Question

1. for the specialized coordinate system given, express the derivatives of a general function f with respect to the new variables in terms of (\frac{partial f}{partial x},\frac{partial f}{partial y},\frac{partial f}{partial z}). (a) cylindrical coordinates: ((r,\theta,z)). cylindrical coordinates are given by: (x = rcos\theta,y = rsin\theta,z = z), and/or (r=sqrt{x^{2}+y^{2}},\theta=arctan(y / x),z = z). (b) spherical coordinates: ((

ho,\theta,phi)). spherical coordinates are given by: (x=
hocos\thetasinphi,y =
hosin\thetasinphi,z=
hocosphi), and/or (
ho=sqrt{x^{2}+y^{2}+z^{2}},\theta=arctan(y / x),phi=arccosleft(\frac{z}{sqrt{x^{2}+y^{2}+z^{2}}}
ight))

Explanation:

Step1: Recall the chain - rule for partial derivatives

The chain - rule for partial derivatives states that if \(f(x,y,z)\) and \(x = x(u,v,w)\), \(y = y(u,v,w)\), \(z = z(u,v,w)\), then \(\frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial u}\).

(a) Cylindrical coordinates \((r,\theta,z)\) where \(x = r\cos\theta\), \(y = r\sin\theta\), \(z = z\)

Step2: Calculate \(\frac{\partial f}{\partial r}\)

Using the chain - rule \(\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial r}\). Since \(\frac{\partial x}{\partial r}=\cos\theta\), \(\frac{\partial y}{\partial r}=\sin\theta\), \(\frac{\partial z}{\partial r} = 0\), we have \(\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta\).

Step3: Calculate \(\frac{\partial f}{\partial\theta}\)

Using the chain - rule \(\frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\theta}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial\theta}\). Since \(\frac{\partial x}{\partial\theta}=-r\sin\theta\), \(\frac{\partial y}{\partial\theta}=r\cos\theta\), \(\frac{\partial z}{\partial\theta} = 0\), we have \(\frac{\partial f}{\partial\theta}=-r\frac{\partial f}{\partial x}\sin\theta + r\frac{\partial f}{\partial y}\cos\theta\).

Step4: Calculate \(\frac{\partial f}{\partial z}\)

Since \(z = z\), \(\frac{\partial f}{\partial z}=\frac{\partial f}{\partial z}\) (no change in the transformation for \(z\)).

(b) Spherical coordinates \((

ho,\theta,\phi)\) where \(x= ho\cos\theta\sin\phi\), \(y = ho\sin\theta\sin\phi\), \(z= ho\cos\phi\)

Step5: Calculate \(\frac{\partial f}{\partial

ho}\)
Using the chain - rule \(\frac{\partial f}{\partial ho}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial ho}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial ho}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial ho}\). Since \(\frac{\partial x}{\partial ho}=\cos\theta\sin\phi\), \(\frac{\partial y}{\partial ho}=\sin\theta\sin\phi\), \(\frac{\partial z}{\partial ho}=\cos\phi\), we have \(\frac{\partial f}{\partial ho}=\frac{\partial f}{\partial x}\cos\theta\sin\phi+\frac{\partial f}{\partial y}\sin\theta\sin\phi+\frac{\partial f}{\partial z}\cos\phi\).

Step6: Calculate \(\frac{\partial f}{\partial\theta}\)

Using the chain - rule \(\frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\theta}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial\theta}\). Since \(\frac{\partial x}{\partial\theta}=- ho\sin\theta\sin\phi\), \(\frac{\partial y}{\partial\theta}= ho\cos\theta\sin\phi\), \(\frac{\partial z}{\partial\theta}=0\), we have \(\frac{\partial f}{\partial\theta}=- ho\frac{\partial f}{\partial x}\sin\theta\sin\phi+ ho\frac{\partial f}{\partial y}\cos\theta\sin\phi\).

Step7: Calculate \(\frac{\partial f}{\partial\phi}\)

Using the chain - rule \(\frac{\partial f}{\partial\phi}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial\phi}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\phi}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial\phi}\). Since \(\frac{\partial x}{\partial\phi}= ho\cos\theta\cos\phi\), \(\frac{\partial y}{\partial\phi}= ho\sin\theta\cos\phi\), \(\frac{\partial z}{\partial\phi}=- ho\sin\phi\), we have \(\frac{\partial f}{\partial\phi}=\frac{\partial f}{\partial x} ho\cos\theta\cos\phi+\frac{\partial f}{\partial y} ho\sin\theta\cos\phi-\frac{\partial f}{\partial z} ho\sin\phi\).

Answer:

(a) Cylindrical coordinates

\(\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta\), \(\frac{\partial f}{\partial\theta}=-r\frac{\partial f}{\partial x}\sin\theta + r\frac{\partial f}{\partial y}\cos\theta\), \(\frac{\partial f}{\partial z}=\frac{\partial f}{\partial z}\)

(b) Spherical coordinates

\(\frac{\partial f}{\partial
ho}=\frac{\partial f}{\partial x}\cos\theta\sin\phi+\frac{\partial f}{\partial y}\sin\theta\sin\phi+\frac{\partial f}{\partial z}\cos\phi\), \(\frac{\partial f}{\partial\theta}=-
ho\frac{\partial f}{\partial x}\sin\theta\sin\phi+
ho\frac{\partial f}{\partial y}\cos\theta\sin\phi\), \(\frac{\partial f}{\partial\phi}=\frac{\partial f}{\partial x}
ho\cos\theta\cos\phi+\frac{\partial f}{\partial y}
ho\sin\theta\cos\phi-\frac{\partial f}{\partial z}
ho\sin\phi\)