How Does Gradient Impact Sphere Rolling Direction on a Surface?

[uzman1243]

Assume perfect sphere lands on a surface given by the function
z = 2x² -3y² at point (2,1,5). I am trying to find a unit vector of the direction in which this perfect sphere will roll.

If I get grad F I'll get a vector field that is perpendicular to the level curves f(x,y) = z = 2x² -3y². This is going to be the steepest ascent. Thus negative grad F should give the steepest descent.

However, this is still a normal to the level surface. How do I find the direction in which it will roll along the surface?

PS. this is not a homework question. I was studying some differential geometry over the holidays and this problem was given in the book.

 

[uzman1243]

How about if I do this:
Find a tangent plane to the surface at point p. Then vector project negative grad f onto the plane?

 

[uzman1243]

How about if I do this:
Find a tangent plane to the surface at point p. Then vector project negative grad f onto the plane?

OK this cannot be done. Any other ideas?

 

[dx]

Use the fact that the magnitude of the gradient is the slope of the function z(x,y).

 

[uzman1243]

Can you please elaborate a bit more?

 

[dx]

Take a vector (a, b, c).

a and b are the x and y components of the gradient. What must c be in order for this vector to point along the surface? You can calculate this from the fact that the rate of change of z along the gradient is √(a² + b²).

I can't give you a complete solution though, as for simple text-book problems like this, the forum rule is that we can only provide help. However, try to use the hint above, and if you can't figure it out, tell me what you tried to do, and I can help you further.

 

[uzman1243]

Take a vector (a, b, c).

a and b are the x and y components of the gradient. What must c be in order for this vector to point along the surface? You can calculate this from the fact that the rate of change of z along the gradient is √(a² + b²).

I can't give you a complete solution though, as for simple text-book problems like this, the forum rule is that we can only provide help. However, try to use the hint above, and if you can't figure it out, tell me what you tried to do, and I can help you further.

f (x,y) = z = 2x^{2} - 3y^{2}

grad f = 4x -6y

Assume those are level curves of the function. Grad F is going to give me a vector that is perpendicular at point a (if substitute point a coordinates to equation)

Thus if I get negative grade f, shouldn't that give me the direction in which it will roll?

 

[dx]

You calculated the divergence, not the gradient. The gradient is a vector, with two components.

grad f = (4x, -6y)

This is a vector in the xy plane, so it doesn't have a component in the z direction. You are looking for a vector that points along the surface z(x,y), i.e. a 3-vector with 3 components.

Now we have a vector already that points in the direction that we want, at least as far as its x and y components are concerned. Now all we need to do is find the z component, to make it a 3-vector.

So, for our problem, we need a vector of the form (4x, -6y, c)

What must c be, to make this vector lie along the tangent plane of f? The tangent plane has a slope which is the magnitude of grad f. So what would c have to be, to make the slope of the vector (4x, -6y, c) come out correctly? i.e. to make it lie along the tangent plane?

 

[uzman1243]

You calculated the divergence, not the gradient. The gradient is a vector, with two components.

Can you explain this part for me? Yes I calculated the divergence. Isn't the divergence a vector in the x-y plane?

What must c be, to make this vector lie along the tangent plane of f? The tangent plane has a slope which is the magnitude of grad f. So what would c have to be, to make the slope of the vector (4x, -6y, c) come out correctly? i.e. to make it lie along the tangent plane?

So the magnitude will be sqrt ( (4x)^2 + (-6y)^2 ) which is equal to c?

Also, I am posting the question and answer as given in the book:

 

[dx]

In your first post, you said you wanted the direction along the surface. If you just want the direction in the xy plane, then that is simply the negative of grad f, as you said. You divide each component by the magnitude to make this a unit vector, as was done in the answer you posted.

I made a mistake in calling 4x - 6y the divergence. That is not true. The divergence is defined for vector fields, not for functions.

However, the gradient 'grad f' is not 4x - 6y, it is 4xi - 6yj as the answer you posted shows.

 

[dx]

So the magnitude will be sqrt ( (4x)^2 + (-6y)^2 ) which is equal to c?

The slope of the vector is c/√((4x)² + (-6y)²), which must be equal to √((4x)² + (-6y)²), so

c = √((4x)² + (-6y)²)√((4x)² + (-6y)²)

= (4x)² + (-6y)²

 

[uzman1243]

In your first post, you said you wanted the direction along the surface. If you just want the direction in the xy plane, then that is simply the negative of grad f, as you said. You divide each component by the magnitude to make this a unit vector, as was done in the answer you posted.

Ah yes. I misread the question. I wanted it along the surface where as the question just wanted the direction it will roll.

I made a mistake in calling 4x - 6y the divergence. That is not true. The divergence is defined for vector fields, not for functions.

However, the gradient 'grad f' is not 4x - 6y, it is 4xi - 6yj as the answer you posted shows.

Isnt grad f = grad . f? It gives the same answer right? Or is this the difference:
Grad f gives a vector
where as grad . f gives scalar?

 

[dx]

Yes, ∇⋅F is a scalar, and ∇f is a vector.

Here F is a vector field and f is a scalar field.

Divergence only applies to vector fields, not scalar fields.

 

[uzman1243]

Yes, ∇⋅F is a scalar, and ∇f is a vector.

Here F is a vector field and f is a scalar field.

Divergence only applies to vector fields, not scalar fields.

Awesome. Thank you so much