Am I even doing this correctly? If not please let me know where mistaken. in the cylidded4.15) EXPRESS THE VECTOR K = (x²^yJay +xyzā₂2Coorddicte system & evelucte it at the pont P(3,3,-4)K-=(x² - y²)²₂ + (xy²)=₂P(3,-3,-4)CosCertusion to cylindiedСозф sno о-Sin Cost OOApАф3A₂A.]]'2²A4 3003²0-p²s ²0$1 JA₂² (peast-2]500 FOJ2Ap= -sin +(cos) (p²cos 4-p³²si² d) + (²cos (0-2)редAp² p²ea²³l p² cos sin g22Q•Ap² Co₂+ 3. + (Op² cos Dsin 0²₂) ECx ³pcool pakx²y²!P/P = K² (7)Z=24² p³2322A₂ = 100) + (p²cos 4-7²5 ~²0) + (1), 3² Cos 05-10₂A₂ = p²cos Dz3k=p²eas²³ D-p²cos($s_²³096+ (p²cos (502) 9₂(35,-4)крKA= [ ( 9 cos(5)) - (9 cos($) s =²(3) < [9(²4) cos(+7) s (7)] 5₂(1.125-3.375).-2,2530+ -15.59cd7x P(0, -2.25 -15.59)

x=ρcos⁡ϕ , y=ρsin⁡ϕ , z=zThe position vector, r→=x ax→+y ay→+z…

Answered: the cylielied 1.15) EXPRESS THE VECTOR…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

100%

Am I even doing this correctly? If not please let me know where mistaken.

in the cylidded
4.15) EXPRESS THE VECTOR K = (x²^yJay +xyzā₂2
Coorddicte system & evelucte it at the pont P(3,3,-4)
K-=(x² - y²)²₂ + (xy²)=₂
P(3,-3,-4)
Cos
Certusion to cylindied
Созф sno о
-Sin Cost O
O
Ap
Аф
3
A₂
A.
]]
'2²
A4 3003²0-p²s ²0
$
1 JA₂² (peast-2]
500 FOJ
2
Ap= -sin +(cos) (p²cos 4-p³²si² d) + (²cos (0-2)
ред
Ap² p²ea²³l p² cos sin g
2
2
Q
•
Ap² Co₂+ 3. + (Op² cos Dsin 0²₂) EC
x ³pcool pakx²y²!
P/P = K² (7)
Z=2
4² p³
23
2
2
A₂ = 100) + (p²cos 4-7²5 ~²0) + (1), 3² Cos 05-10₂
A₂ = p²cos Dz
3
k=p²eas²³ D-p²cos($s_²³096+ (p²cos (502) 9₂
(35,-4)
кр
KA= [ ( 9 cos(5)) - (9 cos($) s =²(3) < [9(²4) cos(+7) s (7)] 5₂
(1.125-3.375).
-2,25
30
+ -15.59cd
7x P(0, -2.25 -15.59)

Expert Solution

Step 1: Transform of the coordinates systems

$x = ρ \cos ϕ, y = ρ \sin ϕ, z = z$

The position vector,

$\vec{r} = x \vec{a_{x}} + y \vec{a_{y}} + z \vec{a_{z}}$

$\vec{r} = ρ \cos ϕ \vec{a_{x}} + ρ \sin ϕ \vec{a_{y}} + z \vec{a_{z}}$

$\frac{\partial \vec{r}}{\partial ρ} = \cos ϕ \vec{a_{x}} + \sin ϕ \vec{a_{y}}$

$\frac{\partial \vec{r}}{\partial ϕ} = - ρ \sin ϕ \vec{a_{x}} + ρ \cos ϕ \vec{a_{y}}$

$\frac{\partial \vec{r}}{\partial z} = \vec{a_{z}}$

Now,

$| \frac{\partial \vec{r}}{\partial ρ} | = 1, | \frac{\partial \vec{r}}{\partial ϕ} | = ρ, | \frac{\partial \vec{r}}{\partial z} | = 1$

The unit vectors

$\vec{a_{ρ}} = \frac{\frac{\partial \vec{r}}{\partial ρ}}{| \frac{\partial \vec{r}}{\partial ρ} |} = \cos ϕ \vec{a_{x}} + \sin ϕ \vec{a_{y}}$

$\vec{a_{ϕ}} = \frac{\frac{\partial \vec{r}}{\partial ϕ}}{| \frac{\partial \vec{r}}{\partial ϕ} |} = - \sin ϕ \vec{a_{x}} + \cos ϕ \vec{a_{y}}$

$\vec{a_{z}} = \vec{a_{z}}$

Therefore

$\vec{a_{x}} = \cos ϕ \vec{a_{ρ}} - \sin ϕ \vec{a_{ϕ}}$

$\vec{a_{y}} = \sin ϕ \vec{a_{ρ}} + \cos ϕ \vec{a_{ϕ}}$

$\vec{a_{z}} = \vec{a_{z}}$

$\begin{aligned} \vec{A} & = A_{x} \vec{a_{x}} + A_{y} \vec{a_{y}} + A_{z} \vec{a_{z}} \\ = A_{x} (\cos ϕ \vec{a_{ρ}} - \sin ϕ \vec{a_{ϕ}}) + A_{y} (\sin ϕ \vec{a_{ρ}} + \cos ϕ \vec{a_{ϕ}}) + A_{z} \vec{a_{z}} \\ = (A_{x} \cos ϕ + A_{y} \sin ϕ) \vec{a_{ρ}} + (- A_{x} \sin ϕ + A_{y} \cos ϕ) \vec{a_{ϕ}} + A_{z} \vec{a_{z}} \end{aligned}$