Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have this:

$$\sqrt{(dx)^2 + (dy)^2}$$

And my book simplified it as:

$$\sqrt{1 + \Big(\frac{dy}{dx}\Big)^2} \times dx$$

I don't have even a close idea how he did it. If it helps, is about path lenght whit integration.

share|improve this question
    
Your last differential in the second part should be outside the radical, not inside it. –  Arturo Magidin Jun 22 '12 at 18:21
add comment

2 Answers

up vote 1 down vote accepted

$$a\sqrt{r} = \sqrt{a^2(r)}\quad\text{if }a\gt 0\text{ and } r\gt 0.$$ So, using changes instead of differentials: $$\begin{align*} \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \Delta x &= \sqrt{\left(\Delta x\right)^2\left(1 + \left(\frac{\Delta y}{\Delta x}\right)^2\right)}\\ &= \sqrt{(\Delta x)^2 + (\Delta y)^2}. \end{align*}$$ Taking limits as $\Delta x\to 0$ converts $\Delta x$ to $dx$, $\Delta y$ to $dy$, and $\frac{\Delta x}{\Delta y}$ to the derivative $\frac{dy}{dx}$.

share|improve this answer
    
Hooo thank you :) I didn't know the first rule :) Yeah, I used dy because I don't know how to put Δ. –  Andres Jun 22 '12 at 18:26
    
@Andres: The first rule is just a consequence of two things: (i) if $a$ and $b$ are both positive, then $\sqrt{ab} = \sqrt{a}\sqrt{b}$; and (ii) $\sqrt{r^2} = |r|$. Putting them both together, if $a$ and $b$ are positive, then $a\sqrt{b} = $\sqrt{a^2}\sqrt{b}=\sqrt{a^2b}$. –  Arturo Magidin Jun 22 '12 at 18:32
    
Thank you :) Where you learn that kind of things ? I am at first year in University. Not EEUU, Argentina :) –  Andres Jun 22 '12 at 18:39
    
@Andres This is standard root manipulation, which I believe is taught in second or third year in high school in Argentina. –  talmid Jun 22 '12 at 18:41
1  
@Andres I suggest khanacademy.org/math/algebra/exponents-radicals, for example. Or pick any third-year high school math book. Properties and tricks of algebraic manipulation like these are ubiquitous in higher-level courses, and people often use them without pointing out what they are doing, like in this case. –  talmid Jun 22 '12 at 18:55
show 6 more comments

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

(Note that we're treating $dx$ and $dy$ as numbers, but that's another issue; see for example this question.)

share|improve this answer
    
It helps if you think about the Pythagorean theorem, first, noting how a right triangle (with legs $dx$ and $dy$) has its hypoteneuse be $\sqrt{(dx)^{2}+(dy)^{2}}$; we want to change this expression into a differential, i.e., something that looks like $f(x)\,dx$. We can do this using the algebraic manipulations shown in this answer... –  Alex Nelson Jun 22 '12 at 18:25
    
Thank you talmid :) Like I said in the other answer i didn't know the rule: $$a \times \sqrt{r+s} = \sqrt{a^2 \times (r+s)}$$ –  Andres Jun 22 '12 at 18:36
    
@Andres Yes, because (assuming $a \geq 0$) $\sqrt{a^2(r+s)} = \sqrt{a^2}\sqrt{r+s} = a \sqrt{r+s}$. –  talmid Jun 22 '12 at 18:39
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.