《高数双语》课件section 9.6.pptx

1、Section 9.6121.Tangent line and normal plane to a space curveP0OxyzL r()()i()j()ktx ty tz t3The tangent line to:r()(),(),()tx ty tz tP0OxyzPPPT000()()P Pr ttr tr 0()r tt 0()r tr 0P Prtt 000limlimttP Prtt 0()r t 0()r t 000(),(),()x ty tz t4The tangent line to The geometric meaning of the derivative o

2、f the direction vector r(t)at t0 is that r(t0)is the direction vector of the tangent to the curve at the corresponding point P0.r(t0)is called the tangent vector to the curve at P0.:r()(),(),()tx ty tz tP0OxyzT0()r t0()r t The Vector equation of the tangent to the curve at P0 is00()()r ttr t 5The eq

3、uation of the tangent line to curve 00()()r ttr t The Vector equation:The Parametric equation:000000()(),()(),()().x txtx ty tyty tz tztz t The Symmetric equation:000000()()()xxyyzzx ty tz t0()0r t 6The tangent line to A curve for which the direction of the tangent varies continuously is called a sm

4、ooth curve.0()0r t 322:r()(,)tttExample1:r()(cos,sin)tttOxy2yOx1piecewise smooth curve7The normal plane to We have seen that for a given space curve if r(t)is derivable at t0 and r(t0)0,then the tangent to at P0 exists and is unique.There is an infinite number of straight lines through the point P0,

5、which are perpendicular to the tangent and lie in the same plane.The plane is called the normal plane to the curve at P0.through the point P0 perpendicular to the tangentthe equation of the normal plane8The normal plane to The equation of the normal plane to the curve at P0,(t=t0)is000000()()()()()(

6、)0 x txx ty tyy tz tzz t Example Find the equations of the tangent line and the normal plane to the following curve at point t=1.22:r()(,2,).tttt9Tangent line and normal plane to a space curveIf the equations of the curve is given in the general form(,)0,:(,)0,F x y zG x y z and the above equations

7、of the curve determine two implicit functions of one variable x,y=y(x)and z=z(x)in the neighbourhood U(P0)and both y(x)and z(x)have continuous derivative.Thenthe symmetric equation of the tangent at P0(x0,y0,z0)is:000001x xx xxxyyzzdydzdxdx10Tangent line and normal plane to a space curveand the equa

8、tion of the normal plane at P0(x0,y0,z0)is:00000()()()0 x xx xdydzxxyyzzdxdx Example Find the equations of the tangent line and the normal plane to the curve at point P0(-2,1,6).22222245,2xyzxyz 11Parametrizing OyzxrAny space point can be imagined thatit lies on a sphere which is centered at(,)P x y

9、 zthe origin and the radius is 222.xyzIf the angle between the projection vector on the xOy plane and the positive OP of direction of x-axis is denoted by,and and the positive direction of z-axisOP the angle between the vector is denoted bysin cos,sin sin,cos,02,0.xryrzrthen the two coordinate syste

10、m are related by,12Parametrizing Oyzx(,)P x y zIf we denote222.xy the surface of the angle between the projection vector OP of on the xOy plane and the positive direction of x-axis is denoted by,and 220.xy Then the coordinate canbe expressed by cos,sin,02.xyzzlies onAnother way to parametrize is ima

11、gine that any point(,)P x y z(,)P x y zis also a point of a space curve or a space surface,then If we can parametrize the equation of the curve or surface.13Tangent Planes and Normal Lines to a SurfaceSuppose that the parametric equation of a surface S is 2rr(,)(,),(,),(,),(,)Ru vx u vy u v z u vu v

12、Dand the partial where r is continuous in D,the point00(,)u vD 0000r(,)r(,).uvu vu v 0000(,)r(,),vu vxyzu vvvv 0000(,)r(,),uu vxyzu vuuu exist,that is,derivatives of r at the point 00(,)u v00(,)u v,then the r(,)u vwe can prove that if is differentiable at the point tangent plane of any smooth curve

13、on the surface through the point r0,00(,)u vwith normal vector must lie in the plane which pass through is called a regular point).and (in this case,00(,)u v0000r(,)r(,)0uvu vu v14Tangent Planes and Normal Lines to a Surface 00r(,)u vrurvrruv xyOzS Therefore,the normal vector is0000r(,)r(,)uvu vu v

14、00(,)(,)(,)(,),(,)(,)(,)u vy zz xx yu vu vu v 00(,)ijkuuuvvvu vxyzxyz def ,.A B CThus the tangent plane is 0000.A xxB yyC zzThe normal line is000.xxyyzzABC15Tangent Planes and Normal Lines to a Surface Example Find the tangent plane and normal line to the right helicoid where the constant cos,sin,(0

15、)xuvyuvazav at the point 2,.4uv SolutionSo,(cos,sin,)ruv uv av(cos,sin,0);(sin,cos,).uvrvvruv uv a Tangent Planes and Normal Lines to a Surface16At the point 2,4uv 22(2,)(,0),(2,)(1,1,).4224uvrra Hence,22(2,)(2,),2.4422uvrraan 22(1)(1)2()0,224a xa yza Thus the equation of the tangent plane at the co

16、rresponding point is:1,1,4a or20.2axayza Tangent Planes and Normal Lines to a Surface17114.2zaxyaa The equation of the normal line at the corresponding point is:1,1,4a Finish.18Tangent Planes and Normal Lines to a Surface derivatives of F are all continuous and the vector ,0,xyzF F F say 0.zF which

17、is determined by(,)zz x y Then,there exists a function,if all the first order partialIf the surface S is expressed by(,)0F x y z Thus,the surface(,)0F x y z and has continuous partial derivative.S can be repressed by r(,)(,(,)x yx y z x y It is easy to see that r(0,1,)(0,1,/),yyyzzFFr(1,0,)(1,0,/),x

18、xxzzFFthen we haven(,).xyzF F F orrr(/,/,1)xyxzyzFF FF19Tangent Planes and Normal Lines to a Surface0(,)A surfaceF x y z 000int,(,),xyzPFor any poPa normal vector at P isFF F 000000000000 0 (,)(,()()()()()(),),.xyzxyzPSo the tangent plane at P to the surface isthF F Fxxyy zF PxxF PyyFatzzsziPThe nor

19、mal line is000000.()()()xyzxxyyzzFPFPF P203.Arc Length:Length of a Smooth CurveEarly in our calculus studies,we learned that the speed is the derivativeUp to now,we considered motion of distance with respect to the time.To study motion along otheroccurring primarily along a straight line.smooth spac

20、e curves,we need to have a measurable length along theThis enables us to locate points along these curves by givingcurve.their directed distance s along the curve from some base point.1 3012sBase pointTime is the natural parameter for describing a moving bodys velocity and acceleration,and s is the

21、naturalparameter for studying the curves shape.21Arc Length:Length of a Smooth CurveDefinition Arc Length:Length of a Smooth Curvethatr()()i()j()k,tf tg th ttThe length of a smooth curve ist is traced exactly once as t increases from t to222.dfdgdhLdtdtdtdt 22Arc Length:Length of a Smooth CurveJust

22、as for plane curves,we can calculate the length of a curve in spacefrom any convenient parametrization that meets the stated conditions.We omit the proof.,is just 222dfdgdhdtdtdtThe square root in the definition,|v|,This enables us to write the formular.ddtthe length of a velocity vector for length

23、a shorter way.Arc Length Formula(Short Form)|v|.Ldt 23Distance Traveled by a GliderExample A glider is soaring upward along the helix r()(cos)i(sin)jk.ttttsec?2t How far does the glider travel along its path from 0t to Solution The path segment during thisrxyzOP/2t 2t time corresponds to one full tu

24、rn of the helix.The length of this portion of the curve is2222020|v|(sin)(cos)222Ldtttt dtdt units of length.times the length of the circle in the2This is xy-plane over which the helix stands.24Arc Length:Length of a Smooth CurveExample Find the arc length of the space curve(),r ttte cost,e sint,eth

25、e arc between the points(1,0,1)and ().220,e,e25Arc Length Parameter with Base Point P(t0)xyzO0()P t()(),(),()P tx ty tz t r()t0r()tBase pointon a smooth curve C parametrized by 0()P tIf we choose a base point on C and a()P tt,each value of t determines a point()s t“directed distance”0()|v()|,tts tdt

26、 measured along C from the base point.We call s(t)an arc length function(弧长函数弧长函数)for the curve.is the negative of the distance.0,()tts t If Each value of s determines a point on Cand this parametrizes C w.r.t.s.26Arc Length Parameter with Base Point P(t0)Arc Length Function with Base Point P(t0)002

27、22()()()()|v()|.tttts txyzddis already given in terms of some parameter t and r()tIf a curve()s tis the arc length function given by the last equation,then we may beThen the curve can be:rr().tt s able to solve for t as a function of reparametrized in terms of s by substituting for:().s tt s 27Findi

28、ng an Arc Length Parametric,the arc length parameter along the helix 00t Example If r()cosisinjk.222ssst sistfrom 0tto 00()|v()|22.ttts tddttSubstituting into the/2.ts Solving this equation for t gives position vector r gives the following arc length parametrizationr()(cos)i(sin)jkttttfor the helix: