5.3: Velocity and Arc Length of General Curves on Surfaces

Given a curve (u(t),v( t)) in the domain of a surface, we have a curve

    X(t) =X(u(t ),v(t))
lying on the surface. By the chain rule for vector functions, we have
    X'( t)=Xu(u(t),v( t))u'(t)
    +Xv( u(t),v(t))v'(t)
(This can be proved directly, or coordinatewise.)

Demonstration 4: Illustration of the Chain Rule
Java not enabled.

This demo allows you to input a surface in the X(u,v) type-in, and a curve in the Curve in UV domain type-in. The curve's domain is defined in the type-in called Domt. Finally, the tapedeck will allow you to move along the curve in the uv-plane and the corresponding curve on the surface. (The tapedeck only goes between 0 and 1 but is scaled to go effectively between the minimum and the maximum of the domain of the curve.) Then Xu(u( t),v(t)) u'(t) and Xv(u( t),v(t)) v'(t) are displayed on the curve in The Surface window.

In this demonstration, we show a point moving along a parametrized curve (u(t),v( t)) in the domain, with velocity vector (u'(t) ,v'(t)) . We show how the velocity vector is the sum of vectors (u'(t),0) and (0,v'(t)) parallel to the coordinate vectors in the plane. We then show the decomposition of the velocity vector X'( t) as a sum of two basis vectors in the tangent plane, Xu(u( t),v(t)) u'(t) and Xv(u( t),v(t)) v'(t) .

To compute the length of the curve X(t) , we must integrate the length of the vector X'( t) over the domain of the curve. The dot product of this vector with itself can be expressed in terms of the dot products of the partial derivative vectors, which we define in terms of the metric coefficients gij(u,v) . We obtain

    |X'(t)| 2=g11(u(t) ,v(t))u '(t)2
    +2g12(u(t) ,v(t))u '(t)v' (t)
    +g22(u(t) ,v(t))v '(t)2
where g12(u,v) =Xu(u,v)·Xv(u,v) and g21(u,v) =Xv(u,v)·Xu(u,v) , and so g12(u,v) =g21(u,v) .

Here we calculate of the lengths of curves on a surface for the simple case of lines in the u,v domain. Metric of lines in the uv-Domain

(u0,v0) will be determined by the type-in [u0,v0] . c now determines the angle that the line makes with the u-axis in the parameter domain.

In this demonstration, we choose a point (u0,v0) in the domain and exhibit the lengths of the curves

    (u(t),v( t))=(u 0+tcos(c),v0 +tsin(c))
where t runs from a to b, chosen so that the curve lies in the domain. For this special case,
    |X'(t)| 2=g11(u(t) ,v(t))cos2(c)
    +2g12(t(t) ,v(t))cos(c)sin(c)
    +g22(u(t) ,v(t))sin2(c)
Next: Linear Independence in Terms of Metric Coefficients