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

Click For Summary
The normal vector to a surface defined by a function f(x, y, z) is indeed the gradient vector, ∇f(x0, y0, z0), at a specific point. This gradient is perpendicular to the tangent plane of the surface where f is constant. The relationship is established through the equation of the tangent plane, which shows that the gradient has a zero dot product with any vector lying in the tangent plane. Therefore, the gradient vector serves as the normal vector, confirming its role in determining the surface's orientation. Understanding this relationship is crucial for grasping concepts in multivariable calculus.
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.
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
12K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
5K