Question

sketch the curve in the xy-plane. Then, for the given point, find the curvature and the radius of curvature. Finally, $$ y=\cos 2 x,\left(\frac{1}{6} \pi, \frac{1}{2}\right) $$

Short answer

Curvature is \( \frac{1}{4} \), radius of curvature is 4.

Step-by-step solution

Understand the Function and Point

We are given the function \( y = \cos(2x) \) and a specific point \( \left(\frac{1}{6}\pi, \frac{1}{2}\right) \). Our goal is to sketch the graph of the function around this point, compute its curvature, and find the radius of curvature.

Sketch the Function

Plot the function \( y = \cos(2x) \) in the xy-plane. This is a cosine wave with a frequency of 2, meaning it completes a full cycle in \( \pi \) rather than in \( 2\pi \). It oscillates between -1 and 1. Sketch the curve, highlighting the point \( \left(\frac{1}{6}\pi, \frac{1}{2}\right) \).

Find the First Derivative

To find the curvature, we first need the first derivative of \( y = \cos(2x) \). Use the chain rule: \( \frac{dy}{dx} = -2\sin(2x) \).

Find the Second Derivative

Next, find the second derivative by differentiating again: \( \frac{d^2y}{dx^2} = -4\cos(2x) \).

Evaluate Derivatives at Given Point

Substitute \( x = \frac{1}{6}\pi \) into the derivatives: \( \frac{dy}{dx} = -2\sin\left(\frac{1}{3}\pi\right) = -\sqrt{3} \), and \( \frac{d^2y}{dx^2} = -4\cos\left(\frac{1}{3}\pi\right) = -4 \times \frac{1}{2} = -2 \).

Calculate the Curvature

The formula for curvature \( \kappa \) is \( \kappa = \frac{|y''|}{(1+(y')^2)^{3/2}} \). Plug in the values: \( \kappa = \frac{|-2|}{(1+(-\sqrt{3})^2)^{3/2}} = \frac{2}{(1+3)^{3/2}} = \frac{2}{8} = \frac{1}{4} \).

Determine the Radius of Curvature

Radius of curvature \( R \) is the reciprocal of curvature: \( R = \frac{1}{\kappa} = \frac{1}{1/4} = 4 \).

Key concepts

Cosine Function

The cosine function, denoted as \( \cos(x) \), is a fundamental trigonometric function that is periodic, oscillating between -1 and 1. In mathematics, a cosine function describes a smooth repetitive oscillation. The function \( y = \cos(2x) \) has a frequency of 2, meaning that its cycle is completed twice as fast compared to the standard cosine function \( y = \cos(x) \), which has a period of \( 2\pi \). The period of \( \cos(2x) \) is reduced to \( \pi \), scaling the wave horizontally.
This property is crucial when sketching the function, as it indicates how often the wave will repeat within a given space. In addition to understanding the basic periodic nature of cosine, it’s important to note its amplitude (which remains 1) and how it affects points on the graph. For example, the given point \( \left(\frac{1}{6}\pi, \frac{1}{2}\right) \) requires us to determine where in the cycle the cosine wave will be at an x-value of \( \frac{1}{6}\pi \). This aids in accurately sketching and analyzing the behavior of the function in problem-solving.

Differentiation

Differentiation is the process of finding the derivative of a function, which quantifies the rate at which a function changes at any point. For trigonometric functions like the cosine function, differentiation involves applying rules such as the product, quotient, or chain rule. In our exercise, we differentiate the function \( y = \cos(2x) \) using the chain rule.
Differentiation Step-by-Step:

• The chain rule helps in finding the derivative of a composite function. Here, the outer function is cosine and the inner function is \( 2x \).
• The first derivative is \( \frac{dy}{dx} = -2\sin(2x) \). This suggests how the slope of the cosine wave advances as x changes.
• To find the second derivative, differentiate again, yielding \( \frac{d^2y}{dx^2} = -4\cos(2x) \). This tells us about the concavity or curvature of the function at any given point.

The process of finding and interpreting these derivatives is vital for understanding the motion and shape of the trigonometric function's graph, as well as solving for curvature.

Radius of Curvature

In calculus, the radius of curvature refers to the radius of the 'best-fit' circle to a curve at a particular point, providing insights into how curved the path is at that locale. For a function \( y = f(x) \), it is inversely related to curvature. To derive it, we use the circular formula for curvature: \( \kappa = \frac{|y''|}{(1+(y')^2)^{3/2}} \).
Through computations, as performed in our exercise:

• Evaluate both the first and second derivatives at the given point. For example, at \( x = \frac{1}{6}\pi \), we calculated \( y'' = -2 \) and \( y' = -\sqrt{3} \).
• Substitute these values to determine \( \kappa \), resulting in \( \frac{1}{4} \).
• The radius of curvature \( R \) is simply the reciprocal, \( R = \frac{1}{\kappa} = 4 \).

This metric is essential when considering applications like motion in physics, where the path of motion is curved, showing the tightness or looseness of the curve at any given segment.

Chain Rule

The chain rule is a vital differentiation technique in calculus used to find the derivative of composite functions. It simplifies the process of differentiating functions that are nested within each other, such as those involving trigonometric functions. Let's break down its utility using our function \( y = \cos(2x) \).
Implementing the Chain Rule includes:

• Recognize the outer and inner functions. For \( \cos(2x) \), \( \cos(u) \) is the outer, and \( u = 2x \) is the inner function.
• Differentiate the outer function \( \cos(u) \) yielding \( -\sin(u) \).
• Differentiate the inner function \( 2x \) to get \( 2 \).
• Combine these using the chain rule: multiply the derivative of the outer function by the derivative of the inner function, resulting in the first derivative \( \frac{dy}{dx} = -2\sin(2x) \).

Understanding how to use the chain rule is critical, as it allows us to handle more complex functions by breaking them down into simpler, manageable parts for differentiation.