Problem 69
Question
A wheel centered at the origin and of radius 10 centimeters is rotating counterclockwise at a rate of 4 revolutions per second. A point \(P\) on the rim is at \((10,0)\) at \(t=0\). (a) What are the coordinates of \(P\) at time \(t\) seconds? (b) At what rate is \(P\) rising (or falling) at time \(t=1\) ?
Step-by-Step Solution
Verified Answer
(a) Coordinates: \((10 \cos(8\pi t), 10 \sin(8\pi t))\); (b) Rising at \(80\pi\) cm/s at \(t=1\).
1Step 1: Determine Angular Velocity
First, calculate the angular velocity \( \omega \) in radians per second. Since the wheel makes 4 revolutions per second, and each revolution is \( 2\pi \) radians, the angular velocity is \( \omega = 4 \times 2\pi = 8\pi \; \text{rad/s}.\)
2Step 2: Parameterize the Point P
Using the angle \( \theta = \omega t \), the coordinates of point \( P \) at time \( t \) are given by \( (x, y) = (10 \cos(\omega t), 10 \sin(\omega t)) \). Substitute the angular velocity to get \( (x, y) = (10 \cos(8\pi t), 10 \sin(8\pi t)) \).
3Step 3: Evaluate the Coordinates at Time t
For any given time \( t \), substitute into the parametric equations: \( x(t) = 10 \cos(8\pi t) \) and \( y(t) = 10 \sin(8\pi t) \) to find the coordinates of \( P \).
4Step 4a: Determine the Rate of Change of y (Step a)
Differentiate the \( y \)-coordinate function to find \( \frac{dy}{dt} \). Since \( y(t) = 10 \sin(8\pi t) \), the derivative is \( \frac{dy}{dt} = 10 \cdot 8\pi \cos(8\pi t) = 80\pi \cos(8\pi t) \).
5Step 4b: Evaluate the Rate of Change at t=1 (Step b)
Substitute \( t = 1 \) into the expression for \( \frac{dy}{dt} \) to find \( \frac{dy}{dt} = 80\pi \cos(8\pi \times 1) = 80\pi \cos(8\pi) = 80\pi \cdot 1 = 80\pi. \) Therefore, at \( t = 1 \), point \( P \) is rising at a rate of \( 80\pi \; \text{cm/s}.\)
Key Concepts
Understanding Angular VelocityExploring Trigonometric Functions in Parametric EquationsCalculating Derivatives for Rate of ChangeAnalyzing Circular Motion
Understanding Angular Velocity
Angular velocity is a measure of how fast an object rotates around a central point. In our exercise, the wheel rotates at 4 revolutions per second.
Each revolution is equivalent to a circle, which contains \(2\pi\) radians. Therefore, to convert the revolutions per second into an angular measurement, we multiply by \(2\pi\).
Consequently, the angular velocity \(\omega\) becomes \(8\pi\; \text{rad/s}\).
Understanding angular velocity helps us know how the position of a point on a rotating object changes over time. As it describes the rotation rate, it aids in connecting time with angles covered in the circular motion.
Each revolution is equivalent to a circle, which contains \(2\pi\) radians. Therefore, to convert the revolutions per second into an angular measurement, we multiply by \(2\pi\).
Consequently, the angular velocity \(\omega\) becomes \(8\pi\; \text{rad/s}\).
Understanding angular velocity helps us know how the position of a point on a rotating object changes over time. As it describes the rotation rate, it aids in connecting time with angles covered in the circular motion.
Exploring Trigonometric Functions in Parametric Equations
Trigonometric functions, sine and cosine, play an essential role when dealing with circular motion.
They allow us to express the position of a point moving around a circle using parametric equations.
In our problem, these trigonometric functions help describe the path of point \(P\).
Sine and cosine functions oscillate between -1 and 1, mirroring the oscillatory nature of circular paths.
With the circle centered at the origin, and given its radius is 10, the coordinates can be parametrized as \(x = 10 \cos(8\pi t)\) and \(y = 10 \sin(8\pi t)\).
This formulation gives us a clear understanding of how \(P\) moves over time, covering the circular trajectory as described by \(t\).
Using these, we can track the precise position of \(P\) for any instance, \(t\), making trigonometric functions indispensable in this context.
They allow us to express the position of a point moving around a circle using parametric equations.
In our problem, these trigonometric functions help describe the path of point \(P\).
Sine and cosine functions oscillate between -1 and 1, mirroring the oscillatory nature of circular paths.
With the circle centered at the origin, and given its radius is 10, the coordinates can be parametrized as \(x = 10 \cos(8\pi t)\) and \(y = 10 \sin(8\pi t)\).
This formulation gives us a clear understanding of how \(P\) moves over time, covering the circular trajectory as described by \(t\).
Using these, we can track the precise position of \(P\) for any instance, \(t\), making trigonometric functions indispensable in this context.
Calculating Derivatives for Rate of Change
Derivatives are a fundamental concept in calculus. They help us determine the rate of change of a function.
In this exercise, we are interested in how quickly the \(y\)-coordinate of point \(P\) changes over time.
This is calculated by taking the derivative of \(y(t) = 10 \sin(8\pi t)\). Differentiating gives us \(\frac{dy}{dt} = 80\pi \cos(8\pi t)\).
This expression tells us the rate at which \(y\) is rising or falling. The derivative essentially provides the slope of the tangent at any point on the curve described by the parametric equation.
Evaluating it at different values of \(t\) gives insights into how fast \(P\) is moving vertically on the circle.
In this exercise, we are interested in how quickly the \(y\)-coordinate of point \(P\) changes over time.
This is calculated by taking the derivative of \(y(t) = 10 \sin(8\pi t)\). Differentiating gives us \(\frac{dy}{dt} = 80\pi \cos(8\pi t)\).
This expression tells us the rate at which \(y\) is rising or falling. The derivative essentially provides the slope of the tangent at any point on the curve described by the parametric equation.
Evaluating it at different values of \(t\) gives insights into how fast \(P\) is moving vertically on the circle.
Analyzing Circular Motion
Circular motion refers to the movement of an object around a circular path.
Key to understanding this is knowing its parameters: radius, speed, and time—elements found in our exercise.
The point \(P\) on the wheel revolves counterclockwise, which is an essential directionality component affecting calculations.
The parameters translate into a parametric setup where both the sine and cosine functions define position relative to the circle's center.
This specific motion requires a keen understanding of angles and radians as they relate to time, expressing how point \(P\) locates itself on this orbit.
Parametrizing through angular velocity and trigonometric functions provides a robust model for such movements, allowing predictions of future locations such as at \(t=1\), and movements between positions on the circular path.
Key to understanding this is knowing its parameters: radius, speed, and time—elements found in our exercise.
The point \(P\) on the wheel revolves counterclockwise, which is an essential directionality component affecting calculations.
The parameters translate into a parametric setup where both the sine and cosine functions define position relative to the circle's center.
This specific motion requires a keen understanding of angles and radians as they relate to time, expressing how point \(P\) locates itself on this orbit.
Parametrizing through angular velocity and trigonometric functions provides a robust model for such movements, allowing predictions of future locations such as at \(t=1\), and movements between positions on the circular path.
Other exercises in this chapter
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