Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the curve in the xy-plane. Then, for the given point, find the curvature and the radius of curvature. Finally, \(y=\sin x,\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)\)

Step-by-Step Solution

Answer

The curvature is \( \frac{2\sqrt{6}}{9} \), and the radius of curvature is \( \frac{3\sqrt{6}}{4} \).

Step 1: Identify the curve equation

The given curve equation is \( y = \sin x \). Our task is to sketch this curve in the xy-plane, which is a sinusoidal wave.

Step 2: Sketch the curve y=sin x

Draw the curve \( y = \sin x \) on the xy-plane. This is a periodic wave that oscillates between -1 and 1 with a period of \( 2\pi \). At \( x = \frac{\pi}{4} \), \( y = \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \), which matches the given point.

Step 3: Find the first derivative of y

To find the curvature, we first need the first derivative of \( y \). Differentiate \( y = \sin x \) with respect to \( x \), giving \( \frac{dy}{dx} = \cos x \).

Step 4: Find the second derivative of y

Next, find the second derivative by differentiating \( \frac{dy}{dx} = \cos x \). This gives \( \frac{d^2y}{dx^2} = -\sin x \).

Step 5: Evaluate derivatives at x=pi/4

Evaluate both \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) at \( x = \frac{\pi}{4} \). Substitute \( x = \frac{\pi}{4} \) into the derivatives: \( \frac{dy}{dx} = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) and \( \frac{d^2y}{dx^2} = -\sin \left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).

Step 6: Calculate curvature formula

The curvature \( \kappa \) is given by the formula: \[ \kappa = \frac{|y''|}{(1+(y')^2)^{3/2}} \]. Substitute the evaluated derivatives: \( y' = \frac{\sqrt{2}}{2} \) and \( y'' = -\frac{\sqrt{2}}{2} \). Hence, \[ \kappa = \frac{| -\frac{\sqrt{2}}{2} |}{(1 + (\frac{\sqrt{2}}{2})^2)^{3/2}} = \frac{\frac{\sqrt{2}}{2}}{\left( 1 + \frac{1}{2} \right)^{3/2}} = \frac{\frac{\sqrt{2}}{2}}{ \frac{3\sqrt{3}}{4} } = \frac{2\sqrt{2}}{3\sqrt{3}}.\]

Step 7: Simplify curvature expression

Further simplify the curvature \( \kappa = \frac{2\sqrt{2}}{3\sqrt{3}} \) by rationalizing the denominator. Multiply the numerator and denominator by \( \sqrt{3} \) to obtain \( \kappa = \frac{2\sqrt{6}}{9} \).

Step 8: Calculate radius of curvature

The radius of curvature \( R \) is the reciprocal of the curvature, \( R = \frac{1}{\kappa} \). Using \( \kappa = \frac{2\sqrt{6}}{9} \), we find \( R = \frac{9}{2\sqrt{6}} \). Rationalize \( R \) by multiplying by \( \sqrt{6} \), yielding \( R = \frac{9\sqrt{6}}{12} = \frac{3\sqrt{6}}{4} \).

Key Concepts

Calculating Derivatives

Derivatives help us understand how functions change. Calculating derivatives involves finding the rate at which one quantity changes with respect to another. In our exercise, we need the derivatives of the function \( y = \sin x \). To find the curvature, we need both the first and second derivatives.

• The first derivative of \( y = \sin x \) is \( \frac{dy}{dx} = \cos x \). It shows the slope of the tangent line to the curve at any given point.
• The second derivative is \( \frac{d^2y}{dx^2} = -\sin x \), which provides information about the curvature or how the direction of the curve's slope changes. This step is crucial as we need these derivatives to find the curvature and radius of curvature at a specific point.

By evaluating these derivatives at \( x = \frac{\pi}{4} \), we identify the behavior of our function at the mentioned point, which is essential for calculating further properties like curvature.

Radius of Curvature

The radius of curvature gives us a measure of how sharply a curve bends at a particular point. It can be thought of as the radius of the circular arc that best approximates the curve at that point.

• The formula to find the curvature \( \kappa \) is \( \kappa = \frac{|y''|}{(1+(y')^2)^{3/2}} \).
• Once we have the curvature, the radius of curvature \( R \) is simply the reciprocal of the curvature, \( R = \frac{1}{\kappa} \).

By substituting the values of the first and second derivatives into this formula, we calculated \( \kappa \) and then \( R \). Rationalizing \( R = \frac{9\sqrt{6}}{12} \), gives us a clear numerical understanding of the curve's "bend" at the point \( (\frac{\pi}{4}, \frac{\sqrt{2}}{2}) \). This shows us how tight or loose the curve is at that point.

Trigonometric Functions

Trigonometric functions like \( \sin \), \( \cos \), and \( \tan \) are fundamental in mathematics, representing periodic phenomena such as waves. In this exercise, the curve \( y = \sin x \) is an example of a sinusoid, which is a type of trigonometric function.

• These functions are periodic, meaning they repeat their values in regular intervals. For \( \sin x \), this period is \( 2\pi \), as it covers all possible values of \( x \) every \( 2\pi \) units.
• Evaluating \( \sin x \) at specific points like \( x = \frac{\pi}{4} \) gives exact values. Here, \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \), which is crucial for finding points on the curve and analyzing its properties.

Understanding these basic properties of trigonometric functions allows us to effectively calculate changes and understand the behavior of periodic curves.

Analytic Geometry

Analytic geometry, or coordinate geometry, uses algebraic equations to represent geometric shapes and analyze their properties. In this context, the function \( y = \sin x \) represents a wave-like curve on the xy-plane.

• Sketching Curves: By plotting functions using x and y coordinates, we can visualize how the curve behaves across different values. The curve \( y = \sin x \) undulates between -1 and 1, forming a classic sinusoidal shape.
• Analyzing Points: By choosing specific points, like \( (\frac{\pi}{4}, \frac{\sqrt{2}}{2}) \), we can explore the curve's behavior at precise locations.

In further steps, tools of analytic geometry enable us to calculate the curvature and radius of curvature, thus exploring the curve's geometric properties more deeply. This ties the abstract algebraic methods and concrete geometric visualizations together beautifully, giving us a comprehensive analysis of the curve.