Image Problem

[niall14]

can somebody give a solution to wat the image of x(u,v)=(cos u*cos v,cos U*sin v, sin u) is? where

x:U ->R^3
u,v is an element of R^2 such that -pi/2 < u < pi/2, -pi < v < pi

help appreciated greatly
thank you

[I like Serena]

Hi niall14, welcome to PF :smile:

can somebody give a solution to wat the image of x(u,v)=(cos u*cos v,cos U*sin v, sin u) is? where

x:U ->R^3
u,v is an element of R^2 such that -pi/2 < u < pi/2, -pi < v < pi

help appreciated greatly
thank you

It is the unit sphere.
Think of coordinates on earth, defined by latitude (u) and longitude (v).

Cheers!

[niall14]

thanks, but how did i show it is a sphere, i can work out it is unit speed?

[I like Serena]

thanks, but how did i show it is a sphere, i can work out it is unit speed?

If you pick any u, and vary v, you'll find you have a circle of the form (r cos v, r sin v, z).
Just like any circle of constant latitude on earth.

Equivalenty you can find the circles through the poles with constant longitude (v).

Btw, with unit sphere I did not mean unit speed.
I meant it's a sphere with radius 1.

[HallsofIvy]

can somebody give a solution to wat the image of x(u,v)=(cos u*cos v,cos U*sin v, sin u) is? where

x:U ->R^3
u,v is an element of R^2 such that -pi/2 < u < pi/2, -pi < v < pi

help appreciated greatly
thank you
If (x, y, z) is in the image then x= cos(u)cos(v), y= cos(u)sin(v), z= sin(v) for some u and v.
x^2+ y^2+ z^2= cos^2(u)cos^2(v)+ cos^2(u)sin^2(v)+ sin^2(v)
= (cos^2(u)+ sin^2(u))cos^2(v)+ sin^2(v)= cos^2(v)+ sin^2(v)= 1

Since x^2+ y^2+ z^2= 1 the image is the surface of the ball with center (0, 0, 0) and radius 1.

As I Like Serena said, this parameterization is just "spherical coordinates",
x= \rho cos(\theta) sin(\phi)
y= \rho sin(\theta) sin(\phi)
[tex]z= \rho cos(\phi)[/itex]
with "u" instead of "\theta", "\pi- v" instead of \phi (to change the cosine to sine and vice-versa) and \rho set equal to 1.

[I like Serena]

Since x^2+ y^2+ z^2= 1 the image is the surface of the ball with center (0, 0, 0) and radius 1.

Good call! :smile: