Problem 69 - Calculus Early Transcendentals: Pearson New International Edition

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

Answer

(a) The coordinates of \(P\) at time \(t\) are \((10\cos(8\pi t), 10\sin(8\pi t))\). (b) At \(t=1\), \(P\) is rising at a rate of \(80\pi\) cm/s.

Step 1: Understand the Rotation

The wheel's radius is 10 cm, and it rotates counterclockwise. One revolution corresponds to a full circle of \(2\pi\) radians. Given 4 revolutions per second, the angular velocity \(\omega\) is \(4 \times 2\pi = 8\pi\) radians per second.

Step 2: Find Position of Point at Time t

The point \(P\) starts at \((10,0)\), which corresponds to an angle of 0 radians. At time \(t\), the angle \(\theta = 8\pi t\). The position coordinates \((x, y)\) of \(P\) can be determined by the parametric equations: \(x = 10\cos(8\pi t)\) and \(y = 10\sin(8\pi t)\).

Step 3: Determine Position at t=1

Substitute \(t = 1\) into the parametric equations: \(x = 10\cos(8\pi \times 1) = 10\cos(8\pi) = 10\) (since\(\cos(8\pi) = 1\)) and \(y = 10\sin(8\pi \times 1) = 10\sin(8\pi) = 0 \) (since \(\sin(8\pi) = 0\)). Therefore, the coordinates of \(P\) at \(t=1\) are \((10, 0)\).

Step 4: Determine Rate of Rising/Falling Using Derivative

The rate of rising (vertical speed) can be found by differentiating the \(y\) position with respect to \(t\). \(y = 10\sin(8\pi t)\), so \(\frac{dy}{dt} = 10 \cdot 8\pi \cos(8\pi t) = 80\pi \cos(8\pi t)\).

Step 5: Calculate Rate at t=1

Substituting \(t = 1\) into \(\frac{dy}{dt} = 80\pi \cos(8\pi t)\), we get \(\frac{dy}{dt} = 80\pi \cos(8\pi \times 1) = 80\pi \times 1 = 80\pi\). Thus, the rate at which \(P\) is rising at \(t = 1\) is \(80\pi \) cm/second.

Key Concepts

Angular Velocity

Angular velocity is a key concept when dealing with rotational motion. It represents the rate of rotation and is often measured in radians per second. For any object moving along a circular path, the angular velocity can be found by considering how many radians the object moves through in a given time. In the case of a wheel, the angular velocity tells us how quickly the wheel is spinning.

• A full circle has an angle of \(2\pi\) radians.
• If a wheel completes four full revolutions in one second, this corresponds to \(4 \times 2\pi = 8\pi\) radians per second.
• Angular velocity, denoted by the symbol \(\omega\), can be calculated using the formula \(\omega = \text{revolutions per second} \times 2\pi\).

Angular velocity helps us understand how fast each point on the wheel is moving. It converts rotational speed into a form that is easier to work with in mathematical calculations.

Parametric Equations

Parametric equations are employed to define a set of related quantities as functions of an independent variable, such as time \(t\). These are especially useful for describing motion along curved paths, like circular motions. Instead of expressing \(y\) directly in terms of \(x\), each coordinate is expressed as a separate equation in terms of a parameter.

• The equations for the coordinates of point \(P\) on the wheel are \(x = 10\cos(8\pi t)\) and \(y = 10\sin(8\pi t)\).
• Here, the parameter \(t\) represents time.
• Such parametric equations capture both the circular path and the changing position of point \(P\).

This concept helps locate different positions of the point as time increases, showing exactly how it moves around the circle.

Differentiation

Differentiation is a fundamental concept in calculus used to determine how a function changes as its input changes. In the context of trigonometric functions and motion, it can be extremely useful for finding rates of change.

• For a point moving in a circular path, differentiation helps calculate the speed in the vertical (or horizontal) direction.
• The vertical speed of point \(P\) at time \(t\) is given by the derivative of the \(y\) position: \( y = 10\sin(8\pi t)\).
• Computing the derivative, we find \( \frac{dy}{dt} = 80\pi \cos(8\pi t) \).

Here, this derivative shows how quickly the point is rising or falling as the wheel spins, which provides valuable insight into the nature of the motion.

Circular Motion

Circular motion refers to the motion of an object along the circumference of a circle. It is characterized by key factors such as radius, rotational speed, and dimensions of motion.

• A point in circular motion traverses a fixed path of constant radius.
• In this exercise, the radius of the circle is 10 cm.
• The point \(P\) on the wheel starts at position \((10, 0)\) and moves according to the parametric equations derived earlier.

Understanding circular motion allows us to describe the trajectory of the point in terms of its position and speed at any given time. This is critical for solving problems involving rotational dynamics and analyzing the behavior of objects moving in circular paths.