/*! 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 30 At time \(t\) seconds, the cente... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 30

At time \(t\) seconds, the center of a bobbing cork is \(3 \sin 2 t\) centimeters above (or below) water level. What is the velocity of the cork at \(t=0, \pi / 2, \pi ?\)

Short Answer

Expert verified
The velocities are 6 cm/s at \( t=0 \), -6 cm/s at \( t=\frac{\pi}{2} \), and 6 cm/s at \( t=\pi \).

Step by step solution

01

Identify the function

The height of the cork above (or below) water level is given by the function \( h(t) = 3 \sin 2t \). This is a sinusoidal function representing the motion of the cork.
02

Differentiate the function

To find the velocity, differentiate the height function \( h(t) \) with respect to time \( t \). The derivative represents the velocity of the cork:\[ h'(t) = \frac{d}{dt}(3 \sin 2t) = 3 \cdot 2 \cos 2t = 6 \cos 2t.\]
03

Evaluate the velocity at \( t = 0 \)

Substitute \( t = 0 \) into the derivative to find the velocity at this time:\[ h'(0) = 6 \cos 2 \times 0 = 6 \cos 0 = 6 \times 1 = 6.\] The velocity of the cork at \( t = 0 \) seconds is 6 cm/s.
04

Evaluate the velocity at \( t = \frac{\pi}{2} \)

Substitute \( t = \frac{\pi}{2} \) into the derivative to find the velocity:\[ h'\left(\frac{\pi}{2}\right) = 6 \cos \left(2 \times \frac{\pi}{2}\right) = 6 \cos \pi = 6 \times (-1) = -6.\] The velocity at \( t = \frac{\pi}{2} \) seconds is -6 cm/s.
05

Evaluate the velocity at \( t = \pi \)

Substitute \( t = \pi \) into the derivative to find the velocity:\[ h'(\pi) = 6 \cos (2 \times \pi) = 6 \cos (2\pi) = 6 \times 1 = 6.\] The velocity at \( t = \pi \) seconds is 6 cm/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Trigonometric Functions
Trigonometric functions, like sine and cosine, are fundamental in understanding wave-like patterns and oscillations. In this exercise, the function representing the cork's motion is sinusoidal: \( h(t) = 3 \sin 2t \). Here, the sine function illustrates how the cork moves up and down over time.
  • **Amplitude**: This is the maximum distance the cork moves from its resting point. In the expression \( 3 \sin 2t \), the amplitude is 3 cm. This means the cork moves 3 cm above or below the water level.
  • **Frequency**: This concerns how many oscillations happen in a given time, influenced by the coefficient inside the sine function. For \( \sin 2t \), the frequency is doubled compared to just \( \sin t \), which means the motion completes its cycle twice as fast.
  • **Period**: The period is the time taken to complete one full cycle of motion. For \( \sin 2t \), the period is \( \frac{2\pi}{2} = \pi \).
Understanding these parameters helps in analyzing how the cork oscillates with regular wave patterns.
Velocity Calculation
When studying the movement of objects, velocity is a critical concept. Velocity refers to the rate of change of the object's position with time. In this context, we've determined the velocity of the cork by differentiating its height function:
The velocity function derived from the height is \( h'(t) = 6 \cos 2t \). Evaluating this derivative at specific times gives the velocity at those moments:
  • At \( t = 0 \): Substitute \( 0 \) into \( h'(t) \) to get \( h'(0) = 6 \cos 0 = 6 \). The velocity is 6 cm/s upward.
  • At \( t = \frac{\pi}{2} \): Substitute \( \frac{\pi}{2} \) to obtain \( h'(\frac{\pi}{2}) = 6 \cos \pi = -6 \). The velocity is 6 cm/s downward, indicating a phase where the cork moves down.
  • At \( t = \pi \): Substitute \( \pi \) to find \( h'(\pi) = 6 \cos 2\pi = 6 \). Again the velocity is 6 cm/s upward, showing a recurring cycle of motion.
These calculations demonstrate how velocity changes at different points in a periodic cycle.
Differentiation Techniques
Differentiation is a mathematical technique used to determine the rate at which a quantity changes. Calculating derivatives is crucial for finding velocities from position functions, as seen in this exercise.
To differentiate the function \( h(t) = 3 \sin 2t \), apply the chain rule, a common differentiation technique:
  • The **chain rule** is used when differentiating composed functions, like sine with an inner function of time \( 2t \).
  • Start by differentiating the outer function, sine, with respect to the angle, yielding \( \cos \, \text{(angle)} \).
  • Then, multiply by the derivative of the inner function \( 2t \), which is 2, to get \( 6 \cos 2t \). This step is where the chain rule is crucial as it accounts for changes in the inner function.
Understanding differentiation and the proper use of techniques like the chain rule help in analyzing motion and change in various contexts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). \(G(x)=\frac{2 x-1}{x-4}\)

Find \(\Delta y\) for the given values of \(x_{1}\) and \(x_{2}\). $$ y=\frac{2}{x+1}, x_{1}=0, x_{2}=0.1 $$

Find \(\Delta y\) for the given values of \(x_{1}\) and \(x_{2}\). $$ y=\frac{1}{x}, x_{1}=1.0, x_{2}=1.2 $$

Find all points on the graph of \(y=9 \sin x \cos x\) where the tangent line is horizontal.

The rate of change of velocity with respect to time is called acceleration. Suppose that the velocity at time \(t\) of a particle is given by \(v(t)=2 t^{2} .\) Find the instantaneous acceleration when \(t=1\) second.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.