Algebraic manipulation of sum of squares.

  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. Simon Bridge

    Simon Bridge 15,279
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    ... check that: does the dx at the end go inside or outside the square-root symbol?
     
  4. 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. HallsofIvy

    HallsofIvy 40,933
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    (1) Factor a "dx" out on the right side.

    (2) Take the square root of both sides.
     
  6. Simon Bridge

    Simon Bridge 15,279
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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.
     
    1 person likes this.
  7. 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. Simon Bridge

    Simon Bridge 15,279
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    Yeah - a step-by-step tidy's things up ;)
    Well done.
     
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