MHB Learn Implicit Differentiation: Solving for r^2 in $y^2 + x^2 = 0

nacho-man
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Can someone please explain how the result is obtained from the first line
$r^2 = y^2 + x^2$
(refer to attached image)

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Given:

$$r^2=x^2+y^2$$

And then differentiating in differential form, we find:

$$2r\,dr=2x\,dx+2y\,dy$$

Divide through by 2:

$$r\,dr=x\,dx+y\,dy$$

And given:

$$\tan(\theta)=\frac{y}{x}$$

And then differentiating in differential form, we find:

$$\sec^2(\theta)\,d\theta=\frac{x\,dy-y\,dx}{x^2}$$

Multiply through by $$x^2=r^2\cos^2(\theta)$$:

$$r^2\,d\theta=x\,dy-y\,dx$$
 
MarkFL said:
Given:

$$r^2=x^2+y^2$$

And then differentiating in differential form, we find:

$$2r\,dr=2x\,dx+2y\,dy$$
Can I confirm what you've done to get the $dr, dx, dy$ out like that?

What is meant by differential form?
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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