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Chapter 11: Problem 93

Convert the equation from polar coordinates into rectangular coordinates. $$ r=-\csc (\theta) \cot (\theta) $$

Short Answer

Expert verified
The equation in rectangular coordinates is \(y^4 x^2 + y^6 = x^2\).

Step by step solution

01

Understand the Relationships

Recall the relationships between polar and rectangular coordinates. You have \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) for conversion. Additionally, \(\csc(\theta) = \frac{1}{\sin(\theta)}\) and \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\).
02

Substitute Trigonometric Identities

Substitute the identities \(\csc(\theta) = \frac{1}{\sin(\theta)}\) and \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\) into the equation: \[r = -\frac{1}{\sin(\theta)} \times \frac{\cos(\theta)}{\sin(\theta)} = -\frac{\cos(\theta)}{\sin^2(\theta)}.\]
03

Express as Rectangular Coordinate Components

Using the relationships \(x = r \cos(\theta)\), \(y = r \sin(\theta)\), and substituting \(\sin(\theta) = \frac{y}{r}\) and \(\cos(\theta) = \frac{x}{r}\) into the equation:\[r = -\frac{\frac{x}{r}}{\left(\frac{y}{r}\right)^2} = -\frac{x}{y^2}.\] Multiply both sides by \(y^2\) to eliminate the fraction:\[y^2 r = -x.\]
04

Confirm Substitution Complete

Now express \(r\) in terms of \(x\) and \(y\). Since \(r = \sqrt{x^2 + y^2}\), substitute \(r = \frac{-x}{y^2}\) in \(y^2 \sqrt{x^2 + y^2} = -x\), providing the equation in consistent terms of \(x\) and \(y\).
05

Eliminate Radical

To simplify further, square both sides to remove the square root, yielding:\[y^4 (x^2 + y^2) = x^2.\] Now expand and simplify, \[y^4 x^2 + y^6 = x^2.\] This is the equation converted into rectangular coordinates.

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Key Concepts

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

Trigonometric Identities
When converting from polar to rectangular coordinates, understanding trigonometric identities is crucial. These identities help in manipulating angles and lengths from one form to another. In this exercise, we are dealing with two specific identities: cosecant and cotangent. The cosecant function, denoted as \(\csc(\theta)\), is the reciprocal of the sine function, represented as \(\csc(\theta) = \frac{1}{\sin(\theta)}\). Similarly, the cotangent function, \(\cot(\theta)\), is the reciprocal of tangent, which can be written using sine and cosine as \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\). Using these identities allows us to express trigonometric functions in terms of basic angles, which simplifies our conversion task. For this exercise, they enable us to rewrite the given polar equation in a form that is easier to transform into rectangular coordinates.
Rectangular Coordinates
Rectangular coordinates are expressed as \((x, y)\). They describe a point on the Cartesian plane using two axes: the x-axis (horizontal) and the y-axis (vertical). To convert polar coordinates to rectangular coordinates, we make use of certain mathematical transformations. Given that we have \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), these equations allow us to express the coordinates \(x\) and \(y\) based on the radius \(r\) and the angle \(\theta\). This exercise illustrates the conversion by substituting polar definitions into the rectangular form. It starts by rearranging polar terms using known trigonometric identities, and then setting these in terms of \(x\) and \(y\). For instance, substituting \(\sin(\theta) = \frac{y}{r}\) and \(\cos(\theta) = \frac{x}{r}\) permits the entire equation to be restated in terms of rectangular coordinates, leading to the final equation \(y^4(x^2 + y^2) = x^2\).
Polar Coordinates
Polar coordinates represent a point in a two-dimensional plane differently than rectangular coordinates. Instead of using horizontal and vertical distances, they utilize the distance from the origin (radius \(r\)) and the angle \(\theta\) from the positive x-axis. The task at hand involves converting these coordinates to a system that relies on a grid-like structure. This conversion captures the intricate dance from curvy paths (polar) to straight lines (rectangular). The original polar equation was \(r = -\csc(\theta) \cot(\theta)\). By converting this to rectangular form, we replace \(r\), \(\sin(\theta)\), and \(\cos(\theta)\) with expressions involving \(x\) and \(y\). Ultimately, this process reflects the relationship between these two coordinate systems, demonstrating that geometrical representations (like polar spirals) can be translated into linear formats.

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Most popular questions from this chapter

Convert the equation from polar coordinates into rectangular coordinates. $$ 12 r=\csc (\theta) $$

In Exercises 47 - 52 , we explore the hyperbolic cosine function, denoted \(\cosh (t)\), and the hyperbolic sine function, denoted \(\sinh (t)\), defined below: $$ \cosh (t)=\frac{e^{t}+e^{-t}}{2} \text { and } \sinh (t)=\frac{e^{t}-e^{-t}}{2} $$ Four other hyperbolic functions are waiting to be defined: the hyperbolic secant \(\operatorname{sech}(t)\) the hyperbolic cosecant \(\operatorname{csch}(t)\), the hyperbolic tangent \(\tanh (t)\) and the hyperbolic cotangent \(\operatorname{coth}(t) .\) Define these functions in terms of \(\cosh (t)\) and \(\sinh (t)\), then convert them to formulas involving \(e^{t}\) and \(e^{-t}\). Consult a suitable reference (a Calculus book, or this entry on the hyperbolic functions) and spend some time reliving the thrills of trigonometry with these 'hyperbolic' functions.

In Exercises \(25-39\), find a parametric description for the given oriented curve. the ellipse \(9 x^{2}+4 y^{2}+24 y=0\), oriented clockwise (Shift the parameter so \(t=0\) corresponds to \((0,0) .)\)

The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line \(y=2 x-4 .\) Let \(\vec{v}_{0}=\langle 0,-4\rangle\) and let \(\vec{s}=\langle 1,2\rangle .\) Let \(t\) be any real number. Show that the vector defined by \(\vec{v}=\vec{v}_{0}+t \vec{s}\), when drawn in standard position, has its terminal point on the line \(y=2 x-4\). (Hint: Show that \(\vec{v}_{0}+t \vec{s}=\langle t, 2 t-4\rangle\) for any real number \(t\).) Now consider the non-vertical line \(y=m x+b\). Repeat the previous analysis with \(\vec{v}_{0}=\langle 0, b\rangle\) and let \(\vec{s}=\langle 1, m\rangle .\) Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of \(\langle 0, b\rangle\) (the position vector of the \(y\) -intercept) and a scalar multiple of the slope vector \(\vec{s}=\langle 1, m\rangle\).

We know that \(|x+y| \leq|x|+|y|\) for all real numbers \(x\) and \(y\) by the Triangle Inequality established in Exercise 36 in Section 2.2. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that \(\|\vec{u}+\vec{v}\| \leq\|\vec{u}\|+\|\vec{v}\|\) for all pairs of vectors \(\vec{u}\) and \(\vec{v}\). (a) (Step 1) Show that \(\|\vec{u}+\vec{v}\|^{2}=\|\vec{u}\|^{2}+2 \vec{u} \cdot \vec{v}+\|\vec{v}\|^{2}\). 6 (b) (Step 2) Show that \(|\vec{u} \cdot \vec{v}| \leq\|\vec{u}\|\|\vec{v}\| .\) This is the celebrated Cauchy-Schwarz Inequality. (Hint: To show this inequality, start with the fact that \(|\vec{u} \cdot \vec{v}|=|\|\vec{u}\|\|\vec{v}\| \cos (\theta)|\) and use the fact that \(|\cos (\theta)| \leq 1\) for all \(\theta\).) (c) (Step 3) Show that \(\|\vec{u}+\vec{v}\|^{2}=\|\vec{u}\|^{2}+2 \vec{u} \cdot \vec{v}+\|\vec{v}\|^{2} \leq\|\vec{u}\|^{2}+2|\vec{u} \cdot \vec{v}|+\|\vec{v}\|^{2} \leq\|\vec{u}\|^{2}+\) \(2\|\vec{u}\|\|\vec{v}\|+\|\vec{v}\|^{2}=(\|\vec{u}\|+\|\vec{v}\|)^{2}\) (d) (Step 4) Use Step 3 to show that \(\|\vec{u}+\vec{v}\| \leq\|\vec{u}\|+\|\vec{v}\|\) for all pairs of vectors \(\vec{u}\) and \(\vec{v}\). (e) As an added bonus, we can now show that the Triangle Inequality \(|z+w| \leq|z|+|w|\) holds for all complex numbers \(z\) and \(w\) as well. Identify the complex number \(z=a+b i\) with the vector \(u=\langle a, b\rangle\) and identify the complex number \(w=c+d i\) with the vector \(v=\langle c, d\rangle\) and just follow your nose!
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