Gradient operator of a function

In summary, the gradient of f(x)=x^3+y^3+z^3-3xyz is (3x^2-3yz, 3y^2-3xz, 3z^2-3xy). The points at which grad(f) is orthogonal to the z-axis and parallel to the x-axis can be found using the dot product test. For (c), the point (0,0,0) satisfies grad(f)=0, but it is not the only solution. Further work is needed to determine any other solutions.
  • #1
bubokribuck
42
0
(1) Let [itex]f(x)=x^3+y^3+z^3-3xyz[/itex], Find grad(f).

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

(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 [itex]grad(f)=(3x^2-3yz, 3y^2-3xz, 3z^2-3xy)=(0,0,0)[/itex], so the points at which grad(f)=0 are (0,0,0). Is that right?
 
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  • #2
bubokribuck said:
(1) Let [itex]f(x)=x^3+y^3+z^3-3xyz[/itex], Find grad(f).

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

(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 [itex]grad(f)=(3x^2-3yz, 3y^2-3xz, 3z^2-3xy)=(0,0,0)[/itex], 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.
 
  • #3
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
[itex]3x^2−3yz=0[/itex]
[itex]3y^2−3xz=0[/itex]
[itex]3z^2−3xy=0[/itex]

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

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

This is what I thought, but I might be wrong though.
 
  • #4
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
[itex]3x^2−3yz=0[/itex]
[itex]3y^2−3xz=0[/itex]
[itex]3z^2−3xy=0[/itex]

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

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.
 

What is the gradient operator of a function?

The gradient operator of a function is a mathematical operator that calculates the rate of change of the function in each direction. It is represented by the symbol ∇ and is also known as the del operator.

How is the gradient operator of a function calculated?

The gradient operator of a function is calculated by taking the partial derivatives of the function with respect to each variable and combining them into a vector. This vector represents the direction and magnitude of the steepest ascent of the function.

What is the significance of the gradient operator in mathematics and science?

The gradient operator is of great significance in mathematics and science as it helps in solving problems related to optimization, vector calculus, and physics. It is also used in fields such as engineering, economics, and computer science.

What is the relationship between the gradient operator and the gradient of a function?

The gradient operator and the gradient of a function are closely related. The gradient of a function is the result of applying the gradient operator to that function. In other words, the gradient of a function is the vector representation of the gradient operator acting on that function.

Can the gradient operator be applied to any type of function?

Yes, the gradient operator can be applied to any type of function, as long as the function is differentiable. This means that the function must have a well-defined slope at every point in its domain.

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