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

Question

Isabella Ackerman · Accepted Answer

The arc length is #4.46# units.

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!

William Abdalla · Answer

undefined To find the length of the curve for $y = \frac{1}{8}(4x^2 - 2\ln(x))$ over the interval $[2, 6]$, you can use the arc length formula:

$$L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$

First, find $\frac{dy}{dx}$ by differentiating $y$ with respect to $x$.

$$\frac{dy}{dx} = \frac{1}{2}x - \frac{1}{4x}$$

Now, plug $\frac{dy}{dx}$ back into the arc length formula and integrate over the interval $[2, 6]$:

$$L = \int_{2}^{6} \sqrt{1 + \left(\frac{1}{2}x - \frac{1}{4x}\right)^2} \, dx$$

After integrating this expression, you will find the length of the curve over the given interval.