Geometric Interpretation of √(x^2+y^2+z^2) and its Derivative

In summary, the equation r = √(x^2+y^2+z^2) can be shown to have the gradient of \vec{r}/r. Geometrically, the direction of the gradient of a scalar function represents the direction of maximum change in the value of that function. In this case, since the scalar function is the distance from the origin, the direction of the gradient is in the direction of the unit vector \vec{u_{r}}, which has a magnitude of 1. This makes sense because the rate of change of r as you move in the direction of r is always 1.
  • #1
aaaa202
1,169
2
Let r = √(x^2+y^2+z^2)
One can easily show that [itex]\nabla[/itex]r= [itex]\vec{r}[/itex]/r.
But I'm having a hard time understanding what this means geometrically - who can help? :)
 
Physics news on Phys.org
  • #2
Not that this is a proof, but intuitively the direction of the gradient of a scalar function gives the direction of maximum change of the value of that function. Since your scalar function is the distance from the origin, and it makes sense that the direction of the gradient is in the direction of [itex]\vec{r}[itex] = [itex]r\vec{u_{r}}[itex].

Grad(r) = [itex]\frac{\vec{r}}{r} = \frac{r\vec{u_{r}}}{r} = \vec{u_{r}}[itex], the unit vector in the direction of the position vector. That makes sense, because the rate of change of r as you move in the direction of r is 1.
 

Related to Geometric Interpretation of √(x^2+y^2+z^2) and its Derivative

What is the geometric interpretation of √(x^2+y^2+z^2)?

The geometric interpretation of √(x^2+y^2+z^2) is the distance from the origin to the point (x,y,z) in three-dimensional space. This is known as the magnitude or norm of the vector (x,y,z).

How is the derivative of √(x^2+y^2+z^2) calculated?

The derivative of √(x^2+y^2+z^2) is calculated using the chain rule. First, the inside of the square root is differentiated, resulting in 2x dx + 2y dy + 2z dz. Then, this expression is divided by 2√(x^2+y^2+z^2) to get the final derivative.

What is the significance of the derivative of √(x^2+y^2+z^2)?

The derivative of √(x^2+y^2+z^2) represents the rate of change of the distance from the origin to the point (x,y,z) in three-dimensional space. It tells us how fast the magnitude of the vector is changing at any given point in space.

How does the geometric interpretation of √(x^2+y^2+z^2) relate to the Pythagorean theorem?

The geometric interpretation of √(x^2+y^2+z^2) is closely related to the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In three-dimensional space, this can be extended to the distance from the origin to a point (x,y,z) using the formula √(x^2+y^2+z^2).

How can the derivative of √(x^2+y^2+z^2) be used in real-world applications?

The derivative of √(x^2+y^2+z^2) has many applications in fields such as physics, engineering, and computer graphics. It can be used to calculate velocities and accelerations of moving objects, determine rates of change in complex systems, and create smooth and realistic 3D animations.

Similar threads

Replies
1
Views
234
  • Calculus
Replies
3
Views
1K
Replies
25
Views
2K
Replies
6
Views
2K
Replies
4
Views
1K
Replies
4
Views
398
Replies
1
Views
1K
Replies
5
Views
1K
  • Calculus
Replies
7
Views
2K
Back
Top