Asked by Arc Length
                Find the length of the arc formed by
y = (1/8)(4x^2-2ln(x)) from x=4 to x=8.
I found the derivative of the function and got y'= x-(1/4x)
Where I'm lost now is after plugging it into the arc length equation: integral of sqrt(1+(x-(1/4x))^2). Squaring the derivative yields me sqrt(1+x^2+1/16x-1/2). Help please.
            
        y = (1/8)(4x^2-2ln(x)) from x=4 to x=8.
I found the derivative of the function and got y'= x-(1/4x)
Where I'm lost now is after plugging it into the arc length equation: integral of sqrt(1+(x-(1/4x))^2). Squaring the derivative yields me sqrt(1+x^2+1/16x-1/2). Help please.
Answers
                    Answered by
            Damon
            
    I got sqrt [ 1 + x^2 -(1/2) + 1/(16x^2) ]
which is
sqrt [ x^2 + (1/2) +1/(16x^2) ]
BUT
x^2 + (1/2) + 1/(16x^2) = [x+ 1/(4x)]^2
ok ?
    
which is
sqrt [ x^2 + (1/2) +1/(16x^2) ]
BUT
x^2 + (1/2) + 1/(16x^2) = [x+ 1/(4x)]^2
ok ?
                    Answered by
            Arc Length
            
    Oh yes, sorry, I merely had a typo. I had the same result as you. What I'm confused about is what to do from that point on. 
    
                    Answered by
            Damon
            
    well, the sqrt of that is just
x + (1/4)(1/x)
integral of that is
(1/2) x^2 + (1/4) ln x
    
x + (1/4)(1/x)
integral of that is
(1/2) x^2 + (1/4) ln x
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.