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.

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Chapter1: Units, Trigonometry. And Vectors
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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.

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)
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
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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