Algebraic manipulation of sum of squares.

In summary, the conversation discusses a confusion with a step in a calculus book where the author goes from (ds)^2=(dx)^2+(dy)^2 to ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx. The individuals in the conversation provide guidance on how to solve the problem, including factoring out a "dx" and taking the square root of both sides. Eventually, it is determined that the right hand side in the first equation becomes what is under the square root in the second equation.
  • #1
tleave2000
8
0
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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  • #2
tleave2000 said:
[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?
 
  • #3
Simon Bridge said:
... 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.
 
  • #4
(1) Factor a "dx" out on the right side.

(2) Take the square root of both sides.
 
  • #5
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.
 
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  • #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]
 
  • #7
Yeah - a step-by-step tidy's things up ;)
Well done.
 

1. What is algebraic manipulation of sum of squares?

Algebraic manipulation of sum of squares is a method used in algebra to simplify expressions that involve sums of squares. It involves using various algebraic identities and properties to transform an expression into a simpler form.

2. Why is algebraic manipulation of sum of squares important?

Algebraic manipulation of sum of squares is important because it allows us to solve complex mathematical problems and equations by simplifying them. It also helps in finding the roots of equations and proving mathematical theorems.

3. What are some common algebraic identities used in manipulating sum of squares?

Some common algebraic identities used in manipulating sum of squares include the difference of squares, perfect square trinomial, and the sum and difference of cubes identities. These identities help in factoring and expanding expressions involving sum of squares.

4. How do you use algebraic manipulation of sum of squares to solve equations?

To solve equations using algebraic manipulation of sum of squares, you need to first identify the expression involving sum of squares and then apply the appropriate algebraic identity to simplify it. This will help in reducing the equation into a simpler form that can be solved easily.

5. Can algebraic manipulation of sum of squares be used in real-life applications?

Yes, algebraic manipulation of sum of squares can be used in various real-life applications such as in engineering, physics, and finance. It can be used to solve problems involving quadratic equations, optimization, and statistical analysis, among others.

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