Question

Multiple Choice A particle is moving around the unit circle (the circle of radius 1 centered at the origin). At the point \((0.6,\) 0.8\()\) the particle has horizontal velocity \(d x / d t=3 .\) What is its vertical velocity \(d y / d t\) at that point? \(\begin{array}{lllll}{\text { (A) }-3.875} & {\text { (B) }-3.75} & {\text { (C) }-2.25} & {\text { (D) } 3.75} & {\text { (E) } 3.875}\end{array}\)

Short answer

The vertical velocity of the particle at the given point is -2.25

Step-by-step solution

Identifying Given Information

The particle is located at the point (0.6, 0.8) on the unit circle, and its horizontal velocity is \(d x / d t=3\). We want to find its vertical velocity \(d y / d t\).

Equation of Motion

In terms of motion on a circle, the position of the particle can be denoted as \((x, y)= (\cos(\theta), \sin(\theta))\), where \(\theta\) is the angle the position vector makes with the +x axis. A unit circle satisfies the equation \(x^{2} + y^{2} = 1\). Differentiating this with respect to time \(t\) gives \(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\). Simplifying this, we get, \(\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}\).

Substituting the Given Values

Substituting \(x = 0.6\), \(y = 0.8\), and \(\frac{dx}{dt} = 3\) into the equation from Step 2 gives \(\frac{dy}{dt} = -\frac{0.6}{0.8} \times 3 = -2.25\) . This is the vertical velocity of the particle at the given point.

Key concepts

Particle Motion

Understanding the concept of particle motion is crucial when examining how particles move in space over time. Particle motion often involves multiple types of motion - like linear, circular, or oscillatory motion. Specifically, when a particle travels around the perimeter of a shape, in this case, a circle, we can describe its position using coordinates. As the particle moves, these coordinates change, and the rate at which they change with respect to time is known as the particle's velocity.

For any point on a unit circle, the particle's movement can be understood in terms of both horizontal and vertical velocities, which correspond to the rate of change of the x and y coordinates, respectively. Importantly, these velocities are not independent but are related due to the constraints of moving along a circle, a relationship that becomes clear through differential calculus.

Differential Calculus

Differential calculus is a powerful mathematical tool that allows us to analyze the rates at which quantities change. Essential in studying motion, it provides a way to calculate velocities as derivatives of position with respect to time. For a particle moving along the unit circle, each coordinate (x,y) is a function of time, thus the horizontal velocity is given by the derivative \(dx/dt\) and the vertical velocity by \(dy/dt\).

In the context of circular motion on a unit circle, we often use the Pythagorean identity \(x^2 + y^2 = 1\) to express the relationship between x and y, as both must satisfy the circle's equation. By taking the derivative of both sides of this equation with respect to time, we involve differential calculus, which gives us a method to relate the horizontal and vertical velocities of the particle.

Unit Circle Equations

The unit circle is a fundamental concept in trigonometry, defined as a circle of radius 1 with its center at the origin of a coordinate system. The equation \(x^2 + y^2 = 1\) represents this circle and any point on it can be expressed as \(x, y) = (\cos(\theta), \sin(\theta))\), where \(\theta\) is the angle of the radius with the positive x-axis.

When considering particle motion on a unit circle, we use these equations to link angles with coordinates. These unit circle equations are essential when we need to transition from linear representations of motion to circular ones, as they provide a bridge between the angle \(\theta\) and the coordinates (x, y). Through these links, complexities like calculating velocities become more manageable.

Motion on a Circle

Motion on a circle, especially a unit circle, is distinctly characterized by the constant radius and the angular movement of the particle. This motion requires us to think about both linear distance traveled along the perimeter and angular displacement, measured in radians or degrees. When analyzing such motion, we can use the relationship between linear and angular velocity, given by \(v = r\omega\), where \(v\) is linear velocity, \(r\) is the radius (1 for the unit circle), and \(\omega\) is angular velocity.

Applying differential calculus to the unit circle's equations allows us to find expressions for both horizontal and vertical velocities. As seen in the exercise, we found the vertical velocity \(dy/dt\) using the given horizontal velocity \(dx/dt\) and the coordinates of the particle on the unit circle. This illustrates how concepts such as particle motion, calculus, and unit circle equations work together to solve problems involving motion on a circle.