Chapter 2, Problem 18P

(a)

To determine

Convert the vector A from cylindrical to rectangular coordinate system and calculate its magnitude at $\left(3,-4,0\right)$.

Explanation of Solution

Given:

  $A=\rho \cos \varphi a_p+\rho z^2\sin \varphi a_z$

Calculation:

Write the given vector A.

  $A=\rho \cos \varphi a_{\rho }+\rho z^2\sin \varphi a_z$

Here, $A_{\rho }=\rho \cos \varphi$, $A_{\varphi }=0$, and $A_z=\rho z^2\sin \varphi$.

Write the expression for the transformation of vector A from $\left(A_x,A_y,A_z\right)\text{to}\left(A_{\rho },A_{\varphi },A_z\right)$.

  $\begin{array}{c}\left[\begin{array}{l}{A}_x\hfill \\ A_y\hfill \\ A_z\hfill \end{array}\right]=\left[\begin{array}{ccc}\cos \varphi & -\sin \varphi & 0\\ \sin \varphi & \cos \varphi & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}{A}_{\rho }\\ A_{\varphi }\\ A_z\end{array}\right]\\ =\left[\begin{array}{c}\rho {\cos }^2\varphi \\ \rho \cos \varphi \sin \varphi \\ \rho z^2\sin \varphi \end{array}\right]\end{array}$        (I)

Write the relationship between cylindrical to rectangular.

  $\rho =\sqrt{x^2+y^2}$

Calculate the $\sin \varphi$ components.

  $\begin{array}{c}\sin \varphi =\frac{y}{\rho }\\ \sin \varphi =\frac{y}{\sqrt{x^2+y^2}}\end{array}$

Calculate the $\cos \varphi$ components.

  $\begin{array}{c}\cos \varphi =\frac{x}{\rho }\\ \cos \varphi =\frac{x}{\sqrt{x^2+y^2}}\end{array}$

Substitute $\rho =\sqrt{x^2+y^2}$, $\cos \varphi =\frac{x}{\sqrt{x^2+y^2}}$ and $\sin \varphi =\frac{y}{\sqrt{x^2+y^2}}$ in Equation (I).

  $\begin{array}{l}\left[\begin{array}{l}{A}_x\hfill \\ A_y\hfill \\ A_z\hfill \end{array}\right]=\left[\begin{array}{c}\sqrt{x^2+y^2}{\left(\frac{x}{\sqrt{x^2+y^2}}\right)}^2\\ \sqrt{x^2+y^2}\left(\frac{x}{\sqrt{x^2+y^2}}\right)\left(\frac{y}{\sqrt{x^2+y^2}}\right)\\ z^2\sqrt{x^2+y^2}\left(\frac{y}{\sqrt{x^2+y^2}}\right)\end{array}\right]\\ \left[\begin{array}{l}{A}_x\hfill \\ A_y\hfill \\ A_z\hfill \end{array}\right]=\left[\begin{array}{c}\frac{x^2}{\sqrt{x^2+y^2}}\\ \frac{xy}{\sqrt{x^2+y^2}}\\ y{z}^2\end{array}\right]\\ A=\frac{x^2}{\sqrt{x^2+y^2}}{a}_x+\frac{xy}{\sqrt{x^2+y^2}}{a}_y+y{z}^2{a}_z\end{array}$

Thus, the vector A in rectangular coordinate system is $\frac{x^2}{\sqrt{x^2+y^2}}{a}_x+\frac{xy}{\sqrt{x^2+y^2}}{a}_y+y{z}^2{a}_z$.

Determine its magnitude at $\left(3,-4,0\right)$.

  $\begin{array}{l}A=\frac{3^2}{\sqrt{3^2+{\left(-4\right)}^2}}{a}_x+\frac{\left(3\times -4\right)}{\sqrt{3^2+{\left(-4\right)}^2}}{a}_y+\left(-4\right){\left(0\right)}^2{a}_z\\ A=1.8a_x-2.4a_y\\ \left|A\right|=\sqrt{{\left(1.8\right)}^2+{\left(-2.4\right)}^2}\\ \left|A\right|=\sqrt{9}\\ \left|A\right|=3\end{array}$

Thus, the magnitude at $\left(3,-4,0\right)$ is $3$.

(b)

To determine

Convert the vector A from cylindrical to spherical coordinate system and calculate its magnitude at $\left(3,-4,0\right)$.

Explanation of Solution

Write the given vector A.

  $A=\rho \cos \varphi a_{\rho }+\rho z^2\sin \varphi a_z$

Here, $A_{\rho }=\rho \cos \varphi$, $A_{\varphi }=0$, and $A_z=\rho z^2\sin \varphi$.

Write the expression for the transformation of vector A from $\left(A_r,A_{\theta },A_{\varphi }\right)\text{to}\left(A_{\rho },A_{\varphi },A_z\right)$.

  $\begin{array}{c}\left[\begin{array}{l}{A}_r\hfill \\ A_{\theta }\hfill \\ A_{\varphi }\hfill \end{array}\right]=\left[\begin{array}{ccc}\sin \theta & 0& \cos \theta \\ \cos \theta & 0& -\sin \theta \\ 0& 1& 0\end{array}\right]\left[\begin{array}{c}{A}_{\rho }\\ A_{\varphi }\\ A_z\end{array}\right]\\ =\left[\begin{array}{c}\rho \cos \varphi \sin \theta +\rho z^2\sin \varphi \cos \theta \\ \rho \cos \varphi \cos \theta -\rho z^2\sin \varphi \sin \theta \\ 0\end{array}\right]\end{array}$        (II)

Write the relationship between cylindrical to rectangular.

  $\rho =\sqrt{x^2+y^2}$

Calculate the $\sin \varphi$ components.

  $\begin{array}{c}\sin \varphi =\frac{y}{\rho }\\ \sin \varphi =\frac{y}{\sqrt{x^2+y^2}}\end{array}$

Calculate the $\cos \varphi$ components.

  $\begin{array}{c}\cos \varphi =\frac{x}{\rho }\\ \cos \varphi =\frac{x}{\sqrt{x^2+y^2}}\end{array}$

Calculate the $\sin \theta$ components.

  $\begin{array}{c}\sin \theta =\frac{\rho }{r}\\ \sin \theta =\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\end{array}$

Calculate the $\cos \theta$ components.

  $\begin{array}{c}\cos \theta =\frac{z}{r}\\ \cos \theta =\frac{z}{\sqrt{x^2+y^2+z^2}}\end{array}$

Substitute $\rho =\sqrt{x^2+y^2}$, $\sin \theta =\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$, $\cos \theta =\frac{z}{\sqrt{x^2+y^2+z^2}}$, $\cos \varphi =\frac{x}{\sqrt{x^2+y^2}}$ and $\sin \varphi =\frac{y}{\sqrt{x^2+y^2}}$ in Equation (II).

  $\begin{array}{l}\left[\begin{array}{l}{A}_r\hfill \\ A_{\theta }\hfill \\ A_{\varphi }\hfill \end{array}\right]=\left[\begin{array}{c}\begin{array}{l}\sqrt{x^2+y^2}\left(\frac{x}{\sqrt{x^2+y^2}}\right)\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\\ +z^2\sqrt{x^2+y^2}\left(\frac{y}{\sqrt{x^2+y^2}}\right)\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\end{array}\\ \begin{array}{l}\sqrt{x^2+y^2}\left(\frac{x}{\sqrt{x^2+y^2}}\right)\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\\ -z^2\sqrt{x^2+y^2}\left(\frac{y}{\sqrt{x^2+y^2}}\right)\left(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\end{array}\\ 0\end{array}\right]\\ \left[\begin{array}{l}{A}_r\hfill \\ A_{\theta }\hfill \\ A_{\varphi }\hfill \end{array}\right]=\left[\begin{array}{c}\left(\frac{x\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)+\left(\frac{y{z}^3}{\sqrt{x^2+y^2+z^2}}\right)\\ \left(\frac{xz}{\sqrt{x^2+y^2+z^2}}\right)-\left(\frac{y{z}^2\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\\ 0\end{array}\right]\\ A=\left\{\begin{array}{l}\left[\left(\frac{x\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)+\left(\frac{y{z}^3}{\sqrt{x^2+y^2+z^2}}\right)\right]a_r\\ +\left[\left(\frac{xz}{\sqrt{x^2+y^2+z^2}}\right)-\left(\frac{y{z}^2\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)\right]a_{\theta }\end{array}\right\}\\ A=\left(\frac{x\sqrt{x^2+y^2}+y{z}^3}{\sqrt{x^2+y^2+z^2}}\right)a_r+\left(\frac{xz-y{z}^2\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)a_{\theta }\end{array}$

Thus, the vector A in spherical coordinate system is $\left(\frac{x\sqrt{x^2+y^2}+y{z}^3}{\sqrt{x^2+y^2+z^2}}\right)a_r+\left(\frac{xz-y{z}^2\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}\right)a_{\theta }$.

Determine its magnitude at $\left(3,-4,0\right)$.

  $\begin{array}{l}A=\frac{3\sqrt{3^2+{\left(-4\right)}^2}+\left(-4\right)\left(0\right)}{\sqrt{3^2+{\left(-4\right)}^2+0^2}}{a}_r+\frac{\left(3\right)\left(0\right)-\left(-4\right)\left(0\right)\sqrt{3^2+{\left(-4\right)}^2}}{\sqrt{3^2+{\left(-4\right)}^2+0^2}}{a}_{\theta }\\ A=3a_r+0a_{\theta }\\ \left|A\right|=\sqrt{{\left(3\right)}^2}\\ \left|A\right|=\sqrt{9}\\ \left|A\right|=3\end{array}$

Thus, the magnitude at $\left(3,-4,0\right)$ is $3$.