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Mathematics 13 Online
OpenStudy (anonymous):

find the arc length of the function y=((x^4)/8)+1/(4x^2) on the interval [1,2]

OpenStudy (anonymous):

what was the formula for arc length.....sqrt(1+dy/dx)?

OpenStudy (anonymous):

\[s=\int\limits_{a}^{b} \sqrt{1+(f'(x))^{2}} dx\] I'm pretty sure that's it?

OpenStudy (anonymous):

k so forget about that formula for now and focus on finding dy/dx. what do you get?

OpenStudy (anonymous):

\[x ^{4}/8 + 1/4x ^{2} \] I this the right equation or is the whole thing being divided by 4x^2?

OpenStudy (anonymous):

\[y'=x ^{3}/2+1/(2x ^{3})\] I got this

OpenStudy (anonymous):

yeah that is the first equation

OpenStudy (anonymous):

got a -1/2x^3

OpenStudy (anonymous):

you forgot a minus sign

OpenStudy (anonymous):

oh yeah thanks

OpenStudy (anonymous):

k, now that you have that lets combine them and then square it

OpenStudy (anonymous):

because the equation is asking for (dy/dx)^2

OpenStudy (anonymous):

ok so when you combine them you get this, right?

OpenStudy (anonymous):

i got (2x^6 - 2)/(4x^3) when i combined them

OpenStudy (anonymous):

\[(x^{6}-1)/4x^{6}\] i got this...

OpenStudy (anonymous):

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

OpenStudy (anonymous):

k im close to the final answer. u still here?

OpenStudy (anonymous):

yeah sorry i am still here. after you square it, why isn't it over 4x^6?

OpenStudy (anonymous):

i still havent squared it. after you do that it should be (x^6-1)^2/4x^6

OpenStudy (anonymous):

oh i see. ok that makes more sense

OpenStudy (anonymous):

now looking back at the formula, you know that you have to add 1. so what do you get?

OpenStudy (anonymous):

the algebra here is a little tricky. let me know what you get and ill chck it

OpenStudy (anonymous):

(x^12+2x^6+1)/(4x^6)

OpenStudy (anonymous):

k now factor the top, can you tell me why that will be useful?

OpenStudy (anonymous):

oh, because it goes under a square root so you want a perfect square if possible

OpenStudy (anonymous):

yup, it works out pretty well so what do you get after taking the square root?

OpenStudy (anonymous):

(x^6+1)/(2x^3)

OpenStudy (anonymous):

thats what i got. now take the integral from 1 to 2

OpenStudy (anonymous):

i got 5/12 as my answer. what did you get?

OpenStudy (anonymous):

ummm...not that. I got 123/16

OpenStudy (anonymous):

I'll check my math hold on

OpenStudy (anonymous):

wolfram says answer is 33/16...so algebra mistake somewhere

OpenStudy (anonymous):

yeah, thats what my answer key says also

OpenStudy (anonymous):

I found my mistake. Thank you so much! This was really helpful :)

OpenStudy (anonymous):

np, but i still havent found my mistake haha. what was your equation after you take the integral?

OpenStudy (anonymous):

oh i just found it

OpenStudy (anonymous):

(x^4)/8-1/(4x^2)

OpenStudy (anonymous):

glad i could help

OpenStudy (anonymous):

me to. again, thank you!!

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