[PDF] MATH 20550 Parametric surfaces Fall 2016 1. Parametric surfaces

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MATH 20550 Parametric surfaces Fall 2016

1.Parametric surfaces

Aparametrized surfaceis, roughly speaking, a vector function of a vector variable r(u;w) =hx(u;w);y(u;w);z(u;w)i:D!R3 dened over some subsetDR2. The \roughly speaking" comes from the fact that we want to impose conditions onrso that the image is 2 dimensional and so really a surface. We will do this below but your informal ideas will suce to get us through some examples. Example.The plane through a pointha;b;cicontaining two vectorsv1andv2. Then r(u;w) =ha;b;ci+uv1+wv2 is one equation for that plane. It is also easy to describe the normal form of the plane. A normal vector is given byN=v1v2and so the equation isNhx;y;zi=Nha;b;ci. Review how to get a plane through 3 points, a plane containing a line and a point, .... Example.The graph ofz=f(x;y) lying overDin thexyplane can be parametrized as r(u;w) =hu;w;f(u;w)i There are similar parametrizations fory=g(x;z),r(u;w) =hu;g(u;w);wiandx=h(y;z), r(u;w) =hu;h(u;w);u;wi.

Example.The sphere of radiusa >0

r(u;w) =hacos(u)sin(w);asin(u)sin(w);acos(w)i06u62; 06w6

This is just spherical coordinates.

Example.Ifhx(t);y(t)ia6t6bis a parametrized curve in thexyplane then r(u;w) =hx(u);y(u);wia6u6b;1< w <1 is the cylinder along that curve. Example.Ifhx(t);y(t)ia6t6bis a parametrized curve in thexyplane then we can rotate about thex-axis to get a surface of revolution r(u;w) =hx(u);y(u)cos(w);y(u)sin(w)ia6u6b; 06w62 We can also rotate about they-axis to get a dierent surface of revolution r(u;w) =hx(u)cos(w);y(u);x(u)sin(w)ia6u6b; 06w62 1 2 Example.Here is a picture of a surface.A surface of Dini r(u;w) =D

2cos(u)sin(w);2sin(u)sin(w);2

cos(w) + ln tanw2 + 0:4uE

26u62; 0:1< w <2

The surface does not determine a parametrized surface any more than a line determines a paramet-

ric equation for it. Nevertheless, in order to do calculations we will need to have a parametrization.

Example.Consider the surfacez=x2+y2+ 1. It can be parametrized as r(u;w) = u;w;u2+w2+ 1 1< u <1;1< w <1 Using cylindrical coordinates it can also be parametrized as r(u;w) = ucos(w);usin(w);u2+ 106u <1; 06w62

Here is a

link to a list of surfaces. One common example of a surface which does not have an obvious parametrization is a level surface,f(x;y;z) =const. There is a theorem, called theImplicit Function Theoremwhich says that a level surfacef(x;y;z) =consthas a parametrization in a neighborhood of a pointha;b;ci

provided one of the partial derivatives is non-zero at the point. This is useful for theoretical work

since in deriving an equation we can usually assume a surface is parametrized even if we can't write a parametrization down.

1.1.Grid lines.In the picture of the Dini surface you can see two colors of greens and some lines

calledgrid lines. These are curves obtained by settingu=const:orw=const: Each grid line is a curve. For example, if we xw=b, then the parametrized curve is (t) =hx(t;b);y(t;b);z(t;b)i It turns out that the tangent lines to the grid lines are important. The calculation is easy. Dene 3 r u(u;w) =@ x@ u ;@ y@ u ;@ z@ u r w(u;w) =@ x@ w ;@ y@ w ;@ z@ w The vectorru(u;w) evaluated at (a;b) is a tangent vector to the grid line (t) =hx(t;b);y(t;b);z(t;b)iatt=a. The vectorrw(u;w) evaluated at (a;b) is a tangent vector to the grid line (t) =hx(a;t);y(a;t);z(a;t)iatt=b. A vector functionr(u;w)is a smooth surfaceprovidedru(u;w)andrw(u;w)are continuous and not parallel at any point inD, except maybe along the boundary of

D. This means that the image ofris 2 dimensional.

One can also speak of a point the image ofr(u;w) as beingsmoothwhich means thatru(u;w) andrw(u;w) are not parallel at the point.

The reason for the name \grid lines" comes from looking at Figure 5 from the book:Figure 5 from the book

The surface in Figure 5 is given by

r(u;v) =h(2 + sinv)cosu;(2 + sinv)sinu;u+ cosvi 4 Each little square in the plane on the left goes over to a piece on the right. Each grid line in the plane goes over to one of the curves on the surface. Most graphics programs draw some grid lines for you. You can see some in the picture of the surface of Dini from above. 5

2.Tangent planes

At a smooth point on a surface, thetangent planeto the surface at the point is the plane which contains the point and the two vectorsru(u;w) andrw(u;w).

Then the normal vector to the tangent plane is

r u(u;w)rw(u;w) so a point is smooth if and only ifru(u;w)rw(u;w)6=h0;0;0iorjru(u;w)rw(u;w)j 6= 0. The parametric equation for the tangent plane at the pointr(a;b) is r(a;b) +uru(a;b) +wrw(a;b) OR Thenormal lineto a surface through a point on the surface is the line perpendicular to the tangent plane. If the point isr(a;b), the equation is r(a;b) +ru(a;b)rw(a;b)t Example.The sphere of radiusa >0 centered at the origin. One description is the level surface x

2+y2+z2=a2. A normal vector is the gradienth2x;2y;2zi. A unit normal vector is

n

1(x;y;z) =Dxa

;ya ;za E A parametrization of the upper half of this sphere is given by r(u;w) =D u;w;pa

2u2w2E

;u2+w26a2 r u=

1;0;upa

2u2w2 r w=

0;1;wpa

2u2w2 r urw= det i j k

1 0upa

2u2w2 0 1 wpa 2u2w2 =upa

2u2w2;wpa

2u2w2;1

jrurwj=apa 2u2w2

A unit normal vector is

n

2(u;w) =ua

;wa ;pa

2u2w2a

This unit normal vector at a given point is equal to the unit normal we got from the gradient. At any point (x0;y0;z0) withz0>0,r(x0;y0) =hx0;y0;z0i. Then n

2(x0;y0) =*

x 0a ;y0a ;pa

2x20y20a

=Dx0a ;y0a ;z0a E =n1(x0;y0;z0) We can parametrize the sphere using spherical coordinates. r(;) =hacos()sin();asin()sin();acos()i; 0662;066 6 r =hasin()sin();acos()sin();0i r =hacos()cos();asin()cos();asin()i r r= i j k asin()sin()acos()sin() 0 acos()cos()asin()cos()asin() a2sin()hcos()sin();sin()sin();cos()i jrrj=a2sin()

A unit normal vector is

n

3(;) =hcos()sin();sin()sin();cos()i

This unit normal vector at a given point is the negative of the unit normal we got from the gra- dient. At any point (x0;y0;z0) nd0and0so thatr(0;0) =hx0;y0;z0i. The parametrization guarantees that this can be done. Then n

3(0;0) =hcos(0)sin(0);sin(0)sin(0);cos(0)i=

acos(0)sin(0)a ;asin(0)sin(0)a ;acos(0)a =Dx0a ;y0a ;z0a E =n1(x0;y0;z0) If we parametrize the sphere using cylindrical coordinates, r(;z) =Dpa

2z2cos();pa

2z2sin();zE

; 0662;a6z6a r =D pa

2z2sin();pa

2z2cos();0E

r z= zpa

2z2cos();zpa

2z2sin();1

r rz= i j k pa

2z2sin()pa

2z2cos() 0

zpa

2z2cos()zpa

2z2sin() 1

=Dpa

2z2cos();pa

2z2sin();zE

jrrzj=a

A unit normal vector is

n

4(;) =

pa 2z2a cos();pa 2z2a sin();za This unit normal vector at a given point is equal to the unit normal we got from the gradient. At any point (x0;y0;z0) nd0so thatr(0;z0) =hx0;y0;z0i. Then n

4(0;z0) =*

pa 2z20a cos(0);pa 2z20a sin(0);z0a =Dx0a ;y0a ;z0a E =n1(x0;y0;z0) 7

3.Orientation

Anorientationfor a surface is a continuous choice of unit normal vector for the surface. Every surface which can be parametrized has two choices of a unit normal vector at every point. The boundary of a solidEin three space has orientations. By@Ewe mean the boundary ofE with the orientation for which the normal vector points out ofE. Example:IfEis the ball of radiusacentered at the origin then@Eis the sphere of radiusa centered at the origin. The gradient vector of the level curvex2+y2+z2=a2, namely 2hx;y;zi points out. Hence the normal vectors we computed in rectangular and cylindrical coordinates point out, and the normal vector we computed in spherical coordinates points inward. An important skill is to be able to decide if a given continuous normal eld on the boundary of a solid is outward or inward. To do this, pick a point on the boundary. The easiest way to do this is to pick a point in the parameter space. Then compute the normal vector you calculated at that point. Then check if it points the way you want. Example:The sphere bounds the ball. Let us use spherical coordinates so r(;) =hacos()sin();asin()sin();acos()i r r=a2sin()hcos()sin();sin()sin();cos()i

Pick a point, say0;2

inspace. Thenr0;2 =ha;0;0i. At this pointrr0;2 a2h1;0;0iso the normal vector points inward. If the outward normal is required, which it often is, just usea2sin()hcos()sin();sin()sin();cos()i. The moral of this last example is to calculate one normal eld and check at a convenient point whether it is the one you need. If it is not, the negative of it will be. 8 Surprisingly, not all surfaces are orientable. The German mathematician Mobius realized that the Mobius strip was such a surface. One parametrization is r(u;w) =D a+bwcosu2 cos(u); a+bwcosu2 sin(u);bwsinu2 E ; 06u62;16w61 witha > b >0. If

A(u) =hcos(u);sin(u);0i

then r(u;w) = a+bwcosu2

A(u) +bwsinu2

k andA(u) hasz-coordinate 0. NoteAhas length 1 (as of course doesk) andAk= 0.a= 1 andb= 0:5 Thezaxis is red, thexaxis is green and theyaxis is yellow. The parametrization is \mostly" one-to-one. Note rst that sincea > b,a+bwcosu2 >0. If r(u1;w1) =r(u0;w0) we may also assumeu06u1. ThenA(u1) =A(u0),w1sinu12 =w0sinu02 anda+bw1cosu12 =a+bw0cosu02 FromA(u1) =A(u0) it follows thatu1=u0oru0= 0 andu1=.

Ifu0= 0,u1=it follows fromw1sinu12

=w0sinu02 thatw1=w0. Hereafter we need only discuss the caseu1=u1. Froma+bw0cosu02 =a+bw1cosu12 it follows thatw1cosu12 =w0cosu02 and sincew1sinu12 =w0sinu02 ,w1=w0.

Henceris one-to-one except for

r(0;w) =r(2;w)

Example.A torus

r(u;w) = a+bcos(w)cos(u);a+bcos(w)sin(u);bsin(w); 06u62;06w62 witha > b >0. 9

Torus isa= 2,b= 1Note

r u(u;w) = a+bcos(w)sin(u);a+bcos(w)cos(u);0 r w(u;w) =hbsin(w)cos(u);bsin(w)sin(u);bcos(w)i r urw= det i j k a+bcos(w)sin(u)a+bcos(w)cos(u) 0 bsin(w)cos(u)bsin(w)sin(u)bcos(w) r jrurwj=ba+bcos(w) The torus is the boundary of a solid and so is orientable. At (0;0) the point isr(0;0) =ha+b;0;0i which is the largest of the four points on thex-axis. (The other three arehab;0;0i,ha+b;0;0i andhab;0;0i.) At (0;0),rurw(0;0) =b(a+b)h1;0;0iand sinceb(a+b)>0, this normal eld points out. There is no need to check additional points but suppose you were asked to check the directionquotesdbs_dbs6.pdfusesText_12

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