Deriving tangent plane equation - assuming 'c' is non-zero

by autodidude
Tags: assuming, deriving, equation, nonzero, plane, tangent
 P: 333 In Stewart's calculus text, the way he derives the tangent plane equation at some point is to divide the general plane equation $$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$ by c This must mean c is always non-zero right? But isn't c is the 'z'-component of the normal vector to the surface at some point? If it's non-zero, does that mean that for some surface in 3D, the normal vector always has a component in the z-direction? The one counter-example I can think of is the case of a sphere of radius 1. At (1,0,0), wouldn't the normal vector be pointing in just the x-direction? But then I'm also not sure that the derivative of z with respect to x is defined at this point...
 Homework Sci Advisor HW Helper Thanks P: 12,953 I suspect you'll find that Stewart is giving a method for finding a non-vertical tangent plane?
 P: 350 It could be that he is referring to the tangent plane of the graph of a function z=f(x,y). Assuming the function is differentiable, then the tangent plane is not vertical.
 P: 333 Deriving tangent plane equation - assuming 'c' is non-zero Ah ok, thanks. So, for a sphere - or a circle in 2D, I know we can take derivatives to figure out the slope at some point but the derivative functions aren't defined at x=r and x=-r, does this mean it's not differentiable?
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 Quote by autodidude In Stewart's calculus text, the way he derives the tangent plane equation at some point is to divide the general plane equation $$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$ by c This must mean c is always non-zero right? But isn't c is the 'z'-component of the normal vector to the surface at some point? If it's non-zero, does that mean that for some surface in 3D, the normal vector always has a component in the z-direction? The one counter-example I can think of is the case of a sphere of radius 1. At (1,0,0), wouldn't the normal vector be pointing in just the x-direction? But then I'm also not sure that the derivative of z with respect to x is defined at this point...
If the problem is to write the equation of the tangent plane (of some surface) in the form z= Ax+ By+ C, then, yes, of course, z must be a function of x and y and that will be true if and only if the normal vector to the surface is NOT parallel to the xy-plane. And that is true if and only if it is possible to write the surface itself in the form z= f(x, y) in some neighborhood of that point.

I don't have a copy of Stewart right here, but I suspect you are misquoting since it does not make sense to talk about "deriving the tangent plane" without talking about the surface. What you are talking about is NOT "deriving the tangent plane" of any surface but deriving it in a specific, limited, form.

The "general plane equation", ax+ by+ cz= d, is itself a perfectly good form for a tangent plane to some surface.
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 Quote by autodidude Ah ok, thanks. So, for a sphere - or a circle in 2D, I know we can take derivatives to figure out the slope at some point but the derivative functions aren't defined at x=r and x=-r, does this mean it's not differentiable?
depends... eg.

In the case of the circle ##x^2+y^2=r^2## - the slope of the tangent to the circle at x=r is what?

What is dy/dx there?

Note, however, it is equally valid to differentiate the other way ... finding dx/dy.
In 3D you have a choice of lines to differentiate against so you need to be more careful with what you mean.
 P: 333 @HallsOfIvy: Yeah, I did misunderstand him (of course I did!) @Simon_Bridge: Undefined? I just looked up the 'differentiable function' in wikipedia and it says that there must be a non-vertical tangent line at every point so I guess circles wouldn't be considered differentiable then (but then they're not functions either are they?). These details weren't emphasised in class, all we had to do was just be able to differentiate different functions and relations and be able to use them to solve applied problems.
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 P: 333 @HallsOfIvy: So in the cartesian coordinate system, if the equation $$x^2+y^2=r^2$$ describes the circle, what would be the functions involved? If it's in terms of x, would it be the two functions describing the top and bottom half of the circle? Also, when we are implicitly differentiating the equation of a circle...are we treating y as a function of x (if differentiating with respect to x)? I'll have to try the differentiating the parametric representation! @Simon_Bridge: Ah, ok....there's lot of subtleties that make it all much more interesting than when I took the course. Yeah, it seems like there are a lot of important points I missed!