/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Sketch the curve in the \(x y\) ... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 23

Sketch the curve in the \(x y\) -plane. Then, for the given point, find the curvature and the radius of curvature. Finally, draw the circle of curvature at the point. Hint: For the curvature, you will use the second formula in Theorem \(A,\) as in Example \(6 .\) $$ y=\tan x,(\pi / 4,1) $$

Short Answer

Expert verified
The curvature at \((\pi/4,1)\) is \(\frac{4}{5\sqrt{5}}\), and the radius is \(\frac{5\sqrt{5}}{4}\).

Step by step solution

01

Understand the Function and Plot the Curve

The function given is \( y = \tan x \). This function represents the tangent curve, which is periodic and has vertical asymptotes at odd multiples of \( \frac{\pi}{2} \). To sketch \( y = \tan x \), plot a few key points such as \( (0, 0) \), \( (\pi/4, 1) \), and understand it's increasing over its period. Highlight vertical asymptotes at \( x = \frac{\pi}{2} + k\pi \), where \( k \) is an integer.
02

Verify Given Point on the Curve

Verify that the point \( (\pi/4, 1) \) lies on the curve \( y = \tan x \). Substitute \( \pi/4 \) into \( \tan x \): \( \tan(\pi/4) = 1 \). Therefore, the point \( (\pi/4, 1) \) is on the curve.
03

Find the First and Second Derivatives

Find the derivatives needed for curvature. The first derivative of \( y = \tan x \) is \( y' = \sec^2 x \). The second derivative is \( y'' = 2\sec^2 x \tan x \). These will be used to find the curvature.
04

Calculate the Curvature

The curvature \( \kappa \) is given by the formula:\[\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\]Substitute \( y' = \sec^2(\pi/4) = 2 \) and \( y'' = 2 \cdot \sec^2(\pi/4) \cdot \tan(\pi/4) = 4 \) into the formula to get:\[\kappa = \frac{|4|}{(1+2^2)^{3/2}} = \frac{4}{5\sqrt{5}}\]
05

Calculate the Radius of Curvature

The radius of curvature \( R \) is the reciprocal of the curvature:\[R = \frac{1}{\kappa} = \frac{5\sqrt{5}}{4}\] This gives the radius of the circle of curvature.
06

Draw the Circle of Curvature

To draw the circle of curvature at the point \( (\pi/4, 1) \), use the calculated radius of \( R = \frac{5\sqrt{5}}{4} \). The center of the osculating circle is along the normal of the curve at \( (\pi/4, 1) \). Since \( y = \tan x \) at \( x = \pi/4 \) has a slope of 1 (the derivative), the normal line is at a slope of -1. Plot the circle with the center found on this normal at the calculated distance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Curve
The function given, \( y = \tan x \), represents the tangent curve. This curve is known for being periodic, meaning it repeats its pattern every \( \pi \) units along the \( x \)-axis. You will notice some key features when examining the tangent curve:
  • Periodicity: It has a repeating pattern with each cycle spanning from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).
  • Vertical Asymptotes: These are found at odd multiples of \( \frac{\pi}{2} \), such as \( x = -\frac{\pi}{2}, \frac{\pi}{2}, \) etc. As \( x \) approaches these points, the tan function goes to infinity or minus infinity, creating discontinuities.
  • Intersections with the X-axis: The curve intersects the \( x \)-axis at multiples of \( \pi \), for instance, the origin \( (0,0) \), creating a repeating pattern.
To sketch \( y = \tan x \), ensure you plot these critical points and respect the asymptotes. Highlight key points like the given \( (\pi/4, 1) \), which lies above the \( x \)-axis, part of its increasing trend in its main period.
Radius of Curvature
The radius of curvature is an essential concept when analyzing curves, particularly how 'bent' or curved a section of the curve is at a point. This radius is the reciprocal of curvature (\(\kappa\)), giving us insight into how tight or broad a curve is around a particular point.
  • Formula: The radius \( R \) is calculated using \( R = \frac{1}{\kappa} \).
  • Interpretation: A smaller radius indicates a sharper curve, similar to a tight turn, while a larger radius suggests a gentler curve.
  • Context in Example: For the tangent curve \( y = \tan x \) at the point \( (\pi/4, 1) \), we've calculated \( R = \frac{5\sqrt{5}}{4} \). This value tells us how large the circle would be at this pinpointed moment, which matches the curve's curvature precisely.
The radius helps in drawing an osculating circle, which is a circle that closely matches the curve at a specific point. Its center can be found by shrinking the normal line of the curve until it equals the radius at the curve point, giving you a tool to understand the curve鈥檚 behavior graphically.
Derivatives in Calculus
Derivatives play a critical role when discussing curvature and the tangent curve.
  • First Derivative (\( y' \)): This derivative indicates the rate of change of the curve or the slope at a particular point. For the function \( y = \tan x \), the first derivative is \( y' = \sec^2 x \). At any point \( x = \pi/4 \), substituting gives \( y' = 2 \), which means it has a slope of 1 at that point in the context of its tangent.
  • Second Derivative (\( y'' \)): This derivative tells us about the curvature and how "curvy" the slope changes. The second derivative for \( y = \tan x \) is \( y'' = 2\sec^2 x \tan x \). At \( x = \pi/4 \), this becomes 4, vital for calculating the curvature value \( \kappa \) used in our context.
  • Connection to Curvature: These derivatives appear in the curvature formula, \( \kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} \), which assesses how quickly the function's direction changes.
Understanding derivatives offers clear insight into how a function behaves, not just at a given point, but how it locally bends, enabling us to look at the curve more intelligently and with a deeper appreciation for its geometry.

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