ASAP Please write neatly, or it is better if it is encoded. Thank you. BELOW ARE EXAMPLES. PLEASE FOLLOW ITS FORMULA AND STEPS. Express the uniform vector field F = 5ax in a) cylindrical components; b) spherical components.
ASAP Please write neatly, or it is better if it is encoded. Thank you. BELOW ARE EXAMPLES. PLEASE FOLLOW ITS FORMULA AND STEPS. Express the uniform vector field F = 5ax in a) cylindrical components; b) spherical components.
An Introduction to Physical Science
14th Edition
ISBN:9781305079137
Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Chapter1: Measurement
Section: Chapter Questions
Problem 12MC
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Question
ASAP
Please write neatly, or it is better if it is encoded. Thank you.
BELOW ARE EXAMPLES. PLEASE FOLLOW ITS FORMULA AND STEPS.
Express the uniform vector field F = 5ax in
a) cylindrical components;
b) spherical components.
Transcribed Image Text:Give the cylindrical coordinates of the point P(x=5, y=2, z=-1).
a) p=√x² + y²
c) z=z
Z=-1
p=√5² +2²
p=5.385
b) p=tan ¹(
❤ = tan¹¹ (²)
❤ = 21.80⁰
Therefore, P(p=5.385, p = 21.80°, z=-1)
![Problem:
Transform the given vector to spherical components at the point specified:
a) D= 4ax - 2ay - 4az @ P(-2,-3,4)
Solution:
a) r= √√x² + y² + z²
r=√(-2)²+(-3)² +4²
r= √29
r= 5.385
0 = cos-1
0 = cos-1
e cos
√x²+y2+
9 = 42.03°
p=tan ¹
p=tan ¹
5.385,
b) D=D.a
5.385.
= cos ¹ (=)
(-2,-3)
p= 56.31°
Since, coordinates (-2,-3) was located in Quadrant III, we will get the angle that
corresponds to the location of the coordinates. Therefore, p = 56.31° +180° =
236.31°.
D₂= D. ae
De (4ax-2ay -4az) - ae
=
De = 4(ax ae) -2(ay ae)- 4(az ae)
180°
56.31°
D= (4ax - 2ay -4az) - ar
D= 4(ax a) -2(ayar) - 4(azar)
D=4(sin cosp) - 2 (sine sinp) - 4(cos6)
D=4[sin(42.03°) cos(236.31°)]- 2[sin(42.03°) sin(236.31°)]-4[cos(42.03°)]
D=-3.34
D₂ = D. ap
Dp (4ax-2ay - 4az). ap
=
D₂ = 4(ax a) -2(ay ap) - 4(az ap)
De = 4(cose cosp) -2(cose sing) - 4(-sine)
De = 4[cos(42.03°) cos(236.31°)]-2[cos (42.03°) sin(236.31°)] +4[sin(42.03°)]
D₂ = 2.27
Dp = 4(-sing) - 2(coso) - 4(0)
Dp = 4[-sin(236.31°)]- 2[cos (236.31°)]
D₂ = 4.44
Therefore, the Spherical Components: D= -3.34a + 2.27ae + 4.44a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bede285-a38f-4732-9564-3dccaeb07e7d%2Fa576e6be-a082-4833-ba7e-582b8a54c2a0%2F58o5a3t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem:
Transform the given vector to spherical components at the point specified:
a) D= 4ax - 2ay - 4az @ P(-2,-3,4)
Solution:
a) r= √√x² + y² + z²
r=√(-2)²+(-3)² +4²
r= √29
r= 5.385
0 = cos-1
0 = cos-1
e cos
√x²+y2+
9 = 42.03°
p=tan ¹
p=tan ¹
5.385,
b) D=D.a
5.385.
= cos ¹ (=)
(-2,-3)
p= 56.31°
Since, coordinates (-2,-3) was located in Quadrant III, we will get the angle that
corresponds to the location of the coordinates. Therefore, p = 56.31° +180° =
236.31°.
D₂= D. ae
De (4ax-2ay -4az) - ae
=
De = 4(ax ae) -2(ay ae)- 4(az ae)
180°
56.31°
D= (4ax - 2ay -4az) - ar
D= 4(ax a) -2(ayar) - 4(azar)
D=4(sin cosp) - 2 (sine sinp) - 4(cos6)
D=4[sin(42.03°) cos(236.31°)]- 2[sin(42.03°) sin(236.31°)]-4[cos(42.03°)]
D=-3.34
D₂ = D. ap
Dp (4ax-2ay - 4az). ap
=
D₂ = 4(ax a) -2(ay ap) - 4(az ap)
De = 4(cose cosp) -2(cose sing) - 4(-sine)
De = 4[cos(42.03°) cos(236.31°)]-2[cos (42.03°) sin(236.31°)] +4[sin(42.03°)]
D₂ = 2.27
Dp = 4(-sing) - 2(coso) - 4(0)
Dp = 4[-sin(236.31°)]- 2[cos (236.31°)]
D₂ = 4.44
Therefore, the Spherical Components: D= -3.34a + 2.27ae + 4.44a
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