Ava-May Nelson

2021-02-05

Given point P(-2, 6, 3) and vector $B=y{a}_{x}+\left(x+z\right){a}_{y}$, express P and B in cylindrical and spherical coordinates. Evaluate A at P in the Cartesian, cylindrical and spherical systems.

dieseisB

Step 1 The objective is to express the point P(−2,6,3) and vector $B=y{a}_{x}+\left(x+z\right){a}_{y}$ in cylindrical and spherical coordinates. Step 2 To convert Cartesian coordinates $\left(x,y,z\right)$ to cylindrical coordinates $\left(r,\theta ,z\right)$ $r=\sqrt{{x}^{2}+{y}^{2}}$
$\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$
$z=z$ The point $P\left(-2,6,3\right)$ in cylindrical coordinates is, $r=\sqrt{4+36}$
$=\sqrt{40}$
$=2\sqrt{10}$ And $\theta ={\mathrm{tan}}^{-1}\left(\frac{6}{-2}\right)$
$=-1.25$ Cylindrical coordinates is $\left(2\sqrt{10},-1.25,3\right)$ To convert Cartesian coordinates $\left(x,y,z\right)$ to spherical coordinates $\left(\rho ,\theta ,\varphi \right)$ $p=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
$\theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$
$\varphi ={\mathrm{tan}}^{-1}\left(\frac{\sqrt{{x}^{2}+{y}^{2}}}{z}\right)$ The point $P\left(-2,6,3\right)$ in spherical coordinates is, $p=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
$\sqrt{4+36+9}$
$=\sqrt{49}$
$=7$ And $\theta ={\mathrm{tan}}^{-1}\left(\frac{6}{-2}\right)$
$=-1.25$ And $\varphi ={\mathrm{tan}}^{-1}\left(\frac{\sqrt{4+36}}{3}\right)$
$={\mathrm{tan}}^{-1}\left(\frac{\sqrt{40}}{3}\right)$
$=1.3034$ Spherical coordinates is $\left(7,-1.25,1.3034\right)$ Step 3 The vector $B=y{a}_{x}+\left(x+z\right){a}_{y}$ in cylindrical and spherical coordinates. In the cartesian system B at P is $B=6{a}_{x}+{a}_{y}$ For vector B, ${B}_{x}=y$
${B}_{y}=x+z$
${B}_{z}=0$ In cylindrical system ${A}_{r}=y\mathrm{cos}\theta +\left(x+z\right)\mathrm{sin}\theta$
${A}_{\theta }=-y\mathrm{sin}\theta +\left(x+z\right)\mathrm{cos}\theta$

fudzisako

To express point P(-2, 6, 3) and vector $𝐁=y{𝐚}_{x}+\left(x+z\right){𝐚}_{y}$ in cylindrical and spherical coordinates, we use the following transformations:
1. Cylindrical Coordinates:
$P:\left(r,\theta ,z\right)\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}r&=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{\left(-2{\right)}^{2}+{6}^{2}}=\sqrt{40}=2\sqrt{10},\phantom{\rule{0ex}{0ex}}\theta &=\mathrm{arctan}\left(\frac{y}{x}\right)=\mathrm{arctan}\left(\frac{6}{-2}\right)=\mathrm{arctan}\left(-3\right)=-\mathrm{arctan}\left(3\right),\phantom{\rule{0ex}{0ex}}z&=z=3.$
$𝐁$ in cylindrical coordinates:
$𝐁:\left({B}_{r},{B}_{\theta },{B}_{z}\right)\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}{B}_{r}&={B}_{x}=0,\phantom{\rule{0ex}{0ex}}{B}_{\theta }&={B}_{y}=x+z=-2+3=1,\phantom{\rule{0ex}{0ex}}{B}_{z}&={B}_{z}=0.$
2. Spherical Coordinates:
$P:\left(r,\theta ,\varphi \right)\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}r&=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}=\sqrt{\left(-2{\right)}^{2}+{6}^{2}+{3}^{2}}=\sqrt{49}=7,\phantom{\rule{0ex}{0ex}}\theta &=\mathrm{arctan}\left(\frac{y}{x}\right)=\mathrm{arctan}\left(\frac{6}{-2}\right)=\mathrm{arctan}\left(-3\right)=-\mathrm{arctan}\left(3\right),\phantom{\rule{0ex}{0ex}}\varphi &=\mathrm{arctan}\left(\frac{\sqrt{{x}^{2}+{y}^{2}}}{z}\right)=\mathrm{arctan}\left(\frac{\sqrt{40}}{3}\right)=\mathrm{arctan}\left(\frac{2\sqrt{10}}{3}\right).$
$𝐁$ in spherical coordinates:
$𝐁:\left({B}_{r},{B}_{\theta },{B}_{\varphi }\right)\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}{B}_{r}&={B}_{x}=0,\phantom{\rule{0ex}{0ex}}{B}_{\theta }&={B}_{y}=x+z=-2+3=1,\phantom{\rule{0ex}{0ex}}{B}_{\varphi }&={B}_{z}=0.$
To evaluate $𝐀$ at point P in different coordinate systems:
1. Cartesian Coordinates:
$𝐀$ at P: $𝐀\left(-2,6,3\right)$.
2. Cylindrical Coordinates:
$𝐀$ at P: $𝐀\left(2\sqrt{10},-\mathrm{arctan}\left(3\right),3\right)$.
3. Spherical Coordinates:
$𝐀$ at P: $𝐀\left(7,-\mathrm{arctan}\left(3\right),\mathrm{arctan}\left(2\sqrt{10}/3\right)\right)$.

xleb123

1. Expressing point P (-2, 6, 3) in cylindrical coordinates:
In cylindrical coordinates, a point is represented as $\left(\rho ,\varphi ,z\right)$, where $\rho$ is the distance from the origin to the point in the xy-plane, $\varphi$ is the angle measured from the positive x-axis to the line segment connecting the origin and the point, and $z$ is the height of the point.
To find $\rho$ and $\varphi$, we can use the following formulas:
$\rho =\sqrt{{x}^{2}+{y}^{2}}$
$\varphi ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$
Given that $P\left(-2,6,3\right)$, we can calculate:
$\rho =\sqrt{\left(-2{\right)}^{2}+{6}^{2}}=\sqrt{40}=2\sqrt{10}$
$\varphi ={\mathrm{tan}}^{-1}\left(\frac{6}{-2}\right)={\mathrm{tan}}^{-1}\left(-3\right)=-1.249$
Therefore, the cylindrical coordinates of point P are $\left(2\sqrt{10},-1.249,3\right)$.
2. Expressing vector B as $B=y{a}_{x}+\left(x+z\right){a}_{y}$ in cylindrical coordinates:
To express vector B in cylindrical coordinates, we need to convert the Cartesian unit vectors ${a}_{x}$ and ${a}_{y}$ into cylindrical unit vectors. The cylindrical unit vectors are denoted as $\left(\stackrel{^}{\rho },\stackrel{^}{\varphi },\stackrel{^}{z}\right)$.
The conversion formulas are:
$\stackrel{^}{\rho }=\mathrm{cos}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}{a}_{x}+\mathrm{sin}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}{a}_{y}$
$\stackrel{^}{\varphi }=-\mathrm{sin}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}{a}_{x}+\mathrm{cos}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}{a}_{y}$
$\stackrel{^}{z}={a}_{z}$
Given that $B=y{a}_{x}+\left(x+z\right){a}_{y}$, we can substitute the conversion formulas:
$B=y\left(\mathrm{cos}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}{a}_{x}+\mathrm{sin}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}{a}_{y}\right)+\left(x+z\right)\left(-\mathrm{sin}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}{a}_{x}+\mathrm{cos}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}{a}_{y}\right)$
Simplifying, we get:
$B=y\mathrm{cos}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\rho }-\left(x+z\right)\mathrm{sin}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\rho }+y\mathrm{sin}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\varphi }+\left(x+z\right)\mathrm{cos}\left(\varphi \right)\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\varphi }$
Therefore, the expression of vector B in cylindrical coordinates is:
$B=\left(y\mathrm{cos}\left(\varphi \right)-\left(x+z\right)\mathrm{sin}\left(\varphi \right)\right)\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\rho }+\left(y\mathrm{sin}\left(\varphi \right)+\left(x+z\right)\mathrm{cos}\left(\varphi \right)\right)\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\varphi }$.
3. Expressing point P (-2, 6, 3) in spherical coordinates:
In spherical coordinates, a point is represented as $\left(r,\theta ,\varphi \right)$, where $r$ is the distance from the origin to the point, $\theta$ is the angle measured from the positive z-axis to the line segment connecting the origin and the point, and $\varphi$ is the angle measured from the positive x-axis to the projection of the line segment on the xy-plane.
To find $r$, $\theta$, and $\varphi$, we can use the following formulas:
$r=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
$\theta ={\mathrm{cos}}^{-1}\left(\frac{z}{r}\right)$
$\varphi ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)$
Given that $
P\left(-2,6,3\right)$
, we can calculate:
$r=\sqrt{\left(-2{\right)}^{2}+{6}^{2}+{3}^{2}}=\sqrt{49}=7$
$\theta ={\mathrm{cos}}^{-1}\left(\frac{3}{7}\right)\approx 1.230$
$\varphi ={\mathrm{tan}}^{-1}\left(\frac{6}{-2}\right)={\mathrm{tan}}^{-1}\left(-3\right)=-1.249$
Therefore, the spherical coordinates of point P are $\left(7,1.230,-1.249\right)$.
4. Evaluating vector A at point P in different coordinate systems:
To evaluate vector A at point P, we can simply substitute the coordinates of P into the expression of vector A in each coordinate system.
a) In Cartesian coordinates:
$A=x\phantom{\rule{0.167em}{0ex}}{a}_{x}+y\phantom{\rule{0.167em}{0ex}}{a}_{y}+z\phantom{\rule{0.167em}{0ex}}{a}_{z}$
Substituting $P\left(-2,6,3\right)$, we get:
$A=-2\phantom{\rule{0.167em}{0ex}}{a}_{x}+6\phantom{\rule{0.167em}{0ex}}{a}_{y}+3\phantom{\rule{0.167em}{0ex}}{a}_{z}$
b) In cylindrical coordinates:
$A={A}_{\rho }\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\rho }+{A}_{\varphi }\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\varphi }+{A}_{z}\phantom{\rule{0.167em}{0ex}}\stackrel{^}{z}$
Substituting the cylindrical coordinates of P $\left(2\sqrt{10},-1.249,3\right)$, we get:
$A={A}_{\rho }\left(2\sqrt{10},-1.249,3\right)+{A}_{\varphi }\phantom{\rule{0.167em}{0ex}}\left(-\mathrm{sin}\left(-1.249\right),\mathrm{cos}\left(-1.249\right),0\right)+{A}_{z}\phantom{\rule{0.167em}{0ex}}\left(0,0,1\right)$
c) In spherical coordinates:
$A={A}_{r}\phantom{\rule{0.167em}{0ex}}\stackrel{^}{r}+{A}_{\theta }\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\theta }+{A}_{\varphi }\phantom{\rule{0.167em}{0ex}}\stackrel{^}{\varphi }$
Substituting the spherical coordinates of P $\left(7,1.230,-1.249\right)$, we get:
$A={A}_{r}\left(7,1.230,-1.249\right)+{A}_{\theta }\phantom{\rule{0.167em}{0ex}}\left(-\mathrm{sin}\left(1.230\right)\mathrm{cos}\left(-1.249\right),-\mathrm{sin}\left(1.230\right)\mathrm{sin}\left(-1.249\right),\mathrm{cos}\left(1.230\right)\right)+{A}_{\varphi }\left(-\mathrm{sin}\left(-1.249\right),\mathrm{cos}\left(-1.249\right),0\right)$.

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