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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

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

Isura Manchanayake
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pdf1234
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  • Why don't you try some simple example like $f(x,y)=x+y$ where you should be able to compute all the terms and see whether what you got is correct? – Gerry Myerson Aug 15 '19 at 02:01

1 Answers1

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

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

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

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

19aksh
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  • this book http://www.math.harvard.edu/~shlomo/docs/Advanced_Calculus.pdf page 172, at the bottom says im wrong. :/ – pdf1234 Aug 15 '19 at 02:13
  • It may be a typo. Your answer is correct. They might have intended $$\bigg(\frac 1r w_\theta \bigg)^2+w_r^2 = w_x^2 + w_y^2$$ – 19aksh Aug 15 '19 at 02:19