- Length of curve to 4 decimal pl

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SUMMARY

The discussion focuses on calculating the length of the curve defined by the equation $y=xe^{-x}$ over the interval $0 \le x \le 2$ to four decimal places. The correct formula for the arc length is established as $L = \int_0^2 \sqrt{1 + [e^{-x}(1-x)]^2} \, dx$. Participants also mention using Desmos for graphing and refer to the TI Nspire calculator for computational assistance. The conversation highlights the importance of understanding the arc length formula for functions expressed in terms of $x$.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the arc length formula for functions.
  • Experience with graphing tools like Desmos.
  • Knowledge of the TI Nspire calculator for computational tasks.
NEXT STEPS
  • Research the application of the arc length formula in different contexts.
  • Learn how to use Desmos for visualizing calculus concepts.
  • Explore advanced integration techniques for complex functions.
  • Familiarize yourself with the TI Nspire programming capabilities for calculus.
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Students and educators in calculus, mathematicians interested in arc length calculations, and anyone utilizing graphing tools for mathematical analysis.

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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