path length of function

[km707]

Find the path length of y= (x[sup:jebgvynu]3[/sup:jebgvynu])/(a[sup:jebgvynu]2[/sup:jebgvynu]) + (a[sup:jebgvynu]2[/sup:jebgvynu])/(12x)

between a/2< x <a

I know I'm supposed to be integrating ds = dx (1+ (dy/dx)[sup:jebgvynu]2[/sup:jebgvynu])[sup:jebgvynu]1/2[/sup:jebgvynu]

but i'm having trouble simplifying the expression inside the square root..

thanks in advance!

 

[soroban]

Hello, km707!

\(\displaystyle \text{Formula: }\;L \;=\;\int^b_a\sqrt{1 + \left(\tfrac{dy}{dx}\right)^2}\,dx\)

\(\displaystyle \text{Find the path length of: }\;y \:=\:\frac{x^3}{a^2} + \frac{a^2}{12x} \quad\text{ for: }\:\tfrac{a}{2} < x < a\)

\(\displaystyle \text{We have: }\:y \;=\;\frac{1}{a^2}\,x^3 + \frac{a^2}{12}\,x^{-1}\)

\(\displaystyle \text{Differentiate: }\;\frac{dy}{dx} \;=\;\frac{3}{a^2}\,x^2 - \frac{a^2}{12}\,x^{-2}\)

\(\displaystyle \text{Square: }\;\left(\frac{dy}{dx}\right)^2 \;=\;\bigg[\frac{3}{a^2}\,x^2 - \frac{a^2}{12}x^{-2}\bigg]^2 \;=\;\frac{9}{a^4}\,x^4 - \frac{1}{2} + \frac{a^4}{144}\,x^{-4}\)

\(\displaystyle \text{Add 1: }\;1 + \left(\frac{dy}{dx}\right)^2 \;=\;1 + \frac{9}{a^4}\,x^4 - \frac{1}{2} + \frac{a^4}{144}\,x^{-4} \;=\; \frac{9}{a^4}\,x^4 + \frac{1}{2} + \frac{a^4}{144}\,x^{-4}\)

\(\displaystyle \text{Factor: }\;1 + \left(\frac{dy}{dx}\right)^2 \;=\;\left(\frac{3}{a^2}\,x^2 + \frac{a^2}{12}\,x^{-2}\right)^2\)

\(\displaystyle \text{Take square root: }\;\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\frac{3}{a^2}\,x^2 + \frac{a^2}{12}\,x^{-2}\)


\(\displaystyle \text{Therefore: }\;L \;=\;\int^a_{\frac{a}{2}}\left(\frac{3}{a^2}\,x^2 + \frac{a^2}{12}\,x^{-2}\right)dx\)


I assume you can finish it . . .

 

[BigGlenntheHeavy]

km707, once you become acclimated in doing Arc Length problems, a good check (if you have a TI-89) is press F3-8 and type in (x^3/a^2+a^2/(12*x),x,a/2,a) and it will return the correct answer, in this case 23a/24.

Note: Only works for Arc Length in Rectangular Coordinates, I think, as I haven't figured out if one can also use this algorithm for Arc Length in parameter or polar form. Any help on this one?