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Algebraic manipulation of sum of squares. 
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#1
Aug2714, 06:26 AM

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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: [tex](ds)^2=(dx)^2+(dy)^2[/tex] to: [tex]ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx[/tex] 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. 


#2
Aug2714, 06:37 AM

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#3
Aug2714, 06:49 AM

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


#4
Aug2714, 07:10 AM

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Algebraic manipulation of sum of squares.
(1) Factor a "dx" out on the right side.
(2) Take the square root of both sides. 


#5
Aug2714, 07:26 AM

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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
Aug2714, 07:56 AM

P: 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.
[tex](ds)^2=(dx)^2+(dy)^2[/tex] [tex](ds)^2=(dx)^2+\frac{(dx)^2(dy)^2}{(dx)^2}[/tex] [tex](ds)^2=(dx)^2(1+\frac{(dy)^2}{(dx)^2})[/tex] [tex]ds=\sqrt{(dx)^2(1+\left(\frac{dy}{dx}\right)^2)}[/tex] [tex]ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx[/tex] 


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