Calc. help please! (Arc Length)?

Question

Find the arc length of the graph of y= (x^4)/8 + 1/(4x^2) over the interval [1,2].

Answer 1

First, recall the formula for arc length:

[a, b]∫√(1+(dy/dx)²) dx

Now, find dy/dx:

dy/dx = x³/2 -1/(2x³)

Squaring this:

(dy/dx)² = x^6/4 - 1/2 + 1/(4x^6)

Plugging into the formula:

[1, 2]∫√(1+x^6/4 - 1/2 + 1/(4x^6)) dx

Simplifying:

[1, 2]∫√(x^6/4 + 1/2 + 1/(4x^6)) dx

Extracting a factor of 1/(2x³):

[1, 2]∫1/(2x³) √(x^12 + 2x^6 + 1) dx

Factoring the expression under the square root:

[1, 2]∫1/(2x³) √((x^6+1)²) dx

Since x^6+1 is positive, √((x^6+1)²) = x^6+1, so we have:

[1, 2]∫(x^6+1)/(2x³) dx

Simplifying:

1/2 [1, 2]∫x³/2 + 1/(2x³) dx

Integrating:

x^4/8 - 1/(4x²) |[1, 2]

Substituting:

(16/8 - 1/16) - (1/8 - 1/4)
32/16 - 1/16 - 2/16 + 4/16
33/16

Answer 2

you need to integrate sqrt(1 + [dy/dx]^2) * dx between the limits

dy/dx = 1/2 * (x^3 - x^-3)
You'll find that
sqrt[1 + (dy/dx)^2] = 1/2 * (x^3 + x^-3)

Now integrating this, we get
x^4 / 8 - 1 / (4x^2)
Apply the limits to get 33/16