14CALCULUSOFVECTOR- VALUEDFUNCTIONS14.1Vector-Valued Functions (ET Section 13.1)Preliminary Questions1. Which one of the following does not parametrize a line?(a) r1(t ) = 8 t, 2t, 3t (b) r2(t ) = t3i 7t3j +t3k(c) r3(t ) = _8 4t3, 2 +5t2, 9t3_SOLUTION(a) This is a parametrization of the line passing through the point (8, 0, 0) in the direction parallel to the vector 1, 2, 3,since:8 t, 2t, 3t= 8, 0, 0 +t 1, 2, 3(b) Using the parameter s = t3we get:_t3i 7t3j +t3k_ = s, 7s, s = s 1, 7, 1This is a parametrization of the line through the origin, with the direction vector v = 1, 7, 1.(c) The parametrization _8 4t3, 2 +5t2, 9t3_ does not parametrize a line. In particular, the points (8, 2, 0) (at t = 0),(4, 7, 9) (at t = 1), and (24, 22, 72) (at t = 2) are not colinear.2. What is the projection of r(t ) = t i +t4j +etk onto the xz-plane?SOLUTION The projection of the path onto the xz-plane is the curve traced by t i + etk = _t, 0, et_. This is the curvez = exin the xz-plane.3. Which projection of cos t, cos 2t, sin tis a circle?SOLUTION The parametric equations arex = cos t, y = cos 2t, z = sin tThe projection onto the xz-plane is cos t, 0, sin t . Since x2+ z2= cos2t +sin2t = 1, the projection is a circle in thexz-plane. The projection onto the xy-plane is traced by the curve cos t, cos 2t, 0. Therefore, x = cos t and y = cos 2t .We express y in terms of x:y = cos 2t = 2 cos2t 1 = 2x21The projection onto the xy-plane is a parabola. The projection onto the yz-plane is the curve 0, cos 2t, sin t . Hencey = cos 2t and z = sin t . We nd y as a function of z:y = cos 2t = 1 2 sin2t = 1 2z2The projection onto the yz-plane is again a parabola.4. What is the center of the circle with parametrizationr(t ) =(2 +cos t )i +2j +(3 sin t )k?SOLUTION The parametric equations arex = 2 +cos t, y = 2, z = 3 sin tTherefore, the curve is contained in the plane y = 2, and the following holds:(x +2)2+(z 3)2= cos2t +sin2t = 1We conclude that the curve r(t ) is the circle of radius 1 in the plane y = 2 centered at the point (2, 2, 3).242 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)5. How do the paths r1(t ) = cos t, sin tand r2(t ) = sin t, cos taround the unit circle differ?SOLUTION The two paths describe the unit circle. However, as t increases from 0 to 2, the point on the path sin t i +cos t j moves in a clockwise direction, whereas the point on the path cos t i +sin t j moves in a counterclockwise direction.6. Which three of the following vector-valued functions parametrize the same space curve?(a) (2 +cos t )i +9j +(3 sin t )k (b) (2 +cos t )i 9j +(3 sin t )k(c) (2 +cos 3t )i +9j +(3 sin 3t )k (d) (2 cos t )i +9j +(3 +sin t )k(e) (2 +cos t )i +9j +(3 +sin t )kSOLUTION All the curves except for (b) lie in the vertical plane y = 9. We identify each one of the curves (a), (c), (d)and (e).(a) The parametric equations are:x = 2 +cos t, y = 9, z = 3 sin tHence,(x +2)2+(z 3)2= (cos t )2+(sin t )2= 1This is the circle of radius 1 in the plane y = 9, centered at (2, 9, 3).(c) The parametric equations are:x = 2 +cos 3t, y = 9, z = 3 sin 3tHence,(x +2)2+(z 3)2= (cos 3t )2+(sin 3t )2= 1This is the circle of radius 1 in the plane y = 9, centered at (2, 9, 3).(d) In this curve we have:x = 2 cos t, y = 9, z = 3 +sin tHence,(x +2)2+(z 3)2= (cos t )2+(sin t )2= 1Again, the circle of radius 1 in the plane y = 9, centered at (2, 9, 3).(e) In this parametrization we have:x = 2 +cos t, y = 9, z = 3 +sin tHence,(x 2)2+(z 3)2= (cos t )2+(sin t )2= 1This is the circle of radius 1 in the plane y = 9, centered at (2, 9, 3).We conclude that (a), (c) and (d) parametrize the same circle whereas (b) and (e) are different curves.Exercises1. What is the domain of r(t ) = eti + 1t j +(t +1)3k?SOLUTION r(t ) is dened for t = 0 and t = 1, hence the domain of r(t ) is:D = {t R : t = 0, t = 1}What is the domain of r(s) = esi +sj +cos sk?3. Find a vector parametrization of the line through P =(3, 5, 7) in the direction v = 3, 0, 1.SOLUTION We use the vector parametrization of the line to obtain:r(t ) = OP +t v = 3, 5, 7 +t 3, 0, 1 = 3 +3t, 5, 7 +t or in the form:r(t ) =(3 +3t )i 5j +(7 +t )kFind a direction vector for the line with parametrization r(t ) = (4 t )i +(2 +5t )j +12t k.S E C T I O N 14.1 Vector-Valued Functions (ET Section 13.1) 2435. Match the space curves in Figure 8 with their projections onto the xy-plane in Figure 9.(A) (B) (C)yxzyxzyxzFIGURE8(i)xy(ii)xy(iii)xyFIGURE9SOLUTION The projection of curve (C) onto the xy-plane is neither a segment nor a periodic wave. Hence, the correctprojection is (iii), rather than the two other graphs. The projection of curve (A) onto the xy-plane is a vertical line, hencethe corresponding projection is (ii). The projection of curve (B) onto the xy-plane is a periodic wave as illustrated in (i).Match the space curves in Figure 8 with the following vector-valued functions:(a) r1(t ) = cos 2t, cos t, sin t (b) r2(t ) = t, cos 2t, sin 2t (c) r3(t ) = 1, t, t 7. Match the vector-valued functions (a)(f) with the space curves (i)(vi) in Figure 10.(a) r(t ) = _t +15, e0.08tcos t, e0.08tsin t_(b) r(t ) = _cos t, sin t, sin 12t_(c) r(t ) =_t, t,25t1 +t2_(d) r(t ) = _cos3t, sin3t, sin 2t_(e) r(t ) = _t, t2, 2t_(f) r(t ) = _cos t, sin t, cos t sin 12t_y(i) (ii) (iii)(iv) (v) (vi)xzyxzyxzyyxxzzyxzFIGURE10SOLUTION(a) (v) (b) (i) (c) (ii)(d) (vi) (e) (iv) (f) (iii)Which of the following curves have the same projection onto the xy-plane?(a) r1(t ) = _t, t2, et_(b) r2(t ) = _et, t2, t_(c) r3(t ) = _t, t2, cos t_9. Match the space curves (A)(C) in Figure 11 with their projections (i)(iii) onto the xy-plane.yyxx(A) (B) (C)(i) (iii) (ii)zyxzyxzzyxzyxzFIGURE11244 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)SOLUTION Observing the curves and the projections onto the xy-plane we conclude that: Projection (i) corresponds tocurve (C); Projection (ii) corresponds to curve (A); Projection (iii) corresponds to curve (B).In Exercises 1013, the function r(t ) traces a circle. Determine the radius, center, and plane containing the circle.r(t ) = (9 cos t )i +(9 sin t )j11. r(t ) = 7i +(12 cos t )j +(12 sin t )kSOLUTION We have:x(t ) = 7, y(t ) = 12 cos t, z(t ) = 12 sin tHence,y(t )2+ z(t )2= 144 cos2t +144 sin2t = 144_cos2t +sin2t_ = 144This is the equation of a circle in the vertical plane x = 7. The circle is centered at the point(7, 0, 0) and its radius is144 = 12.r(t ) = sin t, 0, 4 +cos t 13. r(t ) = 6 +3 sin t, 9, 4 +3 cos t SOLUTION Since y(t ) = 9 the curve is contained in the vertical plane y = 9. By the given equations, x(t ) = 6 +3 sin tand z = 4 +3 cos t , hence:_x 63_2+_z 43_2= sin2t +cos2t = 1We conclude that the function traces a circle in the vertical plane y = 9, centered at the point (6, 9, 4) and with radius 3.Describe the projections of the circle r(t ) = sin t, 0, 4 +cos tonto the coordinate planes.15. Do either of P = (4, 11, 20) or Q = (1, 6, 16) lie on the curve r(t ) = _1 +t, 2 +t2, t4_?SOLUTION The point P =(4, 11, 20) lies on the curve r(t ) = _1 + t, 2 + t2, t4_ if there exists a value of t such thatOP = r(t ). That is,_4, 11, 20_ = _1 +t, 2 +t2, t4_Equating like components we get:1 +t = 42 +t2= 11t4= 20The rst equation implies that t = 3, but this value does not satisfy the third equation. We conclude that P does not lieon the curve. The point Q = (1, 6, 16) lies on the curve if there exists a value of t such that:1, 6, 16 = _1 +t, 2 +t2, t4_or equivalently:1 +t = 12 +t2= 6t4= 16These equations have the solution t = 2, hence Q = (1, 6, 16) lies on the curve.(a) Describe the curve r(t ) = t cos t, t sin t, tand its projections onto the xy- and xz-planes.(b) Plot r(t ) with a computer algebra system if you have one.17. Find the points where the path r(t ) = sin t, cos t, sin t cos 2tintersects the xy-plane.SOLUTION The curve intersects the xy-plane at the points where z = 0. That is, sin t cos 2t = 0 and so either sin t = 0or cos 2t = 0. The solutions are, thus:t = k or t =4 + k2, k = 0, 1, 2, . . .The values t = k yield the points: (sin k, cos k, 0) =_0, (1)k, 0_. The values t =4 +k2yield the points:k = 0 :_sin 4, cos 4, 0_ =_12,12, 0_k = 1 :_sin 34, cos 34, 0_ =_12, 12, 0_k = 2 :_sin 54, cos 54, 0_ =_12, 12, 0_S E C T I O N 14.1 Vector-Valued Functions (ET Section 13.1) 245k = 3 :_sin 74, cos 74, 0_ =_12,12, 0_(Other values of k do not provide new points). We conclude that the curve intersects the xy-plane at the following points:(0, 1, 0), (0, 1, 0),_12,12, 0_,_12, 12, 0_,_12, 12, 0_,_12,12, 0_Parametrize the intersection of the surfacesy2 z2= x 2, y2+ z2= 9using t = y as the parameter (two vector functions are needed as in Example 2).19. Find a parametrization of the curve in Exercise 18 using trigonometric functions.SOLUTION The curve in Exercise 18 is the intersection of the surfaces y2 z2=x 2, y2+ z2=9. The circley2+ z2=9 is parametrized byy=3 cos t , z=3 sin t . Substituting in the rst equation and using the identitycos2t sin2t = cos 2t , gives:x = 2 + y2 z2= 2 +(3 cos t )2(3 sin t )2= 2 +9_cos2t sin2t_ = 2 +9 cos 2tWe obtain the following trigonometric parametrization:r(t ) = 2 +9 cos 2t, 3 cos t, 3 sin t Vivianis Curve C is the intersection of the surfaces x2+ y2= z2, y = z2(Figure 12).(a) Parametrize each of the two parts of C corresponding to x 0 and x 0 taking t = z as parameter.(b) Describe the projection of C onto the xy-plane.(c) Show that C lies on the sphere of radius 1 with center(0, 1, 0). This curve looks like a gure eight lying on asphere [Figure 12(B)].21. Show that any point on x2+ y2= z2can be written in the form(z cos , z sin , z) for some . Use this to nd aparametrization of Vivianis curve (Exercise 20) with as parameter.SOLUTION We rst verify that x = z cos , y = z sin , and z = z satisfy the equation of the surface:x2+ y2= z2cos2 + z2sin2 = z2_cos2 +sin2_ = z2We now show that if(x, y, z) satises x2+ y2= z2, then there exists a value of such that x = z cos , y = z sin .Since x2+ y2= z2, we have |x| |z| and |y| |z|. If z = 0, then also x = y = 0 and any value of is adequate. Ifz = 0 then xz 1 and yz 1, hence there exists 0 such thatxz = cos 0. Hence,yz = _z2 x2z2= _1 _xz_2= _1 cos20 = sin 0Ifxzandyzare both positive, we choose 0such that 0 0 andyz 0 we choose 0 suchthat2< 0< . In either case we can represent the points on the surface as required. Vivianis curve is the intersectionof the surfaces x2+ y2= z2and x = z2. The points on these surfaces are of the form:x2+ y2= z2: (z cos , z sin , z)x = z2: (z2, y, z)(1)The points (x, y, z) on the intersection curve must satisfy the following equations:_z2= z cos y = z sin The rst equation implies that z = 0 or z = cos . The second equation implies that y = 0 or y = cos sin =12 sin 2.The x coordinate is obtained by substituting z = cos in x = z cos (or in x = z2). That is, x = cos2. We obtain thefollowing vector parametrization of the curve:r(t ) =_cos2,12 sin 2, cos _Use sine and cosine to parametrize the intersection of the cylinders x2+ y2=1 and x2+ z2=1 (use twovector-valued functions). Then describe the projections of this curve on the three coordinate planes.23. Use sine and cosine to parametrize the intersection of the surfaces x2+ y2= 1 and z = 4x2, and plot thiscurve using a CAS (Figure 13).xyzFIGURE13Intersection of the surfaces x2+ y2= 1 and z = 4x2.246 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)SOLUTION The points on the cylinder x2+ y2= 1 and on the parabolic cylinder z = 4x2can be written in the form:x2+ y2= 1: (cos t, sin t, z)z = 4x2: (x, y, 4x2)The points (x, y, z) on the intersection curve must satisfy the following equations:x = cos ty = sin tz = 4x2 x = cos t, y = sin t,z = 4 cos2tWe obtain the vector parametrization:r(t ) = _cos t, sin t, 4 cos2t_Using the CAS we obtain the following curve:z2xy1 124211r(t ) = _cos t, sin t, 4 cos2t_Use hyperbolic functions to parametrize the intersection of the surfaces x2 y2= 4, z = xy.In Exercises 2534, nd a parametrization of the curve.25. The vertical line passing through the point (3, 2, 0)SOLUTION The points of the vertical line passing through the point (3, 2, 0) can be written as (3, 2, z). Using z = t asparameter we get the following parametrization:r(t ) = 3, 2, t, < t < The line passing through (1, 0, 4) and (4, 1, 2)27. The line through the origin whose projection on the xy-plane is a line of slope 3 and on the yz-plane is a line ofslope 5 (i.e., y/z = 5)SOLUTION We denote by (x, y, z) the points on the line. The projection of the line on the xy-plane is the line throughthe origin having slope 3, that is the line y = 3x in the xy-plane. The projection of the line on the yz-plane is the linethrough the origin with slope 5, that is the line z = 5y. Thus, the points on the desired line satisfy the following equalities:y = 3xz = 5y y = 3x,z = 5 3x = 15xWe conclude that the points on the line are all the points in the form (x, 3x, 15x). Using x = t as parameter we obtainthe following parametrization:r(t ) = t, 3t, 15t.The horizontal circle of radius 1 with center (2, 1, 4)29. The circle of radius 2 with center (1, 2, 5) in a plane parallel to the yz-planeSOLUTION The circle is parallel to theyz-plane and centered at (1, 2, 5), hence the x-coordinates of the points onthe circle are x = 1. The projection of the circle on the yz-plane is a circle of radius 2 centered at(2, 5). This circle isparametrized by:y = 2 +2 cos t, z = 5 +2 sin tWe conclude that the points on the required circle can be written as (1, 2 +2 cos t, 5 +2 sin t ). This gives the followingparametrization:r(t ) = 1, 2 +2 cos t, 5 +2 sin t.The ellipse_x2_2+_y3_2= 1 in the xy-plane, translated to have center (9, 4, 0)31. The intersection of the plane y =12 with the sphere x2+ y2+ z2= 1S E C T I O N 14.1 Vector-Valued Functions (ET Section 13.1) 247SOLUTION Substituting y =12 in the equation of the sphere gives:x2+_12_2+ z2= 1 x2+ z2=34This circle in the horizontal plane y =12has the parametrization x =32cos t , z =32sin t . Therefore, the points onthe intersection of the plane y =12 and the sphere x2+ y2+z2= 1, can be written in the form_32cos t, 12,32sin t_,yielding the following parametrization:r(t ) =_32cos t,12,32sin t_.The intersection of the surfacesz = x2 y2and z = x2+ xy 133. The ellipse_x2_2+_z3_2= 1 in the xz-plane, translated to have center (3, 1, 5) [Figure 14(A)](A)31(B)yxz zyx31FIGURE14The ellipses described in Exercise 33 and 34.SOLUTION The translated ellipse is in the vertical plane y = 1, hence the y-coordinate of the points on this ellipse isy = 1. The x and z coordinates satisfy the equation of the ellipse:_x 32_2+_z 53_2= 1.This ellipse is parametrized by the following equations:x = 3 +2 cos t, z = 5 +3 sin t.Therefore, the points on the translated ellipse can be written as(3 +2 cos t, 1, 5 +3 sin t ). This gives the followingparametrization:r(t ) = 3 +2 cos t, 1, 5 +3 sin t.The ellipse_y2_2+_z3_2= 1, translated to have center (3, 1, 5) [Figure 14(B)]In Exercises 3537, assume that two paths r1(t ) and r2(t ) intersect if there is a point P lying on both curves. We say thatr1(t ) and r2(t ) collide if r1(t0) = r2(t0) at some time t0.35. Which of the following are true?(a) If r1 and r2 intersect, then they collide.(b) If r1 and r2 collide, then they intersect.(c) Intersectiondependsonlyontheunderlyingcurvestracedbyr1andr2but collisiondependsontheactualparametrizations.SOLUTION(a) This statement is wrong.r1(t ) andr2(t ) may intersect but the point of intersection may correspond to differentvalues of the parameters in the two curves, as illustrated in the following example:r1(t ) = cos t, sin t (the unit circle)r2(s) = s, 1(the horizontal line y = 1)x(0, 1)y248 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)The point of intersection (0, 1) corresponds to t =2and s = 0.(b) This statement is true. If r1(t0) = r2(t0), then the head of the vector r1(t0) (or r2(t0)) is a point of intersection ofthe two curves.(c) The statement is true. Intersection is a geometric property of the curves and it is independent of the parametrizationwe choose for the curves. Collision depends on the actual parametrization. Notice that if we parametrize the line y = 1in the example given in part (a) by r3(s) = _s 2, 1_, then r1_2_ = r3_2_ hence the two paths collide.Determine whether r1 and r2 collide or intersect:r1(t ) = _t2+3, t +1, 6t1_r2(t ) = _4t, 2t 2, t27_37. Determine whether r1 and r2 collide or intersect:r1(t ) = _t, t2, t3_, r2(t ) = _4t +6, 4t2, 7 t_SOLUTION The two paths collide if there exists a value of t such that:_t, t2, t3_ = _4t +6, 4t2, 7 t_Equating corresponding components we obtain the following equations:t = 4t +6t2= 4t2t3= 7 tThe second equation implies that t = 0, but this value does not satisfy the other equations. Therefore, the equations haveno solution, which means that the paths do not collide. The two paths intersect if there exist values of t and s such that:_t, t2, t3_ = _4s +6, 4s2, 7 s_Or equivalently:t = 4s +6t2= 4s2(1)t3= 7 sThe second equation implies that t1 = 2s or t2 = 2s. Substituting t1 = 2s and t2 = 2s in the rst equation gives:t1 = 2s : 2s = 4s +6 2s = 6 s1 = 3t2 = 2s : 2s = 4s +6 6s = 6 s2 = 1The solutions of the rst two equations are thus(t1, s1) = (6, 3); (t2, s2) =(2, 1)(t1, s1) does not satisfy the third equation whereas(t2, s2) does. We conclude that the equations in (1) have a solutiont = 2, s = 1, hence the two paths intersect.Further Insights and ChallengesSketch the curve parametrized by r(t ) = |t | +t, |t | t .39. Find the maximum height above the xy-plane of a point on r(t ) = _et, sin t, t (4 t )_.SOLUTION The height of a point is the value of the z-coordinate of the point. Therefore we need to maximize thefunction z =t (4 t ). z(t ) is a quadratic function having the roots t=0 and t=4, hence the maximum value isobtained at the midpoint of the interval 0 t 4, that is at t = 2. The corresponding value of z is:z max = z(2) = 2 (4 2) = 4The point of maximum height is, thus,(e2, sin 2, 4) (7.39, 0.91, 4)Let C be the curve obtained by intersecting a cylinder of radius rand a plane. Insert two spheres of radius rinto the cylinder above and below the plane, and let F1 and F2 be the points where the plane is tangent to the sphere[Figure 15(A)]. Let Kbe the vertical distance between the equators of the two spheres. Rediscover Archimedessproof that C is an ellipse by showing that every point P on C satisesPF1 + PF2 = KHint: If two lines through a point P are tangent to a sphere and intersect the sphere at Q1 and Q2 as in Figure 15(B),then the segments PQ1 and PQ2 have equal length. Use this to show that PF1 = PR1 and PF2 = PR2.41. Now reprove the result of Exercise 40 using vector geometry. Assume that the cylinder has equation x2+y2= r2and the plane has equation z = ax +by.(a) Show that the upper and lower spheres in Figure 15 have centersC1 =_0, 0, r_a2+b2+1_C2 =_0, 0, r_a2+b2+1_S E C T I O N 14.2 Calculus of Vector-Valued Functions (ET Section 13.2) 249(b) Show that the points where the plane is tangent to the sphere areF1 =r_a2+b2+1_a, b, a2+b2_F2 =r_a2+b2+1_a, b, a2+b2_Hint: Show that C1F1 and C2F2 have length r and are orthogonal to the plane.(c) Verify, with the aid of a computer algebra system, that Eq. (2) holds with K = 2r_a2+b2+1. To simplify thealgebra, observe that since a and b are arbitrary, it sufces to verify Eq. (2) for the point P = (r, 0, ar).SOLUTION(a)and(b) SinceF1is the tangency point of the sphere and the plane, the radius toF1is orthogonal to the plane.Therefore to show that the center of the sphere is at C1 and the tangency point is the given point we must show that:

C1F1 = r (1)C1F1 is orthogonal to the plane. (2)We compute the vectorC1F1:C1F1 =_ra_a2+b2+1,rb_a2+b2+1,r(a2+b2)_a2+b2+1r_a2+b2+1_ =r_a2+b2+1a, b, 1Hence,

C1F1 =r_a2+b2+1 a, b, 1=r_a2+b2+1_a2+b2+(1)2= rWe, thus, proved that (1) is satised. To show (2) we must show that C1F1 is parallel to the normal vector a, b, 1 tothe plane z = ax + by (i.e., ax + by z = 0). The two vectors are parallel since by (1) C1F1 is a constant multiple ofa, b, 1. In a similar manner one can show (1) and (2) for the vectorC2F2.(c) This is an extremely challenging problem. As suggested in the book, we useP =(r, 0, ar), and we also use theexpressions for F1 and F2 as given above. This gives us:PF1 =__1 +2 a2+b22 a_1 +a2+b2_ r2PF2 =__1 +2 a2+b2+2 a_1 +a2+b2_ r2Their sum is not very inspiring:PF1 + PF2 =__1 +2 a2+b22 a_1 +a2+b2_ r2+__1 +2 a2+b2+2 a_1 +a2+b2_ r2Let us look, instead, at(PF1 + PF2)2, and show that this is equal to K2. Since everything is positive, this will implythat PF1 + PF2 = K, as desired.(PF1 + PF2)2= 2 r2+4 a2r2+2 b2r2+2_r4+2b2r4+b4r4= 2 r2+4 a2r2+2 b2r2+2 (1 +b2)r2= 4r2(1 +a2+b2) = K214.2Calculus of Vector-Valued Functions (ET Section 13.2)Preliminary Questions1. State the three forms of the Product Rule for vector-valued functions.SOLUTION The Product Rule for scalar multiplef (t ) of a vector-valued function r(t ) states that:ddtf (t )r(t ) =f (t )r

(t ) +f

(t )r(t )The Product Rule for dot products states that:ddt r1(t ) r2(t ) = r1(t ) r

2(t ) +r

1(t ) r2(t )250 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)Finally, the Product Rule for cross product isddt r1(t ) r2(t ) = r1(t ) r

2(t ) +r

1(t ) r2(t ).In Questions 26, indicate whether true or false and if false, provide a correct statement.2. The derivative of a vector-valued function is dened as the limit of the difference quotient, just as in the scalar-valuedcase.SOLUTION The statement is true. The derivative of a vector-valued function r(t ) is dened a limit of the differencequotient:r

(t ) =limt 0r (t +h) r(t )hin the same way as in the scalar-valued case.3. There are two Chain Rules for vector-valued functions, one for the composite of two vector-valued functions andone for the composite of a vector-valued and scalar-valued function.SOLUTION This statement is false. A vector-valued function r(t ) is a function whose domain is a set of real numbersand whose range consists of position vectors. Therefore, if r1(t ) and r2(t ) are vector-valued functions, the composition(r1 r2)(t ) =r1(r2(t )) has no meaning since r2(t ) is a vector and not a real number. However, for a scalar-valuedfunctionf (t ), the composition r( f (t )) has a meaning, and there is a Chain Rule for differentiability of this vector-valuedfunction.4. The terms velocity vector and tangent vector for a path r(t ) mean one and the same thing.SOLUTION This statement is true.5. The derivative of a vector-valued function is the slope of the tangent line, just as in the scalar case.SOLUTION The statement is false. The derivative of a vector-valued function is again a vector-valued function, henceit cannot be the slope of the tangent line (which is a scalar). However, the derivative, r

(t0) is the direction vector of thetangent line to the curve traced by r(t ), at r(t0).6. The derivative of the cross product is the cross product of the derivatives.SOLUTION The statement is false, since usually,ddt r1(t ) r2(t ) = r

1(t ) r

2(t )The correct statement is the Product Rule for Cross Products. That is,ddt r1(t ) r2(t ) = r1(t ) r

2(t ) +r

1(t ) r2(t )7. State whether the following derivatives of vector-valued functions r1(t ) and r2(t ) are scalars or vectors:(a)ddt r1(t ) (b)ddt_r1(t ) r2(t )_(c)ddt_r1(t ) r2(t )_SOLUTION (a) vector, (b) scalar, (c) vector.ExercisesIn Exercises 14, evaluate the limit.1. limt 3_t2, 4t,1t_SOLUTION By the theorem on vector-valued limits we have:limt 3_t2, 4t,1t_ =_limt 3t2, limt 34t, limt 31t_ =_9, 12,13_.limt sin 2t i +cos t j +tan 4t k3. limt 0e2ti +ln(t +1)j +4kS E C T I O N 14.2 Calculus of Vector-Valued Functions (ET Section 13.2) 251SOLUTION Computing the limit of each component, we obtain:limt 0_e2ti +ln (t +1) j +4k_ =_limt 0e2t_i +_limt 0ln(t +1)_j +_limt 04_k = e0i +(ln 1)j +4k = i +4klimt 0_1t +1,et1t, 4t_5. Evaluatelimh0r(t +h) r(t )hfor r(t ) =_t1, sin t, 4_.SOLUTION This limit is the derivativedrdt . Using componentwise differentiation yields:limh0r (t +h) r(t )h=drdt =_ ddt_t1_,ddt(sin t ) ,ddt (4)_ =_ 1t2 , cos t, 0_.Evaluatelimt 0r(t )tfor r(t ) = sin t, 1 cos t, 2t .In Exercises 714, compute the derivative.7. r(t ) = _t, t2, t3_SOLUTION Using componentwise differentiation we get:drdt =_ ddt (t ),ddt (t2),ddt (t3)_ =_1, 2t, 3t2_v(t ) = _sin 3t, cos 3t_9. w(s) = _es, e2s_SOLUTION Componentwise differentiation gives:w

(s) = __es_

,_e2s_

_ = _es, 2e2s_r() = _tan , 4 2, sin _11. r(t ) = _t t1, 4t2, 8_SOLUTION We compute the derivative of each component to obtain:r

(t ) = _(t t1)

, (4t2)

, (8)

_ = _1 +t2, 8t, 0_c(t ) = t1i e2tk13. a() = (cos 2)i +(sin 2)j +(sin 4)kSOLUTION Using componentwise differentiation yields:a

() = (cos 2)

i +(sin 2)

j +(sin 4)

k = (2 sin 2) i +(2 cos 2) j +(4 cos 4) kb(t ) = _e4t 3, sin(t2), (4t +3)1_ 15. Calculate r

(t ) and r

(t ) for r(t ) = _t, t2, t3_.SOLUTION We perform the differentiation componentwise to obtain:r

(t ) = _(t )

, (t2)

, (t3)

_ = _1, 2t, 3t2_We now differentiate the derivative vector to nd the second derivative:r

(t ) =ddt_1, 2t, 3t2_ = 0, 2, 6t.Sketch the curve r(t ) = _1 t2, t_ for 1 t 1. Compute the tangent vector at t = 1 and add it to the sketch.17. Sketch the curve r1(t ) = _t, t2_ together with its tangent vector at t = 1. Then do the same for r2(t ) = _t3, t6_.SOLUTION Note that r1

(t ) = 1, 2tand so r1

(1) = 1, 2. The graph of r1(t ) satises y = x2. Likewise, r2

(t ) =_3t2, 6t5_ and so r2

(1) = 3, 6. The graph of r2(t ) also satises y =x2. Both graphs and tangent vectors are givenhere.2r2(t )1r1(t )Sketch the cycloid r(t ) = _t sin t, 1 cos t_ together with its tangent vectors at t =3and34 .In Exercises 1922, use the appropriate Product Rule to evaluate the derivative, wherer1(t ) = _8t, 4, t3_, r2(t ) = _0, et, 6_252 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)19.ddt_r1(t ) r2(t )_SOLUTION By the Product Rule for dot products we have:ddt r1 r2 = r1 r2

+r1

r2We compute the derivatives of r1 and r2:r

1 =ddt_8t, 4, t3_ = _8, 0, 3t2_r

2 =ddt_0, et, 6_ = _0, et, 0_By (1) we have:ddt r1(t ) r2(t ) = _8t, 4, t3__0, et, 0_+_8, 0, 3t2__0, et, 6_ = 4et+18t2ddt_t4r1(t )_ 21.ddt_r1(t ) r2(t )_SOLUTION We use the Product Rule for cross products:ddt r1 r2 = r1 r

2 +r

1 r2 = _8t, 4, t3__0, et, 0_+_8, 0, 3t2__0, et, 6_=i j k8t 4 t30 et0+i j k8 0 3t20 et6= t3eti +8t etk +3t2eti +48j +8etk= t2et(t +3)i +48j +8et(t +1)k = _t2et(t +3), 48, 8et(t +1)_ddt_r1(t ) r3(t )_t =5, assuming that r3(5) = 3, 1, 2 and r

3(5) = 1, 2, 7.In Exercises 2325, letr1(t ) = _t2, t3, 4t_, r2(t ) = _t1, 1 +t, 2_23. Let F(t ) = r1(t ) r2(t ).(a) Calculate F

(t ) using the Product Rule.(b) Expand the product r1(t ) r2(t ) and differentiate. Compare with part (a).SOLUTION(a) By the Product Rule for dot products we have:F

(t ) = r1(t ) r

2(t ) +r

1(t ) r2(t )We compute the derivatives of r1(t ) and r2(t ):r

1(t ) =ddt_t2, t3, 4t_ = _2t, 3t2, 4_r

2(t ) =ddt_t1, 1 +t, 2_ = _t2, 1, 0_Thus,F

(t ) = _t2, t3, 4t__t2, 1, 0_+_2t, 3t2, 4__t1, 1 +t, 2_= _1 +t3+0_+_2 +3t2+3t3+8_ = 4t3+3t2+9That is,F

(t ) = 4t3+3t2+9 (1)(b) We now rst compute the product r1(t ) r2(t ) and then differentiate the resulting function. This gives:F(t ) = r1(t ) r2(t ) = _t2, t3, 4t__t1, 1 +t, 2_ = t +t3(1 +t ) +8t = t4+t3+9tDifferentiating F(t ) gives:F

(t ) = 4t3+3t2+9 (2)The derivatives in (1) and (2) are the same, as expected.S E C T I O N 14.2 Calculus of Vector-Valued Functions (ET Section 13.2) 253Let G(t ) = r1(t ) r2(t ).(a) Calculate G

(t ) using the Product Rule.(b) Expand the cross product r1(t ) r2(t ) and differentiate. Compare with part (a).25. Find the rate of change of the angle between r1(t ) and r2(t ) at t = 2, assuming that t is measured in seconds.SOLUTION Recall the formula for the dot product:r1(t ) r2(t ) = r1(t )r2(t ) cos Thus,cos =r1(t ) r2(t )r1(t )r2(t ) =t +t3+t4+8t_t4+t6+16t2_t2+5 +2t +t2=t4+t3+9t_t2+t4+16_1 +5t2+2t3+t4= (t4+t3+9t )_(t2+t4+16)(1 +5t2+2t3+t4)_1/2Taking the derivative, we nd that (after a lot of work)ddt cos =144 +48 t280 t336 t4+264 t56 t624 t8+5 t9__16 +t2+t4_ _1 +5 t2+2 t3+t4__ 32So, at t = 2, we get (after a lot of work)ddt cos t =2=5015953On the other hand, using the chain rule,ddt cos = sin ddt So, we have thatddt =1sin ddt cos Thus, at t = 2,ddt t =2=1sin ddt cos t =2We need only calculate sin at t = 2. From above, we know thatcos = (t4+t3+9t )_(t2+t4+16)(1 +5t2+2t3+t4)_1/2so at t = 2, cos =753. Since sin2 +cos2 = 1, we get that sin =253. We can now conclude thatddt t =2= 5325015953 = 25159In Exercises 2629, evaluateddt r(g(t )) using the Chain Rule.r(t ) = _t2, 2t, 4_, g(t ) = et27. r(t ) = _et, e2t, 4_, g(t ) = 4t +9SOLUTION We rst differentiate the two functions:r

(t ) =ddt_et, e2t, 4_ = _et, 2e2t, 0_g

(t ) =ddt(4t +9) = 4Using the Chain Rule we get:ddt r (g(t )) = g

(t )r

(g(t )) = 4_e4t +9, 2e2(4t +9), 0_ = _4e4t +9, 8e8t +18, 0_r(t ) = 4 sin 2t, 6 cos 2t , g(t ) = t229. r(t ) = _3t, tan1t_, g(t ) = sin t254 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)SOLUTION We rst compute the derivatives of the two functions:r

(t ) =ddt_3t, tan1t_ =_3tln 3,11 +t2_g

(t ) = cos tWe now differentiate the composition function r(g(t )) using the Chain Rule:ddt r(g(t )) = g

(t )r

(g(t )) = cos t_3sin tln 3,11 +sin2t_ =_3sin tcos t ln 3,cos t1 +sin2t_Let v(s) = s2i +2sj +9s2k. Evaluateddsv(g(s)) at s = 4, assuming that g(4) = 3 and g

(4) = 9.31. Letr(t ) = _t2, 1 t, 4t_. Calculate the derivative ofr(t ) a(t ) at t =2, assuming thata(2) = 1, 3, 3 anda

(2) = 1, 4, 1.SOLUTION By the Product Rule for dot products we haveddt r(t ) a(t ) = r(t ) a

(t ) +r

(t ) a(t )At t = 2 we haveddt r(t ) a(t )t =2= r(2) a

(2) +r

(2) a(2) (1)We compute the derivative r

(2):r

(t ) =ddt_t2, 1 t, 4t_ = 2t, 1, 4 r

(2) = 4, 1, 4 (2)Also, r(2) = _22, 1 2, 4 2_ = 4, 1, 8. Substituting the vectors in the equation above, we obtain:ddt r(t ) a(t )t =2= 4, 1, 8 1, 4, 1 +4, 1, 4 1, 3, 3 = (4 4 +8) +(4 3 +12) = 13The derivative of r(t ) a(t ) at t = 2 is 13.Let r(t ) = _t2, t3, et_. Use Example 4 to calculateddt (r r

).In Exercises 3337, nd a parametrization of the tangent line at the point indicated.33. r(t ) = _1 t2, 5t, 2t3_, t = 2SOLUTION The tangent line is parametrized by:(t ) = r(2) +t r

(2) (1)We compute the vectors in the above parametrization:r(2) = _1 22, 5 2, 2 23_ = 3, 10, 16r

(t ) =ddt_1 t2, 5t, 2t3_ = _2t, 5, 6t2_ r

(2) = 4, 5, 24Substituting the vectors in (1) we obtain the following parametrization:(t ) = 3, 10, 16 +t 4, 5, 24 = 3 4t, 10 +5t, 16 +24t r(t ) = _cos 2t, sin 3t, sin 4t_, t =435. r(s) = 4s1i 8s3k, s = 2SOLUTION The tangent line has the following parametrization:(s) = r(2) +sr

(2) (1)We compute the vectors r(2) and r

(2):r(2) = 4 21i 8 23k = 2i kr

(s) =dds_4s1i 8s3k_ = 4s2i +24s4k r

(2) = 4 22i +24 24k = i + 32kS E C T I O N 14.2 Calculus of Vector-Valued Functions (ET Section 13.2) 255Substituting in (1) gives the following parametrization:(t ) = 2i k +s_i + 32k_ =(2 s) i +_32s 1_kor in scalar form:x = 2 s, y = 0, z =32s 1.r(t ) = _t2, t4_, t = 137. r(s) = ln si +s1j +9sk, s = 1SOLUTION The tangent line has the following parametrization:(s) = r(1) +sr

(1) (1)We compute the vectors r(1) and r

(1):r(1) = ln 1i +11j +9 1k = j +9kr

(s) =dds(ln si +s1j +9sk) =1s i s2j +9k r

(1) = i j +9kWe substitute the vectors in (1) to obtain the following parametrization:(s) = j +9k +s(i j +9k) = si +(1 s)j +(9 +9s)kor in scalar form:x = s, y = 1 s, z = 9 +9s.Let r(t ) = sin 2t cos t, sin 2t sin t, cos 2t . Show that r(t ) is constant and conclude using Example 6 that r(t )and r

(t ) are orthogonal. Then compute r

(t ) and verify directly that it is orthogonal to r(t ).39. Show, by nding a counterexample, that in general r

(t ) need not equal r(t )

.SOLUTION Let r(t ) = 1, 1, t . Then r(t ) =_12+12+t2=_2 +t2, hence:r(t )

=ddt__2 +t2_ =2t2_2 +t2 =t_2 +t2On the other hand, we have r

(t ) = 0, 0, 1, hence:r

(t ) =_02+02+12= 1.We see that r

(t ) = r(t )

.In Exercises 4045, evaluate the integrals._102t, 4t, cos 3tdt41._41_t1i +4t j 8t3/2k_dtSOLUTION We perform the integration componentwise. Computing the integral of each component we get:_41t1dt = ln t41= ln 4 ln 1 = ln 4_414t dt = 4 23t3/241=83_43/21_ =563_418t3/2dt = 165 t5/241= 165_45/21_ = 4965Hence,_41_t1i +4t j 8t3/2k_ dt = (ln 4) i + 563j 4965k_10_t et2, t ln(t2+1)_dt43._22_u3i +u5j +u7k_duSOLUTION We perform componentwise integration, but before doing so we notice that u3, u5and u7are all oddfunctions, so their integrals over this symmetric region will all be zero! Thus, the answer is 0i +0j +0k._10_11 +s2 ,s1 +s2_ ds256 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)45._t0_3si +6s2j +9k_dsSOLUTION We rst compute the integral of each component:_t03s ds =32s2t0=32t2_t06s2ds =63s3t0= 2t3_t09 ds = 9st0= 9tHence,_t0_3si +6s2j +9k_ dt =__t03s ds_i +__t06s2ds_j +__t09 ds_k =_32t2_i +_2t3_j +(9t )kIn Exercises 4653, nd the general solutionr(t ) of the differential equation and the solution with the given initialcondition.drdt = 1 2t, 4t , r(0) = 3, 147. r

(t ) = i j, r(0) = 2i +3kSOLUTION The general solution is obtained by integrating r

(t ):r(t ) =_(i j) dt =__1 dt_i __1 dt_j = t i t j +c (1)Hence,r(0) = 0i 0j +c = cThe solution with the initial condition r(0) = 2i +3k must satisfy:r(0) = c = 2i +3kSubstituting in (1) yields the solution:r(t ) = t i t j +2i +3k = (t +2) i t j +3kr

(t ) = t2i +5t j +k, r(0) = j +2k49. r

(t ) = sin 3t, sin 3t, t , r(0) = 0, 1, 8SOLUTION We rst integrate the vector r

(t ) to nd the general solution:r(t ) =_sin 3t, sin 3t, tdt =__sin 3t dt,_sin 3t dt,_t dt_ =_13 cos 3t, 13 cos 3t,12t2_+c (1)Substituting the initial condition we obtain:r(0) =_13 cos 0, 13 cos 0,12 02_+c = 0, 1, 8 =_13, 13, 0_+c = 0, 1, 8Hence,c = 0, 1, 8 _13, 13, 0_ =_13,43, 8_Substituting in (1) we obtain the solution:r(t ) =_13 cos 3t, 13 cos 3t,12t2_+_13,43, 8_ =_13(1 cos 3t ) ,13(4 cos 3t ) , 8 + 12t2_drdt = _e2t, et, e2t_, r(0) = 4, 2, 351. r

(t ) = 16k, r(0) = 1, 0, 0, r

(0) = 0, 1, 0SOLUTION To nd the general solution we rst nd r

(t ) by integrating r

(t ):r

(t ) =_r

(t ) dt =_16kdt =(16t ) k +c1(1)S E C T I O N 14.2 Calculus of Vector-Valued Functions (ET Section 13.2) 257We now integrate r

(t ) to nd the general solution r(t ):r(t ) =_r

(t ) dt =_((16t ) k +c1) dt =__16(t ) dt_k +c1t +c2 = (8t2)k +c1t +c2(2)We substitute the initial conditions in (1) and (2). This gives:r

(0) = c1 = 0, 1, 0 = jr(0) = 0k +c1 0 +c2 = 1, 0, 0 c2 = 1, 0, 0 = iCombining with (2) we obtain the following solution:r(t ) = (8t2)k +t j +i = i +t j +(8t2)kr

(t ) = 0, 0, 1, r(0) = 2, 1, 1, r

(0) = 3, 1, 153. r

(t ) = _et, sin t, cos t_, r(0) = 1, 0, 1, r

(0) = 0, 2, 2SOLUTION We perform integration componentwise on r

(t ) to obtain:r

(t ) =_ _et, sin t, cos t_ dt = _et, cos t, sin t_+c1(1)We now integrate r

(t ) to obtain the general solution:r(t ) =_ __et, cos t, sin t_+c1_dt = _et, sin t, cos t_+c1t +c2(2)Now, we substitute the initial conditions r(0) = 1, 0, 1 and r

(0) = 0, 2, 2 into (1) and (2) and solve for the vectorsc1 and c2. We obtain:r

(0) = 1, 1, 0 +c1 = 0, 2, 2 c1 = 1, 3, 2r(0) = 1, 0, 1 +c2 = 1, 0, 1 c2 = 0, 0, 2Finally we combine the above to obtain the solution:r(t ) = _et, sin t, cos t_+1, 3, 2 t +0, 0, 2 = _ett, sin t +3t, cos t +2t +2_Show that w(t ) = sin(3t +4), sin(3t 2), cos 3tsatises the differential equation w

(t ) = 9w(t ).55. The path r(t ) of a particle satises drdt = _8, 5 3t, 4t2_. Where is the particle located at t = 4 if r(0) = 1, 6, 0?SOLUTION We rst nd the general solution by integrating the vectordrdtcomponentwise. This gives:r(t ) =_ _8, 5 3t, 4t2_dt =_8t, 5t 32t2,43t3_+c (1)Substituting t = 0 we get:r(0) =_8 0, 5 0 32 02,43 03_+c = cThe initial condition r(0) = 1, 6, 0 gives c = 1, 6, 0. Combining with (1) we obtain the following solution:r(t ) =_8t, 5t 32t2,43t3_+1, 6, 0 =_1 +8t, 6 +5t 32t2,43t3_To nd the particles position at t = 4, we substitute t = 4 in r(t ) obtaining:r(4) =_1 +8 4, 6 +5 4 32 42,43 43_ =_33, 2,2563_A ghter plane, which can only shoot bullets straight ahead, travels along the path r(t ) = _5 t, 21 t2, 3 t3/27_. Show that there is precisely one time t at which the pilot can hit a target located at the origin.57. Find all solutions to r

(t ) = v, where v is a constant vector in R3.SOLUTION We denote the components of the constant vector v by v = v1, v2, v3 and integrate to nd the generalsolution. This gives:r(t ) =_v dt =_v1, v2, v3 dt =__v1 dt,_v2 dt,_v3 dt_= v1t +c1, v2t +c2, v3t +c3 = t v1, v2, v3 +c1, c2, c3

We let c = c1, c2, c3 and obtain:r(t ) = t v +c = c +t vNotice that the solutions are the vector parametrizations of all the lines with direction vector v.258 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)Let u be a constant vector in R3. Find the solution of r

(t ) = (sin t )u satisfying r

(0) = 0.59. Find all solutions to r

(t ) = 2r(t ) where r(t ) is a vector-valued function in three-space.SOLUTION We denote the components of r(t ) by r(t ) = x(t ), y(t ), z(t ). Then, r

(t ) = _x

(t ), y

(t ), z

(t )_. Substi-tuting in the differential equation we get:_x

(t ), y

(t ), z

(t )_ = 2 x(t ), y(t ), z(t )Equating corresponding components gives:x

(t ) = 2x(t )y

(t ) = 2y(t )z

(t ) = 2z(t )x(t ) = c1e2ty(t ) = c2e2tz(t ) = c3e2tWe denote the constant vector by c = c1, c2, c3 and obtain the following solutions:r(t ) = _c1e2t, c2e2t, c3e2t_ = e2tc1, c2, c3 = e2tcShow thatddt (a r) = a r

for any constant vector a.61. Prove thatr(t0) andr

(t0) are orthogonal at values t =t0where r(t ) takes on a local minimum ormaximum value. Explain how this result is related to Figure 7. Hint: In the gure, r(t0) is a minimum and the pathr(t ) intersects the sphere of radius r(t0) in a single point (and hence is tangent at that point).zyxr(t0)r(t0)r(t)FIGURE7SOLUTION Suppose that r(t ) takes on a minimum or maximum value at t=t0. Hence, r(t )2also takes on aminimum or maximum value at t = t0, thereforeddtr(t )2t =t0 = 0. Using the Product Rule for dot products we getddtr(t )2t =t0=ddt r(t ) r(t )t =t0= r(t0) r

(t0) +r

(t0) r(t0) = 2r(t0) r

(t0) = 0Thus r(t0) r

(t0) = 0, which implies the orthogonality of r(t0) and r

(t0). In Figure 7, r(t0) is a minimum and thepath intersects the sphere of radius r(t0) at a single point. Therefore, the point of intersection is a tangency point whichimplies that r

(t0) is tangent to the sphere at t0. We conclude that r(t0) and r

(t0) are orthogonal.Newtons Second Law of Motion in vector form states that F =dpdtwhere F is the force acting on an object ofmass m and p = mr

(t ) is the objects momentum. The analogs of force and momentum for rotational motion arethe torque = r F and angular momentum J = r(t ) p(t ). Use the Second Law to prove that =dJdt . This isa rotational version of the Second Law.Further Insights and Challenges63. In this exercise, we verify that the denition of the tangent line using vector-valued functions agrees with the usualdenition in terms of the scalar derivative in the case of a plane curve. Suppose that r(t ) = x(t ), y(t ) traces a planecurve C.(a) Show thatdydx =y

(t )x

(t )at any point such that x

(t ) = 0. Hint: By the Chain Rule, dydt =dydxdxdt .(b) Show that if x

(t0) = 0, then the line L(t ) = r(t0) +t r

(t0) passes through r(t0) and has slopedydxt =t0.SOLUTION(a) By the Chain Rule we havedydt =dydx dxdtHence, at the points wheredxdt = 0 we have:dydx =dydtdxdt=y

(t )x

(t )(b) The line (t ) = a, b +t r

(t0) passes through (a, b) at t = 0. It holds that:(0) = a, b +0r

(t0) = a, bThat is,(a, b) is the terminal point of the vector(0), hence the line passes through(a, b). The line has the directionvector r

(t0) = _x

(t0), y

(t0)_, therefore the slope of the line isy

(t0)x

(t0) which is equal todydxt =t0by part (a).S E C T I O N 14.2 Calculus of Vector-Valued Functions (ET Section 13.2) 259Verify the Sum and Product Rules for derivatives of vector-valued functions.65. Verify the Chain Rule for vector-valued functions.SOLUTION Let g(t ) and r(t ) = x(t ), y(t ), z(t ) be differentiable scalar and vector valued functions respectively. Wemust show that:ddt r (g(t )) = g

(t )r

(g(t )) .We haver (g(t )) = x (g(t )) , y (g(t )) , z (g(t ))We differentiate the vector componentwise, using the Chain Rule for scalar functions. This gives:ddt r (g(t )) =_ ddt(x (g(t ))) ,ddt(y (g(t ))) ,ddt(z (g(t )))_ = _g

(t )x

(g(t )) , g

(t )y

(g(t )) , g

(t )z

(g(t ))_= g

(t )_x

(g(t )) , y

(g(t )) , z

(g(t ))_ = g

(t )r

(g(t ))Verify the Product Rule for cross products [Eq. (6)]. 67. Prove thatddt (r (r

r

)) = r (r

r

)SOLUTION We use the Product Rule for dot products to obtain:ddt_r _r

r

__ = r ddt_r

r

_+r

_r

r

_(1)By the Product Rule for cross products and properties of cross products, we have:ddt_r

r

_ = r

r

+r

r

= r

r

+0 = r

r

(2)Substituting (2) into (1) yields:ddt_r _r

r

__ = r _r

r

_+r

_r

r

_(3)Since r

r

is orthogonal to r

, the dot product r

_r

r

_ = 0. So (3) gives:ddt_r _r

r

__ = r _r

r

_+0 = r _r

r

_Exercises 6871 establish additional properties of vector-valued integrals. Assume that all functions are integrable.Prove the linearity properties_cr(t ) dt = c_r(t ) dt (c any constant)_ _r1(t ) +r2(t )_dt =_r1(t ) dt +_r2(t ) dt69. Prove the Substitution Rule [where g(t ) is a differentiable scalar function]:_bar(g(t ))g

(t ) dt =_g1(b)g1(a)r(u) duSOLUTION (Note that an early edition of the textbook had the integral limits as g(a) and g(b); they should actuallybe g1(a) and g1(b).) We denote the components of the vector-valued function by r(t ) dt = x(t ), y(t ), z(t ). Usingcomponentwise integration we have:_bar(t ) dt =__bax(t ) dt,_bay(t ) dt,_baz(t ) dt_Write_bax(t ) dt as_bax(s) ds. Let s = g(t ), so ds = g

(t ) dt . The substitution gives us_g1(b)g1(a)x(g(t ))g

(t ) dt . Asimilar procedure for the other two integrals gives us:_bar(t ) dt =__g1(b)g1(a)x (g(t )) g

(t ) dt,_g1(b)g1(a)y (g(t )) g

(t ) dt,_g1(b)g1(a)z (g(t )) g

(t ) dt_=_g1(b)g1(a)_x (g(t )) g

(t ), y (g(t )) g

(t ), z (g(t )) g

(t )_ dt=_g1(b)g1(a)x (g(t )) , y (g(t )) , z (g(t )) g

(t ) dt =_g1(b)g1(a)r (g(t )) g

(t ) dtFormulate and verify a version of Integration by Parts for vector-valued integrals.260 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)71. Show that if r(t ) K for t [a, b], then______bar(t ) dt_____ K(b a)SOLUTION Think of r(t ) as a velocity vector. Then,_bar(t ) dt gives the displacement vector from the location at timet = a to the time t = b, and so______bar(t ) dt_____ gives the length of this displacement vector. But, since speed is r(t )which is less than or equal to K, then in the interval a t b, the object can move a total distance not more thanK(b a). Thus, the length of the displacement vector is K(b a), which gives us______bar(t ) dt_____ K(b a), asdesired.14.3Arc Length and Speed (ET Section 13.3)Preliminary Questions1. At a given instant, a car on a roller coaster has velocity vector r

= 25, 35, 10 (in miles per hour). What wouldthe velocity vector be if the speed were doubled? What would it be if the cars direction were reversed but its speedremained unchanged?SOLUTION The speed is doubled but the direction is unchanged, hence the new velocity vector has the form:r

= 25, 35, 10for > 0We use = 2, and so the new velocity vector is 50, 70, 20. If the direction is reversed but the speed is unchanged,the new velocity vector is:r

= 25, 35, 10 .2. Two cars travel in the same direction along the same roller coaster (at different times). Which of the followingstatements about their velocity vectors at a given point P on the roller coaster are true?(a) The velocity vectors are identical.(b) The velocity vectors point in the same direction but may have different lengths.(c) The velocity vectors may point in opposite directions.SOLUTION(a) The length of the velocity vector is the speed of the particle. Therefore, if the speeds of the cars are different thevelocities are not identical. The statement is false.(b) The velocity vector is tangent to the curve. Since the cars travel in the same direction, their velocity vectors point inthe same direction. The statement is true.(c) Since the cars travel in the same direction, the velocity vectors point in the same direction. The statement is false.3. A mosquito ies along a parabola with speed v(t ) = t2. Let L(t ) be the total distance traveled at time t .(a) How fast is L(t ) changing at t = 2?(b) Is L(t ) equal to the mosquitos distance from the origin?SOLUTION(a) By the Arc Length Formula, we have:L(t ) =_tt0r

(t ) dt =_tt0v(t ) dtTherefore,L

(t ) =v(t )To nd the rate of change of L(t ) at t = 2 we compute the derivative of L(t ) at t = 2, that is,L

(2) =v(2) = 22= 4(b) L(t ) is the distance along the path traveled by the mosquito. This distance is usually different from the mosquitosdistance from the origin, which is the length of r(t ).S E C T I O N 14.3 Arc Length and Speed (ET Section 13.3) 261r(t )Distance L(t)Distance fromthe origint0t4. What is the length of the path traced by r(t ) for 4 t 10 if r(t ) is an arc length parametrization?SOLUTION Since r(t ) is an arc length parametrization, the length of the path for 4 t 10 is equal to the length ofthe time interval 4 t 10, which is 6.ExercisesIn Exercises 16, compute the length of the curve over the given interval.1. r(t ) = 3t, 4t 3, 6t +1, 0 t 3SOLUTION We have x(t ) = 3t , y(t ) = 4t 3, z(t ) = 6t +1 hencex

(t ) = 3, y

(t ) = 4, z

(t ) = 6.We use the Arc Length Formula to obtain:L =_30r

(t ) dt =_30_x

(t )2+ y

(t )2+ z

(t )2dt =_30_32+42+62dt = 361r(t ) = 2t i 3t k, 11 t 153. r(t ) = _2t, ln t, t2_, 1 t 4SOLUTION The derivative of r(t ) is r

(t ) =_2, 1t , 2t_. We use the Arc Length Formula to obtain:L =_41r

(t ) dt =_41_22+_1t_2+(2t )2dt =_41_4t2+4 +1t2 dt =_41__2t + 1t_2dt=_41_2t + 1t_ dt = t2+ln t41=(16 +ln 4) (1 +ln 1) = 15 +ln 4r(t ) = _2t2+1, 2t21, t3_, 0 t 45. r(t ) = t i +2t j +(t23)k, 0 t 2. Hint:_ _t2+a2dt =12t_t2+a2+ 12a2ln_t +_t2+a2_SOLUTION The derivative of r(t ) is r

(t ) = i +2j +2t k. Using the Arc Length Formula we get:L =_20r

(t ) dt =_20_12+(2)2+(2t )2dt =_20_4t2+5 dtWe substitute u = 2t , du = 2 dt and use the given integration formula. This gives:L =12_40_u2+5 du =14u_u2+5 + 14 5 ln_u +_u2+5_40=14 4_42+5 + 54 ln_4 +_42+5_ 54 ln5 =21 + 54 ln_4 +21_ 54 ln5=21 + 54 ln 4 +215 6.26r(t ) = t cos t, t sin t, 3t , 0 t 2 7. Compute s(t ) =_t0r

(u) du for r(t ) = _t2, 2t2, t3_.SOLUTION The derivative of r(t ) is r

(t ) = _2t, 4t, 3t2_. Hence,r

(t ) =_(2t )2+(4t )2+(3t2)2=_4t2+16t2+9t4=_20 +9t2t262 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)Hence,s(t ) =_t0r

(u) du =_t0_20 +9u2u duWe compute the integral using the substitution v = 20 +9u2, dv = 18u du. This gives:s(t ) =118_20+9t220v1/2dv =118 23v3/220+9t220=127_(20 +9t2)3/2203/2_.In Exercises 811, nd the speed at the given value of t .r(t ) = 2t +3, 4t 3, 5 t , t = 49. r(t ) = _et 3, 12, 3t1_, t = 3SOLUTION The velocity vector is r

(t ) = _et 3, 0, 3t2_ and at t = 3, r

(3) = _e33, 0, 3 32_ = _1, 0, 13_.The speed is the magnitude of the velocity vector, that is,v(3) = r

(3) =_12+02+_13_2=_109 1.05r(t ) = sin 3t, cos 4t, cos 5t , t =211. r(t ) = cosh t, sinh t, t , t = 0SOLUTION The velocity vector is r

(t ) = sinh t, cosh t, 1. At t = 0 the velocity is r

(0) = sinh(0), cosh(0), 1 =0, 1, 1, hence the speed isv(0) = r

(0) =_02+12+12=2.What is the velocity vector of a particle traveling to the right along the hyperbola y = x1with constant speed5 cm/s when the particles location is (2, 12)?13. A bee with velocity vector r

(t ) starts out at the origin at t = 0 and ies around for Tseconds. Where is the beelocated at time T if_T0r

(u) du = 0? What does the quantity_T0r

(u) du represent?SOLUTION By the Fundamental Theorem for vector-valued functions,_T0r

(u) du =r(T) r(0), hence by thegiven information r(T) = r(0). It follows that at time T the bee is located at the starting point which is at the origin. Theintegral_T0r

(u) du is the length of the path traveled by the bee in the time interval 0 t T. Notice that there is adifference between the displacement and the actual length traveled.Which of the following is an arc length parametrization of a circle of radius 4 centered at the origin?(a) r1(t ) = 4 sin t, 4 cos t (b) r2(t ) = 4 sin 4t, 4 cos 4t (c) r3(t ) = _4 sint4, 4 cost4_15. Let r(t ) = 3t +1, 4t 5, 2t .(a) Calculate s(t ) = _t0 r

(u) du as a function of t .(b) Find the inverse (s) = t (s) and show that r1(s) = r((s)) is an arc length parametrization.SOLUTION(a) We differentiate r(t ) componentwise and then compute the norm of the derivative vector. This gives:r

(t ) = 3, 4, 2r

(t ) =_32+42+22=29We compute s(t ):s(t ) =_t0r

(u) du =_t029 du =29 ut0=29t(b) We nd the inverse (s) = t (s) by solving s = 29t for t . We obtain:s =29t t = (s) =s29We obtain the following arc length parametrization:r1(s) = r_s29_ =_3s29 +1,4s29 5,2s29_To verify that r1(s) is an arc length parametrization we must show that r

1(s) = 1. We compute r

1(s):r

1(s) =dds_3s29 +1,4s29 5,2s29_ =_329,429,229_ =129 3, 4, 2Thus,r

1(s) =129 3, 4, 2=129_32+42+22=129 29 = 1S E C T I O N 14.3 Arc Length and Speed (ET Section 13.3) 263Find an arc length parametrization of the circle in the plane z = 9 with radius 4 and center (1, 4, 9).17. Find a path that traces the circle in the plane y = 10 with radius 4 and center (2, 10, 3) with constant speed 8.SOLUTION We start with the following parametrization of the circle:r(t ) = 2, 10, 3 +4 cos t, 0, sin t= 2 +4 cos t, 10, 3 +4 sin t We need to reparametrize the curve by making a substitution t = (s), so that the new parametrization r1(s) = r_(s)_satises r

1(s) = 8 for all s. We nd r

1(s) using the Chain Rule:r

1(s) =ddsr_(s)_ =

(s)r

_(s)_(1)Next, we differentiate r(t ) and then replace t by (s):r

(t ) = 4 sin t, 0, 4 cos t r

_(s)_ = _4 sin (s), 0, 4 cos (s)_Substituting in (1) we get:r

1(s) =

(s)_4 sin (s), 0, 4 cos (s)_ = 4

(s)_sin (s), 0, cos (s)_Hence,r

1(s) = 4|

(s)|__sin (s)_2+_cos (s)_2= 4|

(s)|To satisfy r

1(s) =8 for all s, we choose

(s) =2. We may take the antiderivative (s) =2 s, and obtain thefollowing parametrization:r1(s) = r_(s)_ = r(2s) = 2 +4 cos(2s), 10, 3 +4 sin(2s) .This is a parametrization of the given circle, with constant speed 8.Show that one arch of the cycloid r(t ) = t sin t, 1 cos thas length 8. Find the value of t in [0, 2] wherethe speed is at a maximum.19. Find an arc length parametrization of r(t ) = _etsin t, etcos t, et_.SOLUTION An arc length parametrization is r1(s) = r_(s)_ where t = (s) is the inverse of the arc length function.We compute the arc length function:s(t ) =_t0r

(u) du (1)Differentiating r(t ) and computing the norm of r

(t ) gives:r

(t ) = _etsin t +etcos t, etcos t etsin t, et_ = etsin t +cos t, cos t sin t, 1r

(t ) = et_(sin t +cos t )2+(cos t sin t )2+12= et(sin2t +2 sin t cos t +cos2t +cos2t 2 sin t cos t +sin2t +1)1/2= et_2(sin2t +cos2t ) +1 = et2 1 +1 =3 et(2)Substituting (2) into (1) gives:s(t ) =_t03 eudu =3 eut0=3(ete0) =3(et1)We nd the inverse function of s(t ) by solving s = 3_et1_ for t . We obtain:s =3(et1)s3 = et1et= 1 +s3 t = (s) = ln_1 +s3_An arc length parametrization for r1(s) = r_(s)_ is:_eln(1+s/(3))sin_ln_1 +s3__, eln(1+s/(3))cos_ln_1 +s3__, eln(1+s/(3))_=_1 +s3__sin_ln_1 +s3__, cos_ln_1 +s3__, 1_264 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)Find an arc length parametrization of r(t ) = _t2, t3_.21. Express the arc length L of y = x3for 0 x 8 as an integral in two ways, using the parametrizations r1(t ) =_t, t3_ and r2(t ) = _t3, t9_. Do not evaluate the integrals, but use substitution to show that they yield the same result.SOLUTION For r1(t ) = _t, t3_ we have r

1(t ) = _1, 3t2_ hence r

1(t ) =_1 +9t4. For r2(t ) = _t3, t9_ we haver

2(t ) = _3t2, 9t8_ hence r

2(t ) =_9t4+81t16. The length L may be computed using the two parametrizations bythe following integrals (notice that in the second parametrization 0 t3 8 hence 0 t 2).L =_80r

1(t ) dt =_80_1 +9t4dt (1)L =_20r

2(t ) dt =_20_1 +9t123t2dt (2)We use the substitution u = t3, du = 3t2dt in the second integral to obtain:_20_1 +9t123t2dt =_80_1 +9u4duThis integral is the same as the integral in (1), in accordance with the well known property: the arc length is independentof the parametrization we choose for the curve.Show that a helix of radius R and height h making N complete turns has the parametrization_R cos_2Nth_, R sin_2Nth_, t_, 0 t h23. Consider the two springs in Figure 5. One has radius 5 cm, height 4 cm, and makes three complete turns. The otherhas height 3 cm, radius 4 cm, and makes ve complete turns.(a) Take a guess as to which spring uses more wire.(b) Compute the lengths of the two springs (use Exercise 22) and compare.4 cm3 turns, radius 5 cm 5 turns, radius 4 cm3 cmFIGURE5Which spring uses more wire?SOLUTION(a) The second wire seems to use more wire than the rst one.(b) Setting R = 5, h = 4 and N = 3 in the parametrization in Exercise 22 gives:r1(t ) =_5 cos 2 3t4, 5 sin 2 3t4, t_ =_5 cos 3t2, 5 sin 3t2, t_, 0 t 4Setting R = 4, h = 3 and N = 5 in this parametrization we get:r2(t ) =_4 cos 2 5t3, 4 sin 2 5t3, t_ =_4 cos 10t3, 4 sin 10t3, t_, 0 t 3We nd the derivatives of the two vectors and their lengths:r

1(t ) =_152sin 3t2,152cos 3t2, 1_ r

1(t ) =_22524+1 =12_2252+4r

2(t ) =_403sin 10t3,403cos 10t3, 1_ r

2(t ) =_1,60029+1 =13_1,6002+9Using the Arc Length Formula we obtain the following lengths:L1 =_4012_2252+4 dt = 2_2252+4 94.3L2 =_3013_1,6002+9 dt =_1,6002+9 125.7We see that the second spring uses more wire than the rst one.Use Exercise 22 to nd a general formula for the length of a helix of radius R and height h that makes N completeturns.25. Evaluate s(t ) =_tr

(u) du for the Bernoulli spiral r(t ) = _etcos 4t, etsin 4t_ (Figure 6). It is convenient totake as the lower limit since s() = 0. Then:S E C T I O N 14.3 Arc Length and Speed (ET Section 13.3) 265(a) Use s to obtain an arc length parametrization of r(t ).(b) Prove that the angle between the position vector and the tangent vector is constant.t = 0t = 210 20xyFIGURE6Bernoulli spiral.SOLUTION(a) We differentiate r(t ) and compute the norm of the derivative vector. This gives:r

(t ) = _etcos 4t 4etsin 4t, etsin 4t +4etcos 4t_ = etcos 4t 4 sin 4t, sin 4t +4 cos 4t r

(t ) = et_(cos 4t 4 sin 4t )2+(sin 4t +4 cos 4t )2= et_cos24t 8 cos 4t sin 4t +16 sin24t +sin24t +8 sin 4t cos 4t +16 cos24t_1/2= et_cos24t +sin24t +16_sin24t +cos24t_ = et1 +16 1 =17etWe now evaluate the improper integral:s(t ) =_tr

(u) du = limR_tR17eudu = limR17eutR = limR17(eteR)=17(et0) =17etAn arc length parametrization of r(t ) is r1(s) = r_(s)_ where t = (s) is the inverse function of s(t ). We nd t = (s)by solving s = 17etfor t :s =17et et=s17 t = (s) = lns17An arc length parametrization of r(t ) is:r1(s) = r_(s)_ =_eln(s/(17))cos_4 lns17_, eln(s/(17))sin_4 lns17__=s17_cos_4 lns17_, sin_4 lns17__(1)(b) The cosine of the angle between the position vector r1(s) and the tangent vector r

1(s) is:cos =r1(s) r

1(s)r1(s)r

1(s)Since for the arc length parametrization r

1(s) = 1, we obtain:cos =r1(s) r

1(s)r1(s)(2)We compute the dot product in (2). We rst compute r

1(s) from (1):r

1(s) =117_cos_4 lns17_, sin_4 lns17__+s17_sin_4 lns17_ 417s117, cos_4 lns17_ 417s117_=117_cos_4 lns17_, sin_4 lns17__+417_sin_4 lns17_, cos_4 lns17__=117_cos_4 lns17_4 sin_4 lns17_, sin_4 lns17_+4 cos_4 lns17__266 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)Thus,r1(s) r

1(s) =s17 117_cos_4 lns17__cos_4 lns17_4 sin_4 lns17__+sin_4 lns17__sin_4 lns17_+4 cos_4 lns17___=s17_cos2_4 lns17_4 cos_4 lns17_sin_4 lns17_+sin2_4 lns17_+4 sin_4 lns17_cos_4 lns17__=s17_cos2_4 lns17_+sin2_4 lns17__ =s17 1 =s17(3)We now compute r1(s) from (1) (Notice that s(t ) > 0 for all t ):r1(s) =|s|17_cos2_4 lns17_+sin2_4 lns17_ =s17 1 =s17(4)Combining (2), (3) and (4) yields:cos =s17s17=117The solution for 0 is = 1.326 rad. Thus, the angle between r1(s) and r

1(s) is constant.Further Insights and ChallengesProve that the length of a curve as computed using the arc length integral does not depend on its parametriza-tion. More precisely, let C be the curve traced by r(t ) for a t b. Let (s) be a differentiable function such that

(s) > 0 and that (c) = a and (d) = b. Then r1(s) = r((s)) parametrizes C for c s d. Verify that_bar

(t ) dt =_dcr

1(s) ds27. Show that path r(t ) =_1 t21 +t2 ,2t1 +t2_parametrizes the unit circle with the point (1, 0) excluded for < t 0 for t > 0, which implies that s(t ) has an inverse function for t 0. Therefore, it sufces to verify that(s(t )) = t . We have:(s(t )) = ln_eln(cosh t )+_e2 ln(cosh t )1_ = ln_cosh t +_cosh2t 1_Since cosh2t 1 = sinh2t we obtain (for t 0): (s(t )) = ln_cosh t +_sinh2t_ = ln (cosh t +sinh t ) = ln_et+et2+ etet2_ = ln_et_ = tWe thus proved that t = (s) is an inverse of s(t ). Therefore, the arc length parametrization is obtained by substitutingt = (s) in r(t ) = t tanh t, sech t . We compute t , tanh t and sech t in terms of s. We have:s = ln (cosh t ) es= cosh t sech t = esAlso:tanh2t = 1 sech2t = 1 e2s tanh t =_1 e2s t = tanh1 _1 e2sSubstituting in r(t ) gives:r1(s) = t tanh t, sech t=_tanh1_1 e2s_1 e2s, es_(c) The tractrix is shown in the following gure:24yx1 268 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)14.4Curvature (ET Section 13.4)Preliminary Questions1. What is the unit tangent vector of a line with direction vector v = 2, 1, 2?SOLUTION A line with direction vector v has the parametrization:r(t ) = OP0 +t vhence, sinceOP0 and v are constant vectors, we have:r

(t ) = vTherefore, since v = 3, the unit tangent vector is:T(t ) =r

(t )r

(t ) =vv = 2/3, 1/3, 2/32. What is the curvature of a circle of radius 4?SOLUTION The curvature of a circle of radius R is1R, hence the curvature of a circle of radius 4 is14.3. Which has larger curvature, a circle of radius 2 or a circle of radius 4?SOLUTION The curvature of a circle of radius 2 is12, and it is larger than the curvature of a circle of radius 4, which is14.4. What is the curvature of r(t ) = 2 +3t, 7t, 5 t ?SOLUTION r(t ) parametrizes the line 2, 0, 5 +t 3, 7, 1, and a line has zero curvature.5. What is the curvature at a point where T

(s) = 1, 2, 3 in an arc length parametrization r(s)?SOLUTION The curvature is given by the formula:(t ) = T

(t )r

(t )In an arc length parametrization, r

(t ) = 1 for all t , hence the curvature is (t ) = T

(t ). Using the given informationwe obtain the following curvature: =1, 2, 3=_12+22+32=146. What is the radius of curvature of a circle of radius 4?SOLUTION The denition of the osculating circle implies that the osculating circles at the points of a circle, is thecircle itself. Therefore, the radius of curvature is the radius of the circle, that is, 4.7. What is the radius of curvature at P if P = 9?SOLUTION The radius of curvature is the reciprocal of the curvature, hence the radius of curvature at P is:R =1P=19ExercisesIn Exercises 16, calculate r

(t ) and T(t ), and evaluate T(1).1. r(t ) = _12t3, 18t2, 9t4_SOLUTION The derivative vector is:r

(t ) = _36t2, 36t, 36t3_ = 36_t2, t, t3_ r

(t ) = 36_t4+t2+t6The unit vector is thus:T(t ) =r

(t )r

(t ) =36t2, t, t3

36_t4+t2+t6 =t t, 1, t2

|t |_1 +t2+t4 =t|t |_1 +t2+t4t, 1, t2

At t = 1 we have:T(1) =1_1 +12+141, 1, 12 =131, 1, 1 =_13,13,13_.S E C T I O N 14.4 Curvature (ET Section 13.4) 269r(t ) = _cos t, sin t, t_ 3. r(t ) = _3 +4t, 3 5t, 9t_SOLUTION We rst nd the vector r

(t ) and its length:r

(t ) = 4, 5, 9 r

(t ) =_42+(5)2+92=122The unit tangent vector is therefore:T(t ) =r

(t )r

(t ) =1122 4, 5, 9 =_4122, 5122,9122_We see that the unit tangent vector is constant, since the curve is a straight line.r(t ) = _1 +2t, t2, 3 t2_ 5. r(t ) = _4t2, 9t_SOLUTION We differentiate r(t ) to obtain:r

(t ) = 8t, 9 r

(t ) =_(8t )2+92=_64t2+81We now nd the unit tangent vector:T(t ) =r

(t )r

(t ) =1_64t2+818t, 9For t = 1 we obtain the vector:T(t ) =164 +81 8, 9 =_8145,9145_.r(t ) = _et, t2_In Exercises 712, use Eq. (3) to calculate (t ).7. r(t ) = _1, et, t_SOLUTION We compute the rst and the second derivatives of r(t ):r

(t ) = _0, et, 1_, r

(t ) = _0, et, 0_.Next, we nd the cross product r

(t ) r

(t ):r

(t ) r

(t ) =i j k0 et10 et0=et1et0i 0 10 0j +0 et0 etk = eti = _et, 0, 0_We need to nd the lengths of the following vectors:r

(t ) r

(t ) = _et, 0, 0_ = etr

(t ) =_02+(et)2+12=_1 +e2tWe now use the formula for curvature to calculate (t ):(t ) = r

(t ) r

(t )r

(t )3=et__1 +e2t_3 =et_1 +e2t_3/2r(t ) = _cos t, t sin t, t_ 9. r(t ) = _4 cos t, t, 4 sin t_SOLUTION By the formula for curvature we have:(t ) = r

(t ) r

(t )r

(t )3(1)First we nd r

(t ) and r

(t ):r

(t ) = 4 sin t, 1, 4 cos t r

(t ) = 4 cos t, 0, 4 sin t We compute the cross product:r

(t ) r

(t ) =i j k4 sin t 1 4 cos t4 cos t 0 4 sin t270 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)=1 4 cos t0 4 sin ti 4 sin t 4 cos t4 cos t 4 sin tj +4 sin t 14 cos t 0k= 4 sin t i _16 sin2t +16 cos2t_j +4 cos t k= 4 sin t i 16j +4 cos t k = 4 sin t, 4, cos t We compute the lengths of the following vectors:r

(t ) r

(t ) = 4_(sin t )2+(4)2+cos2t = 4_sin2t +16 +cos2t = 417r

(t )=_(4 sin t )2+12+(4 cos t )2=_16 sin2t +1 +16 cos2t =17Substituting in (1) gives the following curvature:(t ) =417_17_3 =4171717 =417We see that this curve has constant curvature.r(t ) = _4t +1, 4t 3, 2t_11. r(t ) = _t1, 1, t_SOLUTION By the formula for curvature we have:(t ) = r

(t ) r

(t )r

(t )3(1)We now nd r

(t ), r

(t ) and their cross product. This gives:r

(t ) = _t2, 0, 1_, r

(t ) = _2t3, 0, 0_r

(t ) r

(t ) = _t2i +k_2t3i = 2t3k i = 2t3jWe compute the lengths of the vector in (1):r

(t ) r

(t ) = 2t3j = 2|t3|r

(t ) =__(t )2_2+02+12=_t4+1Substituting in (1) we obtain the following curvature:(t ) =2|t |3__t4+1_3 =2|t |3_t4+1_3/2We multiply through by |t |43/2= |t |6to obtain:(t ) =2|t |3_1 +|t |4_3/2r(t ) = _cosh t, sinh t, t_In Exercises 1316, nd the curvature of the plane curve at the point indicated.13. y = et, t = 3SOLUTION We use the curvature of a graph in the plane:(t ) =| f

(t )|_1 +f

(t )2_3/2In our casef (t ) = et, hencef

(t ) =f

(t ) = etand we obtain:(t ) =et_1 +e2t_3/2 (3) =e3_1 +e6_3/2 0.0025y = cos x, x = 015. y = t4, t = 2S E C T I O N 14.4 Curvature (ET Section 13.4) 271SOLUTION By the curvature of a graph in the plane, we have:(t ) =| f

(t )|_1 +f

(t )2_3/2In this casef (t ) = t4, f

(t ) = 4t3, f

(t ) = 12t2. Hence,(t ) =12t2_1 +_4t3_2_3/2 =12t2_1 +16t6_3/2At t = 2 we obtain the following curvature:(2) =12 22(1 +16 26)3/2 =48(1,025)3/2 0.0015.y = tn, t = 117. Find the curvature of r(t ) = 2 sin t, cos 3t, tat t =3and t =2(Figure 15).yxz 3t =FIGURE15The curve r(t ) = 2 sin t, cos 3t, t .SOLUTION By the formula for curvature we have:(t ) = r

(t ) r

(t )r

(t )3(1)We compute the rst and second derivatives:r

(t ) = 2 cos t, 3 sin 3t, 1 , r

(t ) = 2 sin t, 9 cos 3t, 0At the points t =3and t =2we have:r

_3_ =_2 cos 3, 3 sin 33, 1_ =_2 cos 3, 3 sin , 1_ = 1, 0, 1r

_3_ =_2 sin 3, 9 cos 33, 0_ =_3, 9, 0_r

_2_ =_2 cos 2, 3 sin 32, 1_ = 0, 3, 1r

_2_ =_2 sin 2, 9 cos 32, 0_ = 2, 0, 0We compute the cross products required to use (1):r

_3_r

_3_ =i j k1 0 13 9 0=0 19 0i 1 13 0j +1 03 9k = 9i 3j +9kr

_2_r

_2_ =i j k0 3 12 0 0=3 10 0i 0 12 0j +0 32 0k = 2j +6kHence,___r

_3_r

_3____ =_(9)2+_3_2+92=165___r

_3____ =_12+02+12=2272 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)At t =2we have:___r

_2_r

_2____ =_(2)2+62=40 = 210___r

_2____ =_02+32+12=10Substituting the values for t =3and t =2in (1) we obtain the following curvatures:_3_ =165_2_3 =16522 4.54_2_ =210_10_3 =2101010 = 0.2Find the curvature function (x) for y = sin x. Use a computer algebra system to plot (x) for 0 x 2.Prove that the curvature takes its maximum at x =2and32 . Hint: To simplify the calculation, nd the maximumof (x)2.19. Show that curvature at an inection point of a plane curve y =f (x) is zero.SOLUTION The curvature of the graph y =f (x) in the plane is the following function:(x) =| f

(x)|_1 +f

(x)2_3/2(1)At an inection point the second derivative changes its sign. Therefore, if f

is continuous at the inection point, it iszero at this point, hence by (1) the curvature at this point is zero.Show that the tractrix r(t ) = t tanh t, sech thas the curvature function (t ) = sech t .21. Find the value of such that the curvature of y = exat x = 0 is as large as possible.SOLUTION Using the curvature of a graph in the plane we have:(x) =|y

(x)|_1 + y

(x)2_3/2(1)In our case y

(x) = ex, y

(x) = 2ex. Substituting in (1) we obtain(x) =2ex_1 +2e2x_3/2The curvature at the origin is thus(0) =2e0_1 +2e20_3/2 =2_1 +2_3/2Since (0) and 2(0) have their maximum values at the same values of , we may maximize the function:g() = 2(0) =4(1 +2)3We nd the stationary points:g

() =43(1 +2)34(3)(1 +2)22(1 +2)6=23(1 +2)2(2 2)(1 +2)6= 0The stationary points are the solutions of the following equation:23(1 +2)2(2 2) = 0 3= 0 or 2 2= 0 = 0 = 2Since g() 0 and g(0) = 0, = 0 is a minimum point. Also, g

() is positive immediately to the left of 2 andnegative to the right. Hence, = 2 is a maximum point. Since g() is an even function, = 2 is a maximumpoint as well. Conclusion: (x) takes its maximum value at the origin when = 2.Find the point of maximum curvature on y = ex. 23. Show that the curvature function of the parametrization r(t ) = a cos t, b sin tof the ellipse_xa_2+_yb_2= 1 is(t ) =ab(b2cos2t +a2sin2t )3/28S E C T I O N 14.4 Curvature (ET Section 13.4) 273SOLUTION The curvature is the following function:(t ) = r

(t ) r

(t )r

(t )3(1)We compute the derivatives and their cross product:r

(t ) = a sin t, b cos t, r

(t ) = a cos t, b sin t r

(t ) r

(t ) =(a sin t i +b cos t j) (a cos t i b sin t j)= ab sin2t k +ab cos2t k = ab_sin2t +cos2t_k = abkThus,r

(t ) r

(t ) = abk = abr

(t ) =_(a sin t )2+(b cos t )2=_a2sin2t +b2cos2tSubstituting in (1) we obtain the following curvature:(t ) =ab__a2sin2t +b2cos2t_3 =ab_a2sin2t +b2cos2t_3/2Use a sketch to predict where the points of minimal and maximal curvature occur on an ellipse. Then use Eq. (8)to conrm or refute your prediction.25. In the notation of Exercise 23, assume that a b. Show that b/a2 (t ) a/b2for all t .SOLUTION In Exercise 23 we showed that the curvature of the ellipse r(t ) = a cos t, b sin tis the following function:(t ) =ab_b2cos2t +a2sin2t_3/2Since a b> 0 the quotient becomes greater if we replace a by b in the denominator, and it becomes smaller if wereplace b by a in the denominator. We use the identity cos2t +sin2t = 1 to obtain:ab_a2cos2t +a2sin2t_3/2 (t ) ab_b2cos2t +b2sin2t_3/2ab_a2_cos2t +sin2t__3/2 (t ) ab_b2_cos2t +sin2t__3/2aba3 =ab(a2)3/2 (t ) ab(b2)3/2 =abb3ba2 (t ) ab2Use Eq. (3) to prove that for a plane curve r(t ) = x(t ), y(t ),(t ) = |x

(t )y

(t ) x

(t )y

(t )|(x

(t )2+ y

(t )2)3/2In Exercises 2730, use Eq. (9) to compute the curvature at the given point.27. _t2, t3_, t = 2SOLUTION For the given parametrization, x(t ) = t2, y(t ) = t3, hencex

(t ) = 2tx

(t ) = 2y

(t ) = 3t2y

(t ) = 6tAt the point t = 2 we havex

(2) = 4, x

(2) = 2, y

(2) = 3 22= 12, y

(2) = 12Substituting in Eq. (9) we get(2) = |x

(2)y

(2) x

(2)y

(2)|_x

(2)2+ y

(2)2_3/2= |4 12 2 12|_42+122_3/2=241603/2 0.012_cosh s, s_, s = 0274 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)29. _t cos t, sin t_, t = SOLUTION We have x(t ) = t cos t and y(t ) = sin t , hence:x

(t ) = cos t t sin t x

() = cos sin = 1x

(t ) = sin t (sin t +t cos t ) = 2 sin t t cos t x

() = 2 sin cos = y

(t ) = cos t y

() = cos = 1y

(t ) = sin t y

() = sin = 0Substituting in Eq. (9) gives the following curvature:() = |x

()y

() x

()y

()|_x

()2+ y

()2_3/2= | 1 0 (1)|_(1)2+(1)2_3/2 =22 1.11_sin 3s, 2 sin 4s_, s =231. Let s(t ) =_tr

(u) du for the Bernoulli spiral r(t ) = _etcos 4t, etsin 4t_ (see Exercise 25 in Section 14.3).Show that the radius of curvature is proportional to s(t ).SOLUTION The radius of curvature is the reciprocal of the curvature:R(t ) =1(t )We compute the curvature using the equality given in Exercise 25 in Section 3:(t ) = |x

(t )y

(t ) x

(t )y

(t )|_x

(t )2+ y

(t )2_3/2(1)In our case, x(t ) = etcos 4t and y(t ) = etsin 4t . Hence:x

(t ) = etcos 4t 4etsin 4t = et(cos 4t 4 sin 4t )x

(t ) = et(cos 4t 4 sin 4t ) +et(4 sin 4t 16 cos 4t ) = et(15 cos 4t +8 sin 4t )y

(t ) = etsin 4t +4etcos 4t = et(sin 4t +4 cos 4t )y

(t ) = et(sin 4t +4 cos 4t ) +et(4 cos 4t 16 sin 4t ) = et(8 cos 4t 15 sin 4t )We compute the numerator in (1):x

(t )y

(t ) x

(t )y

(t ) = e2t(cos 4t 4 sin 4t ) (8 cos 4t 15 sin 4t ) +e2t(15 cos 4t +8 sin 4t ) (sin 4t +4 cos 4t )= e2t_68 cos24t +68 sin24t_ = 68e2tWe compute the denominator in (1):x

(t )2+ y

(t )2= e2t(cos 4t 4 sin 4t )2+e2t(sin 4t +4 cos 4t )2= e2t_cos24t 8 cos 4t sin 4t +16 sin24t +sin24t +8 sin 4t cos 4t +16 cos24t_= e2t_cos24t +sin24t +16_sin24t +cos24t__= e2t(1 +16 1) = 17e2t(2)Hence_x

(t )2+ y

(t )2_3/2= 173/2e3tSubstituting in (2) we have(t ) =68e2t173/2e3t =417et R =174et(3)On the other hand, by the Fundamental Theorem and (2) we haves

(t ) = r

(t ) =_x

(t )2+ y

(t )2=_17e2t=17etWe integrate to obtains(t ) =_17 etdt =17 et+C (4)S E C T I O N 14.4 Curvature (ET Section 13.4) 275Since s(t ) =_tr

(u) du, we have limt s(t ) = 0, hence by (4):0 = limt _17et+C_ = 0 +C = C.Substituting C = 0 in (4) we get:s(t ) =17et(5)Combining (3) and (5) gives:R(t ) =14s(t )which means that the radius of curvature is proportional to s(t ).The Cornu spiral is the plane curve r(t ) = x(t ), y(t ), wherex(t ) =_t0sin u22du, y(t ) =_t0cos u22duVerify that (t ) = |t |. Since the curvature increases linearly, the Cornu spiral is used in highway design to createtransitions between straight and curved road segments (Figure 16).33. Plot and compute the curvature (t ) of the clothoid r(t ) = x(t ), y(t ), wherex(t ) =_t0sin u33du, y(t ) =_t0cos u33duSOLUTION We use the following formula for the curvature (given earlier):(t ) = |x

(t )y

(t ) x

(t )y

(t )|_x

(t )2+ y

(t )2_3/2(1)We compute the rst and second derivatives of x(t ) and y(t ). Using the Fundamental Theorem and the Chain Rule weget:x

(t ) = sin t33x

(t ) =3t23cos t33= t2cos t33y

(t ) = cos t33y

(t ) =3t23_sin t33_ = t2sin t33Substituting in (1) gives the following curvature function:(t ) =sin t33_t2sin t33_t2cos t33cos t33__sin t33_2+_cos t33_2_3/2=t2 _sin2t33 +cos2t33_13/2= t2That is, (t ) = t2. Here is a plot of the curvature as a function of t :t(t ) = t2Find the unit normal vector N() to r() =R cos , sin , the circle of radius R. Does N() point inside oroutside the circle? Draw N() at =4with R = 4.35. Find the unit normal vector N(t ) to r(t ) = 4, sin 2t, cos 2t .SOLUTION We rst nd the unit tangent vector:T(t ) =r

(t )r

(t )(1)276 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)We haver

(t ) =ddt 4, sin 2t, cos 2t= 0, 2 cos 2t, 2 sin 2t= 2 0, cos 2t, sin 2t r

(t ) = 2_02+cos22t +(sin 2t )2= 20 +1 = 2Substituting in (1) gives:T(t ) =2 0, cos 2t, sin 2t 2= 0, cos 2t, sin 2t The normal vector is the following vector:N(t ) =T

(t )T

(t )(2)We compute the derivative of the unit tangent vector and its length:T

(t ) =ddt 0, cos 2t, sin 2t= 0, 2 sin 2t, 2 cos 2t= 2 0, sin 2t, cos 2t T

(t ) = 2_02+sin22t +cos22t = 20 +1 = 2Substituting in (2) we obtain:N(t ) = 2 0, sin 2t, cos 2t 2= 0, sin 2t, cos 2t Sketch the graph of r(t ) = _t, t3_. Since r

(t ) = _1, 3t2_, the unit normal N(t ) points in one of the two directions_3t2, 1_. Which sign is correct at t = 1? Which is correct at t = 1?37. Find the normal vectors to r(t ) = t, cos tat t =4and t =34 .SOLUTION The normal vector to r(t ) = t, cos tis T

(t ), where T(t ) =r

(t )r

(t ) is the unit tangent vector. We haver

(t ) = 1, sin t r

(t ) =_12+(sin t )2=_1 +sin2tHence,T(t ) =1_1 +sin2t1, sin t We compute the derivative of T(t ) to nd the normal vector.We use the Product Rule and the Chain Rule to obtain:T

(t ) =1_1 +sin2tddt 1, sin t+_1_1 +sin2t_

1, sin t =1_1 +sin2t0, cos t11 +sin2t 2 sin t cos t2_1 +sin2t1, sin t =1_1 +sin2t0, cos tsin 2t2_1 +sin2t_3/2 1, sin t At t =4we obtain the normal vector:T

_4_ =1_1 +12_0, 12_12_1 +12_3/2_1, 12_ =_0, 13__ 233,133_ =_233 ,233_At t =34we obtain:T

_34_ =1_1 +12_0,12_12_1 +12_3/2_1, 12_ =_0,13_+_ 233,133_ =_ 233,233_Find the unit normal to the Cornu spiral (Exercise 32) at t = .39. Find the unit normal to the clothoid (Exercise 33) at t = 1/3.S E C T I O N 14.4 Curvature (ET Section 13.4) 277SOLUTION The Clothoid is the plane curve r(t ) = x(t ), y(t ) withx(t ) =_t0sin u33du, y(t ) =_t0cos u33duThe unit normal is the following vector:N(t ) =T

(t )T

(t )(1)We rst nd the unit tangent vector T(t ) =r

(t )r

(t ). By the Fundamental Theorem we haver

(t ) =_sin t33, cos t33_ r

(t ) =_sin2 t33 +cos2 t33=1 = 1Hence,T(t ) =_sin t33, cos t33_We now differentiate T(t ) using the Chain Rule to obtain:T

(t ) =_3t23cos t33,3t23sin t33_ = t2_cos t33, sin t33_Hence,T

(t ) = t2_cos2 t33 +_sin t33_2= t2Substituting in (1) we obtain the following unit normal:N(t ) =_cos t33, sin t33_At the point T = 1/3the unit normal isN(1/3) =_cos(1/3)33, sin(1/3)33_ =_cos 3, sin 3_ =_12, 32_Method for Computing N Let v(t ) = r

(t ). Show thatN(t ) =r

(t ) v

(t )T(t )r

(t ) v

(t )T(t )Hint: Differentiate r

(t ) =v(t )T(t ) and note that N is the unit vector in the direction T

.In Exercises 4146, use Eq. (10) to nd N at the point indicated.41. _1 +t2, 2t, t3_, t = 1SOLUTION We compute the values in formula (10). In our caser(t ) = _1 +t2, 2t, t3_Hence,r

(t ) = _2t, 2, 3t2_r

(t ) = 2, 0, 6t v(t ) = r

(t ) =_(2t )2+22+(3t2)2=_4t2+4 +9t4=_4 +4t2+9t4v

(t ) =8t +36t32_4 +4t2+9t4 =4t +18t3_4 +4t2+9t4At the point t = 1, we haver

(1) = 2, 0, 6 , v

(1) =2217, T(1) =r

(1)r

(1) = 2, 2, 317Hence,r

(1) v

(1)T(1) = 2, 0, 6 2217 117 2, 2, 3 = 2, 0, 6 _4417,4417,6617_278 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)=_1017, 4417,3617_ =117 10, 44, 36r

(1) v

(1)T(1) =117_(10)2+(44)2+362=117_3,332 =141717=1417Substituting in equation (10) we getN(1) =r

(1) v

(1)T(1)r

(1) v

(1)T(1) =117 10, 44, 361417=1717 5, 22, 18_t, et, t_, t = 243. _t sin t, 1 cos t_, t = SOLUTION We use the following equality:N(t ) =r

(t ) v

(t )T(t )r

(t ) v

(t )T(t )(1)We compute the vectors in the above equality. For r(t ) = t sin t, 1 cos twe haver

(t ) = 1 cos t, sin t r

(t ) = sin t, cos t v(t ) = r

(t ) =_(1 cos t )2+sin2t =_1 2 cos t +cos2t +sin2t= 1 2 cos t +1 =_2(1 cos t ) =_2 2 sin2t2 = 2sin t2For 0 t 2, sint2 0, hence v(t ) = 2 sint2. Therefore,v

(t ) = 2 12 cos t2 = cos t2, 0 t 2At the point t = we haver

() = sin , cos = 0, 1v

() = cos 2 = 0r

() = 1 cos , sin = 2, 0T() =r

()r

() = 2, 02= 1, 0We now substitute these values in (1) to obtain the following unit normal:N() =0, 1 0 1, 0 0, 1 0 1, 0 = 0, 11= 0, 1_t2, t3_, t = 145. _t1, t, t2_, t = 1SOLUTION We use the equalityN(t ) =r

(t ) v

(t )T(t )r

(t ) v

(t )T(t )(1)We compute the vectors in the above equality. For r(t ) = _t1, t, t2_ we haver

(t ) = _t2, 1, 2t_r

(t ) = _2t3, 0, 2_v(t ) = r

(t ) =_t4+1 +4t2v

(t ) =4t5+8t2_t4+1 +4t2 =2t5+4t_t4+1 +4t2At the point t = 1 we getr

(1) = 1, 1, 2 , r

(1) = 2, 0, 2 , v

(1) =2 41 +1 +4 = 26,S E C T I O N 14.4 Curvature (ET Section 13.4) 279T(1) =r

(1)r

(1) = 1, 1, 26Hence,r

(1) v

(1)T(1) = 2, 0, 2 +26 16 1, 1, 2 =_73,13,43_ =13 7, 1, 4r

(1) v

(1)T(1) =13_(7)2+12+42=1366Substituting in (1) gives the following unit normal:N(1) =13 7, 1, 41366=166 7, 1, 4_cosh t, sinh t, t_, t = 047. Let f (x) = x2. Show that the center of the osculating circle at (x0, x20) is given by_4x30,12 +3x20_.SOLUTION We parametrize the curve by r(x) = _x, x2_. The centerQof the osculating circle at x =x0has theposition vectorOQ = r(x0) +(x0)1N(x0) (1)We rst nd the curvature, using the formula for the curvature of a graph in the plane. We have f

(x) =2xandf

(x) = 2, hence,(x) =| f

(x)|(1 +f

(x)2)3/2 =2(1 +4x2)3/2 (x0)1=12(1 +4x20)3/2To nd the unit normal vector N(x0) we use the following considerations:The tangent vector is r

(x0) = 1, 2x0, hence the vector 2x0, 1 is orthogonal to r

(x0) (since their dot productis zero). Hence N(x0) is one of the two unit vectors 1_1+4x202x0, 1.The graph of f (x) = x2shows that the unit normal vector points in the positive y-direction, hence, the appropriatechoice is:N(x0) =1_1 +4x202x0, 1 (2)yxf (x) = x2We now substitute (2), (3), and r(x0) = _x0, x20_ in (1) to obtainOQ = _x0, x20_+ 12_1 +4x20_3/21_1 +4x20_2x0, 1_ = _x0, x20_+ 12_1 +4x20_2x0, 1= _x0, x20_+_x0 4x30,12_1 +4x20__ =_4x30,12 +3x20_The center of the osculating circle is the terminal point ofOQ, that is,Q =_4x30,12 +3x20_Find a parametrization of the osculating circle to y = x2at x = 1.49. Find a parametrization of the osculating circle to y = sin x at x =2.280 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)SOLUTION We use the parametrization r(x) = x, sin x. The radius of the osculating circle is the radius of curvatureR =1_2_ and the center is the terminal point of the following vector:OQ = r_2_+ RN_2_We rst compute the curvature. Since y

(x) = cos x and y

(x) = sin x, we have:(x) =|y

(x)|(1 + y

(x)2)3/2 =| sin x|(1 +cos2x)3/2 _2_ =sin 2_1 +cos22_3/2 =11 = 1We compute the unit normal vector N(x). N(x) is a unit vector orthogonal to the tangent vector r

(x) = 1, cos x. Weobserve that cos x, 1 is orthogonal to r

(x), since their dot product is zero. Therefore, N(x) is the unit vector in thedirection of either cos x, 1 or cos x, 1, depending on the graph. Considering the accompanying gure, we seethat the unit normal vector at x = /2 points to the negative y-direction. Thus,N(x) =cos x, 1 cos x, 1 =cos x, 1_cos2x +(1)2 N_2_ = 0, 1yxP( , 1)Q( , 0)Wenowndthecenteroftheosculatingcircle. WesubstituteR=1_2_=1, N_2_=0, 1, andr_2_ =_2, sin 2_ = _2, 1_ into (1) to obtainOQ =_2, 1_+1 0, 1 =_2, 0_The osculating circle is the circle with center at the point _2, 0_ and radius 1, so it has the following parametrization:c(t ) =_2, 0_+1 cos t, sin t=_2, 0_+cos t, sin t Use Eq. (7) to nd the center of curvature to r(t ) = _t2, t3_ at t = 1.In Exercises 5155, nd a parametrization of the osculating circle at the point indicated.51. _cos t, sin t_, t =4SOLUTION The curve r(t ) = cos t, sin tis the unit circle. By the denition of the osculating circle, it follows that theosculating circle at each point of the circle is the circle itself. Therefore the osculating circle to the unit circle at t =4isthe unit circle itself._sin t, cos t_, t = 053. _t sin t, 1 cos t_, t = (use Exercise 43)SOLUTIONStep 1. Find and N. In Exercise 43 we found that:N() = 0, 1 (1)To nd we use the formula for curvature:() = r

() r

() r

() 3(2)For r(t ) = t sin t, 1 cos twe have:r

(t ) = 1 cos t, sin t r

() = 1 cos , sin = 2, 0r

(t ) = sin t, cos t r

() = sin , cos = 0, 1Hence,r

() r

() = 2i (j) = 2kS E C T I O N 14.4 Curvature (ET Section 13.4) 281r

() r

()=2k = 2 and r

()=2, 0= 2Substituting in (2) we get: () =223 =14(3)Step 2. Find the center of the osculating circle. The center Q of the osculating circle at r () = , 2 has position vectorOQ = r () +()1N()Substituting (1), (3) and r () = , 2 we get:OQ = , 2 +_14_10, 1 = , 2 +0, 4 = , 2Step 3. Parametrize the osculating circle. The osculating circle has radius R =1() and it is centered at (, 2), henceit has the following parametrization:c(t ) = , 2 +4 cos t, sin t _1 +t2, 2t, t3_, t = 1 (use Exercise 41)55. r(t ) = _cosh t, sinh t, t_, t = 0 (use Exercise 12)SOLUTIONStep 1. Find and N. In Exercise 12 we found that:(t ) =12 cosh2t (0) =12 cosh20 =12(1)We now must nd the unit normal N. We have:r

(t ) = sinh t, cosh t, 1r

(t ) =_sinh2t +cosh2t +1 =_cosh2t 1 +cosh2t +1 =_2 cosh2t =2 cosh tT(t ) =r

(t )r

(t ) =12 cosh t sinh t, cosh t, 1 =12 tanh t, 1, sech t T

(t ) =12_sech2t, 0, sech t tanh t_We compute the length of T

(t ). Using the identity tanh2t +sech2t = 1 we get:T

(t ) =12_sech4t +0 +sech2t tanh2t =12_sech2t_tanh2t +sech2t_ =12_sech2t 1 =sech t2Hence,N(t ) =T

(t )T

(t ) =2sech t12 _sech2t, 0, sech t tanh t_ =12 sech t, 0, tanh t At the point t = 0 we have sech 0 = 1, tanh 0 = 0, henceN(0) = 1, 0, 0 (2)Step 2. Find the center of the osculating circle. The center Q of the osculating circle at r(0) = 1, 0, 0 has positionvector:OQ = r(0) +(0)1N(0)Substituting (1), (2) and r(0) = 1, 0, 0 we get:OQ = 1, 0, 0 +2 1, 0, 0 = 3, 0, 0Step 3. Parametrize the osculating circle. The osculating circle is centered at Q = (3, 0, 0) and has radius R =1(0) = 2,hence it has the following parametrization:c(t ) = 3, 0, 0 +2Ncos t +2Tsin t = 3, 0, 0 +21, 0, 0 cos t +220, 1, 1 sin tFind the curvature and unit normal vector to the helix r(t ) = _cos t, sin t, t_ at t = 0. Then use Eq. (7) to nd thecenter of curvature and a parametrization of the osculating circle at t = 0.282 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)57. Figure 17 shows the graph of the half-ellipse y = _2r x px2, where r and p are positive constants. Show thatthe radius of curvature at the origin is equal to r. Hint: One way of proceeding is to write the ellipse in the form ofExercise 23 and apply Eq. (8).xyrrFIGURE17The curve y =_2r x px2and the osculating circle at the origin.SOLUTION The radius of curvature is the reciprocal of the curvature. We thus must nd the curvature at the origin. Weuse the following simple variant of the formula for the curvature of a graph in the plane:(y) =|x

(y)|_1 + x

(y)2_3/2(1)(The traditional formula of (x) =|y

(x)|_1+y

(x)2_3/2is inappropriate for this problem, as y

(x) is undened at x = 0.) Wend x in terms of y:y =_2r x px2y2= 2r x px2px22r x + y2= 0We solve for x and obtain:x = 1p_r2 py2+rp, y 0.We nd x

and x

:x

= 2py2p_r2 py2 = y_r2 py2x

= 1 _r2 py2 y pyr2py2r2 py2= r2 py2+ py2_r2 py2_3/2= r2_r2 py2_3/2At the origin we get:x

(0) = 0, x

(0) =r2(r2)3/2 = 1rSubstituting in (1) gives the following curvature at the origin:(0) =|x

(0)|(1 + x

(0)2)3/2 =|1r |(1 +0)3/2 =1|r| =1rWe conclude that the radius of curvature at the origin isR =1(0)= rIn a recent study of laser eye surgery by Gatinel, Hoang-Xuan, and Azar, a vertical cross section of the cornea ismodeled by the half-ellipse of Exercise 57. Show that the half-ellipse can be written in the form x =f (y), wheref (y) = p1_r _r2 py2_. During surgery, tissue is removed to a depth t (y) at height y for S y S, wheret (y) is given by Munnerlyns equation (for some R> r):t (y) =_R2 S2_R2 y2_r2 S2+_r2 y2After surgery, the cross section of the cornea has the shape x =f (y) + t (y) (Figure 18). Show that after surgery,the radius of curvature at the point P (where y = 0) is R.59. The angle of inclination of a plane curve with parametrization r(t ) is dened as the angle (t ) between the unittangent vector T(t ) and the x-axis (Figure 19). Show that T

(t ) = |

(t )| and conclude that if r(s) is a parametrizationby arc length, then(s) =dds11Hint: Observe that T(t ) = cos (t ), sin (t ).S E C T I O N 14.4 Curvature (ET Section 13.4) 283yxr(t)(t)T(t) = cos (t), sin (t)FIGURE19The curvature is the rate of change of (t ).SOLUTION Since T(t ) is a unit vector that makes an angle (t ) with the positive x-axis, we haveT(t ) = cos (t ), sin (t ) .Differentiating this vector using the Chain Rule gives:T

(t ) = _

(t ) sin (t ),

(t ) cos (t )_ =

(t ) sin (t ), cos (t )We compute the norm of the vector T

(t ):T

(t ) =

(t ) sin (t ), cos (t )= |

(t )|_(sin (t ))2+(cos (t ))2= |

(t )| 1 = |

(t )|When r(s) is a parametrization by arc length we have:(s) =____dTds____ =____dTdt____dtddds =

(t )1|

(t )|dds =ddsas desired.A particle moves along the path y =x3with unit speed. How fast is the tangent turning (i.e., how fast is theangle of inclination changing) when the particle passes through the point (2, 8)?61. Verify Eq. (11) for a circle of radius R. Suppose that a particle traverses the circle at unit speed. Show that the changein the angle during an interval of length t is = t /R and conclude that

(s) = 1/R.SOLUTION The particle traverses the circle at unit speed hence the parametrization is the arc length parametrization ofthe circle. That is,r(s) = R_cossR, sinsR_The angle is (s) =sR, hence the change in the angle during an interval of length s is: = (s +s) (s) =s +sRsR =sRTherefore,s =1Rand we obtain the following derivative:

(s) = lims0s = lims01R =1RThe curvature of a circle of radius R is1R, therefore, we have:dds =1R.This equality veries Eq. (11), for this case.Let (x) be the angle of inclination at a point on the graph y =f (x) (see Exercise 59).(a) Use the relationf

(x) = tan to prove that ddx =f

(x)(1 +f

(x)2).(b) Use the arc length integral to show thatdsdx =_1 +f

(x)2.(c) Now give a proof of Eq. (5) using Eq. (11).63. Use the parametrization r() =f () cos , f () sin to show that a curve r =f () in polar coordinates hascurvature() = | f ()2+2 f

()22 f () f

()|( f ()2+f

()2)3/212SOLUTION By the formula for curvature we have() = r

() r

()r

()3(1)We differentiate r() and r

():r

() = _ f

() cos f () sin , f

() sin +f () cos _r

() = _ f

() cos f

() sin f

() sin f () cos ,f

() sin +f

() cos +f

() cos f () sin _= __ f

() f ()_cos 2 f

() sin ,_f

() f ()_sin +2 f

() cos _284 C HA P T E R 14 CALCULUSOFVECTOR- VALUEDFUNCTI ONS (ET CHAPTER 13)Hence,r

() r

() = _ f

() cos f () sin ___ f

() f ()_sin +2 f

() cos _k_ f

() sin +f () cos ___f

() f ()_cos 2 f

() sin _k=_ f

()_f

() f ()_cos sin f ()_f

() f ()_sin2 +2 f2() cos22 f () f

() sin cos _f

()_f

() f ()_sin cos f ()_ f

() f ()_cos2+2 f

()2sin2 +2 f () f

() cos sin __k=_f ()_f

() f ()_ _sin2 +cos2_+2 f2()_cos2 +sin2__k=_f ()_f

() f ()_+2 f2()_k=_f () f

() +f 2() +2 f2()_kThe length of the cross product is:r

() r

() = | f 2() +2 f

2() f () f

()| (2)We compute the length of r

():r

()2= _f

() cos f () sin _2+_f

() sin +f () cos _2=f

2() cos2 2 f

() f () cos sin +f 2() sin2 +f

2() sin2+2 f

() f () sin cos +f 2() cos2=f

2()_cos2 +sin2_+f 2()_sin2 +cos2_ =f

2() +f 2()Hence,r

() =_ f

2() +f 2() (3)Substituting (2) and (3) in (1) gives:() = | f 2() +2 f