How Do You Solve Vector Velocity in 3D?

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SUMMARY

The discussion focuses on solving vector velocity in 3D using the equations provided. The acceleration vector is defined as a(t) = i + (30t^4)j + (2e^(-t)ln(e) - 2e^(-t)(1/e))k, while the position vector is r(t) = (2 + 6t)i + (2 + t^6 + 2t)j + (2 + 2e^(-t))k. Participants confirm the correctness of the i and j components but identify an error in the constant for the k component. The discussion emphasizes the importance of recognizing constants in the equations.

PREREQUISITES
  • Understanding of vector calculus
  • Familiarity with 3D coordinate systems
  • Knowledge of exponential functions and their derivatives
  • Proficiency in solving differential equations
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  • Study vector calculus applications in physics
  • Learn about the properties of exponential functions in calculus
  • Explore the concept of acceleration in 3D motion
  • Review techniques for solving differential equations
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Students in physics or engineering, educators teaching vector calculus, and anyone interested in advanced mathematical modeling of motion in three dimensions.

jmagic
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Homework Statement



problem2.png


Homework Equations


i+j+k=1, <1,1,1>


The Attempt at a Solution



a(t)=i+(30t^(4))j+(2e^(-t)ln(e)-2e^(-t)(1/e))k
r(t)=(2+6t)i+(2+t^6+2t)j+(2+2e^-t)k




[

1. The problem statement, a
 
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Welcome to PF!

Hi jmagic! Welcome to PF! :smile:
jmagic said:
r(t)=(2+6t)i+(2+t^6+2t)j+(2+2e^-t)k

i and j are correct, but the constant in k is wrong …

hint: e0 = … ? :smile:
a(t)=i+(30t^(4))j+(2e^(-t)ln(e)-2e^(-t)(1/e))k

Hint: 6 and ln(e) are constants, aren't they? :wink:
 

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