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Chapter 12: Problem 52

Convert the following equations to polar coondinates. $$(x-1)^{2}+y^{2}=1$$

Short Answer

Expert verified
Question: Convert the Cartesian equation \((x-1)^{2} + y^{2} = 1\) to polar coordinates. Answer: The equivalent polar coordinates equation for the given Cartesian equation is \(r = 2\cos\theta\).

Step by step solution

01

Replace x and y with polar coordinates

Replace x with \(r \cos\theta\) and y with \(r \sin\theta\) in the given equation: $$(r\cos\theta -1)^{2} + (r\sin\theta)^{2} = 1$$
02

Expand and simplify the equation

Expand and group the terms: $$r^2\cos^2\theta -2r\cos\theta +1 + r^2\sin^2\theta = 1$$ Combine the \(r^2\cos^2\theta\) and \(r^2\sin^2\theta\) terms: $$r^2(\cos^2\theta + \sin^2\theta) -2r\cos\theta +1 = 1$$ Since \(\cos^2\theta +\sin^2\theta = 1\), we can simplify the equation to: $$r^2 - 2r\cos\theta + 1 = 1$$
03

Solve for r

Now, the goal is to find an equation of the form \(r=f(\theta)\), so subtract 1 from both sides of the equation: $$r^2 -2r\cos\theta = 0$$ Factor out r: $$r(r-2\cos\theta) = 0$$ Since r can鈥檛 be 0, the equation should be: $$r = 2\cos\theta$$ Therefore, the polar coordinates equation for the given Cartesian equation is: $$r = 2\cos\theta$$

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Key Concepts

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

Cartesian coordinates and their representation
Cartesian coordinates are responsible for describing the position of a point on a plane using two values: the x-coordinate and the y-coordinate. These coordinates are based on the Cartesian plane, which is a two-dimensional coordinate system with a horizontal axis (x-axis) and a vertical axis (y-axis). This system enables the representation of geometric figures, equations, and relationships between points in a straightforward manner.
For example:
  • The point (3, 2) means that you move three units along the x-axis and two units along the y-axis from the origin.
  • The equation \(x^2 + y^2 = 1\) represents a circle centered at the origin with a radius of one.
These coordinates form the basis for numerous mathematical concepts and have a wide range of applications, from basic geometry to advanced calculus.
Understanding trigonometric identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables involved. They play a critical role in simplifying expressions and solving equations in various fields, particularly in coordinate conversion.
In the context of polar and Cartesian coordinates, an essential trigonometric identity is \( \cos^2\theta + \sin^2\theta = 1\). This identity helps to relate polar coordinates, defined in terms of \(r\) (radius) and \(\theta\) (angle), with Cartesian coordinates, defined in terms of \(x\) and \(y\).
To convert Cartesian coordinates into polar form based on an equation:
  • Replace \(x\) with \(r \cos \theta\).
  • Replace \(y\) with \(r \sin \theta\).
  • Use trigonometric identities to simplify the equation further.
These steps enable us to thoroughly understand and handle coordinate transformations seamlessly.
Conversion of equations from Cartesian to polar coordinates
Converting equations from Cartesian to polar coordinates involves substituting the Cartesian variables \(x\) and \(y\) with polar variables involving \(r\) and \(\theta\). It allows representing and analyzing equations in a form better suited for calculations involving circles, spirals, and other shapes where radial symmetry is apparent.
To convert equations:
  • Identify the initial equation in Cartesian form. For example, \( (x-1)^2 + y^2 = 1\).
  • Substitute \(x = r\cos\theta\) and \(y = r\sin\theta\) into the equation.
  • Simplify using trigonometric identities such as \( \cos^2 \theta + \sin^2 \theta = 1\).
  • Rearrange to solve for \( r \) in terms of \( \theta \), as in the polar equation \( r = f(\theta)\).
This method results in an equation that provides valuable insights into the shape and behavior of curves in polar coordinates, making it useful in various mathematical and engineering applications.

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Most popular questions from this chapter

Suppose the function \(y=h(x)\) is nonnegative and continuous on \([\alpha, \beta],\) which implies that the area bounded by the graph of h and the x-axis on \([\alpha, \beta]\) equals \(\int_{\alpha}^{\beta} h(x) d x\) or \(\int_{\alpha}^{\beta} y d x .\) If the graph of \(y=h(x)\) on \([\alpha, \beta]\) is traced exactly once by the parametric equations \(x=f(t), y=g(t),\) for \(a \leq t \leq b,\) then it follows by substitution that the area bounded by h is $$\begin{array}{l}\int_{\alpha}^{\beta} h(x) d x=\int_{\alpha}^{\beta} y d x=\int_{a}^{b} g(t) f^{\prime}(t) d t \text { if } \alpha=f(a) \text { and } \beta=f(b) \\\\\left(\text { or } \int_{\alpha}^{\beta} h(x) d x=\int_{b}^{a} g(t) f^{\prime}(t) d t \text { if } \alpha=f(b) \text { and } \beta=f(a)\right)\end{array}$$. Find the area of the region bounded by the astroid \(x=\cos ^{3} t, y=\sin ^{3} t,\) for \(0 \leq t \leq 2 \pi\) (see Example 8 Figure 12.17 ).

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