Question

Find the exact length of the curve.

Answer 1

\[x = \frac{ y^4 }{ 8 } - \frac{ 1 }{ 4y^2 }\] [1,2]

Answer 2

Here is what I have so far.
\[x ^{'} = \frac{ y^3 }{ 2 }-\frac{ 1 }{ 2y^3 }\]
\[\int\limits_{1}^{2} \sqrt{1+(\frac{ y^6 }{ 4 }-\frac{ 1 }{ 2 }-\frac{ 1 }{ 4y^6 })}\]
\[\int\limits_{1}^{2} \sqrt{1+(\frac{ y^3 }{ 2 }-\frac{ 1 }{ 2y^3 })^2}\]
\[\int\limits_{1}^{2}\sqrt{\frac{ y^6 }{ 4 }+\frac{ 1 }{ 2 }-\frac{ 1 }{ 4y^6 }}\]

Answer 3

But Idk what to do from there.

Answer 4

Could I factor the inside? My professor did something similar to that. And canceled the square root that way.

Answer 5

Yes, notice when you had the -1/2 in the middle, it was a perfect square.
So with the +1/2 in the middle, it's probably some other perfect square.

Answer 6

Not seeing it yet? :o

Answer 7

Not really. Its harder to do with fractions :(

Answer 8

Your derivative is wrong... missed a sign

Answer 9

Oh true :U

Answer 10

\[x ^{'}=\frac{ 4y^{4-1} }{ 8 } + \frac{ -2y^{-2-1} }{ 4 } = \frac{ y^3 }{ 2 } - \frac{ y^{-3} }{ 2 }\]

Answer 11

?

Answer 12

I guess maybe you posted the original problem incorrectly.
It shows subtraction in the middle, not addition.

I'm not sure which one is correct.

Answer 13

Oh... Yea I did post the original incorrectly. I apologize..... -sigh-

Answer 14

Ah good :)
Otherwise this problem doesn't work out very nicely lol

Answer 15

But I am stuck on how to factor this bad boi.

Answer 16

\[\large\rm \left(\frac{y^3}{2}-\frac{1}{2y^3}\right)^2=\frac{y^6}{4}\color{orangered}{-\frac12}+\frac{1}{4y^6}\]

Answer 17

Ok fine I'll just ruin the fun for you then :P\[\large\rm \left(\frac{y^3}{2}+\frac{1}{2y^3}\right)^2=\frac{y^6}{4}\color{orangered}{+\frac12}+\frac{1}{4y^6}\]

Answer 18

think of all the work that they put in to making a question you can actually answer (i.e \(1+f'^2\) is a perfect square )
i tried it once, it is not that easy

Answer 19

Ya it's kinda funny how perfectly chosen these functions are :D lol

Answer 20

ikr... i've always thought "how the hell else would you integrate this, were it not so damned convenient"

Answer 21

@myininaya (who i dearly miss) showed me once how to do it with polynomials

Answer 22

whom*

Answer 23

oooo snap

Answer 24

LOL

Answer 25

you just got got son

Answer 26

LOL @zepdrix

Answer 27

XD

Answer 28

Also major LOL to @Seratul

Answer 29

That's a minor faux pas...
It really triggers me though when people use the wrong your/you're XD lol

Like someone will be like "your welcome!"
and I'm thinking in my head.. "my welcome..?"

Answer 30

Maybe they were complimenting your welcome mat at your front door, and were just so blown away by its magnificence, that they got stunned into silence, mid-sentence.

Answer 31

lol :D

Answer 32

Hm, I think it might be who. Oh well.

Answer 33

@Seratul LMAO

Answer 34

@zepdrix @agent0smith Oh I got what you had, its just I forgot to make the it \[+\frac{ 1 }{ 4y^6 }\] If you look at my work. So I didnt think it was correct. Thanks for you're help.

Answer 35

@zepdrix Your the best! :)

Answer 36

LOL

Imagine if someone was talking one of their parents
"My mother, who I dearly miss..."
"*whom" - seratul

Answer 37

All this SAT practice has me confused now. Sorry @satellite73 lmao.

Answer 38

Awesome thread :D

Answer 39

.

Answer 40

my the best? -_- oh boy he triggered me .. i see what he did there..

Answer 41

lmao x'd

Answer 42

@zepdrix Hehe.

Answer 43

I have a problem tho.
I got the wrong answer...

Answer 44

So...

\[\int\limits_{1}^{2}\frac{ y^3 }{ 2 }+\frac{ 1 }{ 2y^3 } = \left[ 2y^4-\frac{ 1 }{ y^2 } \right]_{1}^{2} = \left[ (32 - \frac{ 1 }{ 3 } ) -(2-1) \right] = \frac{ 123 }{ 4 }\]

Answer 45

Thats what I got after the cancellation.

Answer 46

oops, i meant 1/4.

Answer 47

I made a mistake somewhere as well... let's see if we can figure this out.

Answer 48

I just want to say...big fan of this thread! Lol

Answer 49

My threads are always fun. 8)

Answer 50

I take full credit.

Answer 51

lol, you were offline for most of the fun XD pshhh
\[\large\rm \int\limits\limits_{1}^{2}\frac12y^3+\frac12y^{-3}dy\]So power rule on each term,\[\large\rm \frac18y^4-\frac14y^{-2}|_1^2\]Oh ok ok ok I see what I did wrong on paper.

Answer 52

Ya I think your coefficients are messed up.

Answer 53

Power rule? Are you taking the derivative while integrating?? Tsk Tsk

Answer 54

Wait.... Im lost

Answer 55

Power rule for integration :o

Answer 56

LMAO im stupid.

Answer 57

Wait but why did it go to the denominator?

Answer 58

I should call it something else :P
That seems to confuse people a lot

Answer 59

Umm

Answer 60

Wanna make that correction...?

Answer 61

\[\large\rm \int\limits y^n~dy=\frac{1}{n+1}y^{n+1}\]You're asking why this new exponent goes to the denominator?

Answer 62

There we go ^_^

Answer 63

That johnny -_- always sassin' me

Answer 64

OH RIGHT. Gooooood. I am the one who ended up mixing it up with Deriving...

Answer 65

Ohh I see, ya you bought the 4 down into the numerator.

Answer 66

I got it now. Arigato Sensei.

Answer 67

noiceee

Answer 68

Thank you for you're time. <3

Answer 69

Wow cloud9 is gettin rekt right now!!

Answer 70

cloud9? League of legends?