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

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The discussion focuses on the process of implicit differentiation applied to the equation r² = y² + x². It explains how to differentiate this equation in differential form, leading to the result 2r dr = 2x dx + 2y dy, which simplifies to r dr = x dx + y dy. The conversation also touches on the differentiation of the tangent function, resulting in the equation sec²(θ) dθ = (x dy - y dx) / x². Participants seek clarification on the steps taken to isolate dr, dx, and dy, as well as the meaning of differential form. The thread emphasizes understanding implicit differentiation and its applications in calculus.
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?
 

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