Gradient of sphere level fxn with 2 parameters inside the parametric e

[clairaut]

Level function
[L(x,y,z)] = (1/r^2) (x^2 + y^2 + z^2) = 1

Vector [N([x(h,g)], [y(h,g)], [z(g)])] = parametric equation to sphere Level function [L(x,y,z)]

The parametric equations have 2 parameters, h and g

[x(h,g)] = (r [sin (a + gv)]) [cos (b + hw)]

[y(h,g)] = (r [sin (a + gv)]) [sin (b + hw)]

[z(g)] = r [cos (a + gv)]

Where r = radius, a = initial degree phi, b = initial theta, g and h are parameters, AND

v = constant not equal to zero
w = constant not equal to zero

v does NOT EQUAL w

a does NOT equal b

What is the directional derivative of this Level function?

 

[clairaut]

Help.

Without using the zero parametric equation,

I took the directional derivative with respect to parameter h and I got zero.

I took the directional derivative with respect to parameter g and I got zero again.

I'm not sure what how to take the zero parametric equation to define the directional derivative if the parametric equation has 2 parameters instead of just h.

 

[clairaut]

Anybody?

This is excellent food for thought.

 

[HallsofIvy]

The problem is, I suspect, that no one can understand what you are talking about. You say, for example
" [L(x,y,z)] = (1/r^2) (x^2 + y^2 + z^2) = 1" but what is "r"? It can't be the usual distance from the origin because then your equation is an identity, not a function of x, y, and z. You then give parametric equations for the surface of a sphere but what does that have to do with your question, "What is the directional derivative of this Level function?" Again, what function are you talking about? Not [L(x,y,z)] because that either contains the undefined "r" or is a constant.

 

[clairaut]

r is a constant, the distance from origin to surface of the sphere, also known as radius

The mechanics is what's important here. I simply wish to find the directional derivative at a random point on surface of sphere along the tangent direction.

I come across a questionable derivative along the steps.

 

[clairaut]

I'm trying to find the equations of tangent planes and normal lines at this surface

 

[clairaut]

Halls,

How would you find the equation of the tangent plane on the surface of this sphere with the above parametric equations?

 

[slider142]

Help.

Without using the zero parametric equation,

I took the directional derivative with respect to parameter h and I got zero.

I took the directional derivative with respect to parameter g and I got zero again.

I'm not sure what how to take the zero parametric equation to define the directional derivative if the parametric equation has 2 parameters instead of just h.

The directional derivative of a function in the direction of a tangent vector to a level set of that function is always 0. Since the level set is by definition a set of points for which the function's value is a constant.

 

[clairaut]

Yes. It sure appears so.

Then how would you find tan plane equation at surface?

 

[slider142]

Yes. It sure appears so.

Then how would you find tan plane equation at surface?

The tangent plane will be different at each point. The gradient vector of the level function is normal to the tangent plane of the function at each point, a corollary of the previous fact. To find the equation of the tangent plane at a specific point, you can use the fact that the dot product of the gradient vector with any vector in the tangent plane must be 0. Thus, if P = (x0, y0, z0) is the point in your level set L(x, y, z) = 0, then (x, y, z) - P = (x - x0, y - y0, z - z0) is an arbitrary displacement vector attached to that point. We only want the vectors whose dot product with the gradient vector DL(x0, y0, z0) equals 0. That dot product equation is thus the equation satisfied only by (x, y, z) coordinates in the tangent plane to the level set L(x, y, z) = 0 at the point (x0, y0, z0).

 

[clairaut]

The tangent plane will be different at each point. The gradient vector of the level function is normal to the tangent plane of the function at each point, a corollary of the previous fact. To find the equation of the tangent plane at a specific point, you can use the fact that the dot product of the gradient vector with any vector in the tangent plane must be 0. Thus, if P = (x0, y0, z0) is the point in your level set L(x, y, z) = 0, then (x, y, z) - P = (x - x0, y - y0, z - z0) is an arbitrary displacement vector attached to that point. We only want the vectors whose dot product with the gradient vector DL(x0, y0, z0) equals 0. That dot product equation is thus the equation satisfied only by (x, y, z) coordinates in the tangent plane to the level set L(x, y, z) = 0 at the point (x0, y0, z0).

Thanks, but I already knew that.

Given the above parametric equations with the 2 parameters h and g, how would one find the equation of the tangent plane for this sphere?

 

[HallsofIvy]

If a surface is given by parametric equation x= f(u,v), y= g(u,v), z= h(u,v), then the corresponding vector equation would be $\vec{r}(u,v)= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}$. Two tangent vectors to the surface are given by $\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}$ and $\vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}$. A vector perpendicular to the surface is given by the cross product f those two vector.

In particular, with a sphere given by
[x(h,g)] = (r [sin (a + gv)]) [cos (b + hw)]
[y(h,g)] = (r [sin (a + gv)]) [sin (b + hw)]
[z(g)] = r [cos (a + gv)]
the two tangent vectors are given by
$$-rh sin(a+ gv)sin(b+ hw)\vec{i}+ rw sin(a+ gv)cos(b+ hw)\vec{j}$$
and
$$rv cos(a+ gv)cosb+ hw)\vec{i}+ rg cos(a+ gv)sin(b+ hw)\vec{j}- rv sin(a+ gv)\vec{k}$$

and a normal vector is given by the cross product of those two vectors. Once you have a normal vector it is easy to write the equation of the tangent plane at a given point.

 

[clairaut]

Thanks Halls.

I get it.

However,

Is there an operation that takes me from the level function straight into the normal vector for this vector fxn with 2 parameters without going through the crossing of derivatives of the tangent vectors ?

I love the directional derivative of the vector level function with 1 parameter because it leads me to the beautiful gradient dot tan vector equals zero.

 

[clairaut]

Haha

The short operation for the normal vector would simply be the initial position vector for this sphere.

However, for an ellipse... I would need either the crossing of tan vectors or another operation to take me straight from level vector fxn of the ellipsoid to the normal vector.

Kewl!

 

[clairaut]

Hey HALLS,

Are all closed surfaces LEVEL FUNCTIONS?

A hyperbolic paraboloid is NOT a closed surface but I can definitely see it as a level function.

 

[clairaut]

Cones, elliptic paraboloids, hyperbolic paraboloids are open surfaces. Ellipsoids, HYPERBOLOIDS of one and two sheets, and CYLINDERS are CLOSED surfaces. Am I correct?

Help?!

 

[clairaut]

Anybody?!