find the arc length of the function y=((x^4)/8)+1/(4x^2) on the interval [1,2]
what was the formula for arc length.....sqrt(1+dy/dx)?
\[s=\int\limits_{a}^{b} \sqrt{1+(f'(x))^{2}} dx\] I'm pretty sure that's it?
k so forget about that formula for now and focus on finding dy/dx. what do you get?
\[x ^{4}/8 + 1/4x ^{2} \] I this the right equation or is the whole thing being divided by 4x^2?
\[y'=x ^{3}/2+1/(2x ^{3})\] I got this
yeah that is the first equation
got a -1/2x^3
you forgot a minus sign
oh yeah thanks
k, now that you have that lets combine them and then square it
because the equation is asking for (dy/dx)^2
ok so when you combine them you get this, right?
i got (2x^6 - 2)/(4x^3) when i combined them
\[(x^{6}-1)/4x^{6}\] i got this...
the way you can check to see if its right is to pick any value for x and see if you get the same answer in both the new equation and the old one. lets say x=2. do they equal in your two equations? my equation is actually not simplified. you should get (x^6-1)/2x^3
k im close to the final answer. u still here?
yeah sorry i am still here. after you square it, why isn't it over 4x^6?
i still havent squared it. after you do that it should be (x^6-1)^2/4x^6
oh i see. ok that makes more sense
now looking back at the formula, you know that you have to add 1. so what do you get?
the algebra here is a little tricky. let me know what you get and ill chck it
(x^12+2x^6+1)/(4x^6)
k now factor the top, can you tell me why that will be useful?
oh, because it goes under a square root so you want a perfect square if possible
yup, it works out pretty well so what do you get after taking the square root?
(x^6+1)/(2x^3)
thats what i got. now take the integral from 1 to 2
i got 5/12 as my answer. what did you get?
ummm...not that. I got 123/16
I'll check my math hold on
wolfram says answer is 33/16...so algebra mistake somewhere
yeah, thats what my answer key says also
I found my mistake. Thank you so much! This was really helpful :)
np, but i still havent found my mistake haha. what was your equation after you take the integral?
oh i just found it
(x^4)/8-1/(4x^2)
glad i could help
me to. again, thank you!!
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