart A small frictionless cart, attached to the wall by a spring, is pulled 10 \(\mathrm{cm}\) from its rest position and released at time \(t=0\) to roll back and forth for 4 sec. Its position at time \(t\) is \(s=10 \cos \pi t.\) \begin{equation}\begin{array}{l}{\text { a. What is the cart's maximum speed? When is the cart moving }} \\ \quad {\text { that fast? Where is it then? What is the magnitude of the ac- }} \\ \quad {\text { celeration then? }} \\\ {\text { b. Where is the cart when the magnitude of the acceleration is }} \\\ \quad {\text { greatest? What is the cart's speed then? }}\end{array}\end{equation}

\begin{equation} \begin{array}{l}{\text { a. How close does the semicircle } y=\sqrt{16-x^{2}} \text { come to the }} \\ {\text { point }(1, \sqrt{3}) ?} \\ {\text { b. Graph the distance function and } y=\sqrt{16-x^{2}} \text { together and }} \\\ {\text { reconcile what you see with your answer in part (a). }}\end{array} \end{equation}
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