/*! 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 33 Find all points (if any) of hori... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 33

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=\sec \theta, \quad y=\tan \theta $$

Short Answer

Expert verified
The point of horizontal tangency to the curve is at (1,0). There are no points of vertical tangency to the curve.

Step by step solution

01

Convert parametric to rectangular

Write the expression for \( x \) in terms of \( y \). Using the trigonometric identity \( \sec^{2}(\theta)=1+\tan^{2}(\theta) \), replace \( \sec \theta \) in the equation \( x=\sec \theta \) by \( x=\sqrt{1+y^{2}} \). Thus, the curve is described by the equation \( x=\sqrt{1+y^{2}} \).
02

Find the derivative

Next, calculate the derivative of \( x \) in terms of \( y \). The derivative of \( x \) with respect to \( y \) is \( dx/dy=y/(y^{2}+1) \). Keep this expression for the subsequent steps.
03

Find points of horizontal tangency

Horizontal tangency points occur where \( dx/dy=0 \). Hence, solve the equation \( y/(y^{2}+1)=0 \). The solution for \( y \) is \( y=0 \). After substituting \( y=0 \) in the rectangular equation \( x=\sqrt{1+y^{2}} \), the corresponding \( x \) value is \( x=1 \). Hence, the point of horizontal tangency is (1,0).
04

Find points of vertical tangency

Vertical tangency occurs where the derivative is undefined. Hence, solve the equation \( y^{2}+1=0 \). However, there is no solution to this equation in the real number system. Hence, there are no points of vertical tangency to the curve.
05

Validate with graphing utility

After plotting the curve \( x=\sqrt{1+y^{2}} \) using a graphing utility, it can be seen that the curve indeed has a flat point at (1,0), confirming our above analysis.

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=6 \cos \theta & 0 \leq \theta \leq \frac{\pi}{2} & \text { Polar axis } \end{array} $$

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=e^{t}, \quad y=e^{3 t}+1 $$

Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter. $$ x=t^{2}+t, \quad y=t^{2}-t $$

Explain why finding points of intersection of polar graphs may require further analysis beyond solving two equations simultaneously

Find two different sets of parametric equations for the rectangular equation. $$ y=3 x-2 $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.