Syrus said:
Homework Statement
I am working in "Intro to PDEs with Applications" on page 6. Gradients come up in discussions of surfaces expressed as F(x,y,z). In discussing such matters, the buildup includes the assumption that grad F is not equal to the zero vector. A later line reads, "Under the assumption [above], the set of points (x,y,z) in the domain which satisfy the equation F(x,y,z) = c for some appropriate value of c, is a surface is the domain.Homework Equations
The Attempt at a Solution
My question is, what precautions does the gradient not being zero entail? The only answer I've been able to come up with so far is that they mean smooth surface when they simply say surface (since points where grad F = 0) seem to correspond to 'corners' or non-differentiate points. Any deeper or more accurate insight?
Look up the "Implicit Function Theorem".
Having a nonzero gradient vector at ##(x_0,y_0,z_0)## means that at least one of the partial derivatives is nonzero. For example, if ##F_z(x_0,y_0,z_0)
\neq 0##, then ##dF = 0 = F_x dx + F_y dy + F_z dz## can be solved for ##dz## in terms of ##dx## and ##dy## (because we are allowed to divide by ##F_z \neq 0##):
$$ dz = -\frac{F_x}{F_z} dx - \frac{F_y}{F_z} dy.$$
That implies that for ##(x,y)## near ##(x_0,y_0)## we have ##z = f(x,y)## for some smooth function ##f(x,y)##. In other words, if ##F_z \neq 0## the equation ##F(x,y,z) = 0## leads to a unique expression for ##z## as a smooth function of ##(x,y)##, and the function is unique, at least near ##(x_0,y_0,z_0)##.
If the gradient vanishes, we may have a non-smooth and non-unique function (that is, two or more functions ##f_1(x,y)## and ##f_2(x,y)## which can also be non-smooth), or sometimes we can have several different smooth functions ##z = f_i(x,y)## that all satisfy ##0 = F(x,y,f_i(x,y))##. Alternatively, we may have no such functions at all; it all depends on details about the nature of the original function ##F##.