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Ch. 11 - Parametric Equations and Polar Coordinates
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 11 - Parametric Equations and Polar CoordinatesProblema 11.1.36
Capitolo 11, Problema 11.1.36

Finding Parametric Equations


In Exercises 31–36, find a parametrization for the curve.


the ray (half line) with initial point (-1,2) that passes through the point (0,0)

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Identify the initial point of the ray as \(P_0 = (-1, 2)\) and the point it passes through as \(P_1 = (0, 0)\).
Find the direction vector \(\vec{d}\) of the ray by subtracting the coordinates of \(P_0\) from \(P_1\): \(\vec{d} = (0 - (-1), 0 - 2) = (1, -2)\).
Write the parametric equations for the ray starting at \(P_0\) and moving in the direction of \(\vec{d}\) using a parameter \(t \geq 0\): \(x(t) = -1 + 1 \cdot t\) and \(y(t) = 2 - 2 \cdot t\).
Specify the domain of the parameter \(t\) to ensure the curve is a ray (half-line), so \(t \geq 0\).
Thus, the parametric equations describe all points starting at \((-1, 2)\) and moving towards \((0, 0)\) and beyond in the direction of the vector \((1, -2)\).

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Parametric Equations of a Line

Parametric equations express the coordinates of points on a line as functions of a parameter, usually t. For a line through a point P with direction vector d, the parametric form is x = x_0 + t * d_x and y = y_0 + t * d_y, where (x_0, y_0) is the initial point.
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Parameterizing Equations

Direction Vector

The direction vector defines the line's orientation and is found by subtracting the initial point coordinates from another point on the line. For the ray starting at (-1, 2) and passing through (0, 0), the direction vector is (0 - (-1), 0 - 2) = (1, -2).
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Finding Limits by Direct Substitution

Parameter Domain for a Ray

A ray is a half-line starting at an initial point and extending infinitely in one direction. To represent a ray parametrically, the parameter t must be restricted to t ≥ 0, ensuring points lie on or beyond the initial point along the direction vector.
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Eliminating the Parameter
Pratica correlata
Domanda del libro di testo

Examples of Polar Equations


[Technology Exercise] Graph the lines and conic sections in Exercises 65–74.


r = −2 cos θ

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Cartesian to Polar Equations


Replace the Cartesian equations in Exercises 53–66 with equivalent polar equations.


x - y = 3

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Domanda del libro di testo

Polar to Cartesian Equations


Replace the polar equations in Exercises 27–52 with equivalent Cartesian equations. Then describe or identify the graph.


r² = 4r sin θ

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Domanda del libro di testo

Finding Parametric Equations


In Exercises 31–36, find a parametrization for the curve.


the line segment with endpoints (-1,3) and (3,-2)

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Domanda del libro di testo

Finding Cartesian from Parametric Equations


Exercises 1–18 give parametric equations and parameter intervals for the motion of a particle in the xy-plane. Identify the particle’s path by finding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direction of motion.


x = 1 + sin t, y = cos t − 2, 0 ≤ t ≤ π

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Domanda del libro di testo

Symmetries and Polar Graphs


Identify the symmetries of the curves in Exercises 1–12. Then sketch the curves in the xy-plane.


r = 1 + 2 sin θ

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