Math Parametric Surfaces

Transcription

1 Math 13 - Parametric Surfaces Peter A. Perry University of Kentucky April 15, 019

2 Homework Homework D is due Wednesday Work on Stewart problems for 16.6: 1-5 odd, 33, odd Read section 16.7 for Wednesday, Aprll 17 Remember that Dr. Perry will be out of the office April Your lecturer will be Mr. Shane Clark

3 Unit IV: Vector Calculus Lecture 36 Lecture 37 Lecture 38 Lecture 39 Lecture 40 Lecture 41 Lecture 4 Curl and Divergence Parametric Surfaces Surface Integrals Stokes Theorem The Divergence Theorem Final Review, Part I Final Review, Part II

4 Goals of the Day This lecture is about parametric surfaces. You ll learn: How to define and visualize parametric surfaces How to find the tangent plane to a parametric surface at a point How to compute the surface area of a parametric surface using double integrals

5 Parametric Curves and Parametric Surfaces Parametric Curve Parametric Surface A parametric curve in R 3 is given by r(t) = x(t)i + y(t)j + z(t)k where a t b There is one parameter, because a curve is a one-dimensional object There are three component functions, because the curve lives in threedimensional space. A parametric surface in R 3 is given by r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k where (u, v) lie in a region D of the uv plane. There are two parameters, because a surface is a two-dimensional object There are three component functions, because the surface lives in threedimensional space.

6 You Are Living on a Parametric Surface Let u be your latitude (in radians, for this course) Let v be your longitude (in radians) Let R be the surface of the Earth Your position is r(u, v) = R cos(v) cos(u)i + R cos(v) sin(u)j + R sin(v)k π/ v 0 π u π/

7 More Parameterized Surfaces: Planes z y Find a parameteric representation for the plane through 1, 0, 1 that contains the vectors, 0, 1 and 0,, 0 x

8 More Parameterized Surfaces: Planes z y Find a parameteric representation for the plane through 1, 0, 1 that contains the vectors, 0, 1 and 0,, 0 Solution: Let r 0 = 1, 0, 1. Any point in the plane is given by r(s, t) = 1, 0, 1 + s, 0, 1 + t 0,, 0 x

9 More Parameterized Surfaces: Planes z y Find a parameteric representation for the plane through 1, 0, 1 that contains the vectors, 0, 1 and 0,, 0 Solution: Let r 0 = 1, 0, 1. Any point in the plane is given by r(s, t) = 1, 0, 1 + s, 0, 1 + t 0,, 0 x Now you try it: Find a parameteric representation for the plane through the point (0, 1, 5) that contains the vectors, 1, 4 and 3,, 5.

10 More Parameterized Surfaces: The Cylinder v r(u, v) = r cos(u)i + r sin(u)j + vk D = {(u, v) : 0 u π, 0 v h} u parameterizes a cylinder of radius r and height h

11 More Parameterized Surfaces: The Cylinder v r(u, v) = r cos(u)i + r sin(u)j + vk D = {(u, v) : 0 u π, 0 v h} u parameterizes a cylinder of radius r and height h If we fix v and vary u over the cylinder, we trace out a circle

12 More Parameterized Surfaces: The Cylinder v r(u, v) = r cos(u)i + r sin(u)j + vk D = {(u, v) : 0 u π, 0 v h} u parameterizes a cylinder of radius r and height h If we fix v and vary u over the cylinder, we trace out a circle If we fix u and vary v, we trace out a vertical line

13 More Parameterized Surfaces: The Cylinder v r(u, v) = r cos(u)i + r sin(u)j + vk D = {(u, v) : 0 u π, 0 v h} u parameterizes a cylinder of radius r and height h If we fix v and vary u over the cylinder, we trace out a circle If we fix u and vary v, we trace out a vertical line Each of these curves has a tangent vector: r u (u, v) = r sin(u)i + r cos(u)j r v (u, v) = k

14 More Parameterized Surfaces: The Cylinder v r(u, v) = r cos(u)i + r sin(u)j + vk D = {(u, v) : 0 u π, 0 v h} u parameterizes a cylinder of radius r and height h The two tangent vectors r u (u, v) = r sin(u)i + r cos(u)j r v (u, v) = k span the tangent plane to the cylinder at the given point

15 The Tangent Vectors r u and r v Suppose r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k (u, v) D is a parameterized surface. At a point P 0 = r(u 0, v 0 ), the vectors r u (u 0, v 0 ) = x u (u 0, v 0 )i + y u (u 0, v 0 )j + z u (u 0, v 0 )k r v (u 0, v 0 ) = x v (u 0, v 0 )i + y v (u 0, v 0 )j + z v (u 0, v 0 )k are both tangent to the surface.

16 The Tangent Plane r u (u 0, v 0 ) = x u (u 0, v 0 )i + y u (u 0, v 0 )j + z u (u 0, v 0 )k r v (u 0, v 0 ) = x v (u 0, v 0 )i + y v (u 0, v 0 )j + z v (u 0, v 0 )k The tangent plane to a parameterized surface at P 0 = r(u 0, v 0 ) is the plane passing through P 0 and perpendicular to r u (u 0, v 0 ) r v (u 0 v 0 ). Find the equation of the tangent plane to the surface at u = 1, v = 0. r(u, v) = u i + u sin vj + u cos vk

17 The Tangent Plane

18 The Tangent Plane r(u, v) = u, u sin v, u cos v r u (u, v) = u, sin v, cos v r v (u, v) = 0, u cos v, u sin v

19 The Tangent Plane r(u, v) = u, u sin v, u cos v r u (u, v) = u, sin v, cos v r v (u, v) = 0, u cos v, u sin v r(1, 0) = 1, 0, 1 r u (1, 0) =, 0, 1 r v (1, 0) = 0,, 0

20 The Tangent Plane The normal to the plane is r(u, v) = u, u sin v, u cos v r u (u, v) = u, sin v, cos v r v (u, v) = 0, u cos v, u sin v r(1, 0) = 1, 0, 1 r u (1, 0) =, 0, 1 r v (1, 0) = 0,, 0 r u r v = 1, 0,

21 The Tangent Plane The normal to the plane is r(u, v) = u, u sin v, u cos v r u (u, v) = u, sin v, cos v r v (u, v) = 0, u cos v, u sin v r(1, 0) = 1, 0, 1 r u (1, 0) =, 0, 1 r v (1, 0) = 0,, 0 r u r v = 1, 0, so the equation of the plane is ( 1)(x 1) + (z 1) = 0 The tangent plane to the surface at (1, 0, 1) is parameterized by 1 + s, t, 1 + s

22 The Sphere Revisited r(u, v) = sin(v) cos(u)i + sin(v) sin(u)j + cos(v)k 0 u π, 0 v π r u = sin(v) sin(u)i + sin(v) cos(u)j r v = cos(v) cos(u)i + cos(v) sin(u)j sin(v)k Find the tangent plane to the sphere at (u, v) = (π/4, π/4)

23 The Sphere Revisited r(u, v) = sin(v) cos(u)i + sin(v) sin(u)j + cos(v)k 0 u π, 0 v π r u = sin(v) sin(u)i + sin(v) cos(u)j r v = cos(v) cos(u)i + cos(v) sin(u)j sin(v)k Find the tangent plane to the sphere at (u, v) = (π/4, π/4) r(π/4, π/4) = 1 i + 1 j + k r u (π/4, π/4) = 1 i + 1 j r v (π/4, π/4) = 1 i + 1 j k

24 The Sphere Revisited r(u, v) = sin(v) cos(u)i + sin(v) sin(u)j + cos(v)k 0 u π, 0 v π r u = sin(v) sin(u)i + sin(v) cos(u)j r v = cos(v) cos(u)i + cos(v) sin(u)j sin(v)k Find the tangent plane to the sphere at (u, v) = (π/4, π/4) r(π/4, π/4) = 1 i + 1 j + k r u (π/4, π/4) = 1 i + 1 j r v (π/4, π/4) = 1 i + 1 j k n = r u r v = 1 ( 1 i + 1 ) j + k 0 = 1 (x 1 ) + 1 (y 1 ) + (z )

25 Sneak Preview Parametric Curves - Arc Length Parametric Surfaces - Area r(t) = x(t)i + y(t)j + z(t)k r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k r (t) = x (t)i + y (t)j + z (t)k r (t) = x (t) + y (t) + z (t) ds = r (t) dt b L = r (t) dt a r u (u, v) = r (u, v) u r v (u, v) = r (u, v) v da = r u r v du dv S = r u r v du dv D

26 Surface Area v v u z u Find the area A of a small patch of surface The map (u, v) r(u, v) takes the square to a parallelogram with sides r u u and r v v The area of the parallelogram is r u u r v v = r u r v u v The area of the surface is approximately A = r u (u i, v i ) r v (u i, v i ) u v i,j x y and exactly D r u(u i, v i ) r v (u i, v i ) du dv

27 Surface Area of a Sphere r(u, v) = a sin(v) cos(u)i + a sin(v) sin(u)j + a cos(v)k 0 u π, 0 v π r u = a sin(v) sin(u)i + a sin(v) cos(u)j r v = a cos(v) cos(u)i + a cos(v) sin(u)j sin(v)k r u r v = a sin (v) cos(u)i + a sin (v) sin(u)j a cos(v) sin(v)k r u r v = a sin (v) Hence π π S = a sin v du dv = 4πa 0 0

28 Surfaces Area of a Graph The graph of a function z = f (x, y) is also a parameterized surface: r(x, y) = xi + yj + f (x, y)k r x (x, y) = i + f x k r y (x, y) = j + f y k r x r y = f x i + f y j + k ( ) f ( ) f r x r y = x y Hence, the surface area of the graph over a domain D in the xy plane is ( ) f ( ) f S = da D x y

29 Surface Area of a Graph The surface area of the graph over a domain D in the xy plane is ( ) f ( ) f S = da D x y Find the area under the graph of z = x + y that lies over the cylinder x + y = 4

30 Curves and Surfaces Parameterization Curves r(t) = x(t)i + y(t)j + z(t)k Tangent r (t) = x (t)i + y (t)j + z (t)k Tangent line at t = a L(s) = r(a) + sr (a) Arc length differential ds = x (t) + y (t) + z (t) dt Parameterization Tangents Normal Surfaces r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k r u (u, v) = r(u, v) u r v (u, v) = r(u, v) v Area Differential n = r u r v da = r u r v du dv