Problem 21 - Calculus

Question

Find the speed \(d s / d t\) on the circle \(x=2 \cos 3 t, y=2 \sin 3 t\).

Step-by-Step Solution

Answer

The speed is 6.

Step 1: Identify the Parametric Equations

The parametric equations for the motion along a circle are given as \( x = 2 \cos 3t \) and \( y = 2 \sin 3t \). These equations represent the x and y coordinates in terms of the parameter \( t \), which usually represents time.

Step 2: Differentiate the Parametric Equations

To find the speed, we first need to find the derivatives of the parametric equations with respect to time \( t \). Compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):\[\frac{dx}{dt} = \frac{d}{dt}(2 \cos 3t) = -6 \sin 3t\]\[\frac{dy}{dt} = \frac{d}{dt}(2 \sin 3t) = 6 \cos 3t\]

Step 3: Use the Formula for Speed

The speed of a particle moving along a path defined by parametric equations is found using the formula:\[s = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\]Substitute the derivatives found in Step 2:\[s = \sqrt{(-6 \sin 3t)^2 + (6 \cos 3t)^2}\]

Step 4: Simplify the Expression

Simplify the expression under the square root:\[s = \sqrt{36 \sin^2 3t + 36 \cos^2 3t}\]Notice that \( \sin^2 \theta + \cos^2 \theta = 1 \), which simplifies the expression to:\[s = \sqrt{36(\sin^2 3t + \cos^2 3t)} = \sqrt{36}\]Thus, \( s = 6 \).

Step 5: Conclusion: Speed of the Particle

The speed of the particle moving along the circle described by the parametric equations is constant and equal to 6.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. When dealing with parametric equations like those describing motion, derivatives help us understand how the position of an object changes over time.

Let's take the equations given:

• For the x-coordinate: \( x = 2 \, \cos \, 3t \)
• For the y-coordinate: \( y = 2 \, \sin \, 3t \)

To find the velocity components, we derive these with respect to time \( t \):

• \( \frac{dx}{dt} = -6 \, \sin \, 3t \)
• \( \frac{dy}{dt} = 6 \, \cos \, 3t \)

These derivatives give the rate at which the x and y positions change, respectively, which directly informs our calculation of speed.

Speed Calculation

Speed in parametric motion is derived from the velocities in the x and y directions. By combining these into a single expression, we accurately represent how fast an object is moving along its path.

The formula for speed \( s \) is:\[s = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\]This formula consolidates the derivatives found earlier into the geographical notion of speed, combining motion in both the x and y directions into one value. Plugging in:

• \( \left(\frac{dx}{dt}\right)^2 = (-6 \sin 3t)^2 = 36 \sin^2 3t \)
• \( \left(\frac{dy}{dt}\right)^2 = (6 \cos 3t)^2 = 36 \cos^2 3t \)

Summing these gives:\[s = \sqrt{36(\sin^2 3t + \cos^2 3t)} = \sqrt{36}\]Simplifying, we find \( s = 6 \), showing a constant speed.

Trigonometric Identities

Trigonometric identities are vital tools in simplifying expressions, especially when dealing with functions involving sine and cosine. They let us transform complex expressions into simpler ones.

One of the most useful identities is:\[\sin^2 \theta + \cos^2 \theta = 1\]In our speed calculation, we make use of this identity to simplify the under-root expression.

• Originally: \( 36 \sin^2 3t + 36 \cos^2 3t \)
• Apply the identity: \( 36(\sin^2 3t + \cos^2 3t) = 36 \)

Thanks to this trigonometric identity, our expression becomes straightforward to compute, directly leading to \( \sqrt{36} = 6 \). This efficiency shows why trig identities are indispensable in calculus.

Particle Motion

The concept of particle motion revolves around how an object travels through space over time under various forces or constraints. Parametric equations provide a versatile framework to describe such complex movements, as they allow each coordinate to be represented as a function of time.

In the original exercise, the motion of the particle is confined to a circle. The parametric equations \( x = 2 \cos 3t \) and \( y = 2 \sin 3t \) describe this circular path with radius 2. With a parameter such as \( t \) describing time, the position changes smoothly in space, modeling real-world dynamics.

The constant speed obtained from our calculations indicates uniform circular motion. This constant speed means the particle travels equal distances in equal time intervals, a fundamental feature of circular motion. Understanding these concepts can be applied to physics, engineering, and more, providing insight into how objects move in real-world scenarios.