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Describe the graph of the equation. r = − 14i + (1 - t²)j + tk

Describe the graph of the equation. r = − 14i + (1 - t²)j + tk

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Describe the graph of the equation. r = − 14i + (1 − ²)j + tk O It is a parabola in the plane z = -14, with vertex at (1, 0, – 14), opening to the 'right' (positive y). = -14, with vertex at (-14, 1, 0), opening up (positive z). -14, with vertex at (-14, 0, 1), opening to the 'right' (positive y). O It is a parabola in the plane x O It is a parabola in the plane x = O It is a parabola in the plane z = O It is a parabola in the plane x = -14, with vertex at (-14, 1, 0), opening to the 'left' (negative y). -14, with vertex at (1, 0, -14), opening to the 'left' (negative y).
Transcribed Image Text:Describe the graph of the equation. r = − 14i + (1 − ²)j + tk O It is a parabola in the plane z = -14, with vertex at (1, 0, – 14), opening to the 'right' (positive y). = -14, with vertex at (-14, 1, 0), opening up (positive z). -14, with vertex at (-14, 0, 1), opening to the 'right' (positive y). O It is a parabola in the plane x O It is a parabola in the plane x = O It is a parabola in the plane z = O It is a parabola in the plane x = -14, with vertex at (-14, 1, 0), opening to the 'left' (negative y). -14, with vertex at (1, 0, -14), opening to the 'left' (negative y).
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