/*! 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 31 Find all points (if any) of hori... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=3 \cos \theta, \quad y=3 \sin \theta $$

Short Answer

Expert verified
The curve has horizontal tangency at points where \(\theta = \pi/2 + n\pi\), and vertical tangency at points where \(\theta = n\pi\). The corresponding x and y coordinates can be obtained by substituting these values in the original parametric equations.

Step by step solution

01

Parametric Derivative

The first step is finding the derivative of \(y\) with respect to \(x\). This is done by differentiating \(y = 3 \sin \theta\) with respect to \(\theta\) to get \(dy/d\theta = 3 \cos \theta\), and differentiating \(x = 3 \cos \theta\) with respect to \(\theta\) to get \(dx/d\theta = -3 \sin \theta\). Next, \(dy/dx = (dy/d\theta) / (dx/d\theta)\). Plug in our previously obtained values to find that \(dy/dx = -\cot \theta\).
02

Points of Horizontal Tangency

Recall that for horizontal tangency, \(dy/dx = 0\). So, equate \(-\cot \theta = 0\). The cotangent function equals zero at \(\pi/2 + n\pi\) for n as integer. Substituting these \(\theta\) into the equations \(x=3 \cos \theta\) and \(y=3 \sin \theta\), the points of horizontal tangency can be obtained.
03

Points of Vertical Tangency

For vertical tangency, \(dy/dx\) is undefined. This means that \(\cot \theta\) is undefined which happens if \(\sin \theta = 0\). This equals zero at \(n\pi\) for n as integer. Substituting these \(\theta\) into the equations \(x=3 \cos \theta\) and \(y=3 \sin \theta\), the points of vertical tangency can be found.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Derivatives in Parametric Equations
When dealing with parametric equations, calculating the derivative requires a slightly different approach compared to standard Cartesian equations. Here, derivatives help determine how the curve behaves as the parameter, in this case \( \theta \), changes.
To find the derivative \( \frac{dy}{dx} \) when \( y = 3 \sin \theta \) and \( x = 3 \cos \theta \), we first find the derivatives with respect to \( \theta \):
  • \( \frac{dy}{d\theta} = 3 \cos \theta \)
  • \( \frac{dx}{d\theta} = -3 \sin \theta \)
The next step is to divide the two derivatives to find \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3 \cos \theta}{-3 \sin \theta} = -\cot \theta \]
This resulting form, \(-\cot \theta\), is crucial for finding points of tangency on the curve because it describes the slope of the tangent line at any point determined by \( \theta \).
Points of Tangency: Horizontal and Vertical
A tangent to a curve is a line that just "touches" the curve at a particular point. Points of tangency are important in understanding the geometry and slopes of a curve. In this exercise, we are particularly focused on horizontal and vertical tangents.
  • **Horizontal Tangency:** For horizontal tangency, the slope \( \frac{dy}{dx} \) of the tangent line should be 0. For the derived equation \(-\cot \theta = 0\), this occurs when \( \theta = \frac{\pi}{2} + n\pi \), where \( n \) is an integer. By substituting these values of \( \theta \) into the original parametric equations, you find the points on the curve where horizontal tangents exist.
  • **Vertical Tangency:** A vertical tangency occurs where \(\frac{dy}{dx} \) becomes undefined, which for \(-\cot \theta \) is when \( \sin \theta = 0 \), or \( \theta = n\pi \). Again, inserting these \( \theta \) values into the equations helps us to locate where on the curve the vertical tangents appear.
These concepts are important for graph analysis as they can help you identify where curves change directions or exhibit asymptotic behavior.
Role of Trigonometric Functions
Trigonometric functions play a vital role in parametric equations, especially when representing curves like circles or ellipses. In the given equations:
  • \( x = 3 \cos \theta \)
  • \( y = 3 \sin \theta \)
these represent a circle in the parametric form because of the intrinsic properties of \( \sin \theta \) and \( \cos \theta \).
The trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \) highlights that any parametric equation of this format maps onto a circle. Here, \( 3 \) acts as a radius, depicting a path where a point moves along the circle with changing \( \theta \).
Understanding these functions is essential because they not only define the shape and nature of the curve but also come into play when calculating derivatives and determining tangency points. They illustrate how the cycle of \( \theta \) affects \( x \) and \( y \) positions, allowing us to find precise points for analysis.

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

(a) Use a graphing utility to graph each set of parametric equations. \(x=t-\sin t \quad x=2 t-\sin (2 t)\) \(y=1-\cos t \quad y=1-\cos (2 t)\) \(0 \leq t \leq 2 \pi \quad 0 \leq t \leq \pi\) (b) Compare the graphs of the two sets of parametric equations in part (a). If the curve represents the motion of a particle and \(t\) is time, what can you infer about the average speeds of the particle on the paths represented by the two sets of parametric equations? (c) Without graphing the curve, determine the time required for a particle to traverse the same path as in parts (a) and (b) if the path is modeled by \(x=\frac{1}{2} t-\sin \left(\frac{1}{2} t\right) \quad\) and \(\quad y=1-\cos \left(\frac{1}{2} t\right)\)

Show that the graphs of the given equations intersect at right angles. $$ r=\frac{e d}{1+\sin \theta} \quad \text { and } \quad r=\frac{e d}{1-\sin \theta} $$

Sketch a graph of a curve defined by the parametric equations \(x=g(t)\) and \(y=f(t)\) such that \(d x / d t>0\) and \(d y / d t<0\) for all real numbers \(t\).

It Use a graphing utility to graph the polar equation \(r=6[1+\cos (\theta-\phi)]\) for (a) \(\phi=0,(\) b) \(\phi=\pi / 4\) and (c) \(\phi=\pi / 2\). Use the graphs to describe the effect of the angle \(\phi\). Write the equation as a function of \(\sin \theta\) for part (c).

Verify that if the curve whose polar equation is \(r=f(\theta)\) is rotated about the pole through an angle \(\phi\), then an equation for the rotated curve is \(r=f(\theta-\phi)\).

See all solutions

Recommended explanations on Math Textbooks

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.