# Algebraic manipulation of sum of squares.

1. Aug 27, 2014

### tleave2000

Hiya.

I got to an interesting bit in a calculus book, but as usual I'm stumped by a (probably simple) algebraic step.
The author goes from:
$$(ds)^2=(dx)^2+(dy)^2$$
to:
$$ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$
I understand moving the square root across, but I don't understand how the right hand side in the first equation turns into what is under the square root in the second equation.

I hope someone can help. Cheers.

Last edited: Aug 27, 2014
2. Aug 27, 2014

### Simon Bridge

... check that: does the dx at the end go inside or outside the square-root symbol?

3. Aug 27, 2014

### tleave2000

Nice one! It should go outside, not inside. I'll correct it in the original post.

However at first glance the manipulation is still mysterious to me.

4. Aug 27, 2014

### HallsofIvy

Staff Emeritus
(1) Factor a "dx" out on the right side.

(2) Take the square root of both sides.

5. Aug 27, 2014

### Simon Bridge

The clue is that you need a dx^2 in the denominator - so dividing by that is usually a good guess.

So: divide both sides by dx^2 (or just factor it out on the RHS as Ivy suggests.)
The rest should be clear.

Or you can try it backwards - start with the final result and try to get back to the start.
Hint: square both sides.

6. Aug 27, 2014

### tleave2000

Thank you both. I got it a little while after your post Simon, but my working was a mess and Ivy's post really clarified what it was that I had done. I latexed up the tidied version.

$$(ds)^2=(dx)^2+(dy)^2$$
$$(ds)^2=(dx)^2+\frac{(dx)^2(dy)^2}{(dx)^2}$$
$$(ds)^2=(dx)^2(1+\frac{(dy)^2}{(dx)^2})$$
$$ds=\sqrt{(dx)^2(1+\left(\frac{dy}{dx}\right)^2)}$$
$$ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$

7. Aug 27, 2014

### Simon Bridge

Yeah - a step-by-step tidy's things up ;)
Well done.