Find the arc length of the curve y=(1/8)(−x^2+8ln (x)) from x=4 to x=5.

Answer 1

Hi Ana,

So the arc length is the distance along the curve. Each infinitesimal increment of this is made up of the hypotenuse of a right triangle consisting of legs length dx and d(f(x)) respectively.

d(f(x)) = (1/8)d((-x^2)+8ln(x))/dx = (1/8)(-2x)dx + (1/x)dx

So the hypotenuse of those 2 legs = ( (-x/4 + (1/x) )^2 + 1^2)^0.5 dx

Expanding: = ( (x/16 - (1/2) + x^(-2) + 1)^0.5 dx

or ( (x/16) + (1/2) + x^(-2) )^0.5 dx

Next you just must integrate that. That's going to involve some major math functions, since you have to make an algebraic form of the argument ( (x/16) + (1/2) + x^(-2)) by scaling up in powers of x. Recommend you look at Wolfram Alpha for simpler forms (i.e., drop one of the three terms under the square root there), before you put in the full function, and examine it! It's messy.

-- Cheers, -- Mr. d.

Answer 2

The formula for arc length is ∫√(1 + (f '(x))²)dx

For y = 1/8(-x² + 8ln(x)) let's first multiply it out to get:

f(x) = -1/8x² + ln(x)

So f '(x) = -1/4x + 1/x

(f '(x))² = 1/16x² + 1/x²

₄∫⁵√(1 + 1/16x² + 1/x²) dx

This is a pretty complex integral so I'm going to choose to let my TI-84 do the integration

Plugging √(1 + 1/16x² + 1/x²) into my Y1, I'll use "math" 9: fnInt" and type in:

and I get 1.523