Question

I need help with a calculus 2 exercise, with a good explanation to what I'm trying to figure out.

The problem involves finding the arc length when y = (x^2)/2 - (lnx/4) given 2≤ x ≤ 4

So I know the formula involves L = Integral from a to b of sqrt(1+(f'(x))^2)dx

And I took the the derivative of y = (x^2)/2 - (lnx/4)
And got = x - (1/4x)
I then kept reducing it to suit the formula
1 + ( x - 1/4x)^2
I added the alike terms
1 + x^2 - 1/2 + 1/16x^2
x^2 + 1/2 + 1/16x^2
AND then comes my question!
So it is supposed to become this afterwards: (x + 1/4x) ^ 2
But how did that happen, I don't understand how I can reduce it to a square, please help me figure it out.

Answer 1

Explanation:

When writing rational expressions, you need to be aware that ...

  1/4x = (1/4)x ≠ 1/(4x)

Parentheses around the denominator are required, unless you're typesetting the expression and can use a fraction bar for grouping.

The derivative of the curve expression is ...

  y' = x - 1/(4x) . . . . . parentheses added to what you wrote

and the expression (1 -(y')^2) can be written ...

  1 -(y')^2 = x^2 +1/2 +1/(16x^2) . . . . . parentheses added to what you wrote

The first and last terms of this trinomial are both perfect squares, so you might suspect the whole trinomial is a perfect square. You recall that ...

  (a +b)^2 = a^2 + 2ab + b^2

This is a good "pattern" to remember. Using it is a matter of pattern recognition, as is the case with a lot of math.

Here, you have ...

  a = x

  b = 1/(4x)

In order for your trinomial to be a perfect square, the product 2ab must equal the middle term of your trinomial. (Spoiler: it does.)

  2ab = 2(x)(1/(4x)) = (2x)/(4x) = 1/2 . . . . . matches the middle term of 1 -(y')^2

Hence your trinomial can be written as the square ...

  1 -(y')^2 = (x +1/(4x))^2

_____

This is convenient because you want to integrate the square root of this. Your integral then becomes ...

$\displaystyle\int\limits_{2}^{4}{\left(x+\frac{1}{4x}\right)\,dx$