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Chapter 7: Problem 86

Information is given about the motion of a particle moving along a coordinate line. (a) Use a CAS to find the position function of the particle for \(t \geq 0\) (b) Graph the position versus time curve. $$ a(t)=e^{-t} \sin 2 t \sin 4 t, \quad v(0)=0, s(0)=10 $$

Short Answer

Expert verified
Integrate acceleration to get velocity, then velocity to get position. Apply initial conditions to solve constants.

Step by step solution

01

Understand the Given Information

We are given an acceleration function \( a(t) = e^{-t} \sin 2t \sin 4t \) and two initial conditions: velocity \( v(0) = 0 \) and position \( s(0) = 10 \). The task is to find the position function \( s(t) \) for \( t \geq 0 \).
02

Integrate Acceleration to Find Velocity

To find the velocity, integrate the acceleration function with respect to \( t \): \[ v(t) = \int e^{-t} \sin 2t \sin 4t \, dt + C \] where \( C \) is the constant of integration. Use the initial condition \( v(0) = 0 \) to solve for \( C \).
03

Apply Initial Condition for Velocity

Integrate the acceleration function and apply \( v(0) = 0 \). This helps us to find the constant of integration \( C \) in the velocity function.Once the integration is performed, substitute \( t = 0 \) and set the expression equal to 0 to solve for \( C \).
04

Integrate Velocity to Find Position

After finding the expression for \( v(t) \), integrate \( v(t) \) to find the position function:\[ s(t) = \int v(t) \, dt + C' \]where \( C' \) is another constant of integration. Use the initial position condition \( s(0) = 10 \) to solve for \( C' \).
05

Apply Initial Condition for Position

Integrate the velocity function and apply \( s(0) = 10 \). This helps us to find the constant \( C' \) in the position function.Once the integration is performed, substitute \( t = 0 \) and set the expression equal to 10 to solve for \( C' \).
06

Graph the Position Function

With the position function \( s(t) \) determined, graph \( s(t) \) versus \( t \) for \( t \geq 0 \). Use software or a graphing tool to plot the function, verifying it aligns with the initial conditions.

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Key Concepts

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

Understanding the Acceleration Function
Acceleration gives us insight into how the velocity of an object changes over time. In this scenario, the acceleration function is represented as \( a(t) = e^{-t} \sin 2t \sin 4t \). It can seem complex, but its components have specific roles.
* \( e^{-t} \) represents exponential decay, indicating that as time \( t \) increases, the influence of this factor diminishes.
* \( \sin 2t \) and \( \sin 4t \) are sinusoidal functions, implying periodic behavior in the acceleration. Together they suggest how the changes in velocity may speed up and slow down in a repetitive manner.
By examining the acceleration function, we are setting the stage for understanding changes in motion. Acceleration affects how quickly the particle changes its velocity, which subsequently affects its trajectory.
Decoding the Velocity Function
The velocity function gives us the actual speed and direction of the particle at any given time. To uncover this from the acceleration, we use integration.
* Integration refers to finding a function which describes how something accumulates over time or space. Think of it like summing up all little bits of acceleration to see how fast something is going at a point in time.
The velocity function \( v(t) \) is found by integrating the acceleration function \( a(t) \): \[ v(t) = \int e^{-t} \sin 2t \sin 4t \, dt + C \] Here, \( C \) is an arbitrary constant.
To determine \( C \), we use known information, such as the initial velocity condition \( v(0) = 0 \). By substituting \( t = 0 \) into the velocity equation, we solve for \( C \). This step assures that our velocity function accurately represents the motion from the start.
Grasping Integration in Calculus
Integration in calculus is crucial for understanding overall change. When studying motion, integration allows us to transition from knowing how fast something is changing (acceleration) to where it goes (position).
* The process involves reversing differentiation, the mathematical operation which finds instantaneous rates of change.* In our exercise, once we determine the velocity function \( v(t) \), we proceed to integrate it to find the position function \( s(t) \): \[ s(t) = \int v(t) \, dt + C' \] where \( C' \) is another constant.
Subsequently, we use the initial condition \( s(0) = 10 \) to solve for \( C' \).
Integrating in calculus thus is like piecing together small changes to build up the entire picture of motion over time. This is how we arrive at the position function, providing a complete view of the particle's journey.

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Most popular questions from this chapter

In electromagnetic theory, the magnetic potential at a point on the axis of a circular coil is given by $$ u=\frac{2 \pi N I r}{k} \int_{a}^{+\infty} \frac{d x}{\left(r^{2}+x^{2}\right)^{3 / 2}} $$ where \(N, I, r, k\), and \(a\) are constants. Find \(u\).

Engineers want to construct a straight and level road \(600 \mathrm{ft}\) long and \(75 \mathrm{ft}\) wide by making a vertical cut through an intervening hill (see the accompanying figure). Heights of the hill above the centerline of the proposed road, as obtained at various points from a contour map of the region, are shown in the accompanying figure. To estimate the construction costs, the engineers need to know the volume of earth that must be removed. Approximate this volume, rounded to the nearest cubic foot. [Hint: First set up an integral for the cross-sectional area of the cut along the centerline of the road, then assume that the height of the hill does not vary between the centerline and edges of the road.] $$ \begin{array}{cc} \hline \begin{array}{l} \text { HORIZONTAL } \\ \text { DISTANCE } x \text { (ft) } \end{array} & \begin{array}{c} \text { HEIGHT } \\ h(\mathrm{ft}) \end{array} \\ \hline 0 & 0 \\ 100 & 7 \\ 200 & 16 \\ 300 & 24 \\ 400 & 25 \\ 500 & 16 \\ 600 & 0 \end{array} $$

Transform the given improper integral into a proper integral by making the stated \(u\) -substitution; then approximate the proper integral by Simpson's rule with \(n=10\) subdivisions. Round your answer to three decimal places. $$ \int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x ; u=\sqrt{x} $$

Make the \(u\) -substitution and evaluate the resulting definite integral. $$ \begin{aligned} &\int_{0}^{+\infty} \frac{e^{-x}}{\sqrt{1-e^{-x}}} d x ; u=1-e^{-x} \\ &{[\text { Note: } u \rightarrow 1 \text { as } x \rightarrow+\infty .]} \end{aligned} $$

The exact value of the given integral is \(\pi\) (verify). Approximate the integral using (a) the midpoint approximation \(M_{10},(\mathrm{~b})\) the trapezoidal approximation \(T_{10}\), and (c) Simpson's rule approximation \(S_{20}\) using Formula (7). Approximate the absolute error and express your answers to at least four decimal places. $$ \int_{0}^{3} \frac{4}{9} \sqrt{9-x^{2}} d x $$
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