/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 A particle moves along the \(x\)... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 88

A particle moves along the \(x\) -axis with velocity \(d x / d t=f(x) .\) Show that the particle's acceleration is \(f(x) f^{\prime}(x)\)

Short Answer

Expert verified
Acceleration is given by \( f(x) f'(x) \).

Step by step solution

01

Understand the Given Conditions

We are given that the velocity of the particle is denoted as \( \frac{dx}{dt} = f(x) \). The objective is to find the acceleration of the particle, which is the derivative of velocity with respect to time, \( \frac{d^2x}{dt^2} \).
02

Differentiate Velocity with Respect to Time

To find acceleration, differentiate the velocity equation \( \frac{dx}{dt} = f(x) \) with respect to time \( t \). Using the chain rule, we have:\[ \frac{d}{dt} \left( \frac{dx}{dt} \right) = \frac{df(x)}{dt} = \frac{df}{dx} \cdot \frac{dx}{dt} \]
03

Substitute Known Values

We know that \( \frac{df}{dx} = f'(x) \) and \( \frac{dx}{dt} = f(x) \) from the given conditions. Substitute these into the expression from Step 2:\[ \frac{df(x)}{dt} = f'(x) \cdot f(x) \]
04

Conclude the Acceleration Formula

From the previous steps, we write the acceleration as:\[ \frac{d^2x}{dt^2} = f'(x) \cdot f(x) \]This shows the acceleration of the particle is \( f(x) f'(x) \).

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Key Concepts

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

Particle Motion
When studying particle motion in calculus, the position, velocity, and acceleration of a particle are key concepts that help us understand its movement. Motion in one-dimensional space, like along the x-axis, often involves these quantities.
  • **Position** (**x**): Indicates where the particle is located in space.
  • **Velocity** (**\(\frac{dx}{dt}\)**): Describes how fast and in what direction the particle鈥檚 position is changing over time.
  • **Acceleration** (**\(\frac{d^2x}{dt^2}\)**): The rate of change of velocity, providing insights into how the particle's speed and direction are changing.
In the provided exercise, a particle moves along the x-axis with its velocity given by the function \(f(x)\). By understanding how velocity changes, we gain insights into the particle's acceleration, illustrating the interplay between these concepts in motion.
Differentiation
Differentiation is a fundamental mathematical process in calculus that allows us to compute the rate of change of a function. It's essential for analyzing particle motion because it gives us velocity and acceleration as derivatives of the position function.
  • To find **velocity**, differentiate the position function \(x(t)\) with respect to time \(t\): **\(\frac{dx}{dt}\)**.
  • To find **acceleration**, take the derivative of velocity with respect to \(t\): **\(\frac{d^2x}{dt^2}\)**.
Differentiation enables us to derive the relationships between position, velocity, and acceleration. In this exercise, differentiation helps us transition from knowing the velocity function \(f(x)\) to discovering the acceleration \(f(x)f'(x)\). Understanding this process is pivotal when moving from descriptive functions to interpreting meaningful physical quantities.
Chain Rule
The chain rule is a crucial tool in differentiation, especially when dealing with functions of other functions. It helps us differentiate composite functions, where one function is nested inside another.If you have a composite function \(g(f(x))\), the chain rule gives us the derivative as:\[\frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x)\]In the context of particle motion, the chain rule is used to differentiate the velocity function \(f(x)\) with respect to time:\[\frac{d}{dt}[f(x)] = \frac{df}{dx} \cdot \frac{dx}{dt}\]By applying this rule, we obtain the acceleration鈥� proving how crucial the chain rule is in calculus. It allows us to handle complexities of changing rates in a structured way, connecting multiple layers of differentiation to illuminate real-world phenomena.

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Most popular questions from this chapter

On a morning of a day when the sun will pass directly overhead, the shadow of an \(24 \mathrm{m}\) building on level ground is \(18 \mathrm{m}\) long. At the moment in question, the angle \(\theta\) the sun makes with the ground is increasing at the rate of \(0.27^{\circ} / \mathrm{min} . \mathrm{At}\) what rate is the shadow decreasing? (Remember to use radians. Express your answer in centimeters per minute, to the nearest tenth.)

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? \(s=60 t-4.9 t^{2}, \quad 0 \leq t \leq 12.5\) (a heavy object fired straight up from Earth's surface at \(60 \mathrm{m} / \mathrm{s}\) )

a. Find the tangent to the curve \(y=2 \tan (\pi x / 4)\) at \(x=1\) b. What is the smallest value the slope of the curve can ever have on the interval \(-2

Assume that a particle's position on the \(x\) -axis is given by $$x=3 \cos t+4 \sin t$$ where \(x\) is measured in meters and \(t\) is measured in seconds. a. Find the particle's position when \(t=0, t=\pi / 2,\) and \(t=\pi\) b. Find the particle's velocity when \(t=0, t=\pi / 2,\) and \(t=\pi\)

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=u^{5}+1, \quad u=g(x)=\sqrt{x}, \quad x=1$$
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