What is the relationship between the gradient and the normal vector?

Click For Summary
SUMMARY

The gradient vector, denoted as ∇f(x₀, y₀, z₀), is definitively the normal vector to the tangent plane of the surface defined by the equation f(x, y, z) = constant. This relationship is established through the equation ∇f(x₀, y₀, z₀) · (x, y, z) = ∇f(x₀, y₀, z₀) · (x₀, y₀, z₀), which indicates that the gradient is perpendicular to any vector lying in the tangent plane. The directional derivative, expressed as ∂f/∂x cos(θ) + ∂f/∂y cos(φ) + ∂f/∂z cos(ψ), confirms that the gradient vector has a zero dot product with any tangent vector on the surface, reinforcing its role as the normal vector.

PREREQUISITES
  • Understanding of gradient vectors in multivariable calculus
  • Familiarity with tangent planes and their definitions
  • Knowledge of directional derivatives and their calculations
  • Basic concepts of dot products in vector calculus
NEXT STEPS
  • Study the properties of gradient vectors in multivariable functions
  • Learn about tangent planes and their geometric interpretations
  • Explore directional derivatives and their applications in optimization
  • Investigate the relationship between gradients and level surfaces
USEFUL FOR

Students and professionals in mathematics, particularly those studying multivariable calculus, as well as engineers and physicists who apply these concepts in practical scenarios.

jaguar7
Messages
41
Reaction score
0
is the normal just grad(f(x0,y0,z0))? If so, how exactly does this work out to be so? Explain? Thanks... :D

& is the calculus section the most appropriate place to put this question? thanks again. :)
 
Physics news on Phys.org
The tangent plane is defined to be the plane of all points (x,y,z) the equation

\nabla f(x_0,y_0,z_0) \cdot (x,y,z)=\nabla f(x_0,y_0,z_0) \cdot (x_0, y_0, z_0)
(you may not have seen it written down this way, but look at what your definition of the tangent plane and make sure you understand that you can write it in this form).

From this equation, what can you say about a vector lying along the tangent plane (this is different from picking a point in the plane) when compared to the gradient?
 
jaguar7 said:
is the normal just grad(f(x0,y0,z0))? If so, how exactly does this work out to be so? Explain? Thanks... :D

& is the calculus section the most appropriate place to put this question? thanks again. :)
This question is poorly phrased. The normal to what? A normal is a vector perpendicular to some surface and just the function, f(x, y, z), does not determine any surface. The gradient vector, of a function, at a given point, is, as Office Shredder says, normal to the tangent plane of the graph of the surface defined by f(x, y, z)= constant.

We can write the "directional derivative", the rate of change of the function f in the direction that makes angles \theta, \phi, and \psi with the positive x, y, and z axes, respectively, as
\frac{\partial f}{\partial x}cos(\theta)+ \frac{\partial f}{\partial y}cos(\phi)+ \frac{\partial f}{\partial z}cos(\psi)
which is exactly the same as the dot product
\nabla f\cdot cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k}
and now cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k} is the unit vector in the given direction.
If f(x,y,z) is a constant on a given surface, the derivative in any direction tangent to that surface must be 0. That is, \nabla f has 0 dot product with any vector tangent to the surface and so is normal to the surface.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
13K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K