Arc Length

[Gladius]

Hi, I am trying to find the arc length of f(x) = (x^3/6) - (1/2x) between x = 1 and x = 3. So far I have found its derivative and plugged it into the equation for arc length, but I am having trouble evaluating the integral of sqrt( 1 + ((0.5x^2) + (0.5x^-2))^2). The textbook says that it can be found exactly. I've tried many substitutions, but none of the substitutions I make seem to resemble any of the known integrals in my table of integrals . Thanks in advance.

 

[Gladius]

It's actually the first one (1/2)x^-1 Sorry. I'm not sure how to type like you do.

 

[Gladius]

I'm a bit confused. Shouldn't the integral be \(\displaystyle \int\sqrt{1+\frac{(x^{4}+1)^{2}}{4x^{4}}}\)

if the derivative of the function is \(\displaystyle \frac{x^{2}}{2}+\frac{x^{-2}}{2}\)

By the way, thanks for telling my about this LaTex. It definately helps

 

[galactus]

I am so sorry. I made a mistake and added instead of subtracting. Well, time to regroup.

 

[soroban]

Hello, Gladius!

I'm 99.99% sure there is a typo in the problem . . .

\(\displaystyle \text{Find the arc length of: }\;f(x) \:= \:\frac{x^3}{6} + \frac{1}{2x}\;\text{ from }x = 1\,\text{ to }\,x = 3\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \uparrow\) Here!

\(\displaystyle \text{We want: }L \;=\;\int^3_1\sqrt{1 + (y')^2}\,dx\)


\(\displaystyle \text{We have: }\;y \;=\;\frac{x^3}{6} + \frac{1}{2}x^{-1}\)

\(\displaystyle \text{Then: }\;y' \;=\;\frac{x^2}{2} - \frac{1}{2}x^{-2} \;=\;\frac{x^2}{2} - \frac{1}{2x^2}\)

\(\displaystyle \text{And: }\;(y')^2 \;=\;\left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2 \;=\;\frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}\)

\(\displaystyle \text{So: }\;1 + (y')^2\;=\;1 + \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4} \;=\;\frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^2} \;=\;\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2\)

\(\displaystyle \text{Hence: }\;\sqrt{1 + (y')^2} \;=\;\sqrt{\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2} \;=\;\frac{x^2}{2} + \frac{1}{2x^2}\)


\(\displaystyle \text{And we have: }\;L \;=\;\int^3_1\left(\frac{x^2}{2} + \frac{1}{2}x^{-2}\right)\,dx \quad\hdots\quad\text{Got it?}\)

 

[Gladius]

I'm starting to think the same thing. Its so strange because in the text it clearly has a subtraction sign, but as noted by you and Galactus, it gets very complicated. Also, all the problems in the text before this one are relatively straight forward, so I would have a hard time believing that this one was meant to be so hard. This is interesting indeed. I'll have to mention it to my professor.

 

[Gladius]

Thanks for all of the help! This was driving me crazy all day.