Proving r^2 = x^2 + y^2 in Polar Coordinates

[mrcleanhands]

Homework Statement

Consider z=f(x,y), where x=rcosθ and y=rsinθ
(This is a multi part multi variable calculus assignment question). I've just derived dz/dr and now I'm asked... to show that $r^{2}=x^{2}+y^{2}, \theta=a\tan(\frac{y}{x})$

Homework Equations

The Attempt at a Solution

I've drawn up a triangle with r as the hypotenuse, x on the x axis, y on the y axis.

and then x=rcosθ=r * x/r = x by trig laws and then $r^{2}=x^{2}+y^{2}$ by pythagoras theorem etc


this seems too easy. is there some other way to show this?

 

[SteamKing]

Not unless you want to prove the Pythagorean Theorem itself.

 

[LCKurtz]

Homework Statement

Consider z=f(x,y), where x=rcosθ and y=rsinθ
(This is a multi part multi variable calculus assignment question). I've just derived dz/dr and now I'm asked... to show that $r^{2}=x^{2}+y^{2}, \theta=a\tan(\frac{y}{x})$

That doesn't have anything to do with $z = f(x,y)$. It's just standard polar coordinate equations.$$
x^2 + y^2 = r^2\cos^2\theta + r^2\sin^2\theta = r^2$$ $$
\frac y x = \frac{r\sin\theta}{r\cos\theta}=\tan\theta$$

 

[mrcleanhands]

yeah that's why I was confused. I don't get how I'm showing that by just plugging it into equations and referring to the polar equations.