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Surface = constant

by Summer2442
Tags: constant, surface
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Jan15-13, 09:34 AM
P: 8

I wanted to ask what it meant if a surface's equation is equal to a constant. and what does that say about the gradient(surface).

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Jan15-13, 12:18 PM
P: 643
A surface's equation can't be a constant. A constant is an expression, not an equation.
Jan15-13, 01:03 PM
P: 3,091
So you mean like f(x,y,z)=constant

f(x,y,z) = x + y + z = 35 describes a plane in 3 space


g(x,y,z) = x^2 + y^2 + z^2 = constant describes a sphere in 3 space

what can you say about the gradients of these two examples?

Jan15-13, 03:42 PM
P: 8
Surface = constant

yes what i mean is f(x, y, z) = constant,

well for f(x,y,z) the gradient is 1x+1y+z (right?), however for g(x, y, z) it is still a function 2xx+2yy+2zz (right?), does that mean it does not matter, that f(x,y,z) = constant has no effect on the gradient but only the function itself.

if that is the case, then why does it say sometimes that when if f(x,y,z) = constant, then the gradient defines the normal to the surface and that it is perpendicular to dr.

Jan15-13, 03:48 PM
P: 643
If we can describe any surface as f(x,y,z)=g(x,y,z) (which we can for most surfaces, regardless of if they can be represented in terms of elementary functions.) This is equivalent to (f-g)(x,y,z)=0, so we can't really say anything about the surface.

I wanted to take the gradient of both sides, but that appears to be invalid.
Jan15-13, 04:08 PM
P: 8
let me be more specific :)

i hope you can see my attachment, i do not understand the relation between a function f(x, y, z) being equal to a constant, and the gradient of that function being perpendicular to a single increment dr? is there are relation, does f(x, y, z) = constant imply anything.

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Screen shot 2013-01-16 at 00.05.39.png  
Jan15-13, 05:13 PM
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PF Gold
P: 39,682
Quote Quote by Summer2442 View Post

I wanted to ask what it meant if a surface's equation is equal to a constant. and what does that say about the gradient(surface).

It doesn't really make sense to say that an equation is equal to a constant, but, as others have pointed out, the equation decribing any surface can be written in the form f(x,y,z)= constant. Nor does it make sense to talk about the gradient of a surface. I presume what you mean is the gradient of the function. If you describe a surface as f(x,y,z)= constant, the the gradient of f is a vector normal to the surface at each point. That's discussed in any Calculus class in which the gradient of a function of several variables is defined, isn't it?
Jan15-13, 05:24 PM
P: 8
I am sorry i am a beginner in calculus, so there is a lot that i lack, yes i guess that's the point of my confusion, because I recall that the gradient is tangent to the function, but i guess this only applies for a function dependent on only one variable. but when the function is of several variables then the gradient of that function is normal to the surface at each point. is that right?
Jan15-13, 06:02 PM
P: 5,462
A general surface is curved and at any point on the surface, X(x1,,y1,z1), there is not a single tangent line, but an infinity of lines forming a tangent plane.

The gradient of a function f(x,y,z) = [itex]{\nabla _{{X_1}}}f[/itex] is perpendicular to this plane.

Do you need a diagram or can you visualise this?
Jan15-13, 06:31 PM
C. Spirit
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P: 5,661
Hi Summer2442! Consider the family of surfaces [itex](S_{t})[/itex] which are levels sets of the smooth map [itex]F:\mathbb{R}^{3} \rightarrow \mathbb{R}[/itex]. So take an element of the family [itex]S_{t_{0}}[/itex] and any [itex]p\in S_{t_{0}}[/itex] and any [itex]v[/itex] tangent to the surface at that point. Consider a regular curve [itex]c:(-\varepsilon ,\varepsilon )\rightarrow S_{t_{0}}[/itex] such that [itex]c(0) = p, \dot{c}(0) = v[/itex]. We have that [itex]F(c(t)) = const, \forall t\in (-\varepsilon ,\varepsilon )[/itex] so [itex]\frac{\mathrm{d} }{\mathrm{d} t}|_{t = 0}F(c(t)) = \triangledown F|_{p}\cdot v = 0[/itex]. Since this was for an arbitrary point and tangent vector, we can say the gradient of this smooth map is a normal field to the aforementioned family of surfaces.
Jan15-13, 06:33 PM
P: 8
No I can visualise, perfect, thanks a lot that really clarified things.

thanks everyone.

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