# Homework Help: Finding Arc Length

1. Jun 23, 2013

### Bashyboy

1. The problem statement, all variables and given/known data
Find the arc length of the graph, on the interval [1/2, 2], of

$y = \frac{x^3}{6} + \frac{1}{2x}$

2. Relevant equations

$s = \int^b_a \sqrt{1 + [f'(x)]^2}dx$

3. The attempt at a solution

I began with $s = \int_{1/2}^2 \sqrt{1 + (\frac{x^2}{2} - \frac{1}{2x^2})^2}dx$

and became stuck at the step $s = \int_{1/2}^2 \sqrt{1/2 + \frac{x^4}{4} - \frac{1}{4x^4}}dx$

I can't a integration technique that would suffice in solving this problem.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 23, 2013

### SteamKing

Staff Emeritus
Check your expansion of the squared term under the square root. You have made an algebra mistake.
After that, add up the fractions under the square root after getting a common denominator. Plug and chug.

3. Jun 23, 2013

### Bashyboy

I redid the work, and came up with $\frac{x^4}{2} +\frac{1}{2x^4} +\frac{1}{2}$

Is there some way of factoring this?

4. Jun 23, 2013

### SteamKing

Staff Emeritus
Nope, not in this form. Like I said, find a common denominator and add these three fractions together.

5. Jun 24, 2013

### Bashyboy

Okay, I found a common denominator, and then placed the new expression under the square symbol:

$\frac{\sqrt{3x^4 + 2}}{2x^2}$ Is there where you wanted me to lead to?

6. Jun 24, 2013

### ehild

Try again to expand and simplify the expression

$1 +(\frac{x^2}{2} - \frac{1}{2x^2})^2$

ehild

7. Jun 24, 2013

### Bashyboy

ehild, if you could please look at post #6 and see if I correctly simplified the expression.

8. Jun 24, 2013

### ehild

It is wrong, that was I asked you to re-do it.

ehild

9. Jun 24, 2013

### Ray Vickson

Hint: your integral is difficult, but the correct integral is easy.

10. Jun 24, 2013

### SteamKing

Staff Emeritus
(a - b)^2 = a^2 - 2ab + b^2

11. Jun 24, 2013

### HallsofIvy

One of the things teachers work hard on is setting up problems that are easy for their students.

Your function is $x^3/6+ x^{-1}/2$. The derivative is $x^2/2- x^{-2}/2$. Notice that the exponents are 2 and -2. The square is $x^4/4- 1/2+ x^{-4}/4$. The "middle" term is a constant, because, of course, 2+ (-2)= 0. Now adding 1 gives $x^4/4+ 1/2+ x^{-4}/4$. Do you not see that this is exactly the same as the previous formula except that "-1/2" has become "+1/2"? Do you not see that this is exactly "$(x^2/2+ x^{-2})^2$"?

12. Jun 24, 2013

### ehild

Not quite, $(x^2/2+ x^{-2}/2)^2$ instead.

ehild