# Algebraic manipulation of sum of squares.

1. ### tleave2000

8
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. ### Simon Bridge

14,658
... check that: does the dx at the end go inside or outside the square-root symbol?

3. ### tleave2000

8
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. ### HallsofIvy

40,308
Staff Emeritus
(1) Factor a "dx" out on the right side.

(2) Take the square root of both sides.

5. ### Simon Bridge

14,658
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.

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6. ### tleave2000

8
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. ### Simon Bridge

14,658
Yeah - a step-by-step tidy's things up ;)
Well done.