Vector equation of a spherical curve

In summary: Oh, right, I just didn't think of assuming a unit sphere centered at the origin - much easier now. And yeah, the curve just follows a spherical surface.Thanks-
  • #1
BilalX
7
0
Given a curve described by the following function:
r(t) = (cos^2(t), sin(t), sin(t)*cos(t)), 0 ≤ t ≤ 2*Pi

How can I prove it describes a spherical shape? I know that the parametric representation is the following, but I'm not sure how to reconcile that with the expression of a sphere.

x = cos^2(t)
y = sin(t)
z = sin(t)*cos(t)

I'd greatly appreciate any insight, thanks.
 
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  • #2
BilalX said:
Given a curve described by the following function:
r(t) = (cos^2(t), sin(t), sin(t)*cos(t)), 0 ≤ t ≤ 2*Pi

How can I prove it describes a spherical shape? I know that the parametric representation is the following, but I'm not sure how to reconcile that with the expression of a sphere.

x = cos^2(t)
y = sin(t)
z = sin(t)*cos(t)

I'd greatly appreciate any insight, thanks.

Well, for a sphere the x^2+y^2+z^2=r^2
I presume in your case r is equal to one. Plug in the values for x, y,z above and apply trigonometric identities. That said, I don't think a curve can fill an area without appealing to the axiom of choice. Thus perhaps your curve is on the surface of a sphere but I don't think it is a sphere.
 
  • #3
John Creighto said:
Well, for a sphere the x^2+y^2+z^2=r^2
I presume in your case r is equal to one. Plug in the values for x, y,z above and apply trigonometric identities. That said, I don't think a curve can fill an area without appealing to the axiom of choice. Thus perhaps your curve is on the surface of a sphere but I don't think it is a sphere.

Oh, right, I just didn't think of assuming a unit sphere centered at the origin - much easier now. And yeah, the curve just follows a spherical surface.

Thanks-
 
  • #4
John Creighto said:
Well, for a sphere the x^2+y^2+z^2=r^2
I presume in your case r is equal to one. Plug in the values for x, y,z above and apply trigonometric identities. That said, I don't think a curve can fill an area without appealing to the axiom of choice. Thus perhaps your curve is on the surface of a sphere but I don't think it is a sphere.
You can construct space-filling curves without the use of Choice. For example, see the construction given at http://en.wikipedia.org/wiki/Space-filling_curve .
 

FAQ: Vector equation of a spherical curve

What is a vector equation of a spherical curve?

A vector equation of a spherical curve is an equation that represents the position of a point on a curve in three-dimensional space using vectors. It is typically written in the form r(t) = x(t)i + y(t)j + z(t)k, where r(t) is the position vector of the point at a particular parameter t, and i, j, and k are unit vectors in the x, y, and z directions, respectively.

How is a vector equation of a spherical curve different from a parametric equation?

A vector equation of a spherical curve is different from a parametric equation in that it uses vectors to represent the position of a point on the curve, while a parametric equation uses scalar parameters. Additionally, a vector equation can represent curves in three-dimensional space, while a parametric equation is typically used for two-dimensional curves.

What are the advantages of using a vector equation of a spherical curve?

Using a vector equation of a spherical curve allows for a more concise and elegant representation of the curve in three-dimensional space. It also makes it easier to calculate important quantities such as velocity, acceleration, and curvature at any point on the curve.

How do you determine the direction of a spherical curve using its vector equation?

The direction of a spherical curve can be determined by taking the derivative of the vector equation with respect to the parameter t. This will give a vector that is tangent to the curve at that particular point. The direction of the curve can then be determined by finding the unit vector in the same direction as the tangent vector.

Can a vector equation of a spherical curve be converted into a parametric equation?

Yes, a vector equation of a spherical curve can be converted into a parametric equation by equating the components of the position vector r(t) to scalar parameters. This will result in a set of three parametric equations that represent the x, y, and z coordinates of the point on the curve at a particular parameter t.

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