# Thread: Perfect squares for arc lenght

1. ## Perfect squares for arc lenght

hi, I'm having problems recognizing what is a perfect square and what is not. Could anyone share so basic stuff i need to know about them , i'm very lost with them and I have no idea what to do with them. For exemple : $\frac{5x^4}{6}-\frac{3x^{-4}}{10}$

Can i get a full demonstration of why this is a perfect square and tip to recognize it thanks alot

2. Hello, Larrioto!

I'm having problems recognizing what is a perfect square and what is not.

For example: . $\frac{5x^4}{6}-\frac{3x^{-4}}{10}$

Can i get a full demonstration of why this is a perfect square and tip to recognize it?
First of all, it is not a perfect square . . . Let's start at the very beginning.

We have a function: . $y \;=\;\frac{1}{6}x^5 + \frac{1}{10}x^{\:\!\text{-}3}$ .
. . . not a perfect square
. . and we want its arc length over some interval (a,b).

We use the formula: . $L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$

So we form the expession under the radical . . .

We have: . $\frac{dy}{dx} \;=\;\frac{5}{6}x^4 - \frac{3}{10}x^{\:\!\text{-}4}$ .
. . . not a perfect square

Then: . $\left(\frac{dy}{dx}\right)^2 \;=\;\left(\frac{5}{6}x^4 - \frac{3}{10}x^{\:\!\text{-}4}\right)^2 \;=\;\frac{25}{36}x^8 - \frac{1}{2} + \frac{9}{100}x^{\:\!\text{-}8}$

$\text{And: }\;1 + \left(\frac{dy}{dx}\right)^2 \;=\;1 + \frac{25}{36}x^8 - \frac{1}{2} + \frac{9}{100}x^{\;\!\text{-}8} \;=\;\underbrace{\frac{25}{36}x^8 + \frac{1}{2} + \frac{9}{100}x^{\;\!\text{-}8}}_{{\color{blue}\text{This is a perfect square!}}}$ . $=\;\left(\frac{5}{6}x^4 + \frac{3}{10}x^{\;\!\text{-}4}\right)^2$

Hence: . $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\left(\frac{5}{6}x^4 + \frac{3}{10}x^{\;\!\text{-}4}\right)^2} \;=\;\frac{5}{6}x^4 + \frac{3}{10}x^{\;\!\text{-}4}$

Therefore: . $L \;=\;\int^b_a\left(\frac{5}{6}x^4 + \frac{3}{10}x^{\;\!\text{-}4}\right)\,dx$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

So they gave us a rather ugly function,
. . but it produces a simple integral.

I've learned to expect this in an Arc Length problem.
But, as far as I know, there's no way to recognize it in advance.