`y = 1/8(-2x^2 + 4 lnx)`

To determine the length of the arc from x=3 to x=8, apply the formula of arc length (S) which is:

`S = int_a^b sqrt(1+ ((dy)/(dx))^2) dx`

So, take the derivative of y.

`(dy)/(dx) = 1/8(-4x + 4/x) = 1/8 * (-4)(x -1/x) = -1/2...

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`y = 1/8(-2x^2 + 4 lnx)`

To determine the length of the arc from x=3 to x=8, apply the formula of arc length (S) which is:

`S = int_a^b sqrt(1+ ((dy)/(dx))^2) dx`

So, take the derivative of y.

`(dy)/(dx) = 1/8(-4x + 4/x) = 1/8 * (-4)(x -1/x) = -1/2 (x - 1/x)= -1/2 ((x^2 - 1)/x)= - (x^2 - 1)/(2x)`

Then, take the square of `(dy)/(dx)` .

`((dy)/(dx))^2 = (-(x^2-1)/(2x))^2 = (x^2-1)/(2x) * (x^2-1)/(2x) = (x^4-2x^2 +1)/(4x^2)`

Then, add 1 to `((dy)/(dx))^2` .

`1 + ((dy)/(dx))^2=1+(x^4 - 2x^2+1)/(4x^2) = (4x^2 +x^4-2x^2 + 1)/(4x^2) =(x^4 + 2x^2 + 1)/(4x^2) `

`= (x^2+1)^2/(4x^2)`

And, take the square root of `1 + ((dy)/(dx))^2` .

`sqrt(1 + ((dy)/(dx))^2) = sqrt((x^2+1)^2/(4x^2)) = (x^2 +1)/(2x) = x^2/(2x) + 1/(2x)= x/2 + 1/(2x)`

Substitute this to the integral above to determine the arc length.

`S= int_a^b sqrt(1 + ((dy)/(dx))^2) dx= int_3^8 (x/2 + 1/(2x))dx = 1/2 int_3^8 (x + 1/x)dx`

`S= 1/2(x^2/2 + lnx) ` `|_3^8` `= (1/4x^2+1/2lnx )` `|_3^8`

`S = (1/4*8^2 + 1/2 ln 8) - (1/4*3^2 - 1/2 ln3) `

`S= 64/4+1/2 ln8 - 9/4 - 1/2ln3 = 55/4 + 1/2ln8 - 1/2 ln3 = 14.24`

**Hence, the length of the arc is 14.24 units.**