Question #138980
Show that the Mercator projection

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/uf1/vf2/uf2/vf3/uf3/v]\begin{bmatrix} \partial f_{1}/\partial u & \partial f_{1}/\partial v \\ \partial f_{2}/\partial u & \partial f_{2}/\partial v\\ \partial f_{3}/\partial u&\partial f_{3}/\partial v \end{bmatrix} = [(sech u tanh u)cosvsech u sinv(sech u tanh u)sinvsech u cosvsech2 u0]\begin{bmatrix} -(sech \ u \ tanh \ u)\cos v & -sech \ u \ \sin v\\ -(sech \ u \ tanh \ u)\sin v & sech \ u \ \cos v\\ sech^2 \ u &0 \end{bmatrix} . Now if we look at the minor deleting the first row we get sech3 ucosv.sech^3 \ u \cos v. Now sech u0sech \ u\neq 0 , hence minor vanishes if cosv=0.\cos v=0. Now taking the minor after deleting 2nd row gives, sech3 usinv.-sech^3 \ u \sin v. This shows sinv=0.\sin v=0. But both sinv,cosv\sin v , \cos v 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 (sech u)2cos2u+(sech u)2sin2u+(tan h)2u=1.(sech \ u)^2 \cos^2u+(sech \ u)^2\sin^2u+(tan \ h)^2u=1.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS