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The position vector of a particle r(t) = ti + 7 In j+k. Find the acceleration 5 +t vector at t = 1 7 a(1) = 20i - 36j - 14k a(1) = 20i +j + 14k %3D a(1) = 20i + j - 14k %3D 7 a(1) = 20i + 6j + 14k

The position vector of a particle r(t) = ti + 7 In j+k. Find the acceleration 5 +t vector at t = 1 7 a(1) = 20i - 36j - 14k a(1) = 20i +j + 14k %3D a(1) = 20i + j - 14k %3D 7 a(1) = 20i + 6j + 14k

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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The position vector of a particle is r(t) = ti + 7 In 5j+k. Find the acceleration 5+1 vector at t = 1 7 a(1) = 20i - 36j - 14k O a(1) = 20i + j + 14k %3D a(1) = 20i + j - 14k %3! 7 a(1) = 20i + j + 14k
Transcribed Image Text:The position vector of a particle is r(t) = ti + 7 In 5j+k. Find the acceleration 5+1 vector at t = 1 7 a(1) = 20i - 36j - 14k O a(1) = 20i + j + 14k %3D a(1) = 20i + j - 14k %3! 7 a(1) = 20i + j + 14k
Find the equations for the line through the point P(-6, 4, 2), and parallel to the vector u x v where u= 3i + 6j – 4k and v--5i - 3j - 7k. Ox = -62t - 6, y = -41t + 4, z = 21t + 2 x -62t - 6, y = 41t + 4, z = 21t + 2 Ox = -62t - 6, y = 41t + 4, z = -2t + 2 Ox = -62t + 6, y = 41t - 4, z = -2t - 2
Transcribed Image Text:Find the equations for the line through the point P(-6, 4, 2), and parallel to the vector u x v where u= 3i + 6j – 4k and v--5i - 3j - 7k. Ox = -62t - 6, y = -41t + 4, z = 21t + 2 x -62t - 6, y = 41t + 4, z = 21t + 2 Ox = -62t - 6, y = 41t + 4, z = -2t + 2 Ox = -62t + 6, y = 41t - 4, z = -2t - 2
Expert Solution
Step 1

Let r(t) be a twice differentiable vector valued function representing the position vector of a particle at time t. Then the acceleration vector is the second derivative of the position vector.

So,

       r(t)= =t5i + 7ln15+tj + 7tk

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