Hey so I got part a for this question but I have no idea how to even approach
Can anyone help

nvm im dumb just set x=0 and do f(0,y)

Need help with this one though

the normal vector to the tangent plane at (x,y,z) is ∇f(x,y,z), where f(x,y,z)= x^2+y^2-z-1 giving us ∇f(x,y,z)=(2x,2y,-1). we know that the vector from a point (x,y,z) from the origin is (x,y,z)

so we want to find all points on the surface that satisfy the equation ∇f(x,y,z)=k(x,y,z), for some constant k. therefore we need to solve (2x,2y,-1)=(kx,ky,kz).

you should consider the cases when x=0, y=0,x≠0, y≠0 seperately

This post was edited by J0nn on Oct 23 2014 06:52pm

Okay so when I have gradient f = k(x,y,z) and I take the case when x,y are not zero I get k=2 and when I take the case when x,y are zero I have k = -1/z

I'm not sure what to do with these now kinda lost lol

ok so you have the equations
2x = kx
2y = ky
-1 = kz
z = x^2 +y^2 -1

when x and y are 0

you get z = (0)^2 + (0)^2 -1
z = -1
so you get the point (0,0,-1)

and you do this for the other cases of x=0, y=0,x≠0, y≠0

So what I have so far is

x and y = 0
(0,0,-1)

x = 0 , y not zero
(0,1/sqrt(2),-1/2)

y = 0, x not zero
(1/sqrt(2),0,-1/2)

I can't seem to get an actual point for when x and y are not zero though

What do I end up doing with these points? Am I just finding the angles to the z-axis with all of them?

Ehhhh actually just disregard what I wrote above lol

Trying it this way, I have:

x(2-k)=0
y(2-k)=0
kz=-1

So if x and y are zero I have again what you stated (0,0,-1)

Now if k=2 I have...
(gunna edit this while I work on it)
I have I think (x,y,-1/2) where x^2+y^2=1/2

Not sure if this is right

This post was edited by Bloo_Guardian on Oct 23 2014 09:26pm

when x, y ≠ 0 is it easy to show x = y and k =2

then that means that z = -1/2

-1/2 = x^2 + y^2 -1
1/2 = x^2 + y^2
1/2 = 2x^2 since x = y
x = +/- 1/2
therefore yours points are

(-1/2,-1/2,-1/2) and (1/2,1/2,-1/2)

by the way, for your points when x=0 y≠0
you could have (0,1/sqrt(2),-1/2) and (0,-1/sqrt(2),-1/2) because you took the square root. similarly it applies for y=0 x≠0

to find the angle, you sub your points into your vector (2x,2y,-1) and dot it with the (0,0,1) vector and solve for the angle

This post was edited by J0nn on Oct 23 2014 09:30pm

Oh my goodness x = y is a godsend, I did not see that lol

So in total I will have 7 points?