If $w=f(x, y),x=r\cos\theta, y=r\sin\theta$, prove $\frac{dw}{d\theta}^2+\left(r \frac{dw}{dr}\right)^2=\frac{dw}{dx}^2+\frac{dw}{dy}^2$

Question

If $w = f(x,y)$, $x = r\cos(\theta)$, and $y = r\sin(\theta)$, show that

$$\left(\frac{\mathrm{d}w}{\mathrm{d}\theta}\right)^2 + \left(r\frac{\mathrm{d}w}{\mathrm{d}r}\right)^2 = \left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2 + \left(\frac{\mathrm{d}w}{\mathrm{d}y}\right)^2$$

Is this correct? I got

$$\left(\frac{\mathrm{d}w}{\mathrm{d}\theta}\right)^2 + \left(r\frac{\mathrm{d}w}{\mathrm{d}r}\right)^2 = r^2\left(\left(\frac{\mathrm{d}w}{\mathrm{d}x}\right)^2 + \left(\frac{\mathrm{d}w}{\mathrm{d}y}\right)^2\right)$$

Answer 1

What you have got is correct.
$$w_\theta = w_xx_\theta+w_yy_\theta = -r\sin\theta w_x+r\cos\theta w_y $$

$$w_r = \cos\theta w_x + \sin\theta w_y$$

$$w_\theta^2+(rw_r)^2 = r^2\big[(\sin^2\theta+\cos^2\theta)(w_x^2+w_y^2) +2\sin\theta\cos\theta w_x w_y-2\sin\theta\cos\theta w_x w_y\big]$$

$$\implies w_\theta^2+(rw_r)^2 = r^2(w_x^2+w_y^2)$$