Answer to Question #224059 in Calculus for Unknown346307

Question

If A~ = 2yˆi - zˆj + 3xkˆ

(i). Find the unit vectors eˆr, eˆθ and eˆz of a cylindrical coordinates in termsof ˆi, ˆj and kˆ.

(ii). Solve for ˆi, ˆj and kˆ in terms of eˆr, eˆθ and eˆz

(iii). Represent the vector A~ in cylindrical coordinates and determine Ar, Aθ

and Az

Answer 1

Solutiom;

a)

The position vector is given by;

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

In cylindrical coordinates;

$\vec{r}=rcos\theta\hat{i}+rsin\theta\hat{j}+z\hat{k}$

Hence,we find ;

$\hat{e}_r=\frac{\frac{d\vec{r}}{dr}}{|\frac{d\vec{r}}{dr}|}$ $=\frac{cos\theta\hat{i}+sin\theta\hat{j}}{\sqrt{cos^2\theta+sin^2\theta}}$ $\hat{e}_r=cos\theta\hat{i}+sin\theta{j}$

$\hat{e}_{\theta}=\frac{\frac{d\vec{r}}{d\theta}}{|\frac{d\vec{r}}{d\theta}|}$ $=\frac{-rsin\theta\hat{i}+rcos\theta\hat{j}}{\sqrt {r^2cos^2\theta+r^2sin^2\theta}}$

$\hat{e}_{\theta}=-sin\theta\hat{i}+cos\theta\hat{j}$

$\hat{e}_z=\frac{\frac{d\vec{r}}{dz}}{|\frac{d\vec{r}}{dz}|}$ $=\frac{\hat{k}}{1}=\hat{k}$

b)

As calculated;

$\hat{e}_r=cos\theta\hat{i}+sin\theta\hat{j}$ ....(i)

$\hat{e}_{\theta}=-sin\theta\hat{i}+cos\theta\hat{j}$ ...(ii)

$\hat{e}_z=\hat{k}$

Multiply (i) by $cos\theta$ and (ii) by $sin\theta$ and subtract;

$cos\theta\hat{e}_r=cos^2\theta\hat{i}+cos\theta sin\theta\hat{j}$

$-sin\theta\hat{e}_\theta=-sin^2\theta\hat{i}+cos\theta sin\theta\hat{j}$

$\overline{cos\theta\hat{e}_r-sin\theta\hat{e}_{\theta}=(cos^2\theta+sin^2\theta)\hat{i}}$

Hence,

$\hat{i}=cos\theta\hat{e}_r-sin\theta\hat e_{\theta}$

Multiply (i) with $sin\theta$ and (ii) with $cos\theta$ and add;

$sin\theta\hat e_r=sin\theta cos\theta\hat i+sin^2\theta\hat j$

$+cos\theta\hat e_{\theta}=-sin\theta cos\theta\hat i+cos^2\theta\hat j$

$\overline{sin\theta \hat e_r+cos\theta\hat e_{\theta}=(sin^2\theta+cos^2\theta)\hat j}$

Hence,

$\hat j=sin\theta\hat e_r+cos\theta\hat e_{\theta}$

From (iii);

$\hat k=\hat e_z$

c)

Given;

$\vec{A}=2y\hat i-z\hat j+3x\hat k$

In cylindrical coordinates;

$\vec{A}=2(rsin\theta)(cos\theta\hat e_r-sin\theta\hat e_{\theta})-z(sin\theta\hat e_r+cos\theta\hat e_{\theta})+3(rcos\theta)\hat e_z$

By distribution;

$\vec{A}=(2rsin\theta cos\theta-zsin\theta)\hat e_r-(2rsin^2\theta+zcos\theta)\hat e_{\theta}+3rcos\theta\hat e_z$

From which;

$A_r=2rcos\theta sin\theta-zsin\theta$

$A_{\theta}=-2rsin^2\theta-zcos\theta$

$A_z=3rcos\theta$