# Homework Help: Crazy Function for Arc Length!

1. Jan 23, 2008

### jen333

Hi! Here's my question on finding arc length. If I've taken the derivative correctly, is there anyway I can simplify it before putting it into the arc length formula?

1. The problem statement, all variables and given/known data
Find the arc length where 0$$\leq$$x$$\leq$$2
y=(x$$^{3}$$/3)+x$$^{2}$$+x+1/(4x+4)

2. Relevant equations
L=$$\int$$ds=$$\sqrt{1+(dy/dx)^{2}}$$

3. The attempt at a solution
I've only taken the derivative so far:
(dy/dx)=x$$^{2}$$+2x+1-4(4x+4)$$^{-2}$$
=(x+1)$$^{2}$$-4(4x+4)$$^{-2}$$

I tried expanding the equation, but that only makes it more complex.

I know to find the arc length I need to square the derivative and place it in the formula (and possibly using substitution), but I'm just wondering how I can simplify the above equation to make it easier to square and calculate!

Last edited: Jan 23, 2008
2. Jan 24, 2008

### Gib Z

Polynomial division!! =]

3. Jan 24, 2008

### Defennder

Yeah you should get this if you do the polynomial division correctly:

$$\int_0^2 (x+1)^2 + \frac{1}{4(x+1)^4} dx$$.

To make things easier, denote (x+1) by A or some other letter, then it'll look easier.

EDIT: Fixed error

Last edited: Jan 24, 2008
4. Jan 24, 2008

### HallsofIvy

I would look at it like this:
$$y'= x^2+ 2x+ 1- \frac{4}{(4x+ 4)^2}= x^2+ 2x+ 1- \frac{1}{4(x+1)^2}$$
$$= (x+ 1)^2- \left(\frac{1}{2(x+1)}\right)^2$$
so
$$(y')^2= (x+1)^4- 1/2+ \left(\frac{1}{2(x+1)}\right)^4$$
$$(y')^2+ 1= (x+1)^4+ 1/2+ \left(\frac{1}{2(x+1)}\right)^4$$
$$= \left((x+1)^2+ \left(\frac{1}{2(x+1)}\right)^2\right)^2$$
so that
$$\sqrt{(y')^2+ 1}= (x+1)^2+ \frac{1}{2(x+2)}$$

5. Jan 24, 2008

### Defennder

There's a small error here, the last term should read 1/4(x+1)^2 instead.