Question

2014/B5 (a) Draw skecthes to illustrate R, 0 and z coordinate curves for the case of cylindrical polar coordinates (b) Show that the gradient of a scalar field, p, can be expressed in terms of curvilinear coordinates u1, u2 and us, of an orthogonal coordinate system as where h, Idr/dul. Hence obtain a formula for Vip in cylindrical polar coordinates. (c) Evaluate dp/ds, the rate of change of φ with distance, for the field φ-R, cost) at the point R 1,0 π/2, 1/2, for the direction eR + Ca

Answer 1

Solution:2014/B5:Let us suppose the transformation equations

$x=f(u_{1},u_{2},u_{3}),y=g(u_{1},u_{2},u_{3}),z=h(u_{1},u_{2},u_{3})~~~~~~~~(i)$

where f,g and h are continuous, have continuous partial derivatives, and have a

single-valued inverse. We now establish a one-one correspondence between

points in an xyz and
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rectangular coordinate system. Equation(i) can be written as

${\bf{r}}=x{\bf{i}}+y{\bf{j}}+z{\bf{k}}=f(u_{1},u_{2},u_{3}){\bf{i}}+g(u_{1},u_{2},u_{3}){\bf{j}}+h(u_{1},u_{2},u_{3}){\bf{k}}~~~~~~~~(ii)$

A point P can be defined by rectangular coordinates (x,y,z) and curvilinear coordinates

(u1, u2, u3)
.

If $u_{2},u_{3}$ are constant then as $u_{1}$ varies, r describes a curve which is called the

$u_{1}$ coordinate curve. Similarly we define $u_{2}$ and $u_{3}$ coordinate curves through P.

From equation (ii), we have

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The vector $\frac{\partial \bf{r}}{\partial u_{1}}$ is tangent to the $u_{1}$ coordinate curve at P. If
ei
is a unit vector at P

in this direction, we can write $\frac{\partial \bf{r}}{\partial u_{1}}=h_{1}e_{1}$ where
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.

Similarly, $\frac{\partial \bf{r}}{\partial u_{2}}=h_{2}e_{2}$ and $\frac{\partial \bf{r}}{\partial u_{3}}=h_{3}e_{3}$ where $h_{2}=\left | \frac{\partial \bf{r}}{\partial u_{2}} \right |$ and $h_{3}=\left | \frac{\partial \bf{r}}{\partial u_{3}} \right |$

Therefore, equation (iii) can be written as

$d{\bf{r}}=h_{1}du_{1}e_{1}+h_{2}du_{2}e_{2}+h_{3}du_{3}e_{3}~~~~~~~~~~(iv)$

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(a) Cylindrical Polar coordinates$(R,\vartheta,z )$:

Transformation equations:

$x=R\cos\vartheta , y=R\sin\vartheta , z=z$

where $R\geq 0, 0\leq \vartheta <2,\pi , -\infty<z<\infty$

R. theta

(b) If $e_{1},e_{2},e_{3}$ are mutually perpendicular at any point P, the curvillinear

coordinates are called orthogonal.

If $\varphi=\varphi(u_{1},u_{2},u_{3})$ is a scalar field, then

$grade \varphi =\triangledown \varphi =\left ( i\frac{\partial }{\partial x}+j\frac{\partial }{\partial y}+k\frac{\partial }{\partial z} \right )\varphi =\left ( i\frac{\partial\varphi }{\partial x}+j\frac{\partial \varphi }{\partial y}+k\frac{\partial \varphi }{\partial z} \right )$

and

$d{\varphi }=\frac{\partial \varphi}{\partial u_{1}}du_{1}+\frac{\partial \varphi}{\partial u_{2}}du_{2}+\frac{\partial \varphi}{\partial u_{3}}du_{3}$

After simplification, we have

$\triangledown {\varphi }=\frac{1}{h_{1}}\frac{\partial \varphi}{\partial u_{1}}e_{1}+\frac{1}{h_{2}}\frac{\partial \varphi}{\partial u_{2}}e_{2}+\frac{1}{h_{2}}\frac{\partial \varphi}{\partial u_{3}}e_{3}$

In cylindrical polar coordinates,

$u_{1}=R,u_{2}=\vartheta ,u_{3}=z, h_{1}=1, h_{2}=R, h_{3}=1$

Therefore, from $\triangledown {\varphi }=\frac{1}{h_{1}}\frac{\partial \varphi}{\partial u_{1}}e_{1}+\frac{1}{h_{2}}\frac{\partial \varphi}{\partial u_{2}}e_{2}+\frac{1}{h_{2}}\frac{\partial \varphi}{\partial u_{3}}e_{3}$ ,

we have

$\triangledown {\varphi }=\frac{1}{1}\frac{\partial \varphi}{\partial R}e_{1}+\frac{1}{R}\frac{\partial \varphi}{\partial \vartheta }e_{2}+\frac{1}{1}\frac{\partial \varphi}{\partial z}e_{3}$

$=\frac{\partial \varphi}{\partial R}e_{1}+\frac{1}{R}\frac{\partial \varphi}{\partial \vartheta }e_{2}+\frac{\partial \varphi}{\partial z}e_{3}$