7/08/2004 IMPLICIT AND PARAMETRIC SURFACES Maths21aThis is part 3 (of 3) of the homework which is due July 13 at the beginning of class.SUMMARY:• g(x, y, z) = 0 parametric surface.• x = r cos(θ)y = r sin(θ)z = zcylindrical coordinates• x = ρ sin(φ) cos(θ)y = ρ sin(φ) sin(θ)z = ρ cos(φ)spherical coordinates• ~r(u, v) = (x(u, v), y(u, v), z(u, v))defines a parametric surface.EXAMPLES:• x2+ y2+ z2= ρ2sphere• r = 1, cylinder.• ρ = 1, sphere.• r = z, cone• ~r(u, v) = P + u~u + v~v plane• ~r(u, v) = (cos(u) sin(v), sin(u) sin(v), cos(v))sphere• ~r(u, v) = (cos(u), sin(u), v) cylinder• ~r(u, v) = (u, v, f(u, v)) graph of fHomework Problems1) (4 points)a) What is the equation for the surface x2+ y2− 5x = z2in cylindrical coordinates?b) Describe in words or draw a sketch of the surface whose equation is ρ = sin(φ) in sphericalcoordinates (ρ, θ, φ).Solution:a) r2− 5r cos(θ) = z2.b) This is a fat bagle.2) (4 points) Plot the surface with the parametrization ~r(u, v) = (v2cos(u), v2sin(u), v), whereu ∈ [0, 2π] and v ∈ R.Solution:It is a surface of revolution, very thin at the origin. The shape is a parabola but it is bentthe other way round as in the paraboloid.3) (4 points) Find a parametrization for the plane which contains the points (3, 4, 1),(1, 2, 1) and(0, 3, 4).Solution:~r(s, t) = (3 − 2s − 3t, 4 − 2s − t, 1 + 3t).4) (4 points) Find two different parametrisations of the lower half of the ellipsoid 2x2+4y2+z2= 1.Hint: for one of the parametrizations assume that the surface is a graph. For the other, dosomething similar than for the sphere.Solution:1) ~r(u, v) = (u, v, −√1 − 2u2− 4v2.2) ~r(θ, φ) = (sin(φ) cos(θ)/√2, sin(φ) sin(θ)/2, cos(φ)).5) (4 points) Find a parametrisation of the torus which is obtained as the set of points whichhave distance 1 from the circle (2 cos(θ), 2 sin(θ), 0), where θ is the angle occuring in cylindricaland spherical coordinates.Hint: Keep u = t as one of the parameters and let r the distance of a point on the torus tothe z-axis. This distance is r = 2 + cos(φ) if φ is the angle you see on Figure 1. You canread off from the same picture also z = sin(φ). To finish the parametrization problem, youhave to translate back from cylindrical coordinates (r, θ, z) = (2+cos(φ), θ, sin(φ)) to Cartesiancoordinates (x, y, z). Write down your result in the form ~r(θ, φ) = (x(θ, φ), y(θ, φ), z(θ, φ)).Solution:~r(θ, φ) = ((2 + cos(φ)) cos(θ), (2 + cos(φ)) sin(θ), sin(φ)).Challenge Problems(Solutions to these problems are not turned in with the homework.)1) Try to graph without computer the surface r = f(θ, φ) = (2 + sin(3θ))(2 + cos(2φ)) (It is agraph in spherical coordinates (r, φ, θ).)Hint. Do it in stages. First graph r = 2 (the sphere), then r = (2 + sin(3θ)), then draw asketch of the final surface.2) How would you design analogues of spherical or cylindrical coordinates in 4 dimensions?3) Sketch the surface r(u, v) = (2 + 2v cos(πu)) cos(2πu), (2 + 2v cos(πu)) sin(2πu), v sin(πu)).4) The torus is obtained by bending and gluing the ends of a cylinder together.The Klein bottle is obtained in the same way, however, the ends are put together withopposite directions. This can not be achieved without self-intersection. Take one end of thetube, bend it, enter the tube first to match the ends in the opposit direction as for the torus.Can you find a parametrisation r(u, v) for this surface? On the handout for this lecture, youfind a parametrization of the same surface but which looks different. The idea is to have aparametrization which produces the
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