Anyone know how to find the arc length of the curve: y=(((x^(4))/(4))+(1/(8x^(2))) on the interval 1less than or equal to x less than or equal to 2

- anonymous

- schrodinger

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- amistre64

ds = sqrt(dx^2 + dt^2) if I recall correctly

- amistre64

since y is a function of "x" dx^2 = 1

- amistre64

dt was supposed to be dy.....

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## More answers

- anonymous

\[L=\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2 dx}\]

- anonymous

a=1
b=2
but what is (dy/dx)?

- amistre64

looks right.... but I think it has a "ds" on the end, not that it matters much :) and make sure your "dx" is out of the radical

- amistre64

dy/dx = the derivative of your function

- anonymous

I have (dy/dx)= x^(3) - (1/(4x^(3)))

- amistre64

the length thing is the square root of the sum off the derivatives of x and y; since y is a function of x; x'=1

- anonymous

Is that right?

- amistre64

id have to try to decipher the original equation you posted and maybe implicit it for y'...

- anonymous

okay hold on let me write it again, sorry for it being sloppy hahah

- amistre64

:)

- anonymous

\[y=(x^{4}/4) + 1/(8x^2)\]

- anonymous

A lil neater :D

- amistre64

neater yes; but make sure that this is the correct equation.... double check it :)

- anonymous

Yes, lol. And I have to find the arc length of the curve from \[1\le x \le 2\]

- amistre64

4x^3 16x
y' = ------ (-) ---------
4 (8x^2)^2
this should be y' = dy/dx

- amistre64

can reduce as needed :)

- anonymous

y' = x^(3) - (1/4x^(3))
reduced?

- amistre64

yep, that looks good :)

- anonymous

haha awesome.

- anonymous

now I have to square that. That is when I got : x^(6) - 1/2 + 1/(16x^(6))

- amistre64

as long as you kept track of all the stuff, ill trust you on that :)

- anonymous

let me double check real quick, I'm not sureee..

- amistre64

can we get that into 1 term? get like denominators be easier?

- anonymous

I think that might make everything a bit more confusing... lol

- anonymous

hold on a sec :)

- amistre64

4x^6 - 1
-------- ??
4x^3

- anonymous

ummmm... I'm lost. Sorry!
What did you do?

- anonymous

nvm, I see what you did haha

- anonymous

hm, maybe.. let me see real quick

- amistre64

y' = x^(3) - (1/4x^(3))
get like denominators.... 4x^3 is common

- anonymous

ahhhh, smart move, that helped. let me make that change real fast

- amistre64

16x^12 -8x^6 +1
------------------ = y'^2
16x^6

- anonymous

Yeah! now I have to put that back into the equation for arc length..

- amistre64

yep...

- amistre64

reduce it if you want :)

- anonymous

\[\int\limits_{1}^{2} x^6 + 1/16x^6 + 1/2 dx\]

- amistre64

(S) sqrt[1 + x^6 + (1/16x^6) - (1/2)] dx

- anonymous

oh thats not too bad

- amistre64

dont forget its the "square root" of 1 + y'^2 ....

- anonymous

ohhhhh, this is where I got caught up...

- anonymous

yeah, thats where I didn't know what to do next. Should I take the square root of each term separately?

- amistre64

cant do that; they aint multiplied together, it just a one lump sum under that radical and you cant break that up....

- anonymous

x^(3) + 1/4x^(3) + 1/sqrt2 ?

- anonymous

oh you cant do that :( darnnnn

- anonymous

are you sure?

- amistre64

watch:
sqrt(4*3) = sqrt(4) * sqrt(3) = 2sqrt(3)
BUT!!
sqrt(4+3) NOT equal sqrt(4) + sqrt(3)

- anonymous

ohhhhh, you're right. soooo... what do I do?

- anonymous

:(

- amistre64

I dont really know..... thats were I usually have to open up the book and re-read what they have in there for that section :)

- amistre64

it looks like you might be able to do some sort of trig substitution, but i just aint confident about my abilities in that yet...

- anonymous

ahhh, is there a quadratic function we can make out of those three terms?

- anonymous

oh yeahhh, trig substitution... ugh

- amistre64

sqrt(1+tan^2) = something or another.... but I always mess it up :)

- anonymous

lol I'll check...

- anonymous

thanks for helping me on such an exhausting problem! :D

- amistre64

lol ..... I hope I learned something :)

- amistre64

if y' = tan
sqrt(1+tan^2) = sqrt(sec^2) = sec...
(S) sec du or something...

- anonymous

hahaha, ummm... I dont know if its a trig substitution. But if you wanted to move to another problem or anything thats fine. I dont want to bother you all night with just this problem

- anonymous

totally your choice :)

- amistre64

well, library is closing soon, which takes my internet with it :) So I will be heading out anywhoos :)
It was fun... Ciao

- anonymous

Thanks again :)

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