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- #1

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- #2

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You also have the wrong expressions for del F in cartesian, cylindrical and spherical coordinates. Look them up.

- #3

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(∂f/∂x)x+(∂f/∂y)y+(∂f/∂z)z (where x, y, and z should be x(hat), y(hat), and z(hat))

- #4

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- #5

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(∂f/∂x)x+(∂f/∂y)y+(∂f/∂z)z (where x, y, and z should be x(hat), y(hat), and z(hat))

You did not write "just del". You wrote del-dot. The dot was wrong. You can write del f or you can write grad f. Same thing. (But del-dot-

Your expression for grad f is correct for Cartesian coordinates x,y,z. I use

When you get to V2 and V3 the expressions for grad V are very different from Cartesian. V2 is using cylindrical and V3 is using spherical.

- #6

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Yes, ∂(-2xz)/∂y = 0.

No, you are not following your formula in post 3 for V1.

Don't keep changing your notation for the unit vectors. Use

- #7

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dang I felt it might have actually been right. was it simply this 6yi+k

- #8

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dang I felt it might have actually been right. was it simply this 6yi+k

Was what 6y

- #9

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yes but it isn't correct I can imagine.

- #10

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Tell you what, start over with your expression for grad V1 from your post 3, use i j k unit vectors, and let's see if we can't figure this one out. Remember to use V1, not f.

- #11

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∇V1=(∂f/∂x)i+(∂f/∂y)j+(∂f/∂z)k= 6y i + k

- #12

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∇V1=(∂f/∂x)i+(∂f/∂y)j+(∂f/∂z)k= 6y i + k

I said use V1, not f.

OK, one term at a time. Try ∂(V1)/∂x again?

- #13

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ok ∂(V1)/∂x= 6y

- #14

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∂(V1)/∂x = 6y i sorry I forgot the "i" in the previous post.

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∂(V1)/∂x = 6y i sorry I forgot the "i" in the previous post.

What about ∂(-2xz)/∂x?

- #16

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6y-2z i

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- #18

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(6y-2z)i + (6x)j + (1-2x)k

- #19

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for V2 (10cos(phi))-z)i-(10sin(phi))j-(ρ)k

- #20

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for V3 ((-2/r^2)cos(phi))i-((2sin(phi))/(r^2sin(theta)))k

- #21

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(6y-2z)i + (6x)j + (1-2x)k

Right!

- #22

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for V2 (10cos(phi))-z)i-(10sin(phi))j-(ρ)k

Right, except you can't use

- #23

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for V3 ((-2/r^2)cos(phi))i-((2sin(phi))/(r^2sin(theta)))k

Problem here.

With V2 I noticed the angle phi was used; I believe for cylindrical coordinates that angle is usually denoted by theta.

So I have to assume phi is the same angle for V3 as it was for V2, i.e. the angle between the x-y plane and the z axis, in which case grad V3 looks wrong. Do you know which angle is the one made with the z axis and which is in the x-y plane? Then I could check grad V3 also.

Anyway, you seem to be with the program here and have done quite well.

- #24

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Theta is with the z and phi is with the xy

- #25

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for V3 ((-2/r^2)cos(phi))i-((2sin(phi))/(r^2sin(theta)))k

Correct, except again you don't use

V2 shoud have used θ instead of φ. It's not good for φ to be the z-axis angle in one coordinate system and the x-y angle in another. I would explain that to your instructor.

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