- Length of curve to 4 decimal pl

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    Curve Length
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Discussion Overview

The discussion revolves around calculating the length of the curve defined by the function \(y=xe^{-x}\) over the interval \(0 \le x \le 2\). Participants explore different approaches to the arc length formula, including potential misunderstandings regarding the application of the formula based on the function's representation.

Discussion Character

  • Mathematical reasoning, Homework-related, Technical explanation

Main Points Raised

  • One participant presents the arc length formula from a textbook, indicating a need to review the calculation after a long time.
  • Another participant points out that the cited equation is appropriate for \(x\) as a function of \(y\), suggesting an alternative formula for the arc length of \(y\) as a function of \(x\).
  • Several participants inquire about the source or method used to generate a related image, indicating a focus on visual representation.
  • A participant mentions a general formula for arc length in three dimensions, which may not directly apply to the current discussion but introduces a broader context.

Areas of Agreement / Disagreement

There is no consensus on the correct application of the arc length formula, as participants present differing views on the appropriate equations to use based on the function's representation.

Contextual Notes

Participants have not resolved the assumptions regarding the function's representation and the corresponding arc length formula, leading to potential confusion in the application of the equations discussed.

karush
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8.1.1
find the length of the curve to four decimal places
$y=xe^{-x},\quad 0 \le x \le 2$
eq from book
$L=\int_c^d\sqrt{1+[g'(y)]^2}\, dy$

ok I haven't done this in about 2 years and only did a few then so trying to review
rare stuff kinda

desmos graph
 
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your cited book equation is for x as a function of y

the following is for the arclength of y as a function of x

$\displaystyle L = \int_0^2 \sqrt{1 + [e^{-x}(1-x)]^2} \, dx$

arclength1.jpg
 
what generated the image?
 
karush said:
what generated the image?

TI Nspire program on my laptop
 
loaned out my TI Inspire
never got it back... :cry:
 
If we are given that x= f(t), y= g(t), and z= h(t), for \[t_1> t> t_0\] then the arclength is \[\int_{t_0}^{t_1} \sqrt{f^{2}(t)+ g^{2}(t)+ h^2(t)}dt\].
 

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