Chapter 14 - 5E-14(pp 884-893)1/18/0611:47 AMPage...

Download Document
Showing page : 1 of 38
This preview has blurred sections. Sign up to view the full version! View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5E-14(pp 884-893)1/18/0611:47 AMPage 884CHAPTER 14The calculus of vectorvalued functions is usedin Section 14.4 to proveKeplers laws. Thesedescribe the motion ofthe planets about the Sunand also apply to the orbitof a satellite about theEarth, such as the HubbleSpace Telescope.V ector Functions5E-14(pp 884-893)1/18/0611:47 AMPage 885The functions that we have been using so far have been realvalued functions. We now study functions whose values arevectors because such functions are needed to describe curvesand surfaces in space. We will also use vector-valued functionsto describe the motion of objects through space. In particular,we will use them to derive Keplers laws of planetary motion.|||| 14.1Vector Functions and Space CurvesIn general, a function is a rule that assigns to each element in the domain an element in therange. A vector-valued function, or vector function, is simply a function whose domainis a set of real numbers and whose range is a set of vectors. We are most interested in vector functions r whose values are three-dimensional vectors. This means that for every number t in the domain of r there is a unique vector in V3 denoted by r t . If f t , t t , and h tare the components of the vector r t , then f , t, and h are real-valued functions called thecomponent functions of r and we can writertf t ,t t ,h tftitt jht kWe use the letter t to denote the independent variable because it represents time in mostapplications of vector functions.EXAMPLE 1 Ift 3, ln 3rtt , stthen the component functions areftt3ttln 3thtstBy our usual convention, the domain of r consists of all values of t for which the expression for r t is dened. The expressions t 3, ln 3 t , and st are all dened when3 t 0 and t 0. Therefore, the domain of r is the interval 0, 3 .The limit of a vector function r is dened by taking the limits of its component functions as follows.1|||| If lim t l a r tL, this denition is equivalent to saying that the length and direction of thevector r t approach the length and direction ofthe vector L.If r tf t , t t , h t , thenlim r ttlalim f t , lim t t , lim h ttlatlatlaprovided the limits of the component functions exist.Equivalently, we could have used an - denition (see Exercise 43). Limits of vectorfunctions obey the same rules as limits of real-valued functions (see Exercise 41).8855E-14(pp 884-893)8861/18/0611:47 AMPage 886CHAPTER 14 VECTOR FUNCTIONSEXAMPLE 2 Find lim r t , where r tt3 i1tl0sin tk.tte t jSOLUTION According to Denition 1, the limit of r is the vector whose components are thelimits of the component functions of r :lim r tt3lim 1tl0tl0ikilim tetjtl0limtl0sin tkt(by Equation 3.5.2)A vector function r is continuous at a iflim r tratlazP { f(t), g(t), h(t)}In view of Denition 1, we see that r is continuous at a if and only if its component functions f , t, and h are continuous at a.There is a close connection between continuous vector functions and space curves.Suppose that f , t, and h are continuous real-valued functions on an interval I . Then the setC of all points x, y, z in space, whereCx20r(t)=k f(t), g(t), h(t)lxyF IGURE 1C is traced out by the tip of a movingposition vector r(t).ftyztthtand t varies throughout the interval I , is called a space curve. The equations in (2) arecalled parametric equations of C and t is called a parameter. We can think of C as beingtraced out by a moving particle whose position at time t is f t , t t , h t . If we now conf t , t t , h t , then r t is the position vector of thesider the vector function r tpoint P f t , t t , h t on C. Thus, any continuous vector function r denes a space curveC that is traced out by the tip of the moving vector r t , as shown in Figure 1.EXAMPLE 3 Describe the curve dened by the vector functionrtVisual 14.1A shows several curves beingtraced out by position vectors, includingthose in Figures 1 and 2.1t, 25t,16tSOLUTION The corresponding parametric equations arex1ty2z5t16twhich we recognize from Equations 13.5.2 as parametric equations of a line passingthrough the point 1, 2, 1 and parallel to the vector 1, 5, 6 . Alternatively, we could1, 2, 1 andobserve that the function can be written as r r0 t v, where r0v1, 5, 6 , and this is the vector equation of a line as given by Equation 13.5.1.Plane curves can also be represented in vector notation. For instance, the curve givenby the parametric equations x t 2 2 t and y t 1 (see Example 1 in Section 11.1)could also be described by the vector equationrtwhere i1, 0 and jt22 t, t1t22t i0, 1 .EXAMPLE 4 Sketch the curve whose vector equation isrtcos t isin t jtkt1j5E-14(pp 884-893)1/18/0611:47 AMPage 887SECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES887S OLUTION The parametric equations for this curve arexcos tyzsin ttSince x 2 y 2 cos 2t sin 2t 1, the curve must lie on the circular cylinderx 2 y 2 1. The point x, y, z lies directly above the point x, y, 0 , which movescounterclockwise around the circle x 2 y 2 1 in the x y-plane. (See Example 2 inSection 11.1.) Since z t, the curve spirals upward around the cylinder as t increases.The curve, shown in Figure 2, is called a helix.z 0, 1, 2 F IGURE 2xy(1, 0, 0)The corkscrew shape of the helix in Example 4 is familiar from its occurrence in coiledsprings. It also occurs in the model of DNA (deoxyribonucleic acid, the genetic materialof living cells). In 1953 James Watson and Francis Crick showed that the structure of theDNA molecule is that of two linked, parallel helices that are intertwined as in Figure 3.In Examples 3 and 4 we were given vector equations of curves and asked for a geometric description or sketch. In the next two examples we are given a geometric description of a curve and are asked to nd parametric equations for the curve.EXAMPLE 5 Find a vector equation and parametric equations for the line segment thatjoins the point P 1, 3, 2 to the point Q 2, 1, 3 .FIGURE 3|||| Figure 4 shows the line segment PQ inExample 5.SOLUTION In Section 13.5 we found a vector equation for the line segment that joins thetip of the vector r 0 to the tip of the vector r 1:rt1t r0tr10t1z(See Equation 13.5.4.) Here we take r 01, 3, 2 and r 1vector equation of the line segment from P to Q:Q(2, _1, 3)rt1t1, 3,1t, 32t 2,4 t,21, 32,1, 3 to obtain a0tt11orP(1, 3, _2)FIGURE 4rtyx5t0The corresponding parametric equations arex1ty34tz25t0t15E-14(pp 884-893)8881/18/0611:47 AMPage 888CHAPTER 14 VECTOR FUNCTIONSEXAMPLE 6 Find a vector function that represents the curve of intersection of the cylinderx2y2z1 and the plane y2.SOLUTION Figure 5 shows how the plane and the cylinder intersect, and Figure 6 shows thecurve of intersection C, which is an ellipse.zzy+z=2(0, _1, 3)(_1, 0, 2)C(1, 0, 2)(0, 1, 1)+=10yxyxFIGURE 5FIGURE 6The projection of C onto the xy-plane is the circle x 2from Example 2 in Section 11.1 that we can writexcos tysin t0y2t1, z0. So we know2From the equation of the plane, we havez2y2sin tSo we can write parametric equations for C asxcos tyzsin t22sin t ksin t0t2The corresponding vector equation isrtcos t isin t j0t2This equation is called a parametrization of the curve C. The arrows in Figure 6 indicatethe direction in which C is traced as the parameter t increases.Using Computers to Draw Space CurvesSpace curves are inherently more difcult to draw by hand than plane curves; for an accurate representation we need to use technology. For instance, Figure 7 shows a computergenerated graph of the curve with parametric equationsx4sin 20 t cos ty4sin 20 t sin tzcos 20 tIts called a toroidal spiral because it lies on a torus. Another interesting curve, the tre-5E-14(pp 884-893)1/18/0611:47 AMPage 889S ECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES889foil knot, with equationsx2cos 1.5t cos ty2cos 1.5t sin tzsin 1.5tis graphed in Figure 8. It wouldnt be easy to plot either of these curves by hand.zzxyyxFIGURE 7 A toroidal spiralFIGURE 8 A trefoil knotEven when a computer is used to draw a space curve, optical illusions make it difcultto get a good impression of what the curve really looks like. (This is especially true inFigure 8. See Exercise 42.) The next example shows how to cope with this problem.t, t 2, t 3 .EXAMPLE 7 Use a computer to sketch the curve with vector equation r tThis curve is called a twisted cubic.SOLUTION We start by using the computer to plot the curve with parametric equationsx t, y t 2, z t 3 for 2 t 2. The result is shown in Figure 9(a), but its hard tosee the true nature of the curve from that graph alone. Most three-dimensional computergraphing programs allow the user to enclose a curve or surface in a box instead of displaying the coordinate axes. When we look at the same curve in a box in Figure 9(b), wehave a much clearer picture of the curve. We can see that it climbs from a lower cornerof the box to the upper corner nearest us, and it twists as it climbs.z6_266z0x2_64z0_602_2yy20x(a)024_6(b)84z0z01_4_423_24_80x2y0x8_1142(c)_20y24(d)FIGURE 9 Views of the twisted cubic210x(e)_1_2_8012y(f )345E-14(pp 884-893)8901/18/0611:47 AMPage 890CHAPTER 14 VECTOR FUNCTIONSIn Visual 14.1B you can rotate the boxin Figure 9 to see the curve from anyviewpoint.zxyWe get an even better idea of the curve when we view it from different vantagepoints. Part (c) shows the result of rotating the box to give another viewpoint. Parts (d),(e), and (f) show the views we get when we look directly at a face of the box. In particular, part (d) shows the view from directly above the box. It is the projection of thecurve on the xy-plane, namely, the parabola y x 2. Part (e) shows the projection on thex z-plane, the cubic curve z x 3. Its now obvious why the given curve is called atwisted cubic.Another method of visualizing a space curve is to draw it on a surface. For instance, thetwisted cubic in Example 7 lies on the parabolic cylinder y x 2. (Eliminate the parameter from the rst two parametric equations, x t and y t 2.) Figure 10 shows both thecylinder and the twisted cubic, and we see that the curve moves upward from the originalong the surface of the cylinder. We also used this method in Example 4 to visualize thehelix lying on the circular cylinder (see Figure 2).A third method for visualizing the twisted cubic is to realize that it also lies on the cylinder z x 3. So it can be viewed as the curve of intersection of the cylinders y x 2 andz x 3. (See Figure 11.)FIGURE 1084zVisual 14.1C shows how curves ariseas intersections of surfaces.0_4_8_1FIGURE 11|||| Some computer algebra systems provide uswith a clearer picture of a space curve by enclosing it in a tube. Such a plot enables us to seewhether one part of a curve passes in front of orbehind another part of the curve. For example,Figure 13 shows the curve of Figure 12(b) as rendered by the tubeplot command in Maple.Bx01024yWe have seen that an interesting space curve, the helix, occurs in the model of DNA.Another notable example of a space curve in science is the trajectory of a positivelycharged particle in orthogonally oriented electric and magnetic elds E and B. Dependingon the initial velocity given the particle at the origin, the path of the particle is either aspace curve whose projection on the horizontal plane is the cycloid we studied in Section11.1 [Figure 12(a)] or a curve whose projection is the trochoid investigated in Exercise 38in Section 11.1 [Figure 12(b)].BEEt(a) r(t) = kt-sin t, 1-cos t, t lt33(b) r(t) = kt- 2 sin t, 1- 2 cos t, t lFIGURE 12Motion of a charged particle in orthogonally oriented electric and magnetic fieldsFIGURE 135E-14(pp 884-893)1/18/0611:47 AMPage 891SECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES891For further details concerning the physics involved and animations of the trajectories ofthe particles, see the following web sites:12Exercises1. r tt 2, st2. r ttt36||||cos t,ysin t,zcos t,ysin t,zln tcos 10 t,yezsin 10t,zI4. limett1 s1,5. lim st6. limtltzII,3xyxtan tkt1j1tIII1ln ttarctan t , e 2t,1ttt23itl1zzyxx||||7. r tt41, t8. r t9. r tt, cos 2 t, sin 2 t10. r t11. r tyIVSketch the curve with the given vector equation. Indicatewith an arrow the direction in which t increases.1zVt 3, t 2t, 3t,yzVItsin t, 3, cos t12. r tti13. r tt2 i14. r tsin t itjxcos t kt4 j17. P 1,1, 2 ,yxsin t js2 cos t k1518 |||| Find a vector equation and parametric equations for theline segment that joins P to Q.15. P 0, 0, 0 ,yt6 kecos t, sin t, t ln ttl0tsin 5t24. xt2 kln 9te23. xtsin t jt22. xFind the limit.tl07141, s52i23. limwww.physics.ucla.edu/plasma-exp/Beam/Find the domain of the vector function.||||www.phy.ntnu.edu.tw/java/emField/emField.html|||| 14.1lompado.uah.edu/Links/CrossedFields.html16. P 1, 0, 1 ,Q 1, 2, 318. PQ 4, 1, 71924|||| Match the parametric equations with the graphs(labeled IVI). Give reasons for your choices.19. xcos 4 t,yt,t,yt 2,ze21. xt,y11t2 ,sin 4 ttzt2t cos t,y 2, and use thisx226. Show that the curve with parametric equations xsin t,cos t, z sin 2t is the curve of intersection of the surfacesx 2 and x 2 y 2 1. Use this fact to help sketch the curve.yz; 2730|||| Use a computer to graph the curve with the given vectorequation. Make sure you choose a parameter domain and viewpoints that reveal the true nature of the curve.27. r tz20. xy t sin t, z t lies on the cone z 2fact to help sketch the curve.1, 225. Show that the curve with parametric equations xQ 2, 3, 12, 4, 0 , Q 6,sin t, cos t, t 229. r tt 2, st30. r t1, s5t428. r tt21, t, t 2tsin t, sin 2 t, sin 3 t5E-14(pp 884-893)8921/18/0611:47 AMPage 892CHAPTER 14 VECTOR FUNCTIONStrajectories of two particles are given by the vector functions; 31. Graph the curve with parametric equationsx1 cos 16 t cos t, y1 cos 16 t sin t,z 1 cos 16 t. Explain the appearance of the graph byshowing that it lies on a cone.for txs1r2 t4t3, t 2, 5t60. Do the particles collide?s10.25 cos 2 10 t sin tzt, t 2, t 3r1 t0.25 cos 2 10 t cos ty0.5 cos 10 tr2 t12 t, 16 t, 114 tDo the particles collide? Do their paths intersect?41. Suppose u and v are vector functions that possess limits asExplain the appearance of the graph by showing that it lies on asphere.t l a and let c be a constant. Prove the following properties oflimits.(a) lim u tvtlim u tlim v ttlat 2,, z 1 t 3 passes through the points (1, 4, 0) andy 1 3t(9, 8, 28) but not through the point (4, 7, 6).33. Show that the curve with parametric equations xtla(b) lim c u ttla(c) lim u ttlaFind a vector function that represents the curve of intersection of the two surfaces.tlac lim u ttlavtlim u ttlalim v ttla||||34. The cylinder x35. The cone z2y2sx 236. The paraboloid z12, t 240. Two particles travel along the space curves; 32. Graph the curve with parametric equations3436t 2, 7tr1 t4 and the surface zy 2 and the plane z4x21x22; 38. Try to sketch by hand the curve of intersection of theparabolic cylinder y x 2 and the top half of the ellipsoidx 2 4y 2 4 z 2 16. Then nd parametric equations forthis curve and use these equations and a computer to graphthe curve.39. If two objects travel through space along two different curves,its often important to know whether they will collide. (Will amissile hit its moving target? Will two aircraft collide?) Thecurves might intersect, but we need to know whether theobjects are in the same position at the same time. Suppose thex2lim v ttlacos 1.5t cos tzcylinder xy4 and the parabolic cylinder z x . Thennd parametric equations for this curve and use these equationsand a computer to graph the curve.|||| 14.2lim u ttlait doesnt reveal the whole story. Use the parametric equations2; 37. Try to sketch by hand the curve of intersection of the circular2vt42. The view of the trefoil knot shown in Figure 8 is accurate, butyy and the parabolic cylinder ytlaxy2(d) lim u ty2cos 1.5t sin tsin 1.5tto sketch the curve by hand as viewed from above, with gapsindicating where the curve passes over itself. Start by showingthat the projection of the curve onto the xy-plane has polarcoordinates r 2 cos 1.5t andt, so r varies between 1and 3. Then show that z has maximum and minimum valueswhen the projection is halfway between r 1 and r 3.; When you have nished your sketch, use a computer to drawthe curve with viewpoint directly above and compare with yoursketch. Then use the computer to draw the curve from severalother viewpoints. You can get a better impression of the curveif you plot a tube with radius 0.2 around the curve. (Use thetubeplot command in Maple.)43. Show that lim t l a r tthere is a number0.tab if and only if for every0whenever0 such that r tbDerivatives and Integrals of Vector FunctionsLater in this chapter we are going to use vector functions to describe the motion of planets and other objects through space. Here we prepare the way by developing the calculusof vector functions.DerivativesThe derivative r of a vector function r is dened in much the same way as for realvalued functions:5E-14(pp 884-893)1/18/0611:47 AMPage 893SECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONSdrdt1rtlimrthhhl0893rtif this limit exists. The geometric signicance of this denition is shown in Figure 1. If thelpoints P and Q have position vectors r t and r t h , then PQ represents the vectorrt hr t , which can therefore be regarded as a secant vector. If h 0, the scalarmultiple 1 h r t hr t has the same direction as r t hr t . As h l 0, itappears that this vector approaches a vector that lies on the tangent line. For this reason,the vector r t is called the tangent vector to the curve dened by r at the point P, provided that r t exists and r t0. The tangent line to C at P is dened to be the linethrough P parallel to the tangent vector r t . We will also have occasion to consider theunit tangent vector, which isrtrtTtzVisual 14.2 shows an animation ofFigure 1.zr (t+h)-r (t)r (t)QPPQr (t)r (t)r (t+h)r (t+h)CC00yxFIGURE 1r (t+h)-r (t)hyx(a) The secant vector(b) The tangent vectorThe following theorem gives us a convenient method for computing the derivative of avector function r; just differentiate each component of r.f t ,t t ,h t2 Theorem If r tare differentiable functions, thenrtftif t ,t t ,h ttt jf tih t k, where f , t, and httjhtkProofrtlim1rttlim1ttl0tl0limfttl0limtl0fttftrtt ,t tttftttftf t ,t t ,h tt ,h t,ttttttt, limtl0tttf t ,t t ,h t,hthtttttt, limtl0htthtt5E-14(pp 894-903)8941/18/0612:40 PMPage 894CHAPTER 14 VECTOR FUNCTIONSEXAMPLE 1(a) Find the derivative of r t1 t 3 i te t j(b) Find the unit tangent vector at the point where tsin 2 t k.0.SOLUTION(a) According to Theorem 2, we differentiate each component of r :rt(b) Since r 0i and r 03t 2 ijj 2ks1 4EXAMPLE 2 For the curve r tst ir 1 and the tangent vector r 1 .y22 cos 2 t k2 k, the unit tangent vector at the point 1, 0, 0 isr0r0T0t e tj121js52ks5t j, nd r t and sketch the position vectorSOLUTION We have(1, 1)r (1)0r (1)1i2 strtx1jand1i2r1jThe curve is a plane curve and elimination of the parameter from the equations x st,y 2 t gives y 2 x 2, x 0. In Figure 2 we draw the position vector r 1ijstarting at the origin and the tangent vector r 1 starting at the corresponding point 1, 1 .FIGURE 2EXAMPLE 3 Find parametric equations for the tangent line to the helix with parametricequations|||| The helix and the tangent line in Example 3are shown in Figure 3.xat the point 0, 1,yrt4_0.5FIGURE 3y00.512_20xt2.8zzsin tSOLUTION The vector equation of the helix is r t120_12 cos t2 cos t, sin t, t , so2 sin t, cos t, 1The parameter value corresponding to the point 0, 1, 2 is t2, so the tangentvector there is r22, 0, 1 . The tangent line is the line through 0, 1, 2parallel to the vector2, 0, 1 , so by Equations 13.5.2 its parametric equations arex2ty1z2tJust as for real-valued functions, the second derivative of a vector function r is thederivative of r , that is, rr . For instance, the second derivative of the function inExample 3 isrt2 cos t,sin t, 0A curve given by a vector function r t on an interval I is called smooth if r is continuous and r t0 (except possibly at any endpoints of I ). For instance, the helix inExample 3 is smooth because r t is never 0.EXAMPLE 4 Determine whether the semicubical parabola r t1t 3, t 2 is smooth.5E-14(pp 894-903)1/18/0612:40 PMPage 895S ECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONSy895S OLUTION Since3 t 2, 2 trtcusp01FIGURE 4The curve r(t)=k1+t #, t @ lis not smooth.xwe have r 00, 00 and, therefore, the curve is not smooth. The point that corresponds to t 0 is (1, 0), and we see from the graph in Figure 4 that there is a sharpcorner, called a cusp, at (1, 0). Any curve with this type of behavioran abrupt changein directionis not smooth.A curve, such as the semicubical parabola, that is made up of a nite number of smoothpieces is called piecewise smooth.Differentiation RulesThe next theorem shows that the differentiation formulas for real-valued functions havetheir counterparts for vector-valued functions.3 Theorem Suppose u and v are differentiable vector functions, c is a scalar, andf is a real-valued function. Thend1.utvtutvtdtdcu tcu t2.dtdf tutf tutftu t3.dtdut vtu t vtut v t4.dtdutvtutvtutvt5.dtduftf tu ft(Chain Rule)6.dtThis theorem can be proved either directly from Denition 1 or by using Theorem 2 andthe corresponding differentiation formulas for real-valued functions. The proofs of Formulas1, 2, 3, 5, and 6 are left as exercises.Proof of Formula 4 Letutf1 t , f2 t , f3 tvtt1 t , t2 t , t3 t3Thenutvtf1 t t1 tf2 t t2 tf3 t t3 tfi t ti ti1so the ordinary Product Rule givesdutdtvtddt33fi t ti ti1i1dfi t ti tdt3f i t ti tfi t t i ti133f i t ti ti1utfi t t i ti1vtutvt5E-14(pp 894-903)8961/18/0612:40 PMPage 896CHAPTER 14 VECTOR FUNCTIONSEXAMPLE 5 Show that if r tc (a constant), then r t is orthogonal to r t for all t .SOLUTION Sincertrt2rtc2and c 2 is a constant, Formula 4 of Theorem 3 givesdrtdt0rtrtrtrtrt2r trt0, which says that r t is orthogonal to r t .Thus, r t r tGeometrically, this result says that if a curve lies on a sphere with center the origin,then the tangent vector r t is always perpendicular to the position vector r t .IntegralsThe denite integral of a continuous vector function r t can be dened in much the sameway as for real-valued functions except that the integral is a vector. But then we canexpress the integral of r in terms of the integrals of its component functions f, t, and h asfollows. (We use the notation of Chapter 5.)ybanr t dtr t*ilimnlti1nnf t*ilimnlnt t*itii1h t*itji1tki1and soybar t dtybayf t dt ibayt t dt jbah t dt kThis means that we can evaluate an integral of a vector function by integrating each component function.We can extend the Fundamental Theorem of Calculus to continuous vector functions asfollows:ybar t dtRtbaRbwhere R is an antiderivative of r, that is, R tindenite integrals (antiderivatives).EXAMPLE 6 If r t2 cos t iyrtdtsin t jy 2 cos t dt2 sin t iRar t . We use the notation2 t k, thenicos t jy sin t dtt2 ky 2 t dtjkCwhere C is a vector constant of integration, andy02r t dt[2 sin t icos t jt2 k0222ij4kxr tdt for5E-14(pp 894-903)1/18/0612:40 PMPage 897S ECTION 14.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS|||| 14.2897Exercises1. The gure shows a curve C given by a vector function r t .(a) Draw the vectors r 4.5(b) Draw the vectorsr 4.5r4r 4 and r 4.2and0.5r4.r 4.2t13. r t2eijln 1a16. r tta3t kc cos 3t kb sin t jt2 ctbbk3at cos 3 t i15. r t(c) Write expressions for r 4 and the unit tangent vector T(4).(d) Draw the vector T(4).t2 js114. r tr40.2tc1720 |||| Find the unit tangent vector T t at the point with thegiven value of the parameter t.yRCsin 1t i12. r t6 t 5, 4 t 3, 2 t , t17. r tr(4.5)218. r tPtjcos t i3t j20. r tQr(4.2)4 st i19. r t12 sin t i1t k, t2 sin 2 t k,2 cos t j1t0tan t k, t4r(4)023. xt225. xe26. xr1t 5, y24. x0.1ln t,Explain why these vectors are so close to each other inlength and direction.4. r t1t, st , t5. r t1ti6. r tteie7. r tet ie 3 t j,8. r t2 sin t i916||||j,ttt 2, 1it, stje 4t kr t.cos t,t 3; (1, 1, 1)t21, zyye2 st,t1;1, 1, 1tze t;1, 0, 1sin t,t 2;z0, 2, 1s2 cos t, z3e , zcos t, ys2 sin t ;2t3e2t1, 3, 3;4, 1, 1t 3, t 4, t 5cos 3t, sin 3t(b) r tt3t, t 4, t 530. (a) Find the point of intersection of the tangent lines to the00t(a) r t(c) r t13 cos t j,, nd T 0 , r 0 , and r t29. Determine whether the curve is smooth.1t;3curve r tsin t, 2 sin t, cos t at the points wheret 0 and t 0.5.(b) Illustrate by graphing the curve and both tangent lines.t, t 2, t 3 and r2 tsin t, sin 2 t, t intersect at the origin. Find their angle of intersection correct to thenearest degree.31. The curves r1 tFind the derivative of the vector function.9. r t11. r ttt 4, z1, yt, y28. x4t 2 j, t27. xtr t.|||| Find parametric equations for the tangent line to thecurve with the given parametric equations at the specied point.Illustrate by graphing both the curve and the tangent line on a common screen.||||cos t, sin t ,, te2t; 2728(a) Sketch the plane curve with the given vector equation.(b) Find r t .(c) Sketch the position vector r t and the tangent vector r t forthe given value of t.3. r te ,e2t2326 |||| Find parametric equations for the tangent line to thecurve with the given parametric equations at the specied point.function r tt 2, t , 0 t 2, and draw the vectorsr(1), r(1.1), and r(1.1) r(1).(b) Draw the vector r 1 starting at (1, 1) and compare it withthe vector382t22. If r t2. (a) Make a large sketch of the curve described by the vectorr 1.1t, t 2, t 3 , nd r t , T 1 , r t , and r t21. If r tx110. r tcos 3t, t, sin 3tt, 1 t, 3 t 2 andr2 s3 s, s 2, s intersect? Find their angle of intersection correct to the nearest degree.32. At what point do the curves r1 t25E-14(pp 894-903)898333833.34.yPage 898143. Prove Formula 5 of Theorem 3.Evaluate the integral.||||16t 3 i9t 2 j44. Prove Formula 6 of Theorem 3.25 t 4 k dti 2 t 2 j 3 t 3 k and v tnd d dt u t v t .45. If u t41t21022tjy36.y37.yet i2t j38.ycosti4te t jst i3 sin t cos t j39. Find r t if r tt2 i40. Find r t if r tsin t i2 k.r0ijd dt2 sin t cos t k dtsin t k ,47. Show that if r is a vector function such that r exists, thendrtdtrt49. If r tt k dt4t 3 jt 2 k and r 0cos t j0, show that[Hint:j.2rtrtrtrtdutdt48. Find an expression forvtwt .drtdtrt1rtrtr t.50. If a curve has the property that the position vector r t isalways perpendicular to the tangent vector r t , show thatthe curve lies on a sphere with center the origin.2 t k and51. If u trtrt42. Prove Formula 3 of Theorem 3.r t , show thatut41. Prove Formula 1 of Theorem 3.|||| 14.3cos t jvt .ut1k dtt2tjti46. If u and v are the vector functions in Exercise 45, ndln t k dtsink dt23 sin t cos t i01t21235.12:41 PMCHAPTER 14 VECTOR FUNCTIONS0y1/18/06rtrtrtArc Length and CurvatureIn Section 11.2 we dened the length of a plane curve with parametric equations x f t ,y t t , a t b, as the limit of lengths of inscribed polygons and, for the case wheref and t are continuous, we arrived at the formula1zLyba2sf t2ttdtydxdtba2dydt2dtThe length of a space curve is dened in exactly the same way (see Figure 1). Supposethat the curve has the vector equation r tf t , t t , h t , a t b, or, equivalently,the parametric equations x f t , y t t , z h t , where f , t , and h are continuous.If the curve is traversed exactly once as t increases from a to b, then it can be shown thatits length is0y2LybaxyF IGURE 1baThe length of a space curve is the limitof lengths of inscribed polygons.sf tdxdt2tt22dydtht22dtdzdt2dtNotice that both of the arc length formulas (1) and (2) can be put into the more compact form3Lybar t dt5E-14(pp 894-903)1/18/0612:41 PMPage 899S ECTION 14.3 ARC LENGTH AND CURVATUREbecause, for plane curves r tf tiwhereas, for space curves r trt|||| Figure 2 shows the arc of the helix whoselength is computed in Example 1.f tisf ttt jtt2tt2h t k,ftittj22ttjhtksf tht2EXAMPLE 1 Find the length of the arc of the circular helix with vector equationzrtcos t it k from the point 1, 0, 0 to the point 1, 0, 2 .sin t jSOLUTION Since r tsin t icos t jrt(1, 0, 2)sk, we havesin t2cos 2 t1s2The arc from 1, 0, 0 to 1, 0, 2 is described by the parameter interval 0and so, from Formula 3, we have(1, 0, 0)x899t t j,ftirtyyLFIGURE 220r t dty20s2 dtt22 s2A single curve C can be represented by more than one vector function. For instance, thetwisted cubicr1 t4t, t 2, t 31t0u2could also be represented by the function|||| Piecewise-smooth curves were introduced onpage 895.zs(t)Cr(a)0xFIGURE 3yln 2where the connection between the parameters t and u is given by t e u. We say thatEquations 4 and 5 are parametrizations of the curve C. If we were to use Equation 3 tocompute the length of C using Equations 4 and 5, we would get the same answer. In general, it can be shown that when Equation 3 is used to compute the length of any piecewisesmooth curve, the arc length is independent of the parametrization that is used.Now we suppose that C is a piecewise-smooth curve given by a vector functionrtf t i t t j h t k, a t b, and C is traversed exactly once as t increasesfrom a to b. We dene its arc length function s by6r(t)e u, e 2u, e 3ur2 u5stytaruduytadxdu2dydu2dzdu2duThus, s t is the length of the part of C between r a and r t . (See Figure 3.) If we differentiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus, weobtaindsrt7dtIt is often useful to parametrize a curve with respect to arc length because arc lengtharises naturally from the shape of the curve and does not depend on a particular coordinatesystem. If a curve r t is already given in terms of a parameter t and s t is the arc lengthfunction given by Equation 6, then we may be able to solve for t as a function of s: t t s .Then the curve can be reparametrized in terms of s by substituting for t : r r t s . Thus,if s 3 for instance, r t 3 is the position vector of the point 3 units of length along thecurve from its starting point.5E-14(pp 894-903)9001/18/0612:41 PMPage 900CHAPTER 14 VECTOR FUNCTIONSE XAMPLE 2 Reparametrize the helix r tcos t i sin t j t k with respect to arclength measured from 1, 0, 0 in the direction of increasing t .SOLUTION The initial point 1, 0, 0 corresponds to the parameter value tExample 1 we havedsdtand sosTherefore, tysttrts2r u du00. Fromyt0s2 dus2 ts s2 and the required reparametrization is obtained by substituting for t :rtscos(s s2 ) isin(s s2 ) j(s s2 ) kCurvatureIf C is a smooth curve dened by the vector function r, then r ttangent vector T t is given byzTt0xCyFIGURE 4Unit tangent vectors at equally spacedpoints on CVisual 14.3A shows animated unit tangent vectors, like those in Figure 4, for avariety of plane curves and space curves.0. Recall that the unitrtrtand indicates the direction of the curve. From Figure 4 you can see that T t changes direction very slowly when C is fairly straight, but it changes direction more quickly when Cbends or twists more sharply.The curvature of C at a given point is a measure of how quickly the curve changes direction at that point. Specically, we dene it to be the magnitude of the rate of change of theunit tangent vector with respect to arc length. (We use arc length so that the curvature willbe independent of the parametrization.)8Definition The curvature of a curve isdTdswhere T is the unit tangent vector.The curvature is easier to compute if it is expressed in terms of the parameter t insteadof s, so we use the Chain Rule (Theorem 14.2.3, Formula 6) to writedTdtBut ds dt9rtd T dsds dtdTdsandd T dtds dtfrom Equation 7, sotTtrtEXAMPLE 3 Show that the curvature of a circle of radius a is 1 a.5E-14(pp 894-903)1/18/0612:41 PMPage 901S ECTION 14.3 ARC LENGTH AND CURVATURE901SOLUTION We can take the circle to have center the origin, and then a parametrization isrtThereforerta cos t ia cos t ja sin t isoa sin t jandrtrtTtandsin t iTtThis gives T tcos t irtacos t jsin t j1, so using Equation 9, we haveTtrtt1aThe result of Example 3 shows that small circles have large curvature and large circleshave small curvature, in accordance with our intuition. We can see directly from the denition of curvature that the curvature of a straight line is always 0 because the tangent vector is constant.Although Formula 9 can be used in all cases to compute the curvature, the formulagiven by the following theorem is often more convenient to apply.10 Theorem The curvature of the curve given by the vector function r isrttProof Since Trr and rrt3rtds dt, we haverdsTdtrTso the Product Rule (Theorem 14.2.3, Formula 3) givesd 2sTdt 2rUsing the fact that TTdsTdt0 (see Example 2 in Section 13.4), we haverdsdtr2TTNow T t1 for all t , so T and T are orthogonal by Example 5 in Section 14.2.Therefore, by Theorem 13.4.6,rThusandr2dsdtTTT2dsdtTrrds dt 2TrrrrrrrT32dsdt2T5E-14(pp 894-903)9021/18/0612:41 PMPage 902CHAPTER 14 VECTOR FUNCTIONSt, t 2, t 3 at a general pointE XAMPLE 4 Find the curvature of the twisted cubic r tand at 0, 0, 0 .SOLUTION We rst compute the required ingredients:1, 2 t, 3 t 2rtrtrtrt4t 2s1i10rt0, 2, 6 t9t 4jk2t 3t 22 6ts36 t 4rtrt6t 2 i36 t 246t j2k2 s9 t 49t 21Theorem 10 then givesrttrtAt the origin the curvature is2 s1 9 t 2 9 t 41 4t 2 9t 4 3 2rt32.0For the special case of a plane curve with equation y f x , we can choose x as theparameter and write r xx i f x j. Then r xi f x j and r xf x j.Since ijk and jj0, we have r xrxf x k. We also haverxf x 2 and so, by Theorem 10,s1x111fxfx2 32y2EXAMPLE 5 Find the curvature of the parabola yy=and 2, 4 .SOLUTION Since yy=k(x)0FIGURE 5The parabola y= and itscurvature functionx 2 at the points 0, 0 , 1, 1 ,12 x and yx2, Formula 11 gives1yy2 32124x 232xThe curvature at 0, 0 is 02. At 1, 1 it is 12 5 3 2 0.18. At 2, 4 it is3222 170.03. Observe from the expression for x or the graph of in Figure 5 that x l 0 as x l. This corresponds to the fact that the parabola appearsto become atter as x l.The Normal and Binormal VectorsAt a given point on a smooth space curve r t , there are many vectors that are orthogonalto the unit tangent vector T t . We single out one by observing that, since T t1 forall t, we have T t T t0 by Example 5 in Section 14.2, so T t is orthogonal to5E-14(pp 894-903)1/18/0612:42 PMPage 903S ECTION 14.3 ARC LENGTH AND CURVATURE|||| We can think of the normal vector as indicating the direction in which the curve is turningat each point.903T t . Note that T t is itself not a unit vector. But if r is also smooth, we can dene theprincipal unit normal vector N t (or simply unit normal) asTtTtNtT(t)B(t)N(t)The vector B tTtN t is called the binormal vector. It is perpendicular to both Tand N and is also a unit vector. (See Figure 6.)EXAMPLE 6 Find the unit normal and binormal vectors for the circular helixFIGURE 6rt|||| Figure 7 illustrates Example 6 by showingthe vectors T, N, and B at two locations on thehelix. In general, the vectors T, N, and B, starting at the various points on a curve, form a set oforthogonal vectors, called the TNB frame, thatmoves along the curve as t varies. This TNBframe plays an important role in the branch ofmathematics known as differential geometry andin its applications to the motion of spacecraft.cos t isin t jtkSOLUTION We rst compute the ingredients needed for the unit normal vector:rtsin t icos t jrtrtTtrt1s2sin t icos t i1s2Ttksin t js2cos t jk1s2TtzTTtTtNtBNTBcos t iBtTtisin tcos t1s2NtyFIGURE 7Visual 14.3B shows how the TNB framemoves along several curves.cos t,sin t, 0This shows that the normal vector at a point on the helix is horizontal and points towardthe z-axis. The binormal vector isNxsin t jjcos tsin tk101sin t,s2cos t, 1The plane determined by the normal and binormal vectors N and B at a point P on acurve C is called the normal plane of C at P. It consists of all lines that are orthogonalto the tangent vector T. The plane determined by the vectors T and N is called the osculating plane of C at P. The name comes from the Latin osculum, meaning kiss. It is theplane that comes closest to containing the part of the curve near P. (For a plane curve, theosculating plane is simply the plane that contains the curve.)The circle that lies in the osculating plane of C at P, has the same tangent as C at P, lieson the concave side of C (toward which N points), and has radius1 (the reciprocalof the curvature) is called the osculating circle (or the circle of curvature) of C at P. It isthe circle that best describes how C behaves near P; it shares the same tangent, normal,and curvature at P.EXAMPLE 7 Find the equations of the normal plane and osculating plane of the helix inExample 6 at the point P 0, 1,2.SOLUTION The normal plane at P has normal vector r1, 0, 1 , so an equation2is1x00y11z20orzx25E-14(pp 904-913)9041/18/0612:33 PMPage 904CHAPTER 14 VECTOR FUNCTIONS|||| Figure 8 shows the helix and the osculatingplane in Example 7.zThe osculating plane at P contains the vectors T and N, so its normal vector isT N B. From Example 6 we have1sin t,s2Btz=_x+2cos t, 1B11, 0,s2s22A simpler normal vector is 1, 0, 1 , so an equation of the osculating plane isPx1xyFIGURE 800y1z102zorxx 2 at the origin.EXAMPLE 8 Find and graph the osculating circle of the parabola ySOLUTION From Example 5 the curvature of the parabola at the origin is 02. So the11radius of the osculating circle at the origin is 12 and its center is (0, 2 ). Its equationis therefore21x2 (y 1)24yy=osculatingcircle2For the graph in Figure 9 we use parametric equations of this circle:12x120y1212sin tWe summarize here the formulas for unit tangent, unit normal and binormal vectors,and curvature.x1cos tFIGURE 9Visual 14.3C shows how the osculatingcircle changes as a point moves along acurve.|||| 14.316||||2t , sin t3. r t4. r tt2 i5. r ti6. r t12 t itjt k, 00tttrtNtrtrt37. Use Simpsons Rule with narc of the twisted cubic xthe point 2, 4, 8 .|||| Reparametrize the curve with respect to arc length measured from the point where t 0 in the direction of increasing t.109. r tt13 sin t i3t je 2 t cos 2 t i11. r tet2t i10. r t113t 2 k, 0tt sin t , 0ln t k, 138t 3 2 j10e t k,2t j2TtrtTt911t cos t, cos tet js2 t iBtExercises2 sin t, 5t, 2 cos t ,2. r tTtTtNtdTdsFind the length of the curve.1. r trtrtTt4t ke 2 t sin 2 t k2j4t j53 cos t k110 to estimate the length of thet, y t 2, z t 3 from the origin to; 8. Graph the curve with parametric equations xcos t,y sin 3 t, z sin t. Find the total length of this curve correctto four decimal places.12. Reparametrize the curvert2t211i2tt21jwith respect to arc length measured from the point (1, 0) in thedirection of increasing t. Express the reparametrization in itssimplest form. What can you conclude about the curve?5E-14(pp 904-913)1/18/0612:33 PMPage 905S ECTION 14.3 ARC LENGTH AND CURVATURE1316||||905(a) Find the unit tangent and unit normal vectors T t and N t .(b) Use Formula 9 to nd the curvature.3233 |||| Two graphs, a and b, are shown. One is a curve yfxand the other is the graph of its curvature function yx . Identify each curve and explain your choices.13. r t32.2 sin t, 5 t, 2 cos t214. r tt , sin t15. r ts2 t, e t, e1719tt sin t , tt2 i18. r tti19. r t3t ibtjt2 k14 sin t jt24. y4t 3 21.metric curve xt t isx y yxx2 y2 3 2where the dots indicate derivatives with respect to t.3738ex27. yln xf t,y4x 5 22627 |||| At what point does the curve have maximum curvature?What happens to the curvature as x l ?26. y36. Use Theorem 10 to show that the curvature of a plane para-25. yt 3 sin t, 1 3 cos t, t is shown in22Figure 12(b) in Section 14.1. Where do you think the curvatureis largest? Use a CAS to nd and graph the curvature function.For which values of t is the curvature largest?t2zcos x35. The graph of r tUse Formula 11 to nd the curvature.x3C ASt, t 2, t 3 at the point (1, 1, 1).and nd the curvature at the point 1, 4,||||sin 3 t, sin 2 t, sin 3 t . At howmany points on the curve does it appear that the curvaturehas a local or absolute maximum?(b) Use a CAS to nd and graph the curvature function. Doesthis graph conrm your conclusion from part (a)?e t cos t, e t sin t, t at theyx34. (a) Graph the curve r t; 22. Graph the curve with parametric equationsxC AS4 cos t k21. Find the curvature of r tbxpoint (1, 0, 0).23. yatk20. Find the curvature of r t2325yat 2, 2 t, ln t16. r t0Use Theorem 10 to nd the curvature.||||17. r tt cos t, cos t33.y||||e t cos t,37. x38. x28. Find an equation of a parabola that has curvature 4 at theUse the formula in Exercise 36 to nd the curvature.t 3,1e t sin tyyt2torigin.29. (a) Is the curvature of the curve C shown in the gure greater3940at P or at Q ? Explain.(b) Estimate the curvature at P and at Q by sketching theosculating circles at those points.39. r t40. r ty41. x0x11, 0, 131. yxt,z3t;2 cos 3 t ;0, ,21, 1, 112x2at the points 0, 0 and (1, 2 ). Graph both osculating circles andthe parabola on the same screen.; 44. Find equations of the osculating circles of the parabola y14t, z9x 2 4y 2 36 at the points 2, 0 and 0, 3 . Use a graphingcalculator or computer to graph the ellipse and both osculatingcircles on the same screen.|||| Use a graphing calculator or computer to graph both thecurve and its curvature function x on the same screen. Is thegraph of what you would expect?xt, y2; 43. Find equations of the osculating circles of the ellipse; 3031xe(1, 2 , 1)3e t, e t sin t, e t cos t ,2 sin 3 t, y42. xQt 2, 2 t 3, t ,34142 |||| Find equations of the normal plane and osculating planeof the curve at the given point.1Find the vectors T, N, and B at the given point.PC30. y||||5E-14(pp 904-913)9061/18/0612:33 PMCHAPTER 14 VECTOR FUNCTIONSt 3, y 3 t, z t 4 is the normal6 y 8 z 1?45. At what point on the curve xplane parallel to the plane 6 xCASPage 906(b) rr(c) r47. Show that the curvatures2Ns3Brr2t, 1 t 2, 1 t 3 .23curve r td ds ,where is the angle between T and i ; that is, is the angle ofinclination of the tangent line. (This shows that the denitionof curvature is consistent with the denition for plane curvesgiven in Exercise 69 in Section 11.2.)54. Find the curvature and torsion of the curve xysinh t,t at the point 0, 1, 0 .Figure 3 on page 887). The radius of each helix is about10 angstroms (1 10 8 cm). Each helix rises about 34 during each complete turn, and there are about 2.9 10 8 complete turns. Estimate the length of each helix.(b) Show that d B ds is perpendicular to T.(c) Deduce from parts (a) and (b) that d B dss N forsome number s called the torsion of the curve. (Thetorsion measures the degree of twisting of a curve.)(d) Show that for a plane curve the torsion is s0.56. Lets consider the problem of designing a railroad track tomake a smooth transition between sections of straight track.Existing track along the negative x-axis is to be joinedsmoothly to a track along the line y 1 for x 1.(a) Find a polynomial P P x of degree 5 such that the function F dened by50. The following formulas, called the Frenet-Serret formulas,are of fundamental importance in differential geometry:1. d T dsN2. d N dsTB3. d B dsN(Formula 1 comes from Exercise 47 and Formula 3 comesfrom Exercise 49.) Use the fact that N B T to deduceFormula 2 from Formulas 1 and 3.Fx51. Use the Frenet-Serret formulas to prove each of the following.(Primes denote derivatives with respect to t. Start as in theproof of Theorem 10.)(a) rsTs 2Ncosh t, z55. The DNA molecule has the shape of a double helix (see49. (a) Show that d B ds is perpendicular to B.;0Px1if xif 0if x0x11is continuous and has continuous slope and continuouscurvature.(b) Use a graphing calculator or computer to draw the graphof F .Motion in Space: Velocity and Accelerationr (t+h)-r (t)hr (t)QzPr (t)r (t+h)In this section we show how the ideas of tangent and normal vectors and curvature can beused in physics to study the motion of an object, including its velocity and acceleration,along a space curve. In particular, we follow in the footsteps of Newton by using thesemethods to derive Keplers First Law of planetary motion.Suppose a particle moves through space so that its position vector at time t is r t .Notice from Figure 1 that, for small values of h, the vectorC1OFIGURE 13 ss53. Use the formula in Exercise 51(d) to nd the torsion of theN48. Show that the curvature of a plane curve isxrrTa cos t, a sin t, b t , wherea and b are positive constants, has constant curvature and constant torsion. [Use the result of Exercise 51(d).]vectors by the equation|||| 14.4352. Show that the circular helix r tis related to the tangent and normaldTdsr(d)lating plane is parallel to the plane x y z 1? (Note: Youwill need a CAS for differentiating, for simplifying, and forcomputing a cross product.)Bs2s46. Is there a point on the curve in Exercise 45 where the oscu-3srthhrtyapproximates the direction of the particle moving along the curve r t . Its magnitude measures the size of the displacement vector per unit time. The vector (1) gives the average5E-14(pp 904-913)1/18/0612:33 PMPage 907S ECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION 907velocity over a time interval of length h and its limit is the velocity vector v t at time t :vt2limrthhhl0rtrtThus, the velocity vector is also the tangent vector and points in the direction of the tangent line.The speed of the particle at time t is the magnitude of the velocity vector, that is, v t .This is appropriate because, from (2) and from Equation 14.3.7, we havevtrtdsdtrate of change of distance with respect to timeAs in the case of one-dimensional motion, the acceleration of the particle is dened as thederivative of the velocity:atvtrtEXAMPLE 1 The position vector of an object moving in a plane is given byrtt 3 i t 2 j. Find its velocity, speed, and acceleration when tgeometrically.y1 and illustrateSOLUTION The velocity and acceleration at time t arev (1)vt(1, 1)rt3t 2 i2t jatrt6t i2ja (1)x0and the speed isFIGURE 2s 3t 2vtVisual 14.4 shows animated velocity andacceleration vectors for objects movingalong various curves.When tz2s9 t 44t 23i2ja16i2jv1EXAMPLE 2 Find the velocity, acceleration, and speed of a particle with position vectort 2, e t, te t .SOLUTIONv (1)vt1yFIGURE 3rt2 t, e t, 1t etatxs13These velocity and acceleration vectors are shown in Figure 2.rta (1)2t1, we havev1|||| Figure 3 shows the path of the particle inExample 2 with the velocity and accelerationvectors when t 1.2vt2, e t, 2t ets4 t 2e 2tt 2e 2tvt1The vector integrals that were introduced in Section 14.2 can be used to nd positionvectors when velocity or acceleration vectors are known, as in the following example.5E-14(pp 904-913)9081/18/0612:33 PMPage 908CHAPTER 14 VECTOR FUNCTIONSEXAMPLE 3 A moving particle starts at an initial position r 01, 0, 0 with initialvelocity v 0i j k. Its acceleration is a t4 t i 6 t j k. Find its velocityand position at time t .SOLUTION Since a tv t , we haveyatvt2t 2 iydt4t i3t 2 j6t jtkk dtCTo determine the value of the constant vector C, we use the fact that v 0The preceding equation gives v 0C, so C i j k and2t 2 ivt3t 2 j|||| The expression for r t that we obtained inExample 3 was used to plot the path of the particle in Figure 4 for 0 t 3.Since v tyvtz400(2t335y101jt1k1i3t 21jtk.1 k dtjkdt2t 2y6(1, 0, 0)3t 2jr t , we havert2i1i2t 2tkit) itj(1t22t3t3tjt) kD0152020xPutting t0, we nd that D(2t33rtF IGURE 4r0ti, so1) i(1t22t) kIn general, vector integrals allow us to recover velocity when acceleration is known andposition when velocity is known:vtv t0ytt0a u durtr t0ytt0v u duIf the force that acts on a particle is known, then the acceleration can be found fromNewtons Second Law of Motion. The vector version of this law states that if, at any timet, a force F t acts on an object of mass m producing an acceleration a t , then|||| The angular speed of the object moving withd dt, where is the angleposition P isshown in Figure 5.Ftma tEXAMPLE 4 An object with mass m that moves in a circular path with constant angularspeed has position vector r ta cos t i a sin t j. Find the force acting on theobject and show that it is directed toward the origin.yPSOLUTIONvtrta sin t iat0vtaa cos t jx2cos t ia2sin t jTherefore, Newtons Second Law gives the force asFIGURE 5Ftma tm2a cos t ia sin t j5E-14(pp 904-913)1/18/0612:33 PMPage 909S ECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION909Notice that F tm 2 r t . This shows that the force acts in the direction opposite tothe radius vector r t and therefore points toward the origin (see Figure 5). Such a forceis called a centripetal (center-seeking) force.yEXAMPLE 5 A projectile is red with angle of elevation and initial velocity v0. (SeeFigure 6.) Assuming that air resistance is negligible and the only external force is due togravity, nd the position function r t of the projectile. What value of maximizes therange (the horizontal distance traveled)?va0xdSOLUTION We set up the axes so that the projectile starts at the origin. Since the force dueto gravity acts downward, we haveFFIGURE 6where tmamt j9.8 m s2 . ThusaaSince v ttja, we havevtwhere Ctt jCv0 . Thereforev0rtvttt jv0t v0DIntegrating again, we obtain12rtBut Dtt 2 j0, so the position vector of the projectile is given byr012rt3If we write v0tt 2 jt v0v0 (the initial speed of the projectile), thenv0v0 cosiv0 sinjand Equation 3 becomesrtv0 cos[ v0 sinti12ttt 2 ] jThe parametric equations of the trajectory are therefore|||| If you eliminate t from Equations 4, you willsee that y is a quadratic function of x. So thepath of the projectile is part of a parabola.x4v0 costyv0 sint12tt 2The horizontal distance d is the value of x when y 0. Setting yor t2v0 sint. The latter value of t then givesdxv0 cos2v0 sintClearly, d has its maximum value when sin 2v 2 2 sin0cos0, we obtain tv 2 sin 20t1, that is,t4.05E-14(pp 904-913)9101/18/0612:33 PMPage 910CHAPTER 14 VECTOR FUNCTIONSE XAMPLE 6 A projectile is red with muzzle speed 150 m s and angle of elevation 45from a position 10 m above ground level. Where does the projectile hit the ground, andwith what speed?SOLUTION If we place the origin at ground level, then the initial position of the projectileis (0, 10) and so we need to adjust Equations 4 by adding 10 to the expression for y.With v 0 150 m s,45 , and t 9.8 m s2, we havex150 cosy104t75 s2 t150 sin124t9.8 t 210Impact occurs when y 0, that is, 4.9 t 2 75 s2 t 10equation (and using only the positive value of t), we gett75 s2s11,2509.84.9 t 275 s2 t1960. Solving this quadratic21.74Then x 75 s2 21.742306 , so the projectile hits the ground about 2306 m away.The velocity of the projectile isvtrt(75 s275 s2 i9.8 t) jSo its speed at impact isv 21.74s(75 s2 )2(75 s29.8 21.74)2151 m sTangential and Normal Components of AccelerationWhen we study the motion of a particle, it is often useful to resolve the acceleration intotwo components, one in the direction of the tangent and the other in the direction of thenormal. If we write vv for the speed of the particle, thenrtrtTtand sovvtvtvvvTIf we differentiate both sides of this equation with respect to t , we geta5vvTvTIf we use the expression for the curvature given by Equation 14.3.9, then we have6TrTsovTThe unit normal vector was dened in the preceding section as NTTNvNand Equation 5 becomes7avTv2 NvTT , so (6) gives5E-14(pp 904-913)1/18/0612:33 PMPage 911SECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION911Writing a T and a N for the tangential and normal components of acceleration, we haveaTaTaaTTaN NandaNwhereNaT8vv2aNFIGURE 7This resolution is illustrated in Figure 7.Lets look at what Formula 7 says. The rst thing to notice is that the binormal vectorB is absent. No matter how an object moves through space, its acceleration always lies inthe plane of T and N (the osculating plane). (Recall that T gives the direction of motionand N points in the direction the curve is turning.) Next we notice that the tangential component of acceleration is v , the rate of change of speed, and the normal component ofacceleration is v 2, the curvature times the square of the speed. This makes sense if wethink of a passenger in a cara sharp turn in a road means a large value of the curvature, so the component of the acceleration perpendicular to the motion is large and the passenger is thrown against a car door. High speed around the turn has the same effect; in fact,if you double your speed, aN is increased by a factor of 4.Although we have expressions for the tangential and normal components of acceleration in Equations 8, its desirable to have expressions that depend only on r, r , and r . Tothis end we take the dot product of v v T with a as given by Equation 7:vavTvv Tv2 NvTv3 TTNvv(since T T1 and T N0)ThereforeaT9vavrtvrtrtUsing the formula for curvature given by Theorem 14.3.10, we have10aNrtv2rtrt3rtrt2EXAMPLE 7 A particle moves with position function r tt 2, t 2, t 3 . Find the tangentialand normal components of acceleration.rtt2 it2 jrt2t i2t jrt2irtSOLUTIONs8t 22jt3 k3t 2 k6t k9t 4Therefore, Equation 9 gives the tangential component asaTrtrtrtrtrt8 t 18 t 3s8 t 2 9 t 45E-14(pp 904-913)9121/18/0612:33 PMPage 912CHAPTER 14 VECTOR FUNCTIONSrtSinceijk2 t 2 t 3t 22 2 6trt6t 2 i6t 2 jEquation 10 gives the normal component asaNrtrtrt6 s2 t 2s8 t 2 9 t 4Kepler s Laws of Planetary MotionWe now describe one of the great accomplishments of calculus by showing how the material of this chapter can be used to prove Keplers laws of planetary motion. After 20 yearsof studying the astronomical observations of the Danish astronomer Tycho Brahe, theGerman mathematician and astronomer Johannes Kepler (15711630) formulated the following three laws.Keplers Laws1. A planet revolves around the Sun in an elliptical orbit with the Sun at one focus.2. The line joining the Sun to a planet sweeps out equal areas in equal times.3. The square of the period of revolution of a planet is proportional to the cube ofthe length of the major axis of its orbit.In his book Principia Mathematica of 1687, Sir Isaac Newton was able to show thatthese three laws are consequences of two of his own laws, the Second Law of Motion andthe Law of Universal Gravitation. In what follows we prove Keplers First Law. The remaining laws are left as exercises (with hints).Since the gravitational force of the Sun on a planet is so much larger than the forcesexerted by other celestial bodies, we can safely ignore all bodies in the universe except theSun and one planet revolving about it. We use a coordinate system with the Sun at the origin and we let r r t be the position vector of the planet. (Equally well, r could be theposition vector of the Moon or a satellite moving around the Earth or a comet movingaround a star.) The velocity vector is v r and the acceleration vector is a r . We usethe following laws of Newton:Second Law of Motion: FLaw of Gravitation:FmaGMmrr3GMmur2where F is the gravitational force on the planet, m and M are the masses of the planet andr , and u1 r r is the unit vector in thethe Sun, G is the gravitational constant, rdirection of r.We rst show that the planet moves in one plane. By equating the expressions for F inNewtons two laws, we nd thataGMrr35E-14(pp 904-913)1/18/0612:33 PMPage 913SECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATIONand so a is parallel to r. It follows that rto writedrdtva9130. We use Formula 5 in Theorem 14.2.3rvrvvvraThereforerv000hwhere h is a constant vector. (We may assume that h 0; that is, r and v are not parallel.)This means that the vector r r t is perpendicular to h for all values of t , so the planetalways lies in the plane through the origin perpendicular to h. Thus, the orbit of the planetis a plane curve.To prove Keplers First Law we rewrite the vector h as follows:hrvrruru2rrurur2 uururuurr uuThenaGMur2hr2uuGM u u uBut u uthat u uand soFIGURE 8vu(by Theorem 13.4.8, Property 6)1, it follows from Example 5 in Section 14.2GM uhvv11yrxhuhahGM uIntegrating both sides of this equation, we gethcu uuu 2 1 and, since u t0. ThereforeazGM uvuhGM ucwhere c is a constant vector.At this point it is convenient to choose the coordinate axes so that the standard basisvector k points in the direction of the vector h. Then the planet moves in the xy-plane.Since both v h and u are perpendicular to h, Equation 11 shows that c lies in thexy-plane. This means that we can choose the x- and y-axes so that the vector i lies in thedirection of c, as shown in Figure 8.If is the angle between c and r, then r, are polar coordinates of the planet. FromEquation 11 we havervhrGM uGMr u ucGM r urc cosrcGMrrc cos5E-14(pp 914-919)9141/18/0612:21 PMPage 914CHAPTER 14 VECTOR FUNCTIONSwhere cc . Thenrwhere evhvhe cosrvhhhh2h2h . SorWriting d1rGM 1c GM . Butrwhere hrvhGM c cosh 2 GM1 e coseh 2 c1 e cosh 2 c, we obtain the equationr121ede cosComparing with Theorem 11.6.6, we see that Equation 12 is the polar equation of a conicsection with focus at the origin and eccentricity e. We know that the orbit of a planet is aclosed curve and so the conic must be an ellipse.This completes the derivation of Keplers First Law. We will guide you through thederivation of the Second and Third Laws in the Applied Project on page 916. The proofsof these three laws show that the methods of this chapter provide a powerful tool fordescribing some of the laws of nature.|||| 14.4Exercises1. The table gives coordinates of a particle moving through spacealong a smooth curve.(a) Find the average velocities over the time intervals [0, 1],[0.5, 1], [1, 2], and [1, 1.5].(b) Estimate the velocity and speed of the particle at t 1.(d) Draw an approximation to the vector v(2) and estimate thespeed of the particle at t 2.yr(2.4)txy00.51.01.52.02.73.54.55.97.39.87.26.06.47.823.73.33.02.82.72. The gure shows the path of a particle that moves with positionvector r t at time t.(a) Draw a vector that represents the average velocity of theparticle over the time interval 2 t 2.4.(b) Draw a vector that represents the average velocity over thetime interval 1.5 t 2.(c) Write an expression for the velocity vector v(2).r(2)1zr(1.5)021x38 |||| Find the velocity, acceleration, and speed of a particle withthe given position function. Sketch the path of the particle anddraw the velocity and acceleration vectors for the specied valueof t.3. r tt24. r t25. r ttei1, t ,t1t, 4 st , tte j,t105E-14(pp 914-919)1/18/0612:21 PMPage 915S ECTION 14.4 MOTION IN SPACE: VELOCITY AND ACCELERATION6. r tsin t i2 cos t j,7. r tsin t itj; 8. r tt2 jtitt 3 k, ttthat can be used to hit a target 800 m away.028. A batter hits a baseball 3 ft above the ground toward the center1eld fence, which is 10 ft high and 400 ft from home plate. Theball leaves the bat with speed 115 ft s at an angle 50 abovethe horizontal. Is it a home run? (In other words, does the ballclear the fence?)914 |||| Find the velocity, acceleration, and speed of a particle withthe given position function.t29. r t10. r t1, t 3, t 2et js2 t i212. r tti13. r tte cos t i14. r tt sin t i1; 29. Water traveling along a straight portion of a river normallyows fastest in the middle, and the speed slows to almost zeroat the banks. Consider a long stretch of river owing north,with parallel banks 40 m apart. If the maximum water speedis 3 m s, we can use a quadratic function as a basic model forthe rate of water ow x units from the west bank:3x.fx400 x 40(a) A boat proceeds at a constant speed of 5 m s from a pointA on the west bank while maintaining a heading perpendicular to the bank. How far down the river on the oppositebank will the boat touch shore? Graph the path of the boat.(b) Suppose we would like to pilot the boat to land at the pointB on the east bank directly opposite A. If we maintain aconstant speed of 5 m s and a constant heading, nd theangle at which the boat should head. Then graph the actualpath the boat follows. Does the path seem realistic?2 cos t, 3 t, 2 sin t11. r te tkln t jtksin t jtkt2 kt cos t j1516 |||| Find the velocity and position vectors of a particle thathas the given acceleration and the given initial velocity andposition.15. a tk,1718i10 k,16. a tv0v0j,r0i0jk, r 02i3j30. Another reasonable model for the water speed of the river in||||Exercise 29 is a sine function: f x3 sin x 40 . If a boaterwould like to cross the river from A to B with constant headingand a constant speed of 5 m s, determine the angle at whichthe boat should head.(a) Find the position vector of a particle that has the given acceleration and the specied initial velocity and position.; (b) Use a computer to graph the path of the particle.17. a t18. a ti2jti2 t k,2tjv00,cos 2 t k,v0r0iikk,r0j19. The position function of a particle is given byrt91527. A gun has muzzle speed 150 m s. Find two angles of elevation6cos t k,t 2, 5 t, t 216 t . When is the speed a minimum?20. What force is required so that a particle of mass m has theposition function r tt3 it2 jt 3 k?21. A force with magnitude 20 N acts directly upward from thexy-plane on an object with mass 4 kg. The object starts at theorigin with initial velocity v 0i j. Find its position function and its speed at time t.22. Show that if a particle moves with constant speed, then thevelocity and acceleration vectors are orthogonal.3136 |||| Find the tangential and normal components of the acceleration vector.31. r t3tt3 i32. r t1ti33. r t3t 2 j2t jsin t jcos t it2tk234. r tti35. r tteis2 t je tk36. r tticos 2t jsin 2t ktj3t kgure to estimate the tangential and normal components of a.y24. Rework Exercise 23 if the projectile is red from a positiona200 m above the ground.25. A ball is thrown at an angle of 45 to the ground. If the balllands 90 m away, what was the initial speed of the ball?26. A gun is red with angle of elevation 30 . What is themuzzle speed if the maximum height of the shell is 500 m?37. The magnitude of the acceleration vector a is 10 cm s2. Use the23. A projectile is red with an initial speed of 500 m s and angleof elevation 30 . Find (a) the range of the projectile, (b) themaximum height reached, and (c) the speed at impact.0x5E-14(pp 914-919)9161/18/0612:21 PMPage 916CHAPTER 14 VECTOR FUNCTIONS38. If a particle with mass m moves with position vector r t , then40. A rocket burning its onboard fuel while moving through spaceits angular momentum is dened as L tmr tv t andits torque as tmr ta t . Show that L tt.Deduce that if t0 for all t, then L t is constant. (This isthe law of conservation of angular momentum.)has velocity v t and mass m t at time t. If the exhaust gasesescape with velocity ve relative to the rocket, it can be deducedfrom Newtons Second Law of Motion thatm39. The position function of a spaceship isrt3ti2ln t j74t21dvdtdmvedtm0ve.mt(b) For the rocket to accelerate in a straight line from rest totwice the speed of its own exhaust gases, what fraction ofits initial mass would the rocket have to burn as fuel?k(a) Show that v tand the coordinates of a space station are 6, 4, 9 . The captainwants the spaceship to coast into the space station. Whenshould the engines be turned off?v0lnAPPLIED PROJECTKeplers LawsJohannes Kepler stated the following three laws of planetary motion on the basis of masses ofdata on the positions of the planets at various times.Keplers Laws1. A planet revolves around the Sun in an elliptical orbit with the Sun at one focus.2. The line joining the Sun to a planet sweeps out equal areas in equal times.3. The square of the period of revolution of a planet is proportional to the cube of thelength of the major axis of its orbit.Kepler formulated these laws because they tted the astronomical data. He wasnt able to seewhy they were true or how they related to each other. But Sir Isaac Newton, in his PrincipiaMathematica of 1687, showed how to deduce Keplers three laws from two of Newtons ownlaws, the Second Law of Motion and the Law of Universal Gravitation. In Section 14.4 weproved Keplers First Law using the calculus of vector functions. In this project we guide youthrough the proofs of Keplers Second and Third Laws and explore some of their consequences.1. Use the following steps to prove Keplers Second Law. The notation is the same as inthe proof of the First Law in Section 14.4. In particular, use polar coordinates so thatrr cos ir sin j.(a) Show that hr2dk.dt(b) Deduce that r 2ddth.(c) If A A t is the area swept out by the radius vector ras in the gure, show thatyr (t)A(t)dAdtr (t)12r2r t in the time interval t0 , tddt(d) Deduce that0xdAdt12hconstantThis says that the rate at which A is swept out is constant and proves Keplers SecondLaw.5E-14(pp 914-919)1/18/0612:21 PMPage 917C HAPTER 14 REVIEW9172. Let T be the period of a planet about the Sun; that is, T is the time required for it to travelonce around its elliptical orbit. Suppose that the lengths of the major and minor axes of theellipse are 2a and 2b.(a) Use part (d) of Problem 1 to show that T2(b) Show thathGM2 a b h.2edb.a(c) Use parts (a) and (b) to show that T 2423a.GMThis proves Keplers Third Law. [Notice that the proportionality constant 4independent of the planet.]2GM is3. The period of the Earths orbit is approximately 365.25 days. Use this fact and KeplersThird Law to nd the length of the major axis of the Earths orbit. You will need the mass ofthe Sun, M 1.99 10 30 kg, and the gravitational constant, G 6.67 10 11 N m 2 kg2 .4. Its possible to place a satellite into orbit about the Earth so that it remains xed above agiven location on the equator. Compute the altitude that is needed for such a satellite. TheEarths mass is 5.98 10 24 kg; its radius is 6.37 10 6 m. (This orbit is called the ClarkeGeosynchronous Orbit after Arthur C. Clarke, who rst proposed the idea in 1948. The rstsuch satellite, Syncom II, was launched in July 1963.)||||14 ReviewCONCEPT CHECK1. What is a vector function? How do you nd its derivative andits integral?2. What is the connection between vector functions and spacecurves?6. (a) What is the denition of curvature?(b) Write a formula for curvature in terms of r t and T t .(c) Write a formula for curvature in terms of r t and r t .(d) Write a formula for the curvature of a plane curve withequation y f x .3. (a) What is a smooth curve?(b) How do you nd the tangent vector to a smooth curve at apoint? How do you nd the tangent line? The unit tangentvector?4. If u and v are differentiable vector functions, c is a scalar, andf is a real-valued function, write the rules for differentiatingthe following vector functions.(a) u t(b) c u t(c) f t u tvt(d) u t v t(e) u t(f) u f tvt7. (a) Write formulas for the unit normal and binormal vectors ofa smooth space curve r t .(b) What is the normal plane of a curve at a point? What is theosculating plane? What is the osculating circle?8. (a) How do you nd the velocity, speed, and acceleration of aparticle that moves along a space curve?(b) Write the acceleration in terms of its tangential and normalcomponents.5. How do you nd the length of a space curve given by avector function r t ?9. State Keplers Laws.5E-14(pp 914-919)9181/18/0612:21 PMPage 918CHAPTER 14 VECTOR FUNCTIONSTRUE-FALSE QUIZ6. If r t is a differentiable vector function, thenDetermine whether the statement is true or false. If it is true, explain why.If it is false, explain why or give an example that disproves the statement.t3 i1. The curve with vector equation r t2t 3 jdrtdt3t 3 kis a line.t, t 3, t 5 is smooth.3. The curve with vector equation r tcos t, t 2, t 4 is smooth.8. The binormal vector is B t10. Different parametrizations of the same curve result in identicalvttangent vectors at a given point on the curve.EXERCISES1. (a) Sketch the curve with vector functionticos t jsin t k2. Let r ttt , t , t , nd (a) the unit tangent vector, (b) the unit normal vector, and (c) the curvature.01.13. Find the curvature of the curve y16 and the plane xz5.x t 2, y t 4, z t 3 at the point 1, 1, 1 . Graph the curveand the tangent line on a common screen.t2 it cos t jsin t k, evaluate x01 r t dt.2 t 3, y 2 t 1,z ln t. Find (a) the point where C intersects the xz-plane,(b) parametric equations of the tangent line at 1, 1, 0 , and(c) an equation of the normal plane to C at 1, 1, 0 .6. Let C be the curve with equations x7. Use Simpsons Rule with n4 to estimate the length of thearc of the curve with equations x st, y 4 t, z t 2 1from 1, 4, 2 to 2, 1, 17 .8. Find the length of the curve r t0t4 sin t at thex 4 at the point 1, 1 .; 14. Find an equation of the osculating circle of the curve; 4. Find parametric equations for the tangent line to the curve5. If r t3 cos t, ypoints 3, 0 and 0, 4 .3. Find a vector function that represents the curve of intersectiony213123212. Find the curvature of the ellipse x1 t, ln ts2 t, e(a) Find the domain of r.(b) Find lim t l 0 r t .(c) Find r t .of the cylinder x 211. For the curve given by r t(b) Find r t and r t .tTt.tangent vector, normal vector, and curvature as C at thatpoint.5. If u t and v t are differentiable vector functions, thenrtNt9. The osculating circle of a curve C at a point has the sametiating each component function.utd T dt .curvature is4. The derivative of a vector function is obtained by differen-vtrt7. If T t is the unit tangent vector of a smooth curve, then the2. The curve with vector equation r tdutdty x 4 x 2 at the origin. Graph both the curve and its osculating circle.15. Find an equation of the osculating plane of the curvexsin 2 t, yt, zcos 2 t at the point 0, , 1 .16. The gure shows the curve C traced by a particle with positionvector r t at time t.(a) Draw a vector that represents the average velocity of theparticle over the time interval 3 t 3.2.(b) Write an expression for the velocity v(3).(c) Write an expression for the unit tangent vector T(3) anddraw it.y2 t 3 2, cos 2 t, sin 2 t ,1.C19. The helix r1 tcos t i sin t j t k intersects the curver2 t1 t i t 2 j t 3 k at the point 1, 0, 0 . Find theangle of intersection of these curves.e t i e t sin t j e t cos t kwith respect to arc length measured from the point 1, 0, 1 inthe direction of increasing t .r(3)r(3.2)10. Reparametrize the curve r t01x5E-14(pp 914-919)1/18/0612:21 PMPage 919C HAPTER 14 REVIEW17. A particle moves with position functionrt18. A particle starts at the origin with initial velocity ij 3 k.6 t k. Find its position12 t 2 j6t i20. Find the tangential and normal components of the accelerationvector of a particle with position functiont2 k2t j21. A disk of radius 1 is rotating in the counterclockwise directioncos t isin t j(a) Show that the velocity v of the particle isvcos t isin t jt vdwhere vd R t is the velocity of a point on the edge ofthe disk.(b) Show that the acceleration a of the particle isa2 vdt adwhere a d R t is the acceleration of a point on the rimof the disk. The extra term 2 vd is called the Coriolis acceleration; it is the result of the interaction of the rotation ofthe disk and the motion of the particle. One can obtain aphysical demonstration of this acceleration by walkingtoward the edge of a moving merry-go-round.1s1s2Fxat a constant angular speed . A particle starts at the center ofthe disk and moves toward the edge along a xed radius so thatits position at time t, t 0, is given by r tt R t , whereRte t sin t jroad tracks, its important to realize that the acceleration of thetrain should be continuous so that the reactive force exerted bythe train on the track is also continuous. Because of the formulas for the components of acceleration in Section 14.4, this willbe the case if the curvature varies continuously.(a) A logical candidate for a transfer curve to join existingtracks given by y 1 for x 0 and y s2 x forx 1 s2 might be the function f xs1 x 2,, whose graph is the arc of the circle shown0 x 1 s2in the gure. It looks reasonable at rst glance. Show thatthe functionat an initial speed of 43 ft s. It leaves his hand 7 ft above theground.(a) Where is the shot 2 seconds later?(b) How high does the shot go?(c) Where does the shot land?tie t cos t i22. In designing transfer curves to connect sections of straight rail-19. An athlete throws a shot at an angle of 45 to the horizontalrt919(c) Determine the Coriolis acceleration of a particle that moveson a rotating disk according to the equationrtt ln t i t j e t k. Find the velocity, speed, andacceleration of the particle.Its acceleration is a tfunction.;if xif 0if xx2x0x 1 s21 s2is continuous and has continuous slope, but does not havecontinuous curvature. Therefore, f is not an appropriatetransfer curve.(b) Find a fth-degree polynomial to serve as a transfer curvebetween the following straight line segments: y 0 forx 0 and y x for x 1. Could this be done with afourth-degree polynomial? Use a graphing calculator orcomputer to sketch the graph of the connected functionand check to see that it looks like the one in the gure.yy1y=xy=F(x)y=0012x0transfer curve1x5E-14(pp 920-921)1/17/0611:42 AMPROBLEMSPLUSPage 9201. A particle P moves with constant angular speedaround a circle whose center is at the originand whose radius is R. The particle is said to be in uniform circular motion. Assume that themotion is counterclockwise and that the particle is at the point R, 0 when t 0. The positionvector at time t 0 is r tR cos t i R sin t j.(a) Find the velocity vector v and show that v r 0. Conclude that v is tangent to the circleand points in the direction of the motion.(b) Show that the speed v of the particle is the constant R. The period T of the particle isthe time required for one complete revolution. Conclude that2RvTyvvtrx2(c) Find the acceleration vector a. Show that it is proportional to r and that it points towardthe origin. An acceleration with this property is called a centripetal acceleration. Showthat the magnitude of the acceleration vector is aR 2.(d) Suppose that the particle has mass m. Show that the magnitude of the force F that isrequired to produce this motion, called a centripetal force, isFmvR22. A circular curve of radius R on a highway is banked at an angleFIGURE FOR PROBLEM 1FFIGURE FOR PROBLEM 2so that a car can safelytraverse the curve without skidding when there is no friction between the road and the tires.The loss of friction could occur, for example, if the road is covered with a lm of water or ice.The rated speed vR of the curve is the maximum speed that a car can attain without skidding.Suppose a car of mass m is traversing the curve at the rated speed vR. Two forces are acting onthe car: the vertical force, m t, due to the weight of the car, and a force F exerted by, andnormal to, the road. (See the gure.)The vertical component of F balances the weight of the car, so that F cosm t. Thehorizontal component of F produces a centripetal force on the car so that, by Newtons Second Law and part (d) of Problem 1,F sinmg2mvRR2(a) Show that vR Rt tan .(b) Find the rated speed of a circular curve with radius 400 ft that is banked at an angle of 12 .(c) Suppose the design engineers want to keep the banking at 12 , but wish to increase therated speed by 50%. What should the radius of the curve be?3. A projectile is red from the origin with angle of elevationand initial speed v0. Assumingthat air resistance is negligible and that the only force acting on the projectile is gravity, t,we showed in Example 5 in Section 14.4 that the position vector of the projectile isv0 cos t i[ v0 sin t 1 t t 2 ] j . We also showed that the maximum horizontalrt2distance of the projectile is achieved when45 and in this case the range is R v 2 t.0(a) At what angle should the projectile be red to achieve maximum height and what is themaximum height?(b) Fix the initial speed v0 and consider the parabola x 2 2 Ry R 2 0, whose graph isshown in the gure. Show that the projectile can hit any target inside or on the boundaryy_R9200yRx0Dx5E-14(pp 920-921)1/17/0611:42 AMPage 921of the region bounded by the parabola and the x-axis, and that it cant hit any target outside this region.(c) Suppose that the gun is elevated to an angle of inclination in order to aim at a target thatis suspended at a height h directly over a point D units downrange. The target is releasedat the instant the gun is red. Show that the projectile always hits the target, regardless ofthe value v0, provided the projectile does not hit the ground before D.y4. (a) A projectile is red from the origin down an inclined plane that makes an anglevaxFIGURE FOR PROBLEM 4with thehorizontal. The angle of elevation of the gun and the initial speed of the projectile areand v0, respectively. Find the position vector of the projectile and the parametric equationsof the path of the projectile as functions of the time t. (Ignore air resistance.)(b) Show that the angle of elevation that will maximize the downhill range is the anglehalfway between the plane and the vertical.(c) Suppose the projectile is red up an inclined plane whose angle of inclination is . Showthat, in order to maximize the (uphill) range, the projectile should be red in the directionhalfway between the plane and the vertical.(d) In a paper presented in 1686, Edmond Halley summarized the laws of gravity and projectilemotion and applied them to gunnery. One problem he posed involved ring a projectile tohit a target a distance R up an inclined plane. Show that the angle at which the projectileshould be red to hit the target but use the least amount of energy is the same as the anglein part (c). (Use the fact that the energy needed to re the projectile is proportional tothe square of the initial speed, so minimizing the energy is equivalent to minimizing theinitial speed.)5. A projectile of mass m is red from the origin at an angle of elevation . In addition to grav-ity, assume that air resistance provides a force that is proportional to the velocity and thatopposes the motion. Then, by Newtons Second Law, the total force acting on the projectilesatises the equation1d 2Rdt 2mmt jkdRdtwhere R is the position vector and k 0 is the constant of proportionality.(a) Show that Equation 1 can be integrated to obtain the equationkRmdRdtv0tt jdR0.dt(b) Multiply both sides of the equation in part (a) by e k m t and show that the left-hand side ofthe resulting equation is the derivative of the product e k m t R t . Then integrate to nd anexpression for the position vector R t .where v0v06. Find the curvature of the curve with parametric equationsxyt0sin ( 12)d2yyt0cos ( 12)d27. A ball rolls off a table with a speed of 2 ft s. The table is 3.5 ft high.3.5 ftFIGURE FOR PROBLEM 7(a) Determine the point at which the ball hits the oor and nd its speed at the instant ofimpact.(b) Find the angle between the path of the ball and the vertical line drawn through the pointof impact. (See the gure.)(c) Suppose the ball rebounds from the oor at the same angle with which it hits the oor, butloses 20% of its speed due to energy absorbed by the ball on impact. Where does the ballstrike the oor on the second bounce?8. A cable has radius r and length L and is wound around a spool with radius R without over-lapping. What is the shortest length along the spool that is covered by the cable?921...
View Full Document

Ask a homework question - tutors are online