Derivation of Integral Arc Length Formula

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Homework Help Overview

The discussion revolves around the derivation of the integral formula for arc length as presented in a textbook. The original poster expresses confusion regarding a specific step in the derivation that introduces a logarithmic term.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the introduction of a logarithmic term in the derivation of the arc length formula. Some participants question the validity of this term and its implications on the subsequent steps of the derivation.

Discussion Status

Participants are actively engaging with the original poster's confusion, with some suggesting that the logarithmic term is incorrect. There is a focus on clarifying the mathematical reasoning behind the derivation, but no consensus has been reached regarding the presence of the log term.

Contextual Notes

There is a request for additional context, such as a scan of the textbook page in question, to better understand the derivation process. The original poster's reference to a specific textbook edition indicates reliance on that material for the derivation.

El Moriana
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Homework Statement



My textbook [Engineering Mathematics, Stroud, 6th Edition, page932] runs through the derivation of the integral formula for arc length. I got confused at one of the steps:

[partial](ds/dx)=sqrt(1+([partial](dy/dx))^2)
if [partial]dx tends to 0,
ds/dx=sqrt(1+log(dy/dx)^2)
s=intab(sqrt(1+(dy/dx)^2))

Where does the log come from and where does it go?

Homework Equations



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The Attempt at a Solution



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Could you post a scan of that page or at least that section with the derivation?

I don't think that log is supposed to be there.

How I learned it was in a curve if you join any two points, the 'x' distance would be Δx and the corresponding 'y' distance would be Δy. The chord length would be related as

(ΔS)2= (Δx)2 +(Δy)2

or

(ΔS/Δx)2 = 1 + (Δy/Δx)2

as Δx→0, Δy/Δx = dy/dx and ΔS/Δx = dS/dx

so

\frac{dS}{dx} = \sqrt{1+ \left( \frac{dy}{dx} \right)}

which you can then integrate.
 
Here we go, apologies for the messy formula earlier, I didnt realize there was a Latex button.
I was expecting exactly what you typed in the derivation, hence the confusion on my part.
 

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I really don't think that 'log' should be there. Else the next step does not make sense.
 
Ok, well thanks for looking it over.
 

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