/*! 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 19 Convert to polar form. $$y^{2}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Convert to polar form. $$y^{2}=x^{3}$$

Short Answer

Expert verified
The polar form is \( r = \frac{\sin^2 \theta}{\cos^3 \theta} \).

Step by step solution

01

Identify Variables for Conversion

To convert the equation to polar form, identify the variables: \( x \) and \( y \). In polar coordinates, these correspond to \( r \cos \theta \) for \( x \) and \( r \sin \theta \) for \( y \).
02

Substitute Polar Equivalents

Substitute \( x = r \cos \theta \) and \( y = r \sin \theta \) into the given equation \( y^2 = x^3 \). This yields: \[ (r \sin \theta)^2 = (r \cos \theta)^3 \]
03

Simplify the Equation

Simplify the equation: \[ r^2 \sin^2 \theta = r^3 \cos^3 \theta \]Divide both sides by \( r^2 \) (assuming \( r eq 0 \)) to get: \[ \sin^2 \theta = r \cos^3 \theta \]
04

Solve for \( r \)

Solve the equation \( \sin^2 \theta = r \cos^3 \theta \) for \( r \): \[ r = \frac{\sin^2 \theta}{\cos^3 \theta} \]

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.

Coordinate Conversion
Coordinate conversion plays a crucial role when transitioning from Cartesian (also known as rectangular) coordinates to polar coordinates. This process involves changing from the standard x and y coordinates of a point to the polar coordinates, which are represented by two values: the distance from the origin, often denoted as \( r \), and the angle \( \theta \), which is the counterclockwise angle measured from the positive x-axis.
The given exercise starts with an equation in the Cartesian coordinate system and transforms it into polar form. The task requires recognizing how x and y relate to \( r \) and \( \theta \):
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
By substituting these relationships into the initial Cartesian equation, you initiate the conversion process. This mathematical maneuver allows the representation of the same curve, but in terms of its position relative to a central point and direction rather than along a grid. Mastering this conversion helps in analyzing and interpreting curves and shapes in a different coordinate system.
Trigonometry
Trigonometry is essential in the process of converting Cartesian equations to their polar forms. In this context, trigonometric functions serve as the bridge linking two coordinate systems. These functions help express the relationship between linear positions (x and y) and angular positions (\( r \) and \( \theta \)).
Substituting into trigonometric terms is straightforward:
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
These expressions are used to rewrite the equation of a curve using polar coordinates effortlessly. For instance, using \( y = r \sin \theta \) and \( x = r \cos \theta \), the original equation \( y^2 = x^3 \) becomes an equation involving \( r \) and \( \theta \). With trigonometry, one can reconcile the different representations of geometric phenomena seamlessly.
Precalculus Mathematics
Precalculus mathematics involves a skillful understanding of various mathematical properties, including functions, geometry, and coordinate systems. This knowledge base is pivotal in handling problems like coordinate conversions. In the example problem, the simplification of the equation derived from substituting polar equivalents is an application of precalculus mathematics.
Simplification here involves dealing with terms like \( r^2 \sin^2 \theta \) and \( r^3 \cos^3 \theta \). It's essential to know:
  • How to manipulate algebraic expressions consistently
  • The rules for dividing expressions by using properties of exponents (as seen where \( r^2 \) is divided out)
This exercise also demonstrates derivative skills that precalculus students heavily rely upon, creating a foundational stepping stone towards calculus. Having a solid grounding in precalculus mathematics enables students to confidently approach problem-solving that involves transitioning between different mathematical domains.

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

The polar equation of a line is given. In each case: (a) specify the perpendicular distance from the origin to the line; (b) determine the polar coordinates of the points on the line corresponding to \(\theta=0\) and \(\theta=\pi / 2 ;\) (c) specify the polar coordinates of the foot of the perpendicular from the origin to the line; (d) use the results in parts (a), (b), and (c) to sketch the line; and (e) find a rectangular form for the equation of the line. $$r \cos \left(\theta-\frac{\pi}{2}\right)=\sqrt{2}$$

We study the dot product of two vectors. Given two vectors \(\mathbf{A}=\left\langle x_{1}, y_{1}\right\rangle\) and \(\mathbf{B}=\left\langle x_{2}, y_{2}\right\rangle,\) we define the dot product \(\mathbf{A} \cdot \mathbf{B}\) as follows: $$\mathbf{A} \cdot \mathbf{B}=x_{1} x_{2}+y_{1} y_{2}$$ For example, if \(\mathbf{A}=\langle 3,4\rangle\) and \(\mathbf{B}=\langle-2,5\rangle,\) then \(\mathbf{A} \cdot \mathbf{B}=(3)(-2)+(4)(5)=14 .\) Notice that the dot product of two vectors is a real number. For this reason, the dot product is also known as the scalar product. For Exercises \(61-63\) the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are defined as follows: $$\mathbf{u}=\langle-4,5\rangle \quad \mathbf{v}=\langle 3,4\rangle \quad \mathbf{w}=\langle 2,-5\rangle$$ (a) Compute \(\mathbf{u} \cdot \mathbf{v}\) and \(\mathbf{v} \cdot \mathbf{u}\) (b) Compute \(\mathbf{v} \cdot \mathbf{w}\) and \(\mathbf{w} \cdot \mathbf{v}\) (c) Show that for any two vectors \(\mathbf{A}\) and \(\mathbf{B}\), we have \(\mathbf{A} \cdot \mathbf{B}=\mathbf{B} \cdot \mathbf{A} .\) That is, show that the dot product is commutative. Hint: Let \(\mathbf{A}=\left\langle x_{1}, y_{1}\right\rangle,\) and let \(\mathbf{B}=\left\langle x_{2}, y_{2}\right\rangle\)

Graph the parametric equations using the given range for the parameter t. In each case, begin with the standard viewing rectangle and then make adjustments, as necessary, so that the graph utilizes as much of the viewing screen as possible. For example, in graphing the circle given by \(x=\cos t\) and \(y=\sin t,\)it would be natural to choose a viewing rectangle extending from -1 to 1 in both the \(x\) - and \(y\) -directions. $$x=3 t^{2}, y=2 t^{3}, \quad-2 \leq t \leq 2(\text {semicubical parabola})$$

Round each answer to one decimal place. A regular pentagon is inscribed in a circle of radius 1 unit. Find the perimeter of the pentagon. Hint: First find the length of a side using the law of cosines.

From a point on ground level, you measure the angle of clevation to the top of a mountain to be \(38^{\circ} .\) Then you walk 200 m farther away from the mountain and find that the angle of elevation is now \(20^{\circ}\). Find the height of the mountain. Round the answer to the nearest meter.

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.