1. ## Multivaraible Calc. vectors

ok..so here is what I am given for this problem:

The vector function r sends a point (u,v) in the plane to a point (vector) in R3.
r(u,v)= (1/2sqrt(v)*cosh U, sqrt(V)*sinhU, V)
or in other words:
x= 1/2sqrt(V)*coshU
y= sqrt(V)*sinhU
z=V
these points satisfy the equation: z= 4x^2-y^2, and it is a hyperbolic parabloid.
Question:
Fix u=u(sub 0) and consider the parametised curve r(sub v), defined as rv(v)=r(u(sub )), v)
Write explicitly r(subv)(v):

What does this mean? does it mean to replace all u with u(sub 0) in the original vector equation? Help..please someone explain this to me.

2. That should be it

3. Excellent. If you are not too busy, can you check if I've done the next part of the problem right:
So the orginal vector function is as follows:
r(u,v)= (1/2sqrtv*coshU, sqrtv*sinhU, V)
then to write v explicitly I am given that r(sub v)(V)=r(u(sub 0),v)
I did what you said and replaced all the u's in the original vector function with u knot and got:
r(u(sub 0),v)= (1/2sqrtV*coshu(sub0), sqrtv*sinhu(sub0), V)
Then I am asked to compute the tangent vector of r(subv)(V), so I first took the derivitive and I got:
r(v) prime= ( 1/4sqrtv*sinhu(sub0), 1/2sqrtv*coshu(sub0), 1)
Then to get the tangent vector I also calculated it magnitude and got:
the magnitude of r(v)prime= sqrt( everything squared)= sqrt (1/16v*sinh^2u(sub0)+ 1/4v*cosh^2u(sub0)+1)
I then factored out a 1/4 v from the first two terms and got:

sqrt( 1/4v(1/4*sinh^2u(sub0)+cosh^2u(sub0)) +1)
Since I can't simplify this anymore the tangent vector would be the derivitive I calculated earlier over this magnitude.

Sorry I know this is confusing to type/read but I would appreciate any corrections.