Arc Length of y = x^3/6 + 1/2x on [1/2, 2]

[Bashyboy]

Homework Statement

Find the arc length of the graph, on the interval [1/2, 2], of

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

Homework Equations

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

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.

 

[SteamKing]

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.

 

[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?

 

[SteamKing]

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

 

[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?

 

[ehild]

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?

Try again to expand and simplify the expression

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

ehild

 

[Bashyboy]

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

 

[ehild]

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

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

ehild

 

[Ray Vickson]

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?

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

 

[SteamKing]

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

 

[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 "$(x^2/2+ x^{-2})^2$"?

 

[ehild]

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 "$(x^2/2+ x^{-2})^2$"?

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

ehild