A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
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
 5 years ago
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

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ds = sqrt(dx^2 + dt^2) if I recall correctly

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since y is a function of "x" dx^2 = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dt was supposed to be dy.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[L=\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2 dx}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a=1 b=2 but what is (dy/dx)?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0looks 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
 5 years ago
Best ResponseYou've already chosen the best response.0dy/dx = the derivative of your function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have (dy/dx)= x^(3)  (1/(4x^(3)))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0id have to try to decipher the original equation you posted and maybe implicit it for y'...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay hold on let me write it again, sorry for it being sloppy hahah

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=(x^{4}/4) + 1/(8x^2)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0neater yes; but make sure that this is the correct equation.... double check it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, lol. And I have to find the arc length of the curve from \[1\le x \le 2\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04x^3 16x y' =  ()  4 (8x^2)^2 this should be y' = dy/dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0can reduce as needed :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y' = x^(3)  (1/4x^(3)) reduced?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yep, that looks good :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now I have to square that. That is when I got : x^(6)  1/2 + 1/(16x^(6))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as long as you kept track of all the stuff, ill trust you on that :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me double check real quick, I'm not sureee..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0can we get that into 1 term? get like denominators be easier?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think that might make everything a bit more confusing... lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04x^6  1  ?? 4x^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ummmm... I'm lost. Sorry! What did you do?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nvm, I see what you did haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hm, maybe.. let me see real quick

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y' = x^(3)  (1/4x^(3)) get like denominators.... 4x^3 is common

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhhh, smart move, that helped. let me make that change real fast

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.016x^12 8x^6 +1  = y'^2 16x^6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah! now I have to put that back into the equation for arc length..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0reduce it if you want :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{2} x^6 + 1/16x^6 + 1/2 dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(S) sqrt[1 + x^6 + (1/16x^6)  (1/2)] dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dont forget its the "square root" of 1 + y'^2 ....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhhh, this is where I got caught up...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, thats where I didn't know what to do next. Should I take the square root of each term separately?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cant do that; they aint multiplied together, it just a one lump sum under that radical and you cant break that up....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^(3) + 1/4x^(3) + 1/sqrt2 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh you cant do that :( darnnnn

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0watch: sqrt(4*3) = sqrt(4) * sqrt(3) = 2sqrt(3) BUT!! sqrt(4+3) NOT equal sqrt(4) + sqrt(3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhhh, you're right. soooo... what do I do?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I dont really know..... thats were I usually have to open up the book and reread what they have in there for that section :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it 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
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh, is there a quadratic function we can make out of those three terms?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yeahhh, trig substitution... ugh

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(1+tan^2) = something or another.... but I always mess it up :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for helping me on such an exhausting problem! :D

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol ..... I hope I learned something :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if y' = tan sqrt(1+tan^2) = sqrt(sec^2) = sec... (S) sec du or something...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hahaha, 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
 5 years ago
Best ResponseYou've already chosen the best response.0totally your choice :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, library is closing soon, which takes my internet with it :) So I will be heading out anywhoos :) It was fun... Ciao
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.