Algebraic manipulation of sum of squares.

  1. Aug 27, 2014 #1
    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.
     
    Last edited: Aug 27, 2014
  2. jcsd
  3. Aug 27, 2014 #2

    Simon Bridge

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    ... check that: does the dx at the end go inside or outside the square-root symbol?
     
  4. Aug 27, 2014 #3
    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.
     
  5. Aug 27, 2014 #4

    HallsofIvy

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

    (2) Take the square root of both sides.
     
  6. Aug 27, 2014 #5

    Simon Bridge

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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.
     
  7. Aug 27, 2014 #6
    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]
     
  8. Aug 27, 2014 #7

    Simon Bridge

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