Question

How do you find the length of the curve for `y= 1/8(4x^2–2ln(x))` for [2, 6]?

Answer 1

The arc length is `4.46` units.

Explanation:

We have:

`A = int_2^6 sqrt(1 + (dy/dx)^2)dx`

And by derivative rules

`y= 1/2x^2 - 1/4lnx`

`y' = x - 1/(4x)`

Hence:

`A = int_2^6 sqrt(1 + (x - 1/(4x))^2)dx`

`A = int_2^6 sqrt(1 + ((4x^2 - 1)/(4x))^2) dx`

`A = int_2^6 sqrt((16x^2 + 16x^4 - 8x^2 + 1)/(4x)^2)dx`

`A = int_2^6 sqrt(16x^4 + 8x^2 + 1)/(4x)dx`

`A = int_2^6 sqrt((4x^2 +1)^2)/(4x^2) dx`

`A = int_2^6 (4x^2 + 1)/(4x) dx`

`A = int_2^6 x + 1/(4x) dx`

`A = [1/2x^2 + 1/4ln|x|]_2^6`

`A = 1/2(6)^2 +1/4ln|6| - 1/2(2)^2 - 1/4ln|2|`

`A =18 + 1/4ln|6| - 2 - 1/4ln|2|`

`A = 16.27`

Hopefully this helps!