σ(u, v) = (sech u cos v,sech u sin v,tanh u)
is a regular surface patch of the unit sphere
1
Expert's answer
2020-10-19T17:56:58-0400
Jacobian= ⎣⎡∂f1/∂u∂f2/∂u∂f3/∂u∂f1/∂v∂f2/∂v∂f3/∂v⎦⎤ = ⎣⎡−(sechutanhu)cosv−(sechutanhu)sinvsech2u−sechusinvsechucosv0⎦⎤ . Now if we look at the minor deleting the first row we get sech3ucosv. Now sechu=0 , hence minor vanishes if cosv=0. Now taking the minor after deleting 2nd row gives, −sech3usinv. This shows sinv=0. But both sinv,cosv cannot be zero hence one of the two minors never vanish showing the Jacobian has rank two. Also, all points lie on the sphere since (sechu)2cos2u+(sechu)2sin2u+(tanh)2u=1.
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