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Chapter 8: Problem 61

Average value A retarding force, symbolized by the dashpot in the accompanying figure, slows the motion of the weighted spring so that the mass's position at time \(t\) is $$\quad y=2 e^{-t} \cos t, \quad t \geq 0$$ Find the average value of \(y\) over the interval \(0 \leq t \leq 2 \pi\)

Short Answer

Expert verified
The average value of \( y = 2 e^{-t} \cos t \) from \( 0 \) to \( 2\pi \) is approximately 0.

Step by step solution

01

Understand the Formula for Average Value

The average value of a continuous function \( y(t) \) over the interval \([a, b]\) is given by the formula \( \bar{y} = \frac{1}{b-a} \int_{a}^{b} y(t) \, dt \). For this problem, \( y(t) = 2 e^{-t} \cos t \) and the interval is \([0, 2\pi] \). This means \( a = 0 \), \( b = 2\pi \).
02

Set Up the Integral

Plug the function \( y = 2 e^{-t} \cos t \) into the formula for the average value. This gives us the integral \( \int_{0}^{2 \pi} 2 e^{-t} \cos t \, dt \). The average value will be \( \bar{y} = \frac{1}{2\pi} \int_{0}^{2\pi} 2 e^{-t} \cos t \, dt \).
03

Solve the Integral

To solve \( \int_{0}^{2 \pi} 2 e^{-t} \cos t \, dt \), use integration by parts. Let \( u = \cos t \) (so \( du = -\sin t \, dt \)) and \( dv = 2e^{-t} \, dt \) (so \( v = -2e^{-t} \)). The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). After applying this method, calculate the definite integral over \([0, 2\pi]\).
04

Evaluate the Definite Integral

After using integration by parts, evaluate the antiderivatives at the bounds \( 0 \) and \( 2\pi \). Notice that because \( e^{-t} \) approaches zero very quickly as \( t \) increases, the value of the function at \( t = 2\pi \) will be very small. After evaluation, you'll find the integral gives a small number approaching zero.
05

Calculate the Average Value

Since \( \int_{0}^{2\pi} 2 e^{-t} \cos t \, dt \approx 0 \), the average value simplifies as follows: \( \bar{y} = \frac{1}{2\pi} (\text{the evaluated integral}) \approx 0 \). Thus, \( \bar{y} \) is approximately \( 0 \), meaning the average value over the interval is almost zero.

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Key Concepts

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

Integration by Parts
Integration by parts is a handy technique for solving certain integrals. It is particularly useful when you have a product of functions. The formula is\[ \int u \, dv = uv - \int v \, du \]Here's how you can use it:
  • Identify parts of the integral: Choose parts of the equation to represent \( u \) and \( dv \).
  • Differentiation and integration: Calculate \( du \) by differentiating \( u \), and find \( v \) by integrating \( dv \).
  • Substitute: Plug \( u \), \( dv \), \( du \), and \( v \) into the integration by parts formula.
  • Solve the remaining integral: Often, you'll need to compute another simpler integral to finish the job.
In this specific problem, we applied integration by parts to the function \( 2e^{-t} \cos t \). We let \( u = \cos t \), making \( du = -\sin t \, dt \), and \( dv = 2e^{-t} \, dt \), so \( v = -2e^{-t} \). This process turns a tricky integral into something manageable by splitting it up!
Definite Integrals
Definite integrals give the accumulation of values across an interval. They are represented as:\[ \int_{a}^{b} f(x) \, dx \]where \( a \) and \( b \) are the limits and \( f(x) \) is the function you're integrating. When working with a definite integral:
  • Set up the integral: Put the function and its limits into an integral form, ready for evaluation.
  • Evaluate the antiderivative: Calculate the antiderivative or "solution" to the integral.
  • Apply the limits: After finding the antiderivative, substitute the upper and lower limits (\( b \) and \( a \)) into the antiderivative, and subtract the latter from the former.
In this example, the definite integral is \( \int_{0}^{2 \pi} 2e^{-t} \cos t \, dt \). Since \( e^{-t} \) decreases rapidly as \( t \) grows, this evaluation results in a value close to zero, especially when bounded and limited within \( 0 \) to \( 2\pi \). The average value, derived from the integral, provides insights into the function's behavior over the interval.
Retarding Force in Physics
In physics, a retarding force is a kind of damping that gradually slows down motion. Often, it is used to model things like air resistance or friction, which oppose the movement of objects.
  • Dashpot Model: The dampening system is often represented using a "dashpot," a device that applies a force opposite to the direction of movement, effectively slowing it.
  • Real-world applications: This concept appears in car shock absorbers, air resistance for projectiles, and vibration isolation systems.
  • Mathematical Representation: In this exercise, the retarding force impacts the position of the weight on a spring, modeled by \( y = 2 e^{-t} \cos t \), indicating that the further the mass moves in time, the more the force reduces its speed.
This function evolves over time as the exponential factor \( e^{-t} \) continuously decreases, showing how the retarding force weakens the motion in this system. Such a force ensures that motion becomes smoother and less abrupt, which can be crucial for systems requiring stability.

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Most popular questions from this chapter

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{\ln 2} x^{-2} e^{-1 / x} d x$$

About the infinite region in the first quadrant between the curve \(y=e^{-x}\) and the \(x\) -axis. Find the centroid of the region.

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Verify that the functions in Exercises \(11-16\) are probability density functions for a continuous random variable \(X\) over the given interval. Determine the specified probability. $$ f(x)=\sin x \text { over }[0, \pi / 2], P\left(\frac{\pi}{6} < X \leq \frac{\pi}{4}\right) $$
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