Chapter 1, Problem 1.14P

(a)

Interpretation Introduction

Interpretation:

The value of $\nabla \varphi$ at point $\left(3,5\right)$ is to be determined.

Concept Introduction:

The parameters used to define a cartesian coordinate system are $x,y,\text{ and }z$ , where x defines the distance on x -axis, y defines the distance on y -axis, and z defines the distance on z -axis.

The parameters used to define a cylindrical coordinate system are $r,\theta ,\text{ and }z$ , where r defines the radial distance from the point of reference, $\theta$ defines the direction of the point from the reference point, and z defines the perpendicular distance.

The relationship between $x,y,z$ and $r,\theta ,z$ are:

  $\begin{array}{c}x=r\cos \theta \\ y=r\sin \theta \\ z=z\end{array}$

The chain rule is used in calculus to calculate the derivative of a composite function. If there are two differentiable functions and $\left(f\circ g\right)$ is a composite function, then its derivative is given as:

  $\left(f\circ g\right)\text{'}=\left(f^{\prime }\circ g\right)\cdot g^{\prime }$

(b)

Interpretation Introduction

Interpretation:

The value of component of $\nabla \varphi$ at point $\left(3,5\right)$ that makes an angle of $-{60}^{\circ }$ with the x axisis to be determined.

Concept Introduction:

The parameters used to define a cartesian coordinate system are $x,y,\text{ and }z$ , where x defines the distance on x -axis, y defines the distance on y -axis, and z defines the distance on z -axis.

The parameters used to define a cylindrical coordinate system are $r,\theta ,\text{ and }z$ , where r defines the radial distance from the point of reference, $\theta$ defines the direction of the point from the reference point, and z defines the perpendicular distance.

The relationship between $x,y,z$ and $r,\theta ,z$ are:

  $\begin{array}{c}x=r\cos \theta \\ y=r\sin \theta \\ z=z\end{array}$

Chain rule is used in calculus to calculate the derivative of a composite function. If there are two differentiable functions and $\left(f\circ g\right)$ is a composite function, then its derivative is given as:

  $\left(f\circ g\right)\text{'}=\left(f^{\prime }\circ g\right)\cdot g^{\prime }$

Gradient can be converted to a scaler product in order to calculate the component value by multiplying the unit vector e .