Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

A point moves around the circle \(x^{2}+y^{2}=25\) at constant angular speed of 6 radians per second starting at \((5,0)\). Find expressions for \(\mathbf{r}(t), \mathbf{v}(t),\|\mathbf{v}(t)\|\), and $\mathbf{a}(t)

Step-by-Step Solution

Answer

\(\mathbf{r}(t) = 5 \cos(6t) \mathbf{i} + 5 \sin(6t) \mathbf{j}\), \(\mathbf{v}(t) = -30 \sin(6t) \mathbf{i} + 30 \cos(6t) \mathbf{j}\), \(\| \mathbf{v}(t) \| = 30\), \(\mathbf{a}(t) = -180 \cos(6t) \mathbf{i} - 180 \sin(6t) \mathbf{j}\).

Step 1: Understanding the Circle Equation

The equation of the circle given is \(x^2 + y^2 = 25\). This represents a circle centered at the origin (0,0) with a radius of 5.

Step 2: Define the Position Vector

The position vector for a point on the circle can be written in terms of a parameter \(t\) (time) as \(\mathbf{r}(t) = 5 \cos(\theta(t)) \mathbf{i} + 5 \sin(\theta(t)) \mathbf{j}\), where \(\theta(t)\) is the angle of rotation. Since the angular speed is 6 radians per second, \(\theta(t) = 6t\). Thus, the position vector is \(\mathbf{r}(t) = 5 \cos(6t) \mathbf{i} + 5 \sin(6t) \mathbf{j}\).

Step 3: Calculate the Velocity Vector

The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\). Thus, \(\mathbf{v}(t) = \frac{d}{dt}(5 \cos(6t) \mathbf{i} + 5 \sin(6t) \mathbf{j}) = -30 \sin(6t) \mathbf{i} + 30 \cos(6t) \mathbf{j}\).

Step 4: Compute the Magnitude of Velocity

The magnitude of the velocity vector \(\| \mathbf{v}(t) \|\) is the square root of the sum of its components squared: \(\| \mathbf{v}(t) \| = \sqrt{(-30 \sin(6t))^2 + (30 \cos(6t))^2}\). Simplifying, we get \(\| \mathbf{v}(t) \| = 30\) since \(\sin^2(6t) + \cos^2(6t) = 1\).

Step 5: Find the Acceleration Vector

The acceleration vector \(\mathbf{a}(t)\) is the derivative of the velocity vector \(\mathbf{v}(t)\) with respect to \(t\): \(\mathbf{a}(t) = \frac{d}{dt}(-30 \sin(6t) \mathbf{i} + 30 \cos(6t) \mathbf{j}) = -180 \cos(6t) \mathbf{i} - 180 \sin(6t) \mathbf{j}\).

Key Concepts

Position Vector

In circular motion, the position vector describes the location of a point moving on the circle at any given time. If a point is moving along a circle defined by the equation \(x^2 + y^2 = 25\), we understand the center is at the origin \((0,0)\) with a radius of 5.
To write the position vector \(\mathbf{r}(t)\), it's essential to recognize that the point's position depends on time \(t\) as it moves around the circle. Using angular motion principles, this position can be captured as:

• \(\mathbf{r}(t) = 5 \cos(\theta(t)) \mathbf{i} + 5 \sin(\theta(t)) \mathbf{j}\)

Here, \(\theta(t)\) represents the angular displacement, which helps determine where the point is located at time \(t\). If the angular speed is constant, as it is in this problem (6 radians per second), then \(\theta(t) = 6t\). Substituting this in gives:

• \(\mathbf{r}(t) = 5 \cos(6t) \mathbf{i} + 5 \sin(6t) \mathbf{j}\)

This equation gives us the precise position of the point on the circle at any time \(t\).

Velocity Vector

The velocity vector \(\mathbf{v}(t)\) captures both the speed and direction of the point as it moves along the circle. To find this vector, differentiate the position vector with respect to time \(t\). The process involves some calculus, focusing on the derivative of trigonometric functions:- Original position vector is \(\mathbf{r}(t) = 5 \cos(6t) \mathbf{i} + 5 \sin(6t) \mathbf{j}\)- The differentiation gives:

• \(\mathbf{v}(t) = \frac{d}{dt}(5 \cos(6t) \mathbf{i} + 5 \sin(6t) \mathbf{j})\)
• \(\mathbf{v}(t) = -30 \sin(6t) \mathbf{i} + 30 \cos(6t) \mathbf{j}\)

This vector showcases how the point's velocity changes as it moves circularly. The terms include sine and cosine, mirroring how motion alternates directions depending on the angle \(6t\), resulting in a constant speed as we learn next.

Acceleration Vector

Acceleration vector \(\mathbf{a}(t)\) tells us about the change in velocity over time for a point in motion. Calculating the acceleration vector involves taking the derivative of the velocity vector:\(\mathbf{v}(t) = -30 \sin(6t) \mathbf{i} + 30 \cos(6t) \mathbf{j}\). Let's differentiate it:- Differentiate the given velocity vector:

• \(\mathbf{a}(t) = \frac{d}{dt}(-30 \sin(6t) \mathbf{i} + 30 \cos(6t) \mathbf{j})\)
• \(\mathbf{a}(t) = -180 \cos(6t) \mathbf{i} - 180 \sin(6t) \mathbf{j}\)

The components \(-180 \cos(6t)\) and \(-180 \sin(6t)\) in the vector signify how acceleration influences the path with respect to time. Note this vector is directed towards the center of the circle (central acceleration), indicating uniform circular motion.

Angular Speed

Angular speed, especially in circular motion, is a critical component in determining how fast the point moves around the circle. It measures the rate at which the angle \(\theta(t)\) changes with respect to time. Here, the angular speed is given as 6 radians per second.
Angular speed \(\omega\) is typically denoted as the rate of change of angular displacement, written mathematically as \(\omega = \frac{d\theta}{dt}\). In this case:

• Since \(\theta(t) = 6t\), then \(\frac{d\theta}{dt} = 6\).

This constant indicates the point moves consistently around the circle, contributing to the uniformity in motion. The choice of 6 radians per second ensures that each complete circle (which is 2\(\pi\) radians) takes just over a second to complete. This quick pace necessitates high precision in accompanying vectors like position, velocity, and acceleration.