User:Egm6322.s09.xyz/hw6 xyz

Derive the Grad in Polar Coordinates (Method #1)

Recall the translated variables from cartesian to polar coordinates:

$$x = r cos\theta$$

$$y = r sin\theta$$

$$z = z$$

The basis vectors in polar coordinates are:

$$\blacktriangleright \hat r = \frac{\vec r}{r} = \frac{x \hat i + y \hat j}{r} = \frac{x}{r} \hat i + \frac{y}{r} \hat j $$


 * $$= \frac{rcos\theta}{r} \hat i + \frac{rsin\theta}{r} \hat j$$

$$\therefore \hat r = cos\theta \hat i + sin\theta \hat j$$

$$\blacktriangleright \hat \Theta = \hat z \times \hat r$$


 * $$=(\hat k) \times (cos\theta \hat i + sin\theta \hat j)$$


 * $$\therefore \hat \Theta = - sin\theta \hat i + cos\theta \hat j$$

$$\blacktriangleright \hat z = \hat z$$

Assumptions:


 * 1. $$u =u(r, \theta, z)$$ is a scalar field
 * 2. $$du $$is proportional to $$d\hat R$$, where $$\hat R$$ is the the displacement vector

$$\therefore du = \frac{\partial u}{\partial r}dr + \frac{\partial u}{\partial \theta}d\theta + \frac{\partial u}{\partial z}dz$$

and

$$du = \vec \nabla u \cdot d\hat R$$

Equate both expressions for du:

$$\frac{\partial u}{\partial r}dr + \frac{\partial u}{\partial \theta}d\theta + \frac{\partial u}{\partial z}dz = (\vec \nabla u)_r dr + (\vec \nabla u)_{\theta} r d\theta + (\vec \nabla u)_z dz$$

By inspection,

$$\blacktriangleright (\vec \nabla u)_r = \frac{\partial u}{\partial r}$$

$$\blacktriangleright r (\vec \nabla u)_{\theta} = \frac{\partial u}{\partial \theta}$$, or, $$(\vec \nabla u)_{\theta} = \frac{1}{r}\frac{\partial u}{\partial \theta} $$

$$\blacktriangleright (\vec \nabla u)_z = \frac{\partial u}{\partial z} $$

Therefore, the expression for the divergence can be written as a function of the babsis vectors:

$$\therefore \vec \nabla = \frac{\partial}{\partial r} \hat r + \frac{1}{r}\frac{\partial u}{\partial \theta} \hat \theta + \frac{\partial u}{\partial z} \hat z $$

Let $$\hat A = A_r \hat r + A_{\theta} \hat \theta + A_z \hat z$$

Then,

$$\vec \nabla \cdot \vec A =\left[ \frac{\partial}{\partial r} \hat r + \frac{1}{r}\frac{\partial u}{\partial \theta} \hat \theta + \frac{\partial u}{\partial z} \hat z \right] \cdot \left[ A_r \hat r + A_{\theta} \hat \theta + A_z \hat z \right] $$


 * $$= \hat r \left \{

\frac{\partial}{\partial r} [A_r \hat r] +\frac{\partial}{\partial r} [A_{\theta} \hat \theta] +\frac{\partial}{\partial r} [A_z \hat z] \right \}$$


 * $$+ \frac{\hat \theta}{r} \left \{

\frac{\partial}{\partial \theta} [A_r \hat r] +\frac{\partial}{\partial \theta} [A_{\theta} \hat \theta] +\frac{\partial}{\partial \theta} [A_z \hat z] \right \}$$


 * $$+ \hat z \left \{

\frac{\partial}{\partial z} [A_r \hat r] +\frac{\partial}{\partial z} [A_{\theta} \hat \theta] +\frac{\partial}{\partial z} [A_z \hat z] \right \}$$


 * $$=\hat r \left \{ \left( A_r \frac{\partial \hat r}{\partial r} + \hat r \frac{\partial A_r}{\partial r} \right)

+ \left( A_{\theta} \frac{\partial \hat \theta}{\partial r} + \hat \theta \frac{\partial A_{\theta}}{\partial r} \right) + \left( A_z \frac{\partial \hat z}{\partial r} + \hat z \frac{\partial A_z}{\partial r} \right) \right\}$$


 * $$+\frac{\hat \theta}{r} \left \{ \left( A_r \frac{\partial \hat r}{\partial \theta} + \hat r \frac{\partial A_r}{\partial \theta} \right)

+ \left( A_{\theta} \frac{\partial \hat \theta}{\partial \theta} + \hat \theta \frac{\partial A_{\theta}}{\partial \theta} \right) + \left( A_z \frac{\partial \hat z}{\partial \theta} + \hat z \frac{\partial A_z}{\partial \theta} \right) \right\}$$


 * $$+\hat r \left \{ \left( A_r \frac{\partial \hat r}{\partial z} + \hat r \frac{\partial A_r}{\partial z} \right)

+ \left( A_{\theta} \frac{\partial \hat \theta}{\partial z} + \hat \theta \frac{\partial A_{\theta}}{\partial z} \right) + \left( A_z \frac{\partial \hat z}{\partial z} + \hat z \frac{\partial A_z}{\partial z} \right) \right\}$$

The respective derivatives are determined from the basis vectors as follows:

$$\blacktriangleright \frac{\partial{\hat z}}{\partial r} = \frac{\partial{\hat z}}{\partial \theta} = \frac{\partial{\hat z}}{\partial z} = 0$$

$$\blacktriangleright \frac{\partial{\hat r}}{\partial r} = \frac{\partial{\hat r}}{\partial z} = 0$$

$$\blacktriangleright \frac{\partial{\hat r}}{\partial \theta} = -sin \theta \hat i + cos\theta \hat j = \hat \theta$$

$$\blacktriangleright \frac{\partial{\hat \theta}}{\partial r} = \frac{\partial{\hat \theta}}{\partial z} = 0$$

$$\blacktriangleright \frac{\partial{\hat \theta}}{\partial \theta} = -cos \theta \hat i - sin\theta \hat j = \hat r$$

Substituting these values yields:

$$\vec \nabla \cdot \vec A = $$


 * $$=\hat r \left \{ \left( \cancel{A_r (0)} + \hat r \frac{\partial A_r}{\partial r} \right)

+ \left( \cancel{A_{\theta} (0)} + \hat \theta \frac{\partial A_{\theta}}{\partial r} \right) + \left( \cancel{A_z (0)} + \hat z \frac{\partial A_z}{\partial r} \right) \right\}$$


 * $$+\frac{\hat \theta}{r} \left \{ \left( A_r (\hat \theta) + \hat r \frac{\partial A_r}{\partial \theta} \right)

+ \left( A_{\theta} (-\hat r) + \hat \theta \frac{\partial A_{\theta}}{\partial \theta} \right) + \left( \cancel{A_z (0)} + \hat z \frac{\partial A_z}{\partial \theta} \right) \right\}$$


 * $$+\hat z \left \{ \left( \cancel{A_r (0)} + \hat r \frac{\partial A_r}{\partial z} \right)

+ \left( \cancel{A_{\theta} (0)} + \hat \theta \frac{\partial A_{\theta}}{\partial z} \right) + \left( \cancel{A_z (0)} + \hat z \frac{\partial A_z}{\partial z} \right) \right\}$$

Taking the respective dot products of the remaining terms yields:

$$\vec \nabla \cdot \vec A = \frac{\partial A_r}{\partial r} + \frac{1}{r} \left[ A_r + \frac{\partial A_{\theta}}{\partial \theta}\right] + \frac{A_z}{\partial z} $$

or

$$\vec \nabla \cdot \vec A = \frac{1}{r} \frac{\partial }{\partial r}(A_r r) + \frac{1}{r} \frac{\partial A_{\theta}}{\partial \theta} + \frac{A_z}{\partial z} $$

Photos of Student Interaction Using Co-operative Learning

=R6: Photos of Student Interaction Using Co-operative Learning Technique=

A main objective of the co-operative learning scheme is to enhance student interaction.

The following photos were taken during a study session amongst students as they completed the respective assignments given in Report 6.

As shown in the photos, students were able to use the co-operative learning framework as a tool to discuss relevant concepts, share homework solutions, and cultivate a sense of camaraderie with each other.