# Calculate the gradient of the function

• DODGEVIPER13

## Homework Statement

(a) V1=6xy-2xz+z
(b) V2=10ρcos(phi)-ρz
(c) V3=(2/r)cos(phi)

## The Attempt at a Solution

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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.

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))

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.

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.

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.

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

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

Was what 6y i + k ?

yes but it isn't correct I can imagine.

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.

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

∇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?

ok ∂(V1)/∂x= 6y

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

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

6y-2z i

6y-2z i

Better.
Now, can you do the rest of grad V1?

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

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

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

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

Right!

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.

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.

Theta is with the z and phi is with the xy

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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