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

[nacho-man]

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

View attachment 3404

 

[MarkFL]

Given:

\(\displaystyle r^2=x^2+y^2\)

And then differentiating in differential form, we find:

\(\displaystyle 2r\,dr=2x\,dx+2y\,dy\)

Divide through by 2:

\(\displaystyle r\,dr=x\,dx+y\,dy\)

And given:

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

And then differentiating in differential form, we find:

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

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

\(\displaystyle r^2\,d\theta=x\,dy-y\,dx\)

 

[nacho-man]

Given:

\(\displaystyle r^2=x^2+y^2\)

And then differentiating in differential form, we find:

\(\displaystyle 2r\,dr=2x\,dx+2y\,dy\)

Can I confirm what you've done to get the $dr, dx, dy$ out like that?

What is meant by differential form?