Gradient operator of a function

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The gradient of the function f(x) = x^3 + y^3 + z^3 - 3xyz is calculated as grad(f) = (3x^2 - 3yz, 3y^2 - 3xz, 3z^2 - 3xy). For part (c), the point where grad(f) equals zero is confirmed to be (0,0,0), but it is noted that this may not be the only solution. The discussion highlights confusion regarding the concepts of orthogonality to the z-axis and parallelism to the x-axis, suggesting a need for clarification on vector relationships. Participants are encouraged to explore the dot product test to determine relationships between vectors. Further exploration of the equations derived from setting grad(f) to zero is necessary to identify all possible solutions.
bubokribuck
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(1) Let f(x)=x^3+y^3+z^3-3xyz, Find grad(f).

grad(f)=(3x^2-3yz, 3y^2-3xz, 3z^2-3xy).

(2) Identify the points at which grad(f) is
a) orthogonal to the z-axis
b) parallel to the x-axis
c) zero.I have managed to solve for (1), but don't have a clue how to solve for the second part. I have not come across about the theory of "orthogonal to z-axis" and "parallel to x-axis", tried to look up on the internet but still quite confused.

However, for (c) I have come up with something like grad(f)=(3x^2-3yz, 3y^2-3xz, 3z^2-3xy)=(0,0,0), so the points at which grad(f)=0 are (0,0,0). Is that right?
 
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bubokribuck said:
(1) Let f(x)=x^3+y^3+z^3-3xyz, Find grad(f).

grad(f)=(3x^2-3yz, 3y^2-3xz, 3z^2-3xy).

(2) Identify the points at which grad(f) is
a) orthogonal to the z-axis
b) parallel to the x-axis
c) zero.


I have managed to solve for (1), but don't have a clue how to solve for the second part. I have not come across about the theory of "orthogonal to z-axis" and "parallel to x-axis", tried to look up on the internet but still quite confused.

Do you know how to test whether two vectors are parallel or perpendicular?

However, for (c) I have come up with something like grad(f)=(3x^2-3yz, 3y^2-3xz, 3z^2-3xy)=(0,0,0), so the points at which grad(f)=0 are (0,0,0). Is that right?

Certainly that point works. But without seeing what you did, I don't know if you know whether or not that is the only point that works.
 
LCKurtz said:
Do you know how to test whether two vectors are parallel or perpendicular?
Yes, I know that if
1) a and b are parallel, axb=0
2) a and b are perpendicular, axb=|a||b|n

But in the past I have been given specific vectors like a=(1,2,3) b=(5,3,2). This time, I've only been given statements such as "orthogonal to z-axix", and I really don't have a clue what that actually means.

Let's say, for example, if grad(f) is vector a, so what will vector b be? This is where I'm stuck.
Certainly that point works. But without seeing what you did, I don't know if you know whether or not that is the only point that works.

I actually didn't quite understand the question, but guessed that if grad(f)=0, then it's pretty obvious that
3x^2−3yz=0
3y^2−3xz=0
3z^2−3xy=0

simplify them and you get
x^2−yz=0
y^2−xz=0
z^2−xy=0

therefore x=0, y=0, z=0.

This is what I thought, but I might be wrong though.
 
bubokribuck said:
Yes, I know that if
1) a and b are parallel, axb=0
2) a and b are perpendicular, axb=|a||b|n

(2) isn't a test for perpendicularity. What is the dot product test?

But in the past I have been given specific vectors like a=(1,2,3) b=(5,3,2). This time, I've only been given statements such as "orthogonal to z-axix", and I really don't have a clue what that actually means.

Can you find a vector in the direction of the z axis? Then use the dot product test and see what works.

I actually didn't quite understand the question, but guessed that if grad(f)=0, then it's pretty obvious that
3x^2−3yz=0
3y^2−3xz=0
3z^2−3xy=0

simplify them and you get
x^2−yz=0
y^2−xz=0
z^2−xy=0

therefore x=0, y=0, z=0.

This is what I thought, but I might be wrong though.

You can't say "therefore x=0, y=0, z=0" unless you know (0,0,0) is the only solution. Just because it obviously works doesn't mean it is the only solution. You have to work with those equations to either show that is the only solution or find any others that there might be. In this case, you will find others.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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