anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
ds = sqrt(dx^2 + dt^2) if I recall correctly
amistre64
  • amistre64
since y is a function of "x" dx^2 = 1
amistre64
  • amistre64
dt was supposed to be dy.....

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anonymous
  • anonymous
\[L=\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2 dx}\]
anonymous
  • anonymous
a=1 b=2 but what is (dy/dx)?
amistre64
  • 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
  • amistre64
dy/dx = the derivative of your function
anonymous
  • anonymous
I have (dy/dx)= x^(3) - (1/(4x^(3)))
amistre64
  • 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
  • anonymous
Is that right?
amistre64
  • amistre64
id have to try to decipher the original equation you posted and maybe implicit it for y'...
anonymous
  • anonymous
okay hold on let me write it again, sorry for it being sloppy hahah
amistre64
  • amistre64
:)
anonymous
  • anonymous
\[y=(x^{4}/4) + 1/(8x^2)\]
anonymous
  • anonymous
A lil neater :D
amistre64
  • amistre64
neater yes; but make sure that this is the correct equation.... double check it :)
anonymous
  • anonymous
Yes, lol. And I have to find the arc length of the curve from \[1\le x \le 2\]
amistre64
  • amistre64
4x^3 16x y' = ------ (-) --------- 4 (8x^2)^2 this should be y' = dy/dx
amistre64
  • amistre64
can reduce as needed :)
anonymous
  • anonymous
y' = x^(3) - (1/4x^(3)) reduced?
amistre64
  • amistre64
yep, that looks good :)
anonymous
  • anonymous
haha awesome.
anonymous
  • anonymous
now I have to square that. That is when I got : x^(6) - 1/2 + 1/(16x^(6))
amistre64
  • amistre64
as long as you kept track of all the stuff, ill trust you on that :)
anonymous
  • anonymous
let me double check real quick, I'm not sureee..
amistre64
  • amistre64
can we get that into 1 term? get like denominators be easier?
anonymous
  • anonymous
I think that might make everything a bit more confusing... lol
anonymous
  • anonymous
hold on a sec :)
amistre64
  • amistre64
4x^6 - 1 -------- ?? 4x^3
anonymous
  • anonymous
ummmm... I'm lost. Sorry! What did you do?
anonymous
  • anonymous
nvm, I see what you did haha
anonymous
  • anonymous
hm, maybe.. let me see real quick
amistre64
  • amistre64
y' = x^(3) - (1/4x^(3)) get like denominators.... 4x^3 is common
anonymous
  • anonymous
ahhhh, smart move, that helped. let me make that change real fast
amistre64
  • amistre64
16x^12 -8x^6 +1 ------------------ = y'^2 16x^6
anonymous
  • anonymous
Yeah! now I have to put that back into the equation for arc length..
amistre64
  • amistre64
yep...
amistre64
  • amistre64
reduce it if you want :)
anonymous
  • anonymous
\[\int\limits_{1}^{2} x^6 + 1/16x^6 + 1/2 dx\]
amistre64
  • amistre64
(S) sqrt[1 + x^6 + (1/16x^6) - (1/2)] dx
anonymous
  • anonymous
oh thats not too bad
amistre64
  • amistre64
dont forget its the "square root" of 1 + y'^2 ....
anonymous
  • anonymous
ohhhhh, this is where I got caught up...
anonymous
  • anonymous
yeah, thats where I didn't know what to do next. Should I take the square root of each term separately?
amistre64
  • amistre64
cant do that; they aint multiplied together, it just a one lump sum under that radical and you cant break that up....
anonymous
  • anonymous
x^(3) + 1/4x^(3) + 1/sqrt2 ?
anonymous
  • anonymous
oh you cant do that :( darnnnn
anonymous
  • anonymous
are you sure?
amistre64
  • amistre64
watch: sqrt(4*3) = sqrt(4) * sqrt(3) = 2sqrt(3) BUT!! sqrt(4+3) NOT equal sqrt(4) + sqrt(3)
anonymous
  • anonymous
ohhhhh, you're right. soooo... what do I do?
anonymous
  • anonymous
:(
amistre64
  • 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
  • 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
  • anonymous
ahhh, is there a quadratic function we can make out of those three terms?
anonymous
  • anonymous
oh yeahhh, trig substitution... ugh
amistre64
  • amistre64
sqrt(1+tan^2) = something or another.... but I always mess it up :)
anonymous
  • anonymous
lol I'll check...
anonymous
  • anonymous
thanks for helping me on such an exhausting problem! :D
amistre64
  • amistre64
lol ..... I hope I learned something :)
amistre64
  • amistre64
if y' = tan sqrt(1+tan^2) = sqrt(sec^2) = sec... (S) sec du or something...
anonymous
  • 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
  • anonymous
totally your choice :)
amistre64
  • amistre64
well, library is closing soon, which takes my internet with it :) So I will be heading out anywhoos :) It was fun... Ciao
anonymous
  • anonymous
Thanks again :)

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