Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 11: Problem 17
sketch the curve in the xy-plane. Then, for the given point, find the curvature and the radius of curvature. Finally, $$ y=\sin x,\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right) $$
Short Answer
Expert verified
Curvature: \(\frac{2\sqrt{2}}{3\sqrt{3}}\), Radius of Curvature: \(\frac{3\sqrt{6}}{4}\).
Step by step solution
01
Understand and Sketch the Curve
The function given is \( y = \sin x \), which is the sine function. The sine function is periodic with a period of \(2\pi\) and oscillates between -1 and 1. At the point \( x = \frac{\pi}{4} \), the value of \( y \) is \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \). Plot the sine curve and mark the point \( \left( \frac{\pi}{4}, \frac{\sqrt{2}}{2} \right) \) on the graph.
02
Calculate the First Derivative
To find the curvature, we need the first derivative of \( y = \sin x \). The first derivative is \( y' = \cos x \). At \( x = \frac{\pi}{4} \), the derivative \( y' \) is \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).
03
Calculate the Second Derivative
The second derivative of \( y = \sin x \) is \( y'' = -\sin x \). At \( x = \frac{\pi}{4} \), the second derivative \( y'' \) is \( -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).
04
Utilize the Curvature Formula
The curvature \( \kappa \) of a curve \( y = f(x) \) is given by the formula: \[ \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}. \] Substitute \( y' = \frac{\sqrt{2}}{2} \) and \( y'' = -\frac{\sqrt{2}}{2} \) into the formula: \[ \kappa = \frac{|-\frac{\sqrt{2}}{2}|}{(1 + (\frac{\sqrt{2}}{2})^2)^{3/2}} = \frac{\frac{\sqrt{2}}{2}}{(1 + \frac{1}{2})^{3/2}}. \] Simplifying further, \( \kappa = \frac{\frac{\sqrt{2}}{2}}{(\frac{3}{2})^{3/2}} \).
05
Simplify the Curvature Expression
Simplify \( (\frac{3}{2})^{3/2} = \sqrt{\frac{27}{8}} = \frac{3\sqrt{3}}{4} \). Thus, the curvature becomes: \[ \kappa = \frac{\frac{\sqrt{2}}{2}}{\frac{3\sqrt{3}}{4}} = \frac{2\sqrt{2}}{3\sqrt{3}} \].
06
Calculate the Radius of Curvature
The radius of curvature \( R \) is the reciprocal of the curvature \( \kappa \). Thus, \[ R = \frac{1}{\kappa} = \frac{3\sqrt{3}}{2\sqrt{2}} \]. Simplifying, \[ R = \frac{3\sqrt{6}}{4} \].
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sine Function
The sine function, expressed as \( y = \sin x \), is a fundamental mathematical function with periodic behavior. Its pattern repeats every \( 2\pi \), making it ideal for describing wave-like phenomena.
When graphing \( y = \sin x \), you'll notice it oscillates between -1 and 1. This suggests that throughout its cycle, the highest point it reaches is 1 and the lowest is -1.
When graphing \( y = \sin x \), you'll notice it oscillates between -1 and 1. This suggests that throughout its cycle, the highest point it reaches is 1 and the lowest is -1.
- The sine function is symmetric around its midpoint, meaning the graph is a mirror image either side of the origin.
- It starts at zero when \( x = 0 \), reaches the peak of 1 at \( x = \pi/2 \), back to zero at \( x = \pi \), down to -1 at \( x = 3\pi/2 \), and finally returns to zero at \( x = 2\pi \).
Radius of Curvature
The radius of curvature describes how sharply a curve is naturally bending at a particular point. Intuitively, it's like finding the size of a circle that best matches the curve exactly at that point.
Mathematically, the radius of curvature \( R \) is calculated as the reciprocal of the curvature \( \kappa \), given by:\[ R = \frac{1}{\kappa} \]At the point \( \left( \frac{\pi}{4}, \frac{\sqrt{2}}{2} \right) \), we follow a few steps to determine this value:
Mathematically, the radius of curvature \( R \) is calculated as the reciprocal of the curvature \( \kappa \), given by:\[ R = \frac{1}{\kappa} \]At the point \( \left( \frac{\pi}{4}, \frac{\sqrt{2}}{2} \right) \), we follow a few steps to determine this value:
- First, we compute the curvature \( \kappa \).
- The radius of curvature \( R \) is then found by inverting the curvature, meaning we take \( \frac{1}{\kappa} \).
- For example, with a calculated curvature \( \kappa = \frac{2\sqrt{2}}{3\sqrt{3}} \), the radius of curvature becomes \( R = \frac{3\sqrt{3}}{2\sqrt{2}} \).
- Simplifying, it results in \( R = \frac{3\sqrt{6}}{4} \).
Derivatives
Derivatives play a central role in understanding how a function behaves at any given point, especially regarding rates of change and the inclination of curves. When dealing with the function \( y = \sin x \), we use derivatives to find the slope and curvature at particular points.
Here are some key steps related to derivatives in this context:
Here are some key steps related to derivatives in this context:
- The **first derivative** of \( y = \sin x \) is \( y' = \cos x \). This derivative tells us the slope of the tangent line to the sine curve at any point \( x \).
- At \( x = \frac{\pi}{4} \), the slope \( y' = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \). The first derivative essentially dictates how steep the curve is at that specific point.
- The **second derivative** \( y'' = -\sin x \) provides insight into how the function is curving based on its concavity at \( x \).
- For example, when \( x = \frac{\pi}{4} \), \( y'' = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} \). This tells us how the slope of \( y' \) is changing, giving information about curvature.
