Question

I'm trying to figure out another Calculus II arc length question that's a little different. (Posted below in a minute.)

Answer 1

\[x = \frac{ y ^{3} }{ 3 }+ \frac{ 1 }{ 4y }\]\[\frac{ dx }{ dy } = y ^{2}-\frac{ 1 }{ 4y ^{2} }\]\[(\frac{ dx }{ dy })^{2} = y ^{4} - \frac{ 1 }{ 2 }+\frac{ 1 }{ 16y ^{2} }\]
Arc Length formula:\[\int\limits_{a}^{b}\sqrt{1 + f'(x)^{2}}dx\]
\[\int\limits_{1}^{3}\sqrt{y ^{4}+\frac{ 1 }{ 2 }+\frac{ 1 }{ 16y ^{4} }}\]
I tried approaching this with an integral solver and looking through the steps, but Wolfram is coming up with an absurdly complicated antiderivative and I'm thinking that there absolutely has to be some intermediate steps of simplifying or manipulating the integrand to lead to a simpler antiderivative. Any help would be appreciated.

Answer 2

(Also, I thought of trying to factor it liek @Psymon suggested with an earlier problem, but I have no idea how to factor polynomials where you have both positive and negative exponents, and I don't actually know if it can be done, at least with a neat, usable result.)

Answer 3

@terenzreignz @ⒶArchie☁✪

Answer 4

Just in case you come back to see this or you have not found and answer yet.
You can Simplify the integral into like this: (I had a similar problem in my Cal II Class, If i remember correctly)
\[\int\limits_{3}^{1}\sqrt{(y^2+\frac{ 1 }{ 4y^2 })^2}\]
It is difficult to see it, since the negative exponent makes it confusing. But look at your First derivative, and you will see that it's the same except for the negative sign. The +1 changes the sign. That's the clue our professor gave us.
Hope it helps or at least points you in the right direction

Answer 5

Oh, thanks. That makes things *way* easier, I see. So then you just remove the square root sign, take the integral of that (y^3/3 -1/4y) and the evaluate it at the upper/lower bounds.

Answer 6

Yes, exactly. Once you remove the square root you can do the rest

Answer 7

Thanks!

Answer 8

But pretty much, is factoring it. Its just hard to see the factors. So just go by the clue i guess. Lol. And no problem :)