Algebraic manipulation of sum of squares.

[tleave2000]

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.

 

[Simon Bridge]

ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2dx}

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

 

[tleave2000]

... 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]

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

(2) Take the square root of both sides.

 

[Simon Bridge]

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]

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

 

[Simon Bridge]

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