an easy Integral (I think)

N

NRS

Junior Member
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Sep 14, 2009
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Okay, I think this will be thelast question I post today.

I'm trying to find the arc length of this equation: (1/12) x^3 + (1/x)

I square the derivative to get (1/16) x^4 + 1/x^4 + 1/2

Then, I put it into the formula to get:

integral( sqrt ( 1 + 1/16)x^4 + 1/ x^4 + 1/2))dx

From here on, I'm stuck. :(
 
Do not ever underestimate the power of a simple-looking arc-length problem to be much more strenuous than it appears.

You have it right. You should not be stuck. Remember your algebra. Add up those three fractions and see if you can factor the numerator. Maybe you'll get lucky. :D (Thought question: What do you need to know before you get rid of the radical?)

And, just for the record, you may wish to notice two things: 1) Symmetry about the Origin (It's an odd function - Why would this be important?) and 2) Discontinuity at x = 0.
 
tkhunny said:
(Thought question: What do you need to know before you get rid of the radical?)
Click to expand...

I honestly can't remember, in all of my calculusing, my algebra suffered very much.

Although, I was able to easily combine the fractions to get:

integral ( sqrt (x^8 + 24 x ^4 + 16) / 4x^2)

I am still suffering from a severe case of mental block... Maybe just a few more hints....
 
You must have wondered off somewhere. You should get a perfect square in the numerator. Just try it again and be more careful.

If your algebra needs to eb better, upgrade it. Calculus will only be more difficult. Do some geometry and trig, too. It will be worth the effort.
 
I very much appreciate you're tutoring, but I need this problem as a model. I didn't sleep well last night and am unable to think through this problem right now, I have been working on math all day. I'm not trying to make you feel bad for me, I'm just telling you that all the tutoring in the world won't help me right now, and I'd love to get some sleep tonight. Could you please just give me the steps to the problem?
 
I would, but you already have them. That doesn't srike me as particularly useful.

There are very few times in life where thinking and learning will not be beneficial. I seriously doubt this is one of those times.

Do it very carefully. Get it right. Then you will have your model. You almost have it.

\(\displaystyle f(x)\;=\;\frac{x^{3}}{12} + \frac{1}{x}\) - You have that.

\(\displaystyle \frac{df}{dx}\;=\;\frac{x^2}{4}-\frac{1}{x^{2}}\) - You have that.

\(\displaystyle \frac{df}{dx}\;=\;\frac{x^4 - 4}{4x^{2}}\) - That's not hard, is it?

\(\displaystyle \left(\frac{df}{dx}\right)^{2}\;=\;\left(\frac{x^4 - 4}{4x^{2}}\right)^{2}\) - That's simple enough.

\(\displaystyle 1 + \left(\frac{df}{dx}\right)^{2}\;=\;1 + \left(\frac{x^4 - 4}{4x^{2}}\right)^{2}\;=\;\frac{(x^{4}+4)^{2}}{16x^{4}}\) - Okay, that took a little algebra.

\(\displaystyle \int\sqrt{\left(\frac{x^{4}+4}{4x^{2}}\right)^{2}}\;dx = ??\) - Really, are you going to make me do ALL the work?
 
\(\displaystyle f(x) \ = \ \frac{x^{3}}{12}+\frac{1}{x}, \ f'(x) \ = \ \frac{x^{2}}{4}-\frac{1}{x^{2}}, \ [f'(x)]^{2} \ = \ \frac{x^{4}}{16}-\frac{1}{2}+\frac{1}{x^{4}}.\)

\(\displaystyle s \ = \ \int_{a}^{b}[1+[f'(x)]^{2}]^{1/2}dx \ = \ \int_{a}^{b}\bigg[1+\frac{x^{4}}{16}-\frac{1}{2}+\frac{1}{x^{4}}\bigg]^{1/2}dx \ = \ \int_{a}^{b}\bigg(\frac{x^{8}+8x^{4}+16}{16x^{4}}\bigg)^{1/2}dx\)

\(\displaystyle = \ \int_{a}^{b}\bigg[\frac{(x^{4}+4)^{2}}{(4x^{2})^{2}}\bigg]^{1/2}dx \ = \ \int_{a}^{b}\frac{x^{4}+4}{4x^{2}}dx \ = \ \frac{1}{4}\int_{a}^{b}\bigg[x^{2}+\frac{4}{x^{2}}\bigg]dx\)

\(\displaystyle = \ \frac{1}{4}\bigg[\frac{x^{3}}{3}-4x^{-1}\bigg]_{a}^{b} \ = \ \bigg[\frac{x^{3}}{12}-x^{-1}\bigg]_{a}^{b}\)
 
Wow.... You guys really saved me from a night of pain and misery :evil: ! Thankyou! :D :D

The lesson I have learned:

GET BETTER AT FACTORING!
I wouldn't have even had to worry about all of this if I could have factored that trinomial. Is there a good webpage for factoring tips and methods?
 
\(\displaystyle NRS, I don't \ wish \ to \ rain \ on \ your \ parade, \ but \ the \ majority \ of \ Arc \ Length \ problems \ don't \ factor,\)

\(\displaystyle \ the \ above \ one \ was \ contrived.\)
 
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