Curves on a Surface

In general, a curve on a coordinate patch parameterized by

r( u,v) = á x(u,v),  y(u,v),  z( u,v) ñ

is of the form r(t) = r( u(t), v(t) ), or equivalently, r(u,v) maps a curve ( u( t) ,v( t) ) in the uv-plane to a curve r( t) on a surface.  The calculus of such curves r( t) is the same as that in chapter 1 - that is, velocity, arclength, acceleration, etcetera, are calculated in the same fashion as before.

EXAMPLE 1    The latitude-longitude parameterization of a sphere of radius 2Ö2 is given by

r( j,q) = 2Ö2 á cos(j) cos( q) ,cos( j) sin( q) ,sin( j) ñ

What is the length of the curve r( t) = r( p/4, t) , t in [ 0,p/2] , which is an arc at a constant 45^° latitude between the points P(2,0,2) and Q( 0,2,2) ?

LiveGraphics3d Applet

Solution: Since j = p/4 and q = t, the curve r(t) is given by

r( t) = 2Ö2 cos æ è  p 4 ö ø cos( t),  cos æ è  p 4 ö ø sin(t),  sin æ è  p 4 ö ø = 2Ö2  1 Ö2 cos( t) ,  1 Ö2 sin( t) ,  1 Ö2 = á 2cos( t) ,2sin( t) ,2 ñ

Thus, r' = á -2sin( t) ,2cos( t) ,0 ñ and

ds dt   = ||r' || =     4sin2(t) +4cos2( t)   = Ö4 = 2

Consequently, the arclength is

L = ó õ p/2 0   ds dt  dt = ó õ p/2 0  2dt = 2 æ è  p 2 ö ø = p

However, it is often desirable to use the chain rule to find the velocity r^¢ of a curve r(t) = r( u( t) ,v( t) ) , since

r' (t) =  ru  du dt   +  rv  dv dt

Equivalently,a curve v = f( u) is mapped to the curve r( u) = r( u,f( u) ) in a coordinate patch r( u,v) and has a velocity of

r' ( u) = ru  du du   +  rv  dv du   =   ru+rv  dv du

 

For example, a line j = mq+b in the qj-plane is mapped to a curve r( q) = r(mq+b,q) on a sphere of radius R by the latitude-longitude map

r( j,q) = R á cos( j) cos( q),  cos( j) sin(q),  sin( j) ñ

(i.e., m and b are constant).  The velocity of r(q) is given by

r'  = rj  dj dq   +  rq = mrj  +  rq

which can be used to point out a deficiency in the latitude-longitude map.  Namely, if m = tan( a) , then the line j = mq+b forms an angle a with each horizontal line in the jq-plane (i.e., the "paper map" the ship's captain uses to plot a course).  However, the corresponding angle on the globe, which is the angle between r^¢ and the horizontal vector r_q, is not constant (and not equal to a).       

EXAMPLE 2    What is the cosine of the angle b between r^¢ and r_q on the latitude-longitude parameterization of the sphere of radius R?

Solution: Since the latitude-longitude parameterization is orthogonal, it follows that

r' · rq   =   m rj·rq  +  rq·rq   =   ||rq||2

Since r' · r_q = ||r' || ||r_q|| cos( b), it follows that

cos( b) =    r'  · rq || r' ||  || rq||   =    ||rq||2 || r' ||  ||rq||   =    || rq ||  || r'  ||

Moreover, orthogonality of the parameterization implies that

|| r' ||2 = m2||rj||2  +  ||rq||2

Since r_j is a unit vector and

rq = á -cos( j) sin(q) ,cos( j) cos( q),0 ñ = cos( j) á -sin(q) ,cos( q) ,0 ñ

implies that ||r_q|| = | cos( j) | , we are led to the following: 

cos( b) =    | cos( j)| m2+cos2( j)

(1)

      

The significance of (1) is that the angle b depends on both the slope m and the latitude j  (except for m = 0 ).  Indeed, as j approaches p/2, the angle b also approaches p/2, which implies that the actual path on the earth increasingly deviates from that on the "paper map" as the path increases in latitude, and this deviation is toward the north pole (in the northern hemisphere).   
In contrast, as we will explore in the exercises, the Mercator projection is conformal, which means that r( j,q) is an orthogonal parameterization with the property

||rj||  =  ||rq||

Conformal means "shape-preserving,"  which in particular means that a path on a "paper map" using the Mercator Projection is "the same" as the image of the path under r( j,q) .