Chapter 3, Problem 18P

(a)

To determine

The gradient of given scalar field and their value at specified point.

Explanation of Solution

Given:

The scalar field $\left(V\left(x,y,z\right)\right)$ is $10xyz-2x^{2}z$.

The specified point $\left(P\right)$ is $\left(-1,4,3\right)$.

Calculation:

Calculate the gradient of scalar field using the relation.

  $\text{grad}V\left(x,y,z\right)=\frac{\partial V\left(x,y,z\right)}{\partial x}{a}_x+\frac{\partial V\left(x,y,z\right)}{\partial y}{a}_y+\frac{\partial V\left(x,y,z\right)}{\partial z}{a}_z$

  $\begin{array}{c}\text{grad}V\left(x,y,z\right)=\frac{\partial \left(10xyz-2x^{2}z\right)}{\partial x}{a}_x+\frac{\partial \left(10xyz-2x^{2}z\right)}{\partial y}{a}_y+\frac{\partial \left(10xyz-2x^{2}z\right)}{\partial z}{a}_z\\ =\left(10yz-2\left(2x\right)z\right)a_x+\left(10xz\right)a_y+\left(10xy-2x^2\right)a_z\\ =\left(10yz-4xz\right)a_x+10xz{a}_y+\left(10xy-2x^2\right)a_z\end{array}$

Calculate the gradient of scalar field at specified point using the relation.

  $\text{grad}V\left(x,y,z\right)=\left(10yz-4xz\right)a_x+10xz{a}_y+\left(10xy-2x^2\right)a_z$

  $\begin{array}{c}\text{grad}V\left(x,y,z\right)=\{\begin{array}{c}\left(10\left(4\right)\left(3\right)-4\left(-1\right)\left(3\right)\right)a_x+10\left(-1\right)\left(3\right)a_y\\ +\left(10\left(-1\right)\left(4\right)-2{\left(-1\right)}^2\right)a_z\end{array}\\ =\left(120+12\right)a_x-30a_y+\left(-40-2\right)a_z\\ =132a_x-30a_y-42a_z\end{array}$

Thus, the gradient of scalar field is $\underset{\_}{\left(10yz-4xz\right)a_x+10xz{a}_y+\left(10xy-2x^2\right)a_z}$ and their value at the point $\left(-1,4,3\right)$ is $\underset{\_}{132a_x-30a_y-42a_z}$.

(b)

To determine

The gradient of given scalar field and their value at specified point.

Explanation of Solution

Given:

The scalar field $\left(U\left(\rho ,\varphi ,z\right)\right)$ is $2\rho \sin \varphi +\rho z$.

The specified point $\left(Q\right)$ is $\left(2,90°,-1\right)$.

Calculation:

Calculate the gradient of scalar field using the relation.

  $\text{grad}U\left(\rho ,\varphi ,z\right)=\frac{\partial U\left(\rho ,\varphi ,z\right)}{\partial \rho }{a}_{\rho }+\frac{\partial U\left(\rho ,\varphi ,z\right)}{\partial \varphi }{a}_{\varphi }+\frac{\partial U\left(\rho ,\varphi ,z\right)}{\partial z}{a}_z$

  $\begin{array}{c}\text{grad}U\left(\rho ,\varphi ,z\right)=\frac{\partial \left(2\rho \sin \varphi +\rho z\right)}{\partial \rho }{a}_{\rho }+\frac{\partial \left(2\rho \sin \varphi +\rho z\right)}{\partial \varphi }{a}_{\varphi }+\frac{\partial \left(2\rho \sin \varphi +\rho z\right)}{\partial z}{a}_z\\ =\left(2\sin \varphi +z\right)a_{\rho }+\left(2\rho \cos \varphi \right)a_{\varphi }+\left(\rho \right)a_z\\ =\left(2\sin \varphi +z\right)a_{\rho }+2\rho \cos \varphi a_{\varphi }+\rho a_z\end{array}$

Calculate the gradient of scalar field at specified point using the relation.

  $\text{grad}U\left(\rho ,\varphi ,z\right)=\left(2\sin \varphi +z\right)a_{\rho }+2\rho \cos \varphi a_{\varphi }+\rho a_z$

  $\begin{array}{c}\text{grad}U\left(\rho ,\varphi ,z\right)=\left(2\sin \left(90°\right)+\left(-1\right)\right)a_{\rho }+2\left(2\right)\cos \left(90°\right)a_{\varphi }+\left(2\right)a_z\\ =\left(2\left(1\right)-1\right)a_{\rho }+\left(4\right)\left(0\right)a_{\varphi }+2a_z\\ =a_{\rho }+2a_z\end{array}$

Thus, the gradient of scalar field is $\underset{\_}{\left(2\sin \varphi +z\right)a_{\rho }+2\rho \cos \varphi a_{\varphi }+\rho a_z}$ and their value at $\left(2,90°,-1\right)$ is $\underset{\_}{a_{\rho }+2a_z}$.

(c)

To determine

The gradient of given scalar field and their value at specified point.

Explanation of Solution

Given:

The scalar field $\left(W\left(r,\theta ,\varphi \right)\right)$ is $\frac{4}{r}\sin \theta \cos \varphi$.

The specified point $\left(R\right)$ is $\left(1,\pi /6,\pi /2\right)$.

Calculation:

Calculate the gradient of scalar field using the relation.

  $\text{grad}W\left(r,\theta ,\varphi \right)=\frac{\partial W\left(r,\theta ,\varphi \right)}{\partial r}{a}_r+\frac{\partial W\left(r,\theta ,\varphi \right)}{\partial \theta }{a}_{\theta }+\frac{\partial W\left(r,\theta ,\varphi \right)}{\partial \varphi }{a}_{\varphi }$

  $\begin{array}{c}\text{grad}W\left(r,\theta ,\varphi \right)=\frac{\partial \left(\frac{4}{r}\sin \theta \cos \varphi \right)}{\partial r}{a}_r+\frac{\partial \left(\frac{4}{r}\sin \theta \cos \varphi \right)}{\partial \theta }{a}_{\theta }+\frac{\partial \left(\frac{4}{r}\sin \theta \cos \varphi \right)}{\partial \varphi }{a}_{\varphi }\\ =\{\begin{array}{c}\left(\left(-\frac{4}{r^2}\right)\sin \theta \cos \varphi \right)a_r+\left(\frac{4}{r}\left(\cos \theta \right)\cos \varphi \right)a_{\theta }\\ +\left(\frac{4}{r}\sin \theta \left(-\sin \varphi \right)\right)a_{\varphi }\end{array}\\ =-\frac{4}{r^2}\sin \theta \cos \varphi a_r+\frac{4}{r}\cos \theta \cos \varphi a_{\theta }-\frac{4}{r}\sin \theta \sin \varphi a_{\varphi }\end{array}$

Calculate the gradient of scalar field at specified point using the relation.

  $\text{grad}W\left(r,\theta ,\varphi \right)=-\frac{4}{r^2}\sin \theta \cos \varphi a_r+\frac{4}{r}\cos \theta \cos \varphi a_{\theta }-\frac{4}{r}\sin \theta \sin \varphi a_{\varphi }$

  $\begin{array}{c}\text{grad}W\left(r,\theta ,\varphi \right)=\{\begin{array}{c}-\frac{4}{1^2}\sin \left(\frac{\pi }{6}\right)\cos \left(\frac{\pi }{2}\right)a_r+\frac{4}{1}\cos \left(\frac{\pi }{6}\right)\cos \left(\frac{\pi }{2}\right)a_{\theta }\\ -\frac{4}{1}\sin \left(\frac{\pi }{6}\right)\sin \left(\frac{\pi }{2}\right)a_{\varphi }\end{array}\\ =-4\left(0.5\right)\left(0\right)a_r+4\left(0.866\right)\left(0\right)a_{\theta }-4\left(0.5\right)\left(1\right)a_{\varphi }\\ =0+0-2a_{\varphi }\\ =-2a_{\varphi }\end{array}$

Thus, the gradient of scalar field is $\underset{\_}{-\frac{4}{r^2}\sin \theta \cos \varphi a_r+\frac{4}{r}\cos \theta \cos \varphi a_{\theta }-\frac{4}{r}\sin \theta \sin \varphi a_{\varphi }}$ and their value at $\left(1,\pi /6,\pi /2\right)$ is $\underset{\_}{-2a_{\varphi }}$.