Quote (2wo1ne @ Mar 6 2016 03:18am)
how do i do that?
1)
First lets compute σ_u and σ_v
σ_u = (1,0,v)
σ_v = (0,1,u)
then (σ_u) X (σ_v) = (-v,-u,1)
Notice the point (1,2,2) corresponds to u=1, v=2
then our normal vector to the tangent plane is (-2,-1,1)
Now we want to find the equation of the tangent plane at (1,2,2). To do this, we want the dot product of the normal vector and a vector in the tangent plane to equal 0.
We see (x-1,y-2,z-3) is a vector in the tangent plane for (x,y,z) a point in the tangent plane. Thus our equation for the plane is (-2,-1,1)·(x-1,y-2,z-2)= 0
Expanding this we get z = 2x + y - 2. From our surface patch, we have z = xy
Substituting gives xy = 2x + y - 2 => (x-1)y = 2(x-1) => x=1 or y=2
if x=1 then we get the vector (1,y,y) = (1,0,0) + y(0,1,1)
if y=2 then we get the vector (x,2,2x) = (0,2,0) + x(1,0,2)
Thus we get 2 lines.
2) What's this question asking? Part of it is cut off. I'm assuming that its asking that it has the same arclength
First compute σ_u and σ_v
σ_u = (cosv,sinv,1)
σ_v = (-usinv, ucosv, -tanv)
Compute E = (cosv,sinv,1) · (cosv,sinv,1), F = (cosv,sinv,1) · (-usinv, ucosv, -tanv), G = (-usinv, ucosv, -tanv) · (-usinv, ucosv, -tanv)
Then your First Fundamental form is I = 2du² - 2tanvdudv + (u² + tan²v)dv²
E = 2, F = -tanv, G = u² + tan²v
let u = t => du/dt = 1
let v = c =>dv/dt = 0
Arclength = ∫(a to
√[E(du/dt)² + 2F(du/dt)(dv/dt) + G(dv/dt)²] dt
=∫(a to
√2 dt
= √2 (b-a) Thus arclength only depends on a,b. ie arclength is constant for all curves
3) Let q=(x,y,z) , n = (0,0,-1), p=(u,v,0)
Since q,n,p are colinear, we have q-n = k(p-n) for some non-zero scalar k
=> (x,y,z) = (0,0,-1) + k(u,v,1) = (ku,kv,k-1)
=> k = z+1 =>u = x/(z+1) , v = y/(z+1)
So we have f(x,y,z) = (x/(z+1), y/(z+1), 0)
To find f^-1, we use the equation of the hyperboloid x²+y²-z² = -1
=> (ku)² + (kv)² - (k-1)² = -1
=> using k is non-zero we get k = 2/(1-u²-v²)
Thus f^-1(u,v) = (2u/(1-u²-v²), 2v/(1-u²-v²),(u²+v²+1)/(1-u²-v²))