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normal line to the surface, is that anyting like a normal vector? perpendicular to the surface?

I think its the unit normal vector

...do we find the normal plane by partial derivatives?

Yes.

i was reading up on some of the vector stuff last night as my eyes were glossing over :)

if i remember correctly, you don't need to actually find the plane, just the gradient

I think its the unit normal vector

how would I find the equation of the tangent plane at that point?

you a good wizard or a bad wizard? I can never get inline equations....

it's all points perpendicular to that vector

dz/dx(x-x0) + dz/dy(y-y0) + -1(z-z0) = 0, where (x0,y0,z0) is your initial point

ok, so the unit tangent vector is the derivative over the magnitude of the derivative?

so the normal vector is just the gradient of the tangent vector?

No. The normal is the gradient of the surface.

use slope intercept form, y = mx + b, or x = mt + b

where m is the x coordinate of the gradient at that point, and b is the x coordinate at that oint

and do the same with y and b

correct

that will give you the slope of the x coordinate for the parametric equation

that will give me x0 or will it give me a?

what is a?

x=x0 +at

the parametric equations will take the form of x = (dz/dx@x0)*t + x0

yes, that will give you a

ok thanks, and x0 is just equal to 0 correct?

since the point is (0,0,1)

correct

thank you very much

no problem

x = mt + b, where m is the gradient of x @ the point, and b is the x coordinate

so be would be 0, for (0,0,1)