Vectors- gradient and normal unit vector- is this correct?

  • Thread starter tomanator
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  • #1
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Homework Statement



For the scalar field f(x, y, z) = x2 − y2 − z find gradf and normal unit
vector to a surface f(x, y, z) = 0 at the point (1, 1, 0).

Homework Equations




The Attempt at a Solution



I calculated gradf= 2xi -2yj -k

at (1,1,0) this is = 2i -2y -k

normal unit vector= [tex]\frac{2i -2y -k}{\sqrt{9}}[/tex]

normal unit vector = [tex]\frac{1}{3}[/tex](2i -2y -k)

is this correct it seems too simple in terms of marks
 

Answers and Replies

  • #2
phyzguy
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I think you've calculated the gradient correctly, but not the normal unit vector. The gradient is tangent to the surface in the direction where the slope is maximum. The normal vector is perpendicular to the tangent plane of the surface. So the normal unit vector will be perpendicular to the gradient vector. You have just calculated the unit gradient vector.
 
  • #3
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I think you've calculated the gradient correctly, but not the normal unit vector. The gradient is tangent to the surface in the direction where the slope is maximum. The normal vector is perpendicular to the tangent plane of the surface. So the normal unit vector will be perpendicular to the gradient vector. You have just calculated the unit gradient vector.
Thanks for you reply, so do you know how I calculate the normal unit vector then?
 
  • #4
ehild
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Tomanator,
your solution is correct. The gradient in a point of a f(x,y,z)=constant surface is normal to the tangent plane of this surface at the given point.

A small change of f can be written as dF=grad f˙dr, the dot product of the gradf vector with the displacement vector dr. You get the highest change of f if the displacement is parallel to the gradient: the gradient vector points in the direction of the steepest slope of the f function.
If dr is tangent to an f(x,y,z) = constant surface dF=grad f˙dr=0. A dot product of two non-zero vector is 0 if they are perpendicular: gradf and dr are perpendicular, gradf is normal vector to the tangent plane.

ehild
 
  • #5
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Tomanator,
your solution is correct. The gradient in a point of a f(x,y,z)=constant surface is normal to the tangent plane of this surface at the given point.

A small change of f can be written as dF=grad f˙dr, the dot product of the gradf vector with the displacement vector dr. You get the highest change of f if the displacement is parallel to the gradient: the gradient vector points in the direction of the steepest slope of the f function.
If dr is tangent to an f(x,y,z) = constant surface dF=grad f˙dr=0. A dot product of two non-zero vector is 0 if they are perpendicular: gradf and dr are perpendicular, gradf is normal vector to the tangent plane.

ehild
Ok thanks ehild, I'm glad I did it correct first time. I have an exam next week so it's promising :)
 
  • #6
but what if dr is not the tangent to the surface ?? how do u calculate the unit normal vector then ?
 
  • #7
ehild
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The gradient vector is normal to the surfaces f(x,y,z)=constant, and it can be determined from the partial derivatives of f(x,y,z). You do not need any dr to determine it. What I explained was that the gradient vector is really normal to the f=constant surfaces. And you get the change df of the function f when moving from r to r+dr by calculating the dot product grad(f )˙dr=df.

ehild
 
  • #8
yes, i get it .. i always keep forgetting dr is the tangent to the plane of surface. thanks ehild. could you help with what a unit normal vector is ?? is it the direction of the gradient vector? if so, how do you calculate it given the gradient and a point on the surface ?
 
  • #9
Ray Vickson
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Homework Statement



For the scalar field f(x, y, z) = x2 − y2 − z find gradf and normal unit
vector to a surface f(x, y, z) = 0 at the point (1, 1, 0).

Homework Equations




The Attempt at a Solution



I calculated gradf= 2xi -2yj -k

at (1,1,0) this is = 2i -2y -k

normal unit vector= [tex]\frac{2i -2y -k}{\sqrt{9}}[/tex]

normal unit vector = [tex]\frac{1}{3}[/tex](2i -2y -k)

is this correct it seems too simple in terms of marks
Your calculations are correct: grad f is the direction of steepest ascent (i.e., the direction of greatest rate of increase in f), and grad(f)/|grad(f)| is the corresponding unit normal.

RGV
 
  • #10
ehild
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yes, i get it .. i always keep forgetting dr is the tangent to the plane of surface. thanks ehild. could you help with what a unit normal vector is ?? is it the direction of the gradient vector? if so, how do you calculate it given the gradient and a point on the surface ?
See the original post.


ehild
 
  • #11
ehild
Homework Helper
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Your calculations are correct: grad f is the direction of steepest ascent (i.e., the direction of greatest rate of increase in f), and grad(f)/|grad(f)| is the corresponding unit normal.

RGV
Hi RGV,
It was not "tomanator"who asked about finding the normal vector now, but "msslowlearner". "tomanator" solved the problem in the OP, but "msslowlearner" did not look at his solution in the post more than one year ago.

ehild
 
  • #12
yes.., i'm new to the physicsforums .. hardly a week here .. :)
 
  • #13
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Hi

I am trying to learn about covariant and contravariant vectors and derivatives. The videos I have been watching talk about displacement vector as the basis for contravariant vectors and gradient as the basis for covariant vectors. Can somone tlel me the difference between displacemement and gradient vectors?
 

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