Vector calculus finding the surface normal

[frixis]

okay i was reading the book "div, grad cur and all that"
i've just started for my exam. but i got stuck in the beginning. I'm attaching the page because i really don't get it .
basically pages 14 and 15. i don't get how they calculate the components of u and v.
if someone explains u i'll get v obviously.
thanks

 

[chiro]

okay i was reading the book "div, grad cur and all that"
i've just started for my exam. but i got stuck in the beginning. I'm attaching the page because i really don't get it .
basically pages 14 and 15. i don't get how they calculate the components of u and v.
if someone explains u i'll get v obviously.
thanks

U is calculated as a tangent vector. Consider how the rate of change impacts the tangent vector. Think about how a steep surface or how a flat surface would impact the definition of u. Its probably best if visualize a normal one dimensional function and the normal tangent vector to that function.

If you relate u vector to the function tangent in the x direction and v vector to that in the y direction you have two vectors that are orthogonal to each other and represent the best information to calculate a normal since the normal is u x v.

On a steep surface we would have a normal that would have a smaller z component than if the surface were flat on the x,y plane.

In terms of calculation they are converting the x and z components which is df/dx * ux or whatever it says in the book by writing it in terms of its components which are in terms of the orthonormal vectors i j and k which correspond to the unit vectors of the x y and z axis respectively. Just reflect on what units everything is in by looking at the diagram and see how the units for x and z correspond to units or i and k respectively.