Question

For the curve y=x^3/12+x^-1 I must find the length from x=1 to x=3.

Answer 1

To be more specific, I don't need any sort of answer, just a bit of help getting past one part of it so I can finish it on my own.
So, the arclength is equal to \[\int\limits_{a}^{b}\sqrt{1+(\frac{ dy }{ dx })^2 dx}\]
I end up getting dy/dx=1/4x^2+1/x^2, which when squared is equal to x^4/16+1/2+1/x^4.
Putting that back in it gives us
\[\int\limits_{a}^{b}\sqrt{\frac{ x^4 }{ 16 }+ \frac{ 3 }{ 2 }+\frac{ 1 }{ x^4 }} dx\]

I don't know how to manipulate it into anything easier (I have to integrate it by hand.)

Answer 2

\[ y=\frac{x^3}{12}+\frac{1}{x}\]
\[y' = \frac{x^2}{4} - \frac{1}{x^2}\]

Answer 3

ie \[y' = \frac{x^2}{4} \color{red}- \frac{1}{x^2}\]

Answer 4

I had that originally but even then it was still a really ugly trinomial.

Answer 5

\[y'^2 = \frac{x^4}{16} - \frac{1}{2} + \frac{1}{x^4}\]
\[y'^2 \color{red}{+ 1 }= \frac{x^4}{16} + \frac{1}{2} + \frac{1}{x^4}\]
\[= (\frac{x^2}{4} + \frac{1}{x^2})^2\]

Answer 6

Yeah, so that's what I had down on my paper but didn't realize it would make a perfect square. Thank you sir.

Answer 7

mp!