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Surface = constant 
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#1
Jan1513, 09:34 AM

P: 8

Hello,
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). Thanks. 


#2
Jan1513, 12:18 PM

P: 643

A surface's equation can't be a constant. A constant is an expression, not an equation.



#3
Jan1513, 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 vs 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? 


#4
Jan1513, 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 1xˆ+1yˆ+zˆ (right?), however for g(x, y, z) it is still a function 2xxˆ+2yyˆ+2zzˆ (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. thanks. 


#5
Jan1513, 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 (fg)(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. 


#6
Jan1513, 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. thanks. 


#7
Jan1513, 05:13 PM

Math
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PF Gold
P: 39,682




#8
Jan1513, 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?



#9
Jan1513, 06:02 PM

P: 5,462

A general surface is curved and at any point on the surface, X(x_{1,},y_{1},z_{1}), 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? 


#10
Jan1513, 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.



#11
Jan1513, 06:33 PM

P: 8

No I can visualise, perfect, thanks a lot that really clarified things.
thanks everyone. 


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