Is This Calculation of Grad \(\psi\) for \(\psi(x,y,z) = (y-1)z^2\) Correct?

[andrey21]

For the following scalar field:

\psi(x,y,z) = (y-1)z²

Find grad \psi


Here is my attempt at:


Multiplying out brackets:

yz² - z²

Therefore grad \psi = 0+Z² J -2ZK

Is this correct??

 

[transphenomen]

Your close, but you didn't compute the full partial with respect to Z for k.

 

[andrey21]

Ok do I have to take into account all the terms when computing full partial with repsect to x,y,z. So would that give me:

0 + Z² j + 2zy - 2z k

 

[Mark44]

Ok do I have to take into account all the terms when computing full partial with repsect to x,y,z. So would that give me:

0 + Z² j + 2zy - 2z k

You should write this as either
0i + z²j + 2(y - 1)zk
or
<0, z², 2(y - 1)z>

The way you wrote it, the 2zy term isn't associated with any of the unit vectors.

 

[andrey21]

Ok thanks Mark44 so other than writing it the wrong way the answer is correct? I have another similar question:

Find grad \psi = x²(y-1)z

Multiply out brackets:

x²yz - x²z

grad \psi = (2xy -2xz)i +x²zj +(x²y -x²)k

Is this correct??

 

[dextercioby]

Not really. Compute the partial derivative wrt x again.

EDIT to your post. Yes, it's correct now.

 

[andrey21]

Ok I think I've seen my error, should it be:

(2xyz -2xz)i