Find parametric equations for the curve of intersection of the circular cylinder x^2 + z^2 = 1 and the hyperbolic paraboloid z = x^2 - y^2.
I have no idea how to do this.
10 points for the best explanation.
Thanks.
Z.
I have no idea how to do this.
10 points for the best explanation.
Thanks.
Z.
-
Locus of R(x,y,z) is points of intersection:
r = |OR|
for cylinder;
x^2 + z^2 = 1
let x = rcos(u), z = rsin(u) ... angle u is in (x,z) plane
[rcos(u)]^2 + [rsin(u)]^2 = 1
==> r^2 =1 i.e. |OR| is of unit measure or r = 1 .. your cylinder is a circle?
for parabaloid;
z = x^2 - y^2
rsin(u) = [rcos(t)]2 - [rsin(t)]^2
rcos(u) = r^2 [cos(t)^2 -sin(t)^2] ...but cos(t)cos(t) -sin(t)sin(t) = cos(2t)
1.cos(u) = 1. cos(2t)
cos(u) = cos(2t) , u = ± 2t
x = cos(t), y = sin(t), z = ± sin(2t)
1^2 = sin(2t)^2 since r^2 =x^2 + y^2 + z^2 in general
2t = π (1± 2n)
t = π/2 (1± 2n)
x = cos{π/2 (1± 2n)} = 0 for all n
y = sin{π/2 (1± 2n)} = { -1, 1}
z = ± sin{π (1± 2n)} = 0 for all n
the cylinder (or circle since r^2 =1 here) intersects the hyperbaloid at two points
(0,-1,0) and (0,1,0)
r = |OR|
for cylinder;
x^2 + z^2 = 1
let x = rcos(u), z = rsin(u) ... angle u is in (x,z) plane
[rcos(u)]^2 + [rsin(u)]^2 = 1
==> r^2 =1 i.e. |OR| is of unit measure or r = 1 .. your cylinder is a circle?
for parabaloid;
z = x^2 - y^2
rsin(u) = [rcos(t)]2 - [rsin(t)]^2
rcos(u) = r^2 [cos(t)^2 -sin(t)^2] ...but cos(t)cos(t) -sin(t)sin(t) = cos(2t)
1.cos(u) = 1. cos(2t)
cos(u) = cos(2t) , u = ± 2t
x = cos(t), y = sin(t), z = ± sin(2t)
1^2 = sin(2t)^2 since r^2 =x^2 + y^2 + z^2 in general
2t = π (1± 2n)
t = π/2 (1± 2n)
x = cos{π/2 (1± 2n)} = 0 for all n
y = sin{π/2 (1± 2n)} = { -1, 1}
z = ± sin{π (1± 2n)} = 0 for all n
the cylinder (or circle since r^2 =1 here) intersects the hyperbaloid at two points
(0,-1,0) and (0,1,0)