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Algebraic manipulation of sum of squares.

by tleave2000
Tags: algebraic, manipulation, squares
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tleave2000
#1
Aug27-14, 06:26 AM
P: 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:
[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.
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Simon Bridge
#2
Aug27-14, 06:37 AM
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Quote Quote by tleave2000 View Post
[tex]ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2dx}[/tex]
... check that: does the dx at the end go inside or outside the square-root symbol?
tleave2000
#3
Aug27-14, 06:49 AM
P: 8
Quote Quote by Simon Bridge View Post
... check that: does the dx at the end go inside or outside the square-root symbol?
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.

HallsofIvy
#4
Aug27-14, 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.
Simon Bridge
#5
Aug27-14, 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.
tleave2000
#6
Aug27-14, 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]
Simon Bridge
#7
Aug27-14, 08:35 AM
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Yeah - a step-by-step tidy's things up ;)
Well done.


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