Calculate the gradient of the function
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The discussion revolves around calculating the gradient of three scalar functions: V1, V2, and V3, while addressing notation issues and coordinate systems. Participants clarify that the gradient should not use "div" notation, as it is applicable only to vector fields, and emphasize the need for correct unit vector notation in Cartesian, cylindrical, and spherical coordinates. The correct expressions for the gradients of V1, V2, and V3 are derived, with specific attention to the use of angles and unit vectors appropriate for each coordinate system. There is a consensus that V2 should use θ instead of φ to avoid confusion between coordinate systems. The conversation concludes with a recommendation to clarify these notational issues with the instructor.
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rude man
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You have the notation wrong for gradient. You wrote div instead. You can't take the div of a scalar. V1, V2 and V3 are all scalars.
You also have the wrong expressions for del F in cartesian, cylindrical and spherical coordinates. Look them up.
You also have the wrong expressions for del F in cartesian, cylindrical and spherical coordinates. Look them up.
DODGEVIPER13
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so what you are saying is that instead of just del I should have used del f for notation?
(∂f/∂x)x+(∂f/∂y)y+(∂f/∂z)z (where x, y, and z should be x(hat), y(hat), and z(hat))
(∂f/∂x)x+(∂f/∂y)y+(∂f/∂z)z (where x, y, and z should be x(hat), y(hat), and z(hat))
DODGEVIPER13
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so if I am right which I probably am not then (a) 6yUx+(-0)+zUz the reason I put 0 is the the partial derivative of -2xz with respect to y is 0 correct since there is no y term.
rude man
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DODGEVIPER13 said:so what you are saying is that instead of just del I should have used del f for notation?
(∂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-F is OK whereas del F or grad F are not.)
Your expression for grad f is correct for Cartesian coordinates x,y,z. I use i, j , and k for the unit vectors. I also use bold type for all vectors.
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.
rude man
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DODGEVIPER13 said:so if I am right which I probably am not then (a) 6yUx+(-0)+zUz the reason I put 0 is the the partial derivative of -2xz with respect to y is 0 correct since there is no y term.
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 i j k why not.
DODGEVIPER13
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dang I felt it might have actually been right. was it simply this 6yi+k
rude man
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DODGEVIPER13 said:dang I felt it might have actually been right. was it simply this 6yi+k
Was what 6y i + k ?
DODGEVIPER13
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yes but it isn't correct I can imagine.
rude man
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I'm getting dizzy. Correct for what?
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.
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.
DODGEVIPER13
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∇V1=(∂f/∂x)i+(∂f/∂y)j+(∂f/∂z)k= 6y i + k
rude man
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DODGEVIPER13 said:∇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?
DODGEVIPER13
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ok ∂(V1)/∂x= 6y
DODGEVIPER13
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∂(V1)/∂x = 6y i sorry I forgot the "i" in the previous post.
rude man
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DODGEVIPER13 said:∂(V1)/∂x = 6y i sorry I forgot the "i" in the previous post.
What about ∂(-2xz)/∂x?
DODGEVIPER13
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6y-2z i
rude man
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DODGEVIPER13 said:6y-2z i
Better.
Now, can you do the rest of grad V1?
DODGEVIPER13
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(6y-2z)i + (6x)j + (1-2x)k
DODGEVIPER13
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for V2 (10cos(phi))-z)i-(10sin(phi))j-(ρ)k
DODGEVIPER13
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for V3 ((-2/r^2)cos(phi))i-((2sin(phi))/(r^2sin(theta)))k
rude man
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DODGEVIPER13 said:(6y-2z)i + (6x)j + (1-2x)k
Right!
rude man
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DODGEVIPER13 said:for V2 (10cos(phi))-z)i-(10sin(phi))j-(ρ)k
Right, except you can't use i and j for the ρ and phi unit vectors. Since I don't know how to put hats over letters I would use 1ρ and 1Φ which is still standard.
rude man
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DODGEVIPER13 said: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.
DODGEVIPER13
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Theta is with the z and phi is with the xy
rude man
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DODGEVIPER13 said:for V3 ((-2/r^2)cos(phi))i-((2sin(phi))/(r^2sin(theta)))k
Correct, except again you don't use i j k as unit vectors for spherical coordinates. Use 1r, 1φ and 1θ.
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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