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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 12, Problem 12.1.15

15–30. Working with parametric equations Consider the following parametric equations.
a. Eliminate the parameter to obtain an equation in x and y.
b. Describe the curve and indicate the positive orientation.


x = 3 + t, y = 1 − t; 0 ≤ t ≤ 1

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1
Identify the given parametric equations: \(x = 3 + t\) and \(y = 1 - t\), with the parameter \(t\) ranging from \(0\) to \(1\).
To eliminate the parameter \(t\), solve one of the equations for \(t\). For example, from \(x = 3 + t\), isolate \(t\) to get \(t = x - 3\).
Substitute the expression for \(t\) into the other equation: replace \(t\) in \(y = 1 - t\) with \(x - 3\), resulting in \(y = 1 - (x - 3)\).
Simplify the equation to express \(y\) solely in terms of \(x\): \(y = 1 - x + 3\), which simplifies further to \(y = 4 - x\).
Interpret the curve: the equation \(y = 4 - x\) represents a straight line. The parameter \(t\) increases from \(0\) to \(1\), so the curve starts at the point when \(t=0\) (which is \((3,1)\)) and ends at \(t=1\) (which is \((4,0)\)). This indicates the positive orientation of the curve is from \((3,1)\) to \((4,0)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter, usually denoted t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves and motions.
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Parameterizing Equations

Eliminating the Parameter

Eliminating the parameter involves manipulating the parametric equations to remove t, resulting in a direct relationship between x and y. This often requires solving one equation for t and substituting into the other, yielding a Cartesian equation of the curve.
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Eliminating the Parameter

Curve Orientation and Interval of Parameter

The orientation of a parametric curve is determined by the direction in which the parameter t increases. The interval for t specifies the portion of the curve traced, and understanding this helps describe the curve's direction and endpoints.
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Finding Area Between Curves that Cross on the Interval
Related Practice
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r = 4 cos 2θ; at the tips of the leaves

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The region inside the limaçon r = 2 + cos θ

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Textbook Question

15–30. Working with parametric equations Consider the following parametric equations.

a. Eliminate the parameter to obtain an equation in x and y.

b. Describe the curve and indicate the positive orientation.


x = 3 cos t, y = 3 sin t; π ≤ t ≤ 2π

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Textbook Question

13–30. Graphing conic sections Determine whether the following equations describe a parabola, an ellipse, or a hyperbola, and then sketch a graph of the curve. For each parabola, specify the location of the focus and the equation of the directrix; for each ellipse, label the coordinates of the vertices and foci, and find the lengths of the major and minor axes; for each hyperbola, label the coordinates of the vertices and foci, and find the equations of the asymptotes.


4x = -y²

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Textbook Question

90–94. Focal chords A focal chord of a conic section is a line through a focus joining two points of the curve. The latus rectum is the focal chord perpendicular to the major axis of the conic. Prove the following properties.

The length of the latus rectum of the parabola y ² =4px or x ² =4py is 4|p|.

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Textbook Question

13–30. Graphing conic sections Determine whether the following equations describe a parabola, an ellipse, or a hyperbola, and then sketch a graph of the curve. For each parabola, specify the location of the focus and the equation of the directrix; for each ellipse, label the coordinates of the vertices and foci, and find the lengths of the major and minor axes; for each hyperbola, label the coordinates of the vertices and foci, and find the equations of the asymptotes.


x² + y²/9 = 1

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