# Multivaraible Calc. vectors

• Oct 13th 2008, 04:07 PM
ScullyX51
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.
• Oct 13th 2008, 04:13 PM
WantMath
That should be it
• Oct 13th 2008, 04:35 PM
ScullyX51
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.