What is the interpretation of Vx'(t)2 + y'(t)² for a particle following the trajectory (x(t), y(t))?

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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What is the interpretation of Vx'(t)2 + y'(t)² for a particle following the trajectory (x(t), y(t))?
Transcribed Image Text:What is the interpretation of Vx'(t)2 + y'(t)² for a particle following the trajectory (x(t), y(t))?
Expert Solution
Step 1

The particle is following the trajectory (x(t), y(t)).

Velocity across x axis = Vx = Rate of change of x = x'(t)

Velocity across y axis = Vy = Rate of change of y = y'(t)

 

Step 2

Hence, Velocity, V = Vxi + Vyj = x'(t)i + y'(t)j

Hence, speed = magnitude of the velocity vector V = [Vx2 + Vy2]1/2 = [x'(t)2 + y'(t)2]1/2

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