Rewrite curve as arclength function

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SUMMARY

The discussion focuses on computing the arclength function s(t) for the curve defined by r = . The correct formula for the arclength function is derived as s(t) = 3^(1/2)e^(3t), but it is emphasized that the initial condition s(0) = 0 must be applied correctly. The integral s(t) - s(0) = ∫_0^t |r'(t)| dt is crucial for obtaining the correct arclength function, as it ensures the initial point is accounted for in the calculation.

PREREQUISITES
  • Understanding of parametric curves in three-dimensional space
  • Knowledge of vector calculus, specifically derivatives of vector functions
  • Familiarity with integral calculus and the concept of arclength
  • Experience with exponential functions and their properties
NEXT STEPS
  • Study the derivation of arclength for parametric curves in three dimensions
  • Learn about the application of initial conditions in calculus problems
  • Explore the properties of exponential functions in relation to parametric equations
  • Practice solving similar problems involving arclength and vector functions
USEFUL FOR

Students studying calculus, particularly those focusing on vector calculus and arclength computations, as well as educators looking for examples of applying initial conditions in mathematical problems.

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Homework Statement


Consider the curve r = <cos(3t)e^(3t),sin(3t)e^(3t),e^(3t)>
compute the arclength function s(t) with the initial point t = 0.


Homework Equations


s = integral |r'(t)|dt


The Attempt at a Solution


Okay so if you work all of this out it turns out it's not as bad as it looks.. it's set up to come out really nicely it appears. I end up with

s = 3^(1/2)e^(3t)

but my online homework program is saying that this is wrong... Do I ever use the information that the initial point is t = 0? I don't understand why they need to tell me that...
 
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When you put ##t=0## you aren't getting ##s(0)=0## like the problem asks. You need to calculate$$
s(t) - s(0) = \int_0^t|r'(t)|dt$$with s(0)=0. I'm guessing you didn't handle the lower limit correctly.
 

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