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Determine all points P at which the tangent line to the curve given parametrically by x(t) t³ - 6t, is parallel to the line (−3t, 2t). = Y t²

Determine all points P at which the tangent line to the curve given parametrically by x(t) t³ - 6t, is parallel to the line (−3t, 2t). = Y t²

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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**Determine all points \( P \) at which the tangent line to the curve given parametrically by:** \[ x(t) = t^3 - 6t, \quad y = t^2 \] **is parallel to the line \((-3t, 2t)\).** 1. \( P = (5, -3), \quad (-4, 3) \) 2. \( P = (5, 3), \quad (-4, -3) \) 3. \( P = (-5, -3), \quad (4, 3) \) 4. \( P = (5, 1), \quad (-4, 4) \) 5. \( P = (-5, 4), \quad (4, 1) \) 6. \( P = (-5, 1), \quad (4, 4) \) **Explanation:** This problem involves finding points on a parametric curve where the tangent line is parallel to a given line. The curve is described by \( x(t) = t^3 - 6t \) and \( y = t^2 \). The condition for parallel lines is that their direction vectors are proportional. Here, the direction vector of the given line is \((-3, 2)\). You need to set up and solve equations based on the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) relative to the direction vector.
Transcribed Image Text:**Determine all points \( P \) at which the tangent line to the curve given parametrically by:** \[ x(t) = t^3 - 6t, \quad y = t^2 \] **is parallel to the line \((-3t, 2t)\).** 1. \( P = (5, -3), \quad (-4, 3) \) 2. \( P = (5, 3), \quad (-4, -3) \) 3. \( P = (-5, -3), \quad (4, 3) \) 4. \( P = (5, 1), \quad (-4, 4) \) 5. \( P = (-5, 4), \quad (4, 1) \) 6. \( P = (-5, 1), \quad (4, 4) \) **Explanation:** This problem involves finding points on a parametric curve where the tangent line is parallel to a given line. The curve is described by \( x(t) = t^3 - 6t \) and \( y = t^2 \). The condition for parallel lines is that their direction vectors are proportional. Here, the direction vector of the given line is \((-3, 2)\). You need to set up and solve equations based on the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) relative to the direction vector.
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