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9.1, y, , INTRODUCTION Recall that a curve C in the xy-plane is simply a set of ordered pairs, , (x(t0), y(t0)), , (x, y). We say that C is a parametric curve if the x- and y-coordinates of a point on the curve are, defined by a pair of functions x f (t), y g(t) that are continuous on some interval a t b., The notion of a parametric curve extends to 3-space as well. A parametric curve in space, or, space curve, is a set of ordered triples (x, y, z) where, , C, , r(t0) = 〈x(t0), y(t0)〉, , x f (t),, , y g(t),, , z h(t),, , (1), , x, , are continuous on an interval defined by a t b. In this section we combine the concepts of, parametric curves with vectors., , (a) 2-space, z, r(t0) = 〈x(t0), y(t0), z(t0)〉, , Vector Functions, , C, , Vector-Valued Functions It is often convenient in science and engineering to introduce, a vector r whose components are functions of a parameter t. We say that, , (x(t0), y(t0), z(t0)), , r(t) f (t), g(t) f (t) i g(t) j, y, , x, (b) 3-space, , FIGURE 9.1.1 Curves defined by vector, functions, , r(t) f (t), g(t), h(t) f (t) i g(t) j h(t) k,, , and, , are vector-valued functions or simply vector functions. As shown in FIGURE 9.1.1, for a given, value of the parameter, say t0, the vector r(t0) is the position vector of a point P on a curve C., In other words, as the parameter t varies, we can envision the curve C being traced out by the, moving arrowhead of r(t)., We have already seen an example of parametric equations, as well as the vector function of, a space curve, in Section 7.5, when we discussed the line in 3-space., , Circular Helix, , EXAMPLE 1, , Graph the curve traced by the vector function, r(t) 2 cos t i 2 sin t j t k,, , t 0., , SOLUTION The parametric equations of the curve are x 2 cos t, y 2 sin t, z t. By, eliminating the parameter t from the first two equations:, x 2 y2 (2 cos t)2 (2 sin t)2 22, we see that a point on the curve lies on the circular cylinder x2 y2 4. As seen in FIGURE 9.1.2, and the accompanying table, as the value of t increases, the curve winds upward in a spiral, or circular helix., t, 0, p/2, p, 3p/2, 2p, 5p/2, 3p, 7p/2, 4p, 9p/2, , x, , y, , z, , 2, 0, 2, 0, 2, 0, 2, 0, 2, 0, , 0, 2, 0, 2, 0, 2, 0, 2, 0, 2, , 0, p/2, p, 3p/2, 2p, 5p/2, 3p, 7p/2, 4p, 9p/2, , z, , cylinder, x2 + y2 = 4, , (, , 7π, 0, –2,, 2, , (, , (, , (, , 3π, 2, , (, , (, , (, , (2, 0, 2 π ), , (2, 0, 0), , FIGURE 9.1.2 Circular helix in Example 1, |, , CHAPTER 9 Vector Calculus, , (, , 0, 2,, , 5π, 2, , (, , (–2, 0, π), , x, , 480, , 9π, 2, , (–2, 0, 3π ), , (2, 0, 4 π ), , 0, –2,, , 0, 2,, , 0, 2,, , π, 2, , (, , y
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The curve in Example 1 is a special case of the vector function, r(t) a cos t i b sin t j ct k,, , a 0,, , b 0,, , c 0,, , which describes an elliptical helix. When a b, the helix is circular. The pitch of a helix is defined, to be the number 2pc. Problems 9 and 10 in Exercises 9.1 illustrate two other kinds of helixes., z, , Circle in a Plane, , EXAMPLE 2, , x2 + y2 = 4, z = 3, , Graph the curve traced by the vector function, r(t) 2 cos t i 2 sin t j 3 k., y, , SOLUTION The parametric equations of this curve are x 2 cos t, y 2 sin t, z 3. As, in Example 1, we see that a point on the curve must also lie on the cylinder x 2 y 2 4., However, since the z-coordinate of any point has the constant value z 3, the vector function, r(t) traces out a circle 3 units above the xy-plane. See FIGURE 9.1.3., , x, , FIGURE 9.1.3 Curve in Example 2, , Curve of Intersection, , EXAMPLE 3, , Find the vector function that describes the curve C of intersection of the plane y 2x and the, paraboloid z 9 x 2 y 2., , z, z = 9 – x2 – y2, , SOLUTION We first parameterize the curve C of intersection by letting x t. It follows, that y 2t and z 9 t 2 (2t)2 9 5t 2. From the parametric equations x t, y 2t,, z 9 5t 2, we see that a vector function describing the trace of the paraboloid in the plane, y 2x is given by r(t) t i 2t j (9 5t 2) k. See FIGURE 9.1.4., , C, , y, y = 2x, , x2 + y2 = 9, x, , FIGURE 9.1.4 Curve in Example 3, , Limits, Continuity, and Derivatives The fundamental notion of the limit of a vector function r(t) f (t), g(t), h(t) is defined in terms of the limits of the component functions., Definition 9.1.1, , Limit of a Vector Function, , If limtSa f (t), limtSa g(t), and limtSa h(t) exist, then, lim r (t) hlim f (t), lim g(t), lim h(t)i., tSa, , tSa, , tSa, , tSa, , The notation t S a in Definition 9.1.1 can, of course, be replaced by t S a, t S a, t S q, or, t S q., As an immediate consequence of Definition 9.1.1, we have the following result., Theorem 9.1.1, , Properties of Limits, , If limtSa r1(t) L1 and limtSa r2(t) L2, then, (i) lim cr1(t) cL1,, tS a, , c a scalar, , (ii) lim [r1(t) r2(t)] L1 L2, tS a, , (iii) lim r1(t) r2(t) L1 L2., tS a, , Definition 9.1.2, , Continuity of a Vector Function, , A vector function r is said to be continuous at t a if, (i) r(a) is defined,, (ii) lim r(t) exists, and, (iii) lim r(t) r(a)., tS a, , tS a, , Equivalently, r(t) is continuous at t a if and only if the component functions f, g, and h are, continuous there., 9.1 Vector Functions, , |, , 481
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Definition 9.1.3, , Derivative of Vector Function, , The derivative of a vector function r is, r (t) lim, , DtS0, , 1, [r(t t) r(t)], Dt, , (2), , for all t for which the limit exists., The derivative of r is also written dr/dt. The next theorem will show that on a practical level, the derivative of a vector function is obtained by simply differentiating its component functions., Theorem 9.1.2, , Differentiation of Components, , If r(t) f (t), g(t), h(t), where f, g, and h are differentiable, then, r (t) f (t), g (t), h (t)., , PROOF: From (2) we have, , tangent, , P, Δr, , r9(t) lim, , r(t), , z, , DtS0, , r(t + Δt), , C, , lim h, , y, , DtS0, , x, , (a), , Δr, , Δt > 0, Δt, , r(t), , z, , r(t + Δt), , Smooth Curves When the component functions of a vector function r have continuous, first derivatives and r (t) 0 for all t in the open interval (a, b), then r is said to be a smooth, function and the curve C traced by r is called a smooth curve., Geometric Interpretation of r (t) If the vector r (t) is not 0 at a point P, then it may, be drawn tangent to the curve at P. As seen in FIGURE 9.1.5, the vectors, , C, , y, x, , r r(t t) r(t), , (b), , FIGURE 9.1.5 Vector r (t) is tangent to, curve C at P, , r′, , ((, , f (t Dt) 2 f (t) g(t Dt) 2 g(t) h(t Dt) 2 h(t), ,, ,, i., Dt, Dt, Dt, , Taking the limit of each component yields the desired result., , tangent, P, , 1, B k f ( t Dt), g(t Dt), h(t Dt)l 2 k f ( t), g(t), h(t)l R, Dt, , Dr, 1, , [r(t t) r(t)], Dt, Dt, , are parallel. If we assume lim tS0 r/ t exists, it seems reasonable to conclude that as t S 0,, r(t) and r(t t) become close and, as a consequence, the limiting position of the vector r/ t, is the tangent line at P. Indeed, the tangent line at P is defined as that line through P parallel to r (t)., , y, , π, 6, , EXAMPLE 4, , ( (, , (1, 0), , Tangent Vectors, , Graph the curve C that is traced by a point P whose position is given by r(t) cos 2t i sin t j,, 0 t 2p. Graph r (0) and r (p/6)., , r′(0), , 1 1, ,, 2 2, , x, , x = 1 – 2y2, , SOLUTION By clearing the parameter from the parametric equations x cos 2t, y sin t,, 0 t 2p, we find that C is the parabola x 1 2y2, 1 x 1. From r (t) 2 sin 2t i cos t j, we find, r (0) j, , FIGURE 9.1.6 Tangent vectors in, Example 4, , 482, , and, , |, , CHAPTER 9 Vector Calculus, , and, , p, !3, r9 a b !3 i , j., 6, 2, , In FIGURE 9.1.6 these vectors are drawn tangent to the curve C at (1, 0) and ( 12 , 12 ), respectively.
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Tangent Line, , EXAMPLE 5, , Find parametric equations of the tangent line to the graph of the curve C whose parametric, equations are x t 2, y t 2 t, z 7t at t 3., SOLUTION The vector function that gives the position of a point P on the curve is given by, r(t) t 2 i (t 2 t) j 7t k. Now,, r (t) 2t i (2t 1) j 7 k, , and so, , r (3) 6 i 5 j 7 k,, , which is tangent to C at the point whose position vector is, r(3) 9 i 6 j 21 k;, that is, P(9, 6, 21). Using the components of r (3), we see that parametric equations of the, tangent line are x 9 6t, y 6 5t, z 21 7t., , Higher-Order Derivatives Higher-order derivatives of a vector function are also, obtained by differentiating its components. In the case of the second derivative, we have, r (t) f (t), g (t), h (t) f (t) i g (t) j h (t) k., , Derivative of a Vector Function, , EXAMPLE 6, , If r(t) (t 2t ) i 4t j et k, then, 3, , 2, , r (t) (3t 2 4t) i 4 j et k, Theorem 9.1.3, , and, , r (t) (6t 4) i et k., , Chain Rule, , If r is a differentiable vector function and s u(t) is a differentiable scalar function, then the, derivative of r(s) with respect to t is, dr, dr ds, , r (s) u (t)., dt, ds dt, , EXAMPLE 7, , Chain Rule, , If r(s) cos 2s i sin 2s j e3s k, where s t 4, then, dr, [2 sin 2s i 2 cos 2s j 3e3s k]4t 3, dt, 4, , 8t 3 sin(2t 4) i 8t 3 cos(2t 4) j 12t 3 e23t k., Details of the proof of the next theorem are left as exercises., Theorem 9.1.4, , Rules of Differentiation, , Let r1 and r2 be differentiable vector functions and u(t) a differentiable scalar function., d, (i), fr (t) r2(t)g r9(t), r9(t), 1, 2, dt 1, d, (ii), fu(t) r1(t)g u(t) r19(t) u9(t) r1(t), dt, d, (iii), fr (t) r2(t)g r1(t) r29(t) r19(t) r2(t), dt 1, d, (iv), fr (t) 3 r2(t)g r1(t) 3 r29(t) r19(t) 3 r2(t)., dt 1, , 9.1 Vector Functions, , |, , 483
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Note of caution., , Since the cross product of two vectors is not commutative, the order in which r1 and r2 appear, in part (iv) of Theorem 9.1.4 must be strictly observed., , Integrals of Vector Functions If f, g, and h are integrable, then the indefinite and, definite integrals of a vector function r(t) f (t) i g(t) j h(t) k are defined, respectively, by, , #r (t) dt B #f (t) dtR i B #g(t) dtR j B #h(t) dtR k, #, , b, , a, , r (t) dt B, , #, , b, , a, , f (t) dtR i B, , #, , b, , a, , b, , g(t) dtR j B, , # h(t) dtR k., a, , The indefinite integral of r is another vector function R c such that R (t) r(t)., , Integral of a Vector Function, , EXAMPLE 8, , If r(t) 6t 2 i 4e2t j 8 cos 4t k,, then, , #r(t) dt B #6t, , 2, , #, , #, , dtR i B 4e 2t dtR j B 8 cos 4t dtR k, , [2t 3 c1] i [2e2t c2] j [2 sin 4t c3] k, 2t 3 i 2e2t j 2 sin 4t k c,, where c c1i c2 j c3 k., , Length of a Space Curve If r(t) f (t) i g(t) j h(t) k is a smooth function, then it, can be shown that the length of the smooth curve traced by r is given by, b, , s, , # "f f 9(t)g, , b, , 2, , a, , 2, , 2, , fg9(t)g fh9(t)g dt , , # ir9(t)i dt., , (3), , a, , Arc Length as a Parameter A curve in the plane or in space can be parameterized in, , terms of the arc length s., EXAMPLE 9, , Example 1 Revisited, , Consider the helix of Example 1. Since i r (t) i !5, it follows from (3) that the length of, the curve from r(0) to an arbitrary point r(t) is, t, , s, , # "5 du "5t,, 0, , where we have used u as a dummy variable of integration. Using t s/ !5, we obtain a vector, equation of the helix as a function of arc length:, r(s) 2 cos, , s, , "5, , i 2 sin, , Parametric equations of the helix are then, f (s) 2 cos, , s, "5, , ,, , g(s) 2 sin, , s, , "5, , j, , s, "5, , ,, , s, , "5, , k., , h(s) , , (4), , s, "5, , ., , The derivative of a vector function r(t) with respect to the parameter t is a tangent vector to the, curve traced by r. However, if the curve is parameterized in terms of arc length s, then r (s) is a, unit tangent vector. To see this, let a curve be described by r(s), where s is arc length. From (3),, the length of the curve from r(0) to r(s) is s 0s i r (u) i du. Differentiation of this last equation, with respect to s then yields i r (s) i 1., 484, , |, , CHAPTER 9 Vector Calculus
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Exercises, , 9.1, , Answers to selected odd-numbered problems begin on page ANS-20., , In Problems 1–10, graph the curve traced by the given vector, function., 1. r(t) 2 sin t i 4 cos t j t k; t 0, 2. r(t) cos t i t j sin t k; t 0, 3. r(t) t i 2t j cos t k; t 0, 4. r(t) 4i 2 cos t j 3 sin t k, 5. r(t) et, e2t , 6. r(t) cosh t i 3 sinh t j, 7. r(t) !2 sin t,!2 sin t, 2 cos t; 0 t p/2, 8. r(t) t i t 3 j t k, 9. r(t) e t cos t i et sin t j e t k, 10. r(t) t cos t, t sin t, t 2 , , d, [r(t) (r (t) r (t))], dt, d, 30., [r (t) (r2(t) r3(t))], dt 1, 29., , d, 1, cr1(2t) r2 a b d, dt, t, d 3 2, 32., ft r(t )g, dt, 31., , In Problems 33–36, evaluate the given integral., 33., , #, , 2, , (t i 3t 2 j 4t 3 k) dt, , 21, , In Problems 11–14, find the vector function that describes the, curve C of intersection between the given surfaces. Sketch the, curve C. Use the indicated parameter., 11. z x 2 y2, y x; x t, 12. x 2 y2 z2 1, y 2x; x t, 13. x 2 y2 9, z 9 x2; x 3 cos t, 14. z x 2 y2, z 1; x sin t, sin 2t, 15. Given that r(t) , i (t 2)5j t ln t k, find, t, lim 1 r( t)., tS 0, , 16. Given that limtSa r1(t) i 2 j k and limtSa r2(t) 2 i , , 5 j 7 k, find:, (a) lim [4r1(t) 3r2(t)], tS a, , (b) lim r1(t) r2(t)., tS a, , In Problems 17–20, find r (t) and r (t) for the given vector function., 17. r(t) ln t i j, t 0, 18. r(t) t cos t sin t, t cos t , 19. r(t) te2t, t 3, 4t 2 t, 20. r(t) t 2 i t 3 j tan1t k, In Problems 21–24, graph the curve C that is described by r and, graph r at the indicated value of t., 21. r(t) 2 cos t i 6 sin t j; t p/6, 22. r(t) t 3 i t 2 j; t 1, 4, 23. r(t) 2 i t j , k; t 1, 1 t2, 24. r(t) 3 cos t i 3 sin t j 2t k; t p/4, In Problems 25 and 26, find parametric equations of the tangent, line to the given curve at the indicated value of t., 1, , 1, , 25. x t, y 2 t 2, z 3 t 3; t 2, 26. x t 3 t, y , , 6t, , z (2t 1)2; t 1, t11, , In Problems 27–32, find the indicated derivative. Assume that all, vector functions are differentiable., d, d, 27., [r(t) r (t)], 28., [r(t) (t r(t))], dt, dt, , 4, , 34., , # ( "2t 1i 2 "t j sin p t k) dt, # (te i e j te k) dt, 1, # 1 t (i t j t k) dt, 0, , 35., 36., , 2t, , t, , t2, , 2, , 2, , In Problems 37–40, find a vector function r that satisfies the, indicated conditions., 37. r (t) 6i 6t j 3t 2 k; r(0) i 2j k, 3, 38. r (t) t sin t 2 i cos 2t j; r(0) 2 i, 39. r (t) 12t i 3t 1/2j 2k; r (1) j, r (1) 2i k, 40. r (t) sec2 t i cos t j sin t k;, r (0) i j k, r(0) j 5k, In Problems 41–44, find the length of the curve traced by the, given vector function on the indicated interval., 41. r(t) a cos t i a sin t j ct k; 0 t 2p, 42. r(t) t i t cos t j t sin t k; 0 t p, 43. r(t) et cos 2t i et sin 2t j et k; 0 t 3p, 2, 44. r(t) 3t i !3t 2 j 3 t 3 k; 0 t 1, 45. Express the vector equation of a circle r(t) a cos t i , a sin t j as a function of arc length s. Verify that r (s) is a, unit vector., 46. If r(s) is the vector function given in (4), verify that r (s) is a, unit vector., 47. Suppose r is a differentiable vector function for which, i r(t) i c for all t. Show that the tangent vector r (t) is, perpendicular to the position vector r(t) for all t., 48. In Problem 47, describe geometrically the kind of curve C for, which i r(t) i c., , Miscellaneous Problems, 49., 50., 51., 52., , Prove Theorem 9.1.4(ii)., Prove Theorem 9.1.4(iii)., Prove Theorem 9.1.4(iv)., If v is a constant vector and r is integrable on [a, b], prove, that ab v r(t) dt v ab r(t) dt., 9.1 Vector Functions, , |, , 485
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9.2, , Motion on a Curve, , INTRODUCTION In the preceding section we saw that the first and second derivatives of the, , vector function r(t) f (t), g(t), h(t) f (t) i g(t) j h(t) k can be obtained by differentiating the component functions f, g, and h. In this section we give a physical interpretation to the, vectors r (t) and r (t)., , Velocity and Acceleration Suppose a particle or body moves along a curve C so that, its position at time t is given by the vector function r(t) f (t) i g(t) j h(t) k. If the component, functions f, g, and h have second derivatives, then the vectors, v(t) r (t) f (t) i g (t) j h (t) k, a(t) r (t) f (t) i g (t) j h (t) k, are called the velocity and acceleration of the particle, respectively. The scalar function i v(t) i, is the speed of the particle. Since, iv(t)i g, , z, , t, , speed is related to arc length s by s (t) i v(t) i . In other words, arc length is given by s et01 i v(t) i dt., It also follows from the discussion of Section 9.1 that if P(x1, y1, z1) is the position of the particle, on C at time t1, then we may draw v(t1) tangent to C at P. Similar remarks hold for curves traced, by the vector function r(t) f (t) i g(t) j., , v(2), C, P(4, 2, 5), a(2), , EXAMPLE 1, , Velocity and Acceleration Vectors, , The position of a moving particle is given by r(t) t 2 i t j 52 t k. Graph the curve defined, by r(t) and the vectors v(2) and a(2)., , y, x = y2, , SOLUTION Since x t 2, y t, the path of the particle is above the parabola x y2. When, t 2 the position vector r(2) 4 i 2 j 5 k indicates that the particle is at the point, P(4, 2, 5). Now,, 5, v(t) r (t) 2t i j k, and, a(t) r (t) 2 i, 2, , (4, 2, 0), , x, , dy 2, dr, dx 2, dz 2, g a b a b a b, dt, Å dt, dt, dt, , FIGURE 9.2.1 Velocity and acceleration, vectors in Example 1, , so that v(2) 4 i j 52 k and a(2) 2 i. These vectors are shown in FIGURE 9.2.1., If a particle moves with a constant speed c, then its acceleration vector is perpendicular to, the velocity vector v. To see this, note that i v i 2 c2 or v v c2. We differentiate both sides, with respect to t and obtain, with the aid of Theorem 9.1.4(iii):, d, dv, dv, dv, (v v) v , , v 2v , 0., dt, dt, dt, dt, Thus,, , EXAMPLE 2, , dv, v0, dt, , or, , a(t) v(t) 0 for all t., , Velocity and Acceleration Vectors, , Suppose the vector function in Example 2 of Section 9.1 represents the position of a particle, moving in a circular orbit. Graph the velocity and acceleration vector at t p/4., SOLUTION Recall that r(t) 2 cos t i 2 sin t j 3 k is the position vector of a particle, moving in a circular orbit of radius 2 in the plane z 3. When t p/4 the particle is at the, point P( !2, !2, 3). Now,, v(t) r (t) 2 sin t i 2 cos t j, , a(t) r (t) 2 cos t i 2 sin t j., 486, , |, , CHAPTER 9 Vector Calculus
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a, , z, , ((, π, 4, , v, , Since the speed is i v(t) i 2 for all time t, it follows from the discussion preceding this, example that a(t) is perpendicular to v(t). (Verify this.) As shown in FIGURE 9.2.2, the vectors, , ((, π, 4, , p, p, p, v a b 2 sin i 2 cos j "2 i "2 j, 4, 4, 4, , P(√2 , √2 , 3), z=3, , p, p, p, a a b 2 cos i 2 2 sin j "2 i 2 "2 j, 4, 4, 4, , y, , are drawn at the point P. The vector v(p/4) is tangent to the circular path and a(p/4) points, along a radius toward the center of the circle., , x, , FIGURE 9.2.2 Velocity and acceleration, vectors in Example 2, , Centripetal Acceleration For circular motion in the plane, described by r(t) r0 cos vt i , r0 sin vt j, r0 and v constants, it is evident that r v2 r. This means that the acceleration, vector a(t) r (t) points in the direction opposite to that of the position vector r(t). We then say, a(t) is centripetal acceleration. See FIGURE 9.2.3. If v i v(t) i and a i a(t) i , we leave it as, an exercise to show that a v2 /r0., , v(t1), v(t2), r(t2), , Curvilinear Motion in the Plane Many important applications of vector functions, occur in regard to curvilinear motion in a plane. For example, planetary and projectile motion, take place in a plane., In analyzing the motion of short-range ballistic projectiles,* we begin with the acceleration, of gravity written in vector form: a(t) g j., If, as shown in FIGURE 9.2.4, a projectile is launched with an initial velocity v0 v0 cos u i , v0 sin u j from an initial height s0 s0 j, then, , a(t1), , FIGURE 9.2.3 Circular motion, , #, , v(t) (g j) dt gt j c1,, , y, v0, (v0 sin θ ) j, s0 j, , where v(0) v0 implies that c1 v0. Therefore,, v(t) (v0 cos u) i (gt v0 sin u) j., , θ, (v0 cos θ )i, , Integrating again and using r(0) s0 yields, x, , FIGURE 9.2.4 Trajectory of a projectile, y, , H, x, , 1, r(t) (v0 cos u)t i c gt 2 (v0 sin u) t s0 d j., 2, , (1), , Hence, parametric equations for the trajectory of the projectile are, 1, x(t) (v0 cos u)t,, y(t) gt 2 (v0 sin u)t s0., (2), 2, We are naturally interested in finding the maximum height H and the range R attained by a, projectile. As shown in FIGURE 9.2.5, these quantities are the maximum values of y(t) and x(t),, respectively., EXAMPLE 3, , Trajectory of a Projectile, , A projectile is launched from ground level with an initial speed v0 768 ft/s at an angle of, elevation u 30. Find (a) the vector function and parametric equations of the projectile’s, trajectory, (b) the maximum altitude attained, (c) the range of the projectile, and (d) the, impact speed., , (a) Maximum height H:, Find t1 for which y′(t1) = 0;, H = ymax = y(t1), y, , SOLUTION (a) The initial height and velocity are, respectively, s0 0 and, , R, , r9(0) v0 (768 cos 308)i (768 sin 308)j 384"3i 384j., x, , (b) Range R:, Find t1 > 0 for which y (t1) = 0;, R = xmax = x(t1), , FIGURE 9.2.5 Maximum height and range, of a projectile, , (3), , Integrating a(t) 32j and using (3) give, v(t) (384"3)i (32t 384)j., , (4), , *A projectile is shot or hurled rather than self-propelled. In the analysis of long-range ballistic motion,, the curvature of the Earth must be taken into consideration., , 9.2 Motion on a Curve, , |, , 487
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Integrating (4) and using r(0) s0 0 then give, r(t) (384"3)t i (16t 2 384t)j., The components of this vector function,, x(t) (384"3)t, y(t) 16t 2 384t,, , (5), , are the parametric equations of the projectile’s trajectory., (b) From (5) we see that y (t) 0 when 32t 384 0 or t 12 s. The maximum, height attained by the projectile is, H y(12) 16(12)2 384(12) 2304 ft., (c) From (5) we see that y(t) 0 when 16t 2 384t 16t(t 2 24) 0. The time, that projectile hits the ground is t 24 s and the corresponding range is, R x(24) 384"3(24) < 15,963 ft., (d) Finally, from (4) we see that v(24) (384"3)i (384)j and so the impact speed is, 7 v(24) 7 "(384!3)2 (384)2 768 ft/s., , In Example 3, note that the impact speed is the same as the launch speed v0 768 ft/s. Verify, that this is still the case if we change the angle of elevation to, say, u 50. See Problem 16 in, Exercises 9.2., , REMARKS, We have seen that the rate of change of arc length ds/dt is the same as the speed i v(t)i i r (t)i ., However, as we shall see in the next section, it does not follow that the scalar acceleration, d 2s/dt 2 is the same as i a(t)i i r (t)i . See Problem 24 in Exercises 9.2., , Exercises, , 9.2, , Answers to selected odd-numbered problems begin on page ANS-21., , In Problems 1–8, r(t) is the position vector of a moving particle., Graph the curve and the velocity and acceleration vectors at the, indicated time. Find the speed at that time., 1, , 1. r(t) t 2 i 4 t 4 j; t 1, , 1, j; t 1, t2, r(t) cosh 2t i sinh 2t j; t 0, r(t) 2 cos t i (1 sin t) j; t p/3, r(t) 2 i (t 1)2 j t k; t 2, r(t) t i t j t 3 k; t 2, r(t) t i t 2 j t 3 k; t 1, r(t) t i t 3 j t k; t 1, Suppose r(t) t 2 i (t 3 2t) j (t 2 5t) k is the position, vector of a moving particle. At what points does the particle, , 2. r(t) t 2 i , 3., 4., 5., 6., 7., 8., 9., , 488, , |, , CHAPTER 9 Vector Calculus, , pass through the xy-plane? What are its velocity and acceleration at these points?, 10. Suppose a particle moves in space so that a(t) 0 for all time t., Describe its path., 11. A shell is fired from ground level with an initial speed of, 480 ft/s at an angle of elevation of 30. Find:, (a) a vector function and parametric equations of the shell’s, trajectory,, (b) the maximum altitude attained,, (c) the range of the shell, and, (d) the speed at impact., 12. Rework Problem 11 if the shell is fired with the same initial, speed and the same angle of elevation but from a cliff 1600 ft, high.
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13. A used car is pushed off an 81-ft-high sheer seaside cliff with, , 19. If a projectile is launched from level ground, the initial con-, , a speed of 4 ft/s. Find the speed at which the car hits the water., 14. A small projectile is launched from ground level with an initial, speed of 98 m/s. Find the possible angles of elevation so that, its range is 490 m., 15. A football quarterback throws a 100-yd “bomb” at an angle, of 45 from the horizontal. What is the initial speed of the, football at the point of release?, 16. If a projectile is launched from level ground, then the initial conditions are r(0) 0, r9(0) v0 (v0 cos u)i (v0 sin u)j and, the vector function (1) becomes, , ditions are r(0) 0, r9(0) v0 (v0 cos u)i (v0 sin u)j,, and if linear air resistance is taken into consideration, the, vector function analogue of (1) is, , r(t) (v0 cos u)t i f12gt 2 (v0 sin u)tg j., , (6), , This function is equivalent to the solution x(t), y(t) of the, system of linear differential equations (8) in Problem 21 of, Exercises 4.6 subject to the same initial conditions. Underlying, that earlier discussion, as well as in the derivation of (1), is the, assumption that the projectile is not subject to air resistance., Use (6) to show that the impact speed of a projectile is the, same as the initial speed 7 r9(0) 7 7 v0 7 v0 for any angle, of elevation u, 0 , u , 90., 17. When air resistance is ignored, we saw in Problem 21, of Exercises 4.6 that a projectile launched at an angle u,, 0 , u , 90 from level ground (r(0) 0), then its horizontal, range and maximum height are, respectively,, R, , v 20, sin 2u, g, , and, , H, , v 20, sin2 u., 2g, , From the first formula, it follows that when the projectile, is launched at distinct complementary angles the horizontal, range is the same and that the maximum range is attained, when the angle of elevation is u 45. But if the projectile is, launched from an initial height s0 . 0 the foregoing formulas, and statements are not valid., (a) With v0 480 ft/s, s0 1600 ft rework Problem 12,, but this time use the complementary angle of elevation, u 60. Compare the maximum height, range, and impact speed of the projectile with the answers to parts (b),, (c) and (d) of Problem 12., (b) Use a graphing utility or CAS to plot the trajectory of the, projectile defined by the parametric equations x(t) and, y(t) in part (a) of Problem 12. Repeat for the parametric, equations in part (a) of this problem. Superimpose both, of these curves on the same coordinate system., 18. As mentioned in Problem 17, if a projectile is launched from, an initial height s0 . 0 the maximum range of the projectile, is not attained using u 45 as the angle of elevation., (a) A projectile is launched from an initial height of v0 480 ft/s,, s0 1600 ft, at an angle of elevation of u 45. Use a, calculator or CAS to find the time the projectile hits the, ground and the corresponding range., (b) If the angle of elevation in part (a) is changed to u 39.76,, show that the range is greater than that in part (a)., (c) Use a graphing utility or CAS to plot the trajectories of, the projectiles in parts (a) and (b). Superimpose both of, these curves on the same coordinate system., , r(t) , , mv0 cos u, (1 2 e bt>m) i, b, mv0 sin u, m2g, mg, ca, 2 b(1 2 e bt>m) 2, td j,, b, b, b, , This function is equivalent to the solution x(t), y(t) of the, system of linear differential equations (11) in Problem 22 of, Exercises 4.6 subject to the same initial conditions. Here b . 0, is a constant of proportionality related to the air resistance or, drag. If m 14 slug, g 32 ft/s 2, b 0.02, v0 300 ft/s,, and u 38, use a calculator or CAS to find the impact speed, of the projectile. See part (b) of Problem 22 in Exercises 4.6., 20. Suppose the angle of elevation of the projectile in Problem 19, is changed to u 52. Do you think that the impact speed of, the projectile is the same, greater than, or less than the value, found in that problem? Prove your assertion., 21. A projectile is fired from a cannon directly at a target that, is dropped from rest simultaneously as the cannon is fired., Show that the projectile will strike the target in midair. See, FIGURE 9.2.6. [Hint: Assume that the origin is at the muzzle of, the cannon and that the angle of elevation is u. If rp and rt are, position vectors of the projectile and target, respectively, is, there a time at which rp rt?], , FIGURE 9.2.6 Cannon in Problem 21, 22. In army field maneuvers sturdy equipment and supply packs, , are simply dropped from planes that fly horizontally at a, slow speed and a low altitude. A supply plane flies horizontally over a target at an altitude of 1024 ft at a constant, speed of 180 mi/h. Use (1) to determine the horizontal distance a supply pack travels relative to the point from which, it was dropped. At what line-of-sight angle a should the, supply pack be released in order to hit the target indicated, in FIGURE 9.2.7?, , α, , supply, pack, , 1024 ft, , target, , FIGURE 9.2.7 Supply plane in Problem 22, , 9.2 Motion on a Curve, , |, , 489
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23. Suppose that r(t) ⫽ r0 cos vt i ⫹ r0 sin vt j is the position, , vector of an object that is moving in a circle of radius r0 in, the xy-plane. If i v(t) i ⫽ v, show that the magnitude of the, centripetal acceleration is a ⫽ i a(t) i ⫽ v2 /r0., 24. The motion of a particle in space is described by, r(t) ⫽ b cos t i ⫹ b sin t j ⫹ ct k,, , t ⱖ 0., , (a) Compute i v(t) i ., (b) Compute s ⫽ 兰0t i v(t) i dt and verify that ds/dt is the same, as the result of part (a)., (c) Verify that d 2s/dt 2 ⫽ i a(t) i ., 25. The effective weight we of mass m at the equator of the Earth, is defined by we ⫽ mg ⫺ ma, where a is the magnitude of the, centripetal acceleration given in Problem 23. Determine the, effective weight of a 192-lb person if the radius of the Earth, is 4000 mi, g ⫽ 32 ft/s2, and v ⫽ 1530 ft/s., 26. Consider a bicyclist riding on a flat circular track of radius, r0. If m is the combined mass of the rider and bicycle, fill in, the blanks in FIGURE 9.2.8. [Hint: Use Problem 23 and force ⫽, mass ⫻ acceleration. Assume that the directions are upward, and to the left.] The resultant vector U gives the direction the, bicyclist must be tipped to avoid falling. Find the angle f from, the vertical at which the bicyclist must be tipped if her speed, is 44 ft/s and the radius of the track is 60 ft., resultant, , 〈0, —〉, , U = 〈—, —〉, , φ, , force exerted, by track =, opposite of, the combined, weight of bike, and person, , 〈—, 0〉, centripetal force, , FIGURE 9.2.8 Bicyclist in Problem 26, 27. The velocity of a particle moving in a fluid is described by, , means of a velocity field v ⫽ v1 i ⫹ v2 j ⫹ v3 k, where the components v1, v2, and v3 are functions of x, y, z, and time t. If the, velocity of the particle is v(t) ⫽ 6t 2 x i ⫺ 4ty2 j ⫹ 2t(z ⫹ 1) k,, find r(t). [Hint: Use separation of variables.], 28. Suppose m is the mass of a moving particle. Newton’s second, law of motion can be written in vector form as, dp, d, F ⫽ m a ⫽ (m v) ⫽, ,, dt, dt, where p ⫽ m v is called linear momentum. The angular, momentum of the particle with respect to the origin is defined, to be L ⫽ r ⫻ p, where r is its position vector. If the torque, of the particle about the origin is t ⫽ r ⫻ F ⫽ r ⫻ d p/dt,, show that t is the time rate of change of angular momentum., 490, , |, , CHAPTER 9 Vector Calculus, , 29. Suppose the Sun is located at the origin. The gravitational force, , F exerted on a planet of mass m by the Sun of mass M is, F 5 2k, , Mm, u., r2, , F is a central force—that is, a force directed along the position vector r of the planet. Here k is the gravitational constant,, r ⫽ i r i , u ⫽ r/r is a unit vector in the direction of r, and the, minus sign indicates that F is an attractive force—that is, a, force directed toward the Sun. See FIGURE 9.2.9., (a) Use Problem 28 to show that the torque acting on the, planet due to this central force is 0., (b) Explain why the angular momentum L of a planet is constant., planet, m, , sun, r, , F, , M, , FIGURE 9.2.9 Force F in Problem 29, , Discussion Problems, 30. In this problem the student will use the properties in Sections, , 7.4 and 9.1 to prove Kepler’s first law of planetary motion:, The orbit of a planet is an ellipse with the Sun at one focus. We, assume that the Sun has mass M and is located at the origin, r, is the position vector of a body of mass m moving under the, gravitational attraction of the Sun, and u ⫽ r/r is a unit vector, in the direction of r., (a) Use Problem 29 and Newton’s second law of motion, F ⫽ m a to show that, d 2r, u, 5 2kM 2 ., dt 2, r, (b) Use part (a) to show that r ⫻ r⬙ ⫽ 0., d, (c) Use part (b) to show that (r ⫻ v) ⫽ 0., dt, (d) It follows from part (c) that r ⫻ v ⫽ c, where c is a constant, vector. Show that c ⫽ r 2 (u ⫻ u⬘)., d, (e) Show that (u ⭈ u) ⫽ 0 and consequently u ⭈ u⬘ ⫽ 0., dt, (f ) Use parts (a), (e), and (d) to show that, d, du, (v ⫻ c) ⫽ kM, ., dt, dt, (g) By integrating the result in part (f) with respect to t, we, get v ⫻ c ⫽ kMu ⫹ d, where d is another constant vector., Dot both sides of this last expression by the vector r ⫽, r u and use Problem 61 in Exercises 7.4 to show that, r⫽, , c 2>kM, , 1 ⫹ (d>kM) cos u, , ,, , where c ⫽ i c i , d ⫽ i d i , and u is the angle between d and r.
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(h) Explain why the result in part (g) proves Kepler’s first law., (i) At perihelion, the point in the orbit where the body is, closest to the Sun, the vectors r and v are perpendicular, and have magnitudes r0 and v0, respectively. Use this, information and parts (d) and (g) to show that c r0v0, and d r0v02 kM., 31. Suppose a projectile is launched from an initial height s0 at, an angle of elevation u. If air resistance is ignored, show that, the horizontal range is given by, , Note that when s0 0 this formula reduces to the range R, given in Problem 17., 32. Consider the formula in Problem 31 as a function of the single, variable u. Show that the range of a projectile launched from, an initial height s0 is a maximum for the angle of inclination, 2s0g v 20, ., Å 2s0g 2v 20, , Note that when s0 0 this formula reduces to the value given, in Problem 17, that is, u p/4 or 45. Use this formula to, verify the angle of inclination used in part (b) of Problem 18., , v0 cos u, R, (v0 sin u "v 20 sin2 u 2s0g )., g, , 9.3, , u cos1, , Curvature and Components of Acceleration, , INTRODUCTION Let C be a smooth curve in either 2- or 3-space traced out by a vector, , function r(t). In this section we are going to consider in greater detail the acceleration vector, a(t) r (t) introduced in the last section. But before doing this, we need to examine a scalar, quantity called the curvature of a curve., , A Definition We know that r (t) is a tangent vector to the curve C, and consequently, T(t) , T, , T, , T, , T, , C, , T, T, , T, , P3, , d r>dt, r9(t), dr, d r ds, dr, T(t)., , and so, , , dt, ds dt, ds, ds>dt, ir9(t)i, , T, , P1, , FIGURE 9.3.1 Unit tangents, , (1), , is a unit tangent. But recall from the end of Section 9.1 that if C is parameterized by arc length s,, then a unit tangent to the curve is also given by dr/ds. The quantity i r (t) i in (1) is related to arc, length s by ds/dt i r (t) i . Since the curve C is smooth, we know from pages 482 and 484 that, ds/dt 0. Hence by the Chain Rule,, , P2, , T, , r9(t), ir9(t)i, , (2), , Now suppose C is as shown in FIGURE 9.3.1. As s increases, T moves along C, changing direction, but not length (it is always of unit length). Along the portion of the curve between P1 and P2 the, vector T varies little in direction; along the curve between P2 and P3, where C obviously bends, more sharply, the change in the direction of the tangent T is more pronounced. We use the rate, at which the unit vector T changes direction with respect to arc length as an indicator of the, curvature of a smooth curve C., Definition 9.3.1, , Curvature, , Let r(t) be a vector function defining a smooth curve C. If s is the arc length parameter and, T d r/ds is the unit tangent vector, then the curvature of C at a point is, k5 g, , dT, g., ds, , (3), , The symbol k in (3) is the Greek letter kappa. Now since curves are generally not parameterized, by arc length, it is convenient to express (3) in terms of a general parameter t. Using the Chain, Rule again, we can write, d T>dt, dT, d T ds, dT, 5, and consequently, 5, ., dt, ds dt, ds, ds>dt, 9.3 Curvature and Components of Acceleration, , |, , 491
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In other words, curvature is given by, k(t) , EXAMPLE 1, , iT9(t)i, ., ir9(t)i, , (4), , Curvature of a Circle, , Find the curvature of a circle of radius a., SOLUTION A circle can be described by the vector function r(t) a cos t i a sin t j. Now, from r (t) a sin t i a cos t j and i r (t) i a, we get, T(t) , large curvature κ, , r9(t), sin t i cos t j, ir9(t)i, , and, , T (t) cos t i sin t j., , Hence, from (4) the curvature is, k(t) , small curvature κ, , iT9(t)i, " cos2 t sin2 t, 1, , ., a, a, ir9(t)i, , (5), , The result in (5) shows that the curvature at a point on a circle is the reciprocal of the radius, of the circle and indicates a fact that is in keeping with our intuition: A circle with a small, radius curves more than one with a large radius. See FIGURE 9.3.2., , FIGURE 9.3.2 Curvature of a circle in, Example 1, , Tangential and Normal Components of Acceleration Suppose a particle moves, in 2- or 3-space on a smooth curve C described by the vector function r(t). Then the velocity of the, particle on C is v(t) r (t), whereas its speed is ds/dt v i v(t) i . Thus, (1) implies v(t) v T., Differentiating this last expression with respect to t gives acceleration:, a(t) v, z, , T, P, N, , is a unit normal to the curve C at P with direction given by d T/dt. The vector N is also called, idT>dti, , it follows from (7) that d T/dt , the principal normal. But since curvature is k , v, kv N. Thus, (6) becomes, , aNN, y, , a(t) kv2 N , , x, C, , By writing (8) as, , FIGURE 9.3.3 Components of, acceleration, z, , P, , N, osculating plane, , C, , FIGURE 9.3.4 Osculating plane, , a(t) aN N aT T,, , |, , (8), (9), , The Binormal A third unit vector defined by, y, , B(t) T(t), , N(t), , is called the binormal. The three unit vectors T, N, and B form a right-handed set of mutually, orthogonal vectors called the moving trihedral. The plane of T and N is called the osculating, plane,* the plane of N and B is said to be the normal plane, and the plane of T and B is the, rectifying plane. See FIGURE 9.3.4., *Literally, this means the “kissing” plane., , 492, , dv, T., dt, , we see that the acceleration vector a of the moving particle is the sum of two orthogonal vectors aN N and aT T. See FIGURE 9.3.3. The scalar functions aT dv/dt and aN kv2 are called the, tangential and normal components of the acceleration, respectively. Note that the tangential, component of the acceleration results from a change in the magnitude of the velocity v, whereas, the normal component of the acceleration results from a change in the direction of v., , B=T×N, T, , x, , (6), , Furthermore, with the help of Theorem 9.1.4(iii), it follows from the differentiation of, T T 1 that T d T/dt 0. Hence, at a point P on C the vectors T and d T/dt are orthogonal., If i d T/dt i 0, the vector, d T>dt, (7), N(t) , id T>dti, , aTT, , a, , dT, dv, , T., dt, dt, , CHAPTER 9 Vector Calculus
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The three mutually orthogonal unit vectors T, N, and B can be thought of as a movable righthanded coordinate system since, B(t) T(t), , N(t) B(t), , N(t),, , T(t) N(t), , T(t),, , B(t)., , This movable coordinate system is referred to as the TNB-frame., , Tangent, Normal, and Binormal Vectors, , EXAMPLE 2, , The position of a moving particle is given by r(t) 2 cos t i 2 sin t j 3t k. Find the vectors, T, N, and B. Find the curvature., SOLUTION Since r (t) 2 sin t i 2 cos t j 3 k, i r (t)i "13, and so from (1) we, see that a unit tangent is, T(t) , Next, we have, , 2, "13, , sin t i , , 2, "13, , cos t j , , dT, 2, 2, , cos t i 2, sin t j and, dt, "13, "13, , Hence, (7) gives the principal normal, , g, , 3, "13, , k., , dT, 2, g , ., dt, "13, , N(t) cos t i sin t j., Now, the binormal is, , i, , B(t) T(t) 3 N(t) 4 , , , , 3, , 2, , j, 2, , sin t, , cos t, "13, sin t, , "13, cos t, , "13, , sin t i 2, , k, 3, , 3, , "13, , "13, 0, , cos t j , , 2, , "13, , 4, , k., , Finally, using i d T/dti 2/ "13 and i r (t)i "13, we obtain from (4) that the curvature, at any point is the constant, k, , 2> "13, "13, , , , 2, ., 13, , The fact that the curvature in Example 2 is constant is not surprising, since the curve defined, by r(t) is a circular helix., EXAMPLE 3, , Osculating, Normal, Rectifying Planes, , At the point corresponding to t p>2 on the circular helix in Example 2, find an equation of, (a) the osculating plane, (b) the normal plane, and (c) the rectifying plane., SOLUTION, , From r(p>2) k0, 2, 3p>2l the point P in question is (0, 2, 3p>2)., , (a) A normal vector to the osculating plane at P is, B(p>2) T(p>2) 3 N(p>2) , , 3, "13, , i, , 2, "13, , k., , To find an equation of a plane we do not require a unit normal, so in lieu of B(p>2) it, is a bit simpler to use k3, 0, 2l. From (11) of Section 7.5 an equation of the osculating, plane is, 3(x 2 0) 0(y 2 2) 2az 2, , 3p, b 0, 2, , or, , 3x 2z 3p., , 9.3 Curvature and Components of Acceleration, , |, , 493
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2y, 0, , 1, (b) At the point P, the vector T(p>2) !13, k2, 0, 3l or k2, 0, 3l is normal to the plane, containing N(p>2) and B(p>2). Hence an equation of the normal plane is, , –2, 20, , 2(x 2 0) 0(y 2 2) 3az 2, , 10, , –10, , –2.5, , 0, x, , 2.5, , 5, , FIGURE 9.3.5 Helix and osculating plane, in Example 3, , 4x 6z 9p., , or, , (c) Finally, at the point P, the vector N(p>2) k0, 1, 0l is normal to the plane containing T(p>2) and B(p>2). An equation of the rectifying plane is, , z, 0, , –5, , 3p, b 0, 2, , 0(x 2 0) (1)(y 2 2) 0az 2, , 3p, b 0, 2, , or, , y 2., , With the help of Mathematica, portions of the helix and the osculating plane in Example 3 are, shown in FIGURE 9.3.5. The point (0, 2, 3p>2) is indicated in the figure by the red dot., , Formulas for aT , aN , and Curvature By dotting, and in turn crossing, the vector, v v T with (9), it is possible to obtain explicit formulas involving r, r , and r for the tangential, and normal components of the acceleration and the curvature. Observe that, v a aN (v T N) aT (v T T) aTv, 0, , 1, , yields the tangential component of acceleration, aT , , r9(t) r0(t), dv, va, , ., , dt, ivi, ir9(t)i, , (10), , On the other hand,, v, , a aN (v T, , N) aT (v T, B, , T) aNv B., 0, , Since i B i 1, it follows that the normal component of acceleration is, aN kv 2 , , ir9(t) 3 r0(t)i, iv 3 ai, , ., ivi, ir9(t)i, , (11), , Solving (11) for the curvature gives, k(t) , , EXAMPLE 4, , iv 3 ai, ivi3, , , , ir9(t) 3 r0(t)i, ir9(t)i3, , ., , (12), , Curvature of Twisted Cubic, , The curve traced by r(t) t i 12 t 2 j 13 t 3 k is said to be a “twisted cubic.” If r(t) is the position, vector of a moving particle, find the tangential and normal components of the acceleration at, any t. Find the curvature., SOLUTION, , v(t) r (t) i t j t 2 k,, , a(t) r (t) j 2t k., , Since v a t 2t 3 and i vi "1 t 2 t 4 , it follows from (10) that, aT , , Now,, , 494, , |, , CHAPTER 9 Vector Calculus, , i, v 3 a 31, 0, , dv, t 2t 3, , ., dt, "1 t 2 t 4, j, t, 1, , k, t 2 3 t 2 i 2 2t j k, 2t
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and i v, , ai "t 4 4t 2 1. Thus, (11) gives, aN kv 2 , , "t 4 4t 2 1, "1 t 2 t 4, , , , t 4 4t 2 1, ., Å t4 t2 1, , From (12) we find that the curvature of the twisted cubic is given by, k(t) , tangent, C, P, , ρ, , FIGURE 9.3.6 Radius of curvature, , (t 4 4t 2 1) 1>2, (t 4 t 2 1) 3>2, , ., , Radius of Curvature The reciprocal of the curvature, r 1/k, is called the radius of, curvature. The radius of curvature at a point P on a curve C is the radius of a circle that “fits”, the curve there better than any other circle. The circle at P is called the circle of curvature and, its center is the center of curvature. The circle of curvature has the same tangent line at P as, the curve C, and its center lies on the concave side of C. For example, a car moving on a curved, track, as shown in FIGURE 9.3.6, can, at any instant, be thought to be moving on a circle of radius r., Hence, the normal component of its acceleration aN kv2 must be the same as the magnitude, of its centripetal acceleration a v2 /r. Therefore, k 1/r and r 1/k. Knowing the radius of, curvature, we can determine the speed v at which a car can negotiate a banked curve without, skidding. (This is essentially the idea in Problem 26 in Exercises 9.2.), , REMARKS, By writing (6) as, a(t) , , ds d T, d 2s, 2 T,, dt dt, dt, , we note that the so-called scalar acceleration d 2s/dt 2, referred to in the last remark, is now, seen to be the tangential component of the acceleration aT., , 9.3, , Exercises, , Answers to selected odd-numbered problems begin on page ANS-21., , In Problems 1 and 2, for the given position function, find the, unit tangent., 1. r(t) (t cos t sin t) i (t sin t cos t) j t 2 k, t 0, 2. r(t) et cos t i et sin t j "2et k, 3. Use the procedure outlined in Example 2 to find T, N, B, and, , k for motion on a general circular helix that is described by, r(t) a cos t i a sin t j ct k., 4. Use the procedure outlined in Example 2 to show on the twisted, cubic of Example 4 that at t 1:, T, , 1, "3, , B, , (i j k), N , 1, "6, , 1, "2, , (i 2 j 2 k), k , , (i 2 k),, "2, ., 3, , In Problems 5 and 6, find an equation of the osculating plane, to the given space curve at the point that corresponds to, the indicated value of t., 5. The circular helix of Example 2; t p/4, 6. The twisted cubic of Example 4; t 1, , In Problems 7–16, r(t) is the position vector of a moving, particle. Find the tangential and normal components of the, acceleration at any t., r(t) i t j t 2 k, r(t) 3 cos t i 2 sin t j t k, r(t) t 2 i (t 2 1) j 2t 2 k, r(t) t 2 i t 3 j t 4 k, r(t) 2t i t 2 j, r(t) tan1 t i 12 ln(1 t 2) j, r(t) 5 cos t i 5 sin t j, r(t) cosh t i sinh t j, r(t) et (i j k), r(t) t i (2t 1) j (4t 2) k, Find the curvature of an elliptical helix that is described by, r(t) a cos t i b sin t j ct k, a 0, b 0, c 0., 18. (a) Find the curvature of an elliptical orbit that is described, by r(t) a cos t i b sin t j c k, a 0, b 0, c 0., (b) Show that when a b, the curvature of a circular orbit, is the constant k 1/a., 19. Show that the curvature of a straight line is the constant k 0., [Hint: Use (2) in Section 7.5.], 7., 8., 9., 10., 11., 12., 13., 14., 15., 16., 17., , 9.3 Curvature and Components of Acceleration, , |, , 495
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20. Find the curvature at t p of the cycloid that is described by, , r(t) a(t sin t) i a(1 cos t) j, a 0., 21. Let C be a plane curve traced by r(t) f (t) i g(t) j, where f, , and g have second derivatives. Show that the curvature at a, point is given by, k, , Z f9(t)g0(t) 2 g9(t) f 0(t)Z, (f f9(t)g 2 fg9(t)g 2)3>2, ZF0(x)Z, f1 (F9(x))2g 3>2, , 23. y x 2; (0, 0), (1, 1), 24. y x 3; (1, 1), ( 12 , 18 ), , Discussion Problems, , ., , 22. Show that if y F(x), the formula for k in Problem 21 reduces to, , k, , In Problems 23 and 24, use the result of Problem 22 to find the, curvature and radius of curvature of the curve at the indicated, points. Decide at which point the curve is “sharper.”, , 25. Discuss the curvature near a point of inflection of y F(x)., 26. Show that i a(t) i 2 a2N a2T ., , ., , 9.4, , Partial Derivatives, , z, (x, y, z), where z = f (x, y), , INTRODUCTION In this section we consider functions of two or more variables and how, to find the instantaneous rate of change—that is, the derivative—of such functions with respect, to each variable., , f (x, y), , y, , (x, y), domain of z = f (x, y), , x, , FIGURE 9.4.1 Function of two variables, , z, z = y2 – x2, , Functions of Two Variables Recall from calculus that a function of two variables, is a rule of correspondence that assigns to each ordered pair of real numbers (x, y) of a subset, of the xy-plane one and only one number z in the set R of real numbers. The set of ordered pairs, (x, y) is called the domain of the function and the set of corresponding values of z is called the, range. A function of two variables is usually written z f (x, y). The variables x and y are called, the independent variables of the function, and z is called the dependent variable. The graph, of a function z f (x, y) is a surface in 3-space. See FIGURE 9.4.1., Level Curves For a function z f (x, y), the curves defined by f (x, y) c, for suitable, c, are called the level curves of f. The word level arises from the fact that we can interpret the, equation f (x, y) c as the projection onto the xy-plane of the curve of intersection, or trace, of, z f (x, y) and the (horizontal or level) plane z c. See FIGURE 9.4.2., , c=1, , z, plane, z=c, , y, , y, , surface, z = f (x, y), , x, (a) Surface, , increasing, values of f, , y, y, , c=0, c=1, x, , c = –1, , FIGURE 9.4.3 Surface and level curves in, Example 1, |, , CHAPTER 9 Vector Calculus, , (b) Level curves, , FIGURE 9.4.2 Surface and level curves, , EXAMPLE 1, (b) Level curves, , f (x, y) = c, (a) Surface, , x, , 496, , x, , Level Curves, , The level curves of the function f (x, y) y2 x 2 are defined by y2 x 2 c. As shown in, FIGURE 9.4.3, when c 0 or c 0, a member of this family of curves is a hyperbola. For c 0,, we obtain the lines y x and y x.
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Functions of Three or More Variables Functions of three or more variables are, defined analogously to functions of two variables. For example, a function of three variables, is a rule of correspondence that assigns to each ordered triple of real numbers (x, y, z) of a subset, of 3-space one and only one number w in the set R of real numbers. We write w F(x, y, z)., , z, c=1, , Level Surfaces Although we cannot draw a graph of a function of three variables, w F(x, y, z), we can draw the surfaces defined by F(x, y, z) c for suitable values of c. These, surfaces are called level surfaces. This is an unfortunate, though standard, choice of words, since, level surfaces are usually not level., , c=2, , y, , Level Surfaces, , EXAMPLE 2, c = –2, x, , Describe the level surfaces of the function F(x, y, z) (x 2 y2)/z., SOLUTION, , 0 the level surfaces are given by, x2 1 y2, c, z, , c = –1, , FIGURE 9.4.4 Level surfaces in Example 2, , For c, , or, , x 2 y 2 cz., , A few members of this family of paraboloids are shown in FIGURE 9.4.4., , Partial Derivatives The derivative of a function of one variable y f (x) is given by, dy, f (x Dx) 2 f (x), lim, ., dx, DxS0, Dx, In exactly the same manner, we can define a derivative of a function of two variables with respect, to each variable. If z f (x, y), then the partial derivative with respect to x is, f (x Dx, y) 2 f (x, y), 0z, lim, 0x, DxS0, Dx, , (1), , and the partial derivative with respect to y is, f (x, y Dy) 2 f (x, y), 0z, lim, ,, 0y, DyS0, Dy, , (2), , provided each limit exists., In (1) the variable y does not change in the limiting process; that is, y is held fixed. Similarly,, in (2) the variable x is held fixed. The two partial derivatives (1) and (2) then represent the rates, of change of f with respect to x and y, respectively. On a practical level:, To compute z / x, use the laws of ordinary differentiation while treating y as a constant., To compute z / y, use the laws of ordinary differentiation while treating x as a constant., EXAMPLE 3, , Partial Derivatives, , If z 4x 3y2 4x 2 y6 1, find z / x and z / y., SOLUTION We hold y fixed and treat constants in the usual manner. Thus,, 0z, 12x 2y2 8x., 0x, By treating x as a constant, we obtain, 0z, 8x 3y 6y5., 0y, , Alternative Symbols The partial derivatives z /x and z /y are often represented by, alternative symbols. If z f (x, y), then, 0f, 0z, 5, 5 zx 5 fx and, 0x, 0x, , 0f, 0z, 5, 5 zy 5 fy ., 0y, 0y, 9.4 Partial Derivatives, , |, , 497
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Higher-Order and Mixed Derivatives For a function of two variables z ⫽ f (x, y),, the partial derivatives ⭸z /⭸x and ⭸z /⭸y are themselves functions of x and y. Consequently, we, can compute second and higher partial derivatives. Indeed, we can find the partial derivative, of ⭸z/⭸x with respect to y, and the partial derivative of ⭸z/⭸y with respect to x. The latter types of, partial derivatives are called mixed partial derivatives. In summary, for z ⫽ f (x, y):, Second-order partial derivatives:, 0 2z, 0 0z, ⫽, a b, 0x 0x, 0x 2, , and, , Third-order partial derivatives:, , 0 3z, 0 0 2z, ⫽, a b, 3, 0x 0x 2, 0x, , and, , Mixed second-order partial derivatives:, 0 2z, 0 0z, ⫽, a b, 0x0y, 0x 0y, , and, , 0 2z, 0 0z, ⫽, a b., 0y 0y, 0y 2, 0 3z, 0 0 2z, ⫽, a b., 3, 0y 0y 2, 0y, 0 2z, 0 0z, ⫽, a b., 0y0x, 0y 0x, , Alternative Symbols The second- and third-order partial derivatives are denoted by fxx ,, fyy , fxxx, and so on. The subscript notation for mixed second partial derivatives is fxy or fyx. Note that, fxy ⫽ ( fx)y ⫽, , 0 0z, 0 2z, 0 2z, a b ⫽, and fyx ⫽, ., 0y 0x, 0y0x, 0x0y, , Although we shall not prove it, if a function f has continuous second partial derivatives, then the, order in which a mixed second partial derivative is done is irrelevant; that is,, fxy ⫽ fyx ., , (3), , Functions of Three or More Variables The rates of change of a function of three variables w ⫽ F(x, y, z) in the x, y, and z directions are ⭸w/⭸x, ⭸w/⭸y, and ⭸w/⭸z, respectively. To compute,, say, ⭸w/⭸x, we differentiate with respect to x in the usual manner while holding both y and z constant., In this manner we extend the process of partial differentiation to functions of any number of variables., Partial Derivatives, , EXAMPLE 4, ⫺3pt, , If F(x, y, t) ⫽ e, , cos 4x sin 6y, then the partial derivatives with respect to x, y, and t are, in turn,, Fx(x, y, t) ⫽ ⫺4e⫺3pt sin 4x sin 6y,, Fy(x, y, t) ⫽ 6e⫺3pt cos 4x cos 6y,, Ft(x, y, t) ⫽ ⫺3pe⫺3pt cos 4x sin 6y., , Chain Rule The Chain Rule for functions of one variable states that if y ⫽ f (u) is a differentiable function of u, and u ⫽ g(x) is a differentiable function of x, then the derivative of the, composite function is, dy du, dy, 5, ., dx, du dx, , (4), , For a composite function of two variables z ⫽ f (u, v), where u ⫽ g(x, y) and v ⫽ h(x, y), we would, naturally expect two formulas analogous to (4), since we can compute both ⭸z /⭸x and ⭸z /⭸y. The, Chain Rule for functions of two variables is summarized as follows:, Theorem 9.4.1, , Chain Rule, , If z ⫽ f (u, v) is differentiable and u ⫽ g(x, y) and v ⫽ h(x, y) have continuous first partial, derivatives, then, 0z, 0z 0u, 0z 0v, ⫽, ⫹, ,, 0x, 0u 0x, 0v 0x, , 498, , |, , CHAPTER 9 Vector Calculus, , 0z, 0z 0u, 0z 0v, ⫽, ⫹, ., 0y, 0u 0y, 0v 0y, , (5)
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REMARKS, If w F(x, y, z) has continuous partial derivatives of any order, then analogous to (3), the, mixed partial derivatives are equal:, Fxyz Fyzx Fzyx,, , Fxxy Fyxx Fxyx ,, , and so on., , Exercises, , 9.4, , Answers to selected odd-numbered problems begin on page ANS-22., , In Problems 1–6, sketch some of the level curves associated, with the given function., 1. f (x, y) x 2y, 2. f (x, y) y2 x, 2, , 4. f (x, y) "36 2 4x 2 2 9y 2, y 2 x2, , 1, , 6. f (x, y) tan ( y x), , In Problems 7–10, describe the level surfaces but do not graph., x2, z2, 7. F(x, y, z) , , 9, 4, 8. F(x, y, z) x 2 y2 z2, 9. F(x, y, z) x 2 3y2 6z2, 10. F(x, y, z) 4y 2z 1, 11. Graph some of the level surfaces associated with F(x, y, z) , x 2 y2 z2 for c 0, c 0, and c 0., 12. Given that, F(x, y, z) , , y2, x2, z2, , ,, 16, 4, 9, , In Problems 13–32, find the first partial derivatives of the given, function., 13. z x 2 xy2 4y5, 14. z x3 6x 2y3 5y2, 4 3, 2 6, 5, 15. z 5x y x y 6x 4y, 16. z tan(x 3y2), 4"x, 3y 2 1, 19. z (x 3 y2)1, 21. z cos2 5x sin2 5y, 3, , 23. f (x, y) xe x y, , 29., 31., 32., , 20. z (x4 7y2 3y)6, 2, , tan 1y2, , 24. f (u, f) f2 sin, , u, f, , xy, (x 2 2 y 2)2, "s, "r, 2, g(u, v) ln(4u2 5v3) 28. h(r, s) , s, r, y/z, w 2 "xy ye, 30. w xy ln(xz), F(u, v, x, t) u2w2 uv3 vw cos(ut 2) (2x 2t)4, 4 5, G( p, q, r, s) ( p2q 3)r s, , 25. f (x, y) , 27., , 3x 2 y, x 2y, , 18. z 4x 3 5x 2 8x, 22. z e x, , 500, , |, , 26. f (x, y) , , CHAPTER 9 Vector Calculus, , 2, , 33. z ln(x2 y2), , 34. z e x, , 2 y2, , cos 2xy, , In Problems 35 and 36 verify that the given function satisfies the, wave equation:, a2, , 0 2u, 0 2u, , ., 0x 2, 0t 2, , 35. u cos at sin x, 36. u cos(x at) sin(x at), 37. The molecular concentration C(x, t) of a liquid is given by, 2, , C(x, t) t 1/2 e x >kt . Verify that this function satisfies the, diffusion equation:, k 0 2C, 0C, , ., 2, 4 0x, 0t, 38. The pressure P exerted by an enclosed ideal gas is given by, , find the x-, y-, and z-intercepts of the level surface that passes, through (4, 2, 3)., , 17. z , , 0 2z, 0 2z, , 0., 0x 2, 0y 2, , 2, , 3. f (x, y) "x 2 y 2 1, 5. f (x, y) e, , In Problems 33 and 34, verify that the given function satisfies, Laplace’s equation:, , P k(T/V), where k is a constant, T is temperature, and V is, volume. Find:, (a) the rate of change of P with respect to V,, (b) the rate of change of V with respect to T, and, (c) the rate of change of T with respect to P., In Problems 39–48, use the Chain Rule to find the indicated, partial derivatives., 0z 0z, 2, 39. z e uv ; u x 3, v x y2;, ,, 0x 0y, 0z 0z, 40. z u2 cos 4v; u x2y3, v x 3 y3;, ,, 0x 0y, 0z 0z, 41. z 4x 5y2; x u4 8v3, y (2u v)2;, ,, 0u 0v, x2y, v 2 0z 0z, u, 42. z , ; x , y ;, ,, v, u 0u 0v, xy, 0w 0w, 43. w (u2 v2)3/2; u et sin u, v et cos u;, ,, 0t 0u, 0w 0w, 44. w tan1 !uv; u r 2 2 s 2, v r 2s 2;, ,, 0r 0s, 0R 0R, 2, 2, 2 2, 45. R rs2t 4; r ue v , s ve u , t e u v ;, ,, 0u 0v, x, x 0Q 0Q, 46. Q ln( pqr); p t 2 sin1 x, q 2 , r tan1 ;, ,, t 0x 0t, t
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47. w "x 2 y 2; x ln (rs tu),, , t, 0w 0w 0w, cosh rs;, , ,, u, 0t 0r 0u, 2, 2, 2, 48. s p q r 4t; p fe3u, q cos(f u), r fu2,, 0s 0s, t 2f 8u;, ,, 0f 0u, y, , In Problems 49–52, use (8) to find the indicated derivative., dz, 49. z ln(u2 v2); u t 2, v t 2;, dt, dz, 3, 4, 5t, 50. z u v uv ; u e , v sec 5t;, dt, p, p dw, 51. w cos(3u 4v); u 2t , v t 2 ;, 2, 2, 4 dt t p, 4, dw, 52. w e xy; x , , y 3t 5;, 2, 2t 1, dt t 0, 53. If u f (x, y) and x r cos u, y r sin u, show that Laplace’s, equation 2u/x 2 2u/y2 0 becomes, 2, , 2, , 0u, 1 0u, 1 0u, , 2 2 0., 2, r, 0r, 0r, r 0u, , y, , FIGURE 9.4.6 Triangle in Problem 57, , 0.08T, 3.6, 2 2., V 2 0.0427, V, , 58. A particle moves in 3-space so that its coordinates at any time, , If dT/dt and dV/dt are rates at which the temperature and volume change, respectively, use the Chain Rule to find dP/dt., 55. The equation of state for a thermodynamic system is, F(P, V, T) 0, where P, V, and T are pressure, volume, and, , 9.5, , z = f(x, y), , y, , rate of change of f, in the i-direction, is ∂ z, ∂x, , are x 4 cos t, y 4 sin t, z 5t, t 0. Use the Chain Rule, to find the rate at which its distance, w "x 2 y 2 z 2, , from the origin is changing at t 5p/2 seconds., , Directional Derivative, , INTRODUCTION We saw in the last section that for a function f of two variables x and y,, , z, , x, , x, , θ, , 54. Van der Waals’ equation of state for the real gas CO2 is, , P, , temperature, respectively. If the equation defines V as a function, of P and T, and also defines T as a function of V and P, show that, 0F, 0T, 1, 0V, , ., 0T, 0F, 0T, 0V, 0V, 56. The voltage across a conductor is increasing at a rate of, 2 volts/min and the resistance is decreasing at a rate of, 1 ohm/min. Use I E/R and the Chain Rule to find the rate, at which the current passing through the conductor is changing, when R 50 ohms and E 60 volts., 57. The length of the side labeled x of the triangle in FIGURE 9.4.6, increases at a rate of 0.3 cm/s, the side labeled y increases, at a rate of 0.5 cm/s, and the included angle u increases at a, rate of 0.1 rad/s. Use the Chain Rule to find the rate at which, the area of the triangle is changing at the instant x 10 cm,, y 8 cm, and u p/6., , rate of change of f, in the j-direction, is ∂ z, u ∂y, What is the, rate of change of, f in the direction, given by the vector u?, , FIGURE 9.5.1 An arbitrary direction is, denoted by the vector u, , the partial derivatives z/x and z/y give the slope of the tangent to the trace, or curve of intersection of the surface defined by z f (x, y) and vertical planes that are, respectively, parallel, to the x- and y-coordinates axes. Equivalently, we can think of the partial derivative z/x as, the rate of change of the function f in the direction given by the vector i, and z/y as the rate of, change of the function f in the j-direction. There is no reason to confine our attention to just two, directions. In this section we shall see how to find the rate of change of a differentiable function, in any direction. See FIGURE 9.5.1., , The Gradient of a Function In this and the next sections it is convenient to introduce, a new vector based on partial differentiation. When the vector differential operator, = 5i, , 0, 0, 0, 0, 0, or = 5 i, 1j, 1j, 1k, 0x, 0y, 0x, 0y, 0z, , is applied to a differentiable function z f (x, y) or w F(x, y, z), we say that the vectors, =f (x, y) , =F(x, y, z) , , 0f, 0f, i, j, 0x, 0y, , (1), , 0F, 0F, 0F, i, j, k, 0x, 0y, 0z, , (2), , 9.5 Directional Derivative, , |, , 501
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are the gradients of the respective functions. The symbol , an inverted capital Greek delta, is, called “del” or “nabla.” The vector f is usually read “grad f.”, , Gradient, , EXAMPLE 1, , Compute f (x, y) for f (x, y) 5y x3y2., SOLUTION, , From (1), f (x, y) , , 0, 0, (5y x3y2)i , (5y x3y2) j; therefore, 0x, 0y, , f (x, y) 3x2y2 i (5 2x3y) j., , Gradient at a Point, , EXAMPLE 2, 2, , If F(x, y, z) xy 3x 2 z3, find F(x, y, z) at (2, 1, 4)., SOLUTION, , From (2), F(x, y, z) ( y 2 6x) i 2xy j 3z2 k and so, F(2, 1, 4) 13 i 4 j 48 k., , A Generalization of Partial Differentiation Suppose u cos u i sin u j is a unit, vector in the xy-plane that makes an angle u with the positive x-axis and is parallel to the vector, v from (x, y, 0) to (x x, y y, 0). If h "(Dx)2 (Dy)2 0, then v h u. Furthermore,, let the plane perpendicular to the xy-plane that contains these points slice the surface z f ( x, y), in a curve C. We ask: What is the slope of the tangent line to C at a point P with coordinates, (x, y, f (x, y)) in the direction given by v? See FIGURE 9.5.2., From the figure we see that x h cos u and y h sin u so that the slope of the indicated, secant line is, f (x Dx, y Dy) 2 f (x, y), f (x h cos u, y h sin u) 2 f (x, y), , ., h, h, , (3), , z, tangent, secant, surface, z = f (x, y), C, f(x + Δx, y + Δy) – f(x, y), , P, h, , u Δx, , θ, x, , (x, y, 0), v = hu, θ, Δy (x + Δx, y + Δy, 0), , y, , FIGURE 9.5.2 C is the curve of intersection of the surface and the plane determined by vector v, , We expect the slope of the tangent at P to be the limit of (3) as h S 0. This slope is the rate of change, of f at P in the direction specified by the unit vector u. This leads us to the following definition:, Definition 9.5.1, , Directional Derivative, , The directional derivative of z f (x, y) in the direction of a unit vector u cos u i sin u j is, Du f (x, y) lim, , hS0, , provided the limit exists., , 502, , |, , CHAPTER 9 Vector Calculus, , f (x h cos u, y h sin u) 2 f (x, y), h, , (4)
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Observe that (4) is truly a generalization of partial differentiation, since, u0, and, , u, , implies that Di f (x, y) lim, , hS0, , f (x h, y) 2 f (x, y), 0z, , h, 0x, , f (x, y h) 2 f (x, y), p, 0z, implies that Dj f (x, y) lim, ., 2, hS0, h, 0y, , Method for Computing the Directional Derivative While (4) could be used, to find Du f (x, y) for a given function, as usual we seek a more efficient procedure. The next, theorem will show how the concept of the gradient of a function plays a key role in computing, a directional derivative., Computing a Directional Derivative, , Theorem 9.5.1, , If z f (x, y) is a differentiable function of x and y and u cos u i sin u j, then, Du f (x, y) f (x, y) u., , (5), , PROOF: Let x, y, and u be fixed so that g(t) f (x t cos u, y t sin u) is a function of one, , variable. We wish to compare the value of g (0), which is found by two different methods. First,, by the definition of a derivative,, g9(0) lim, , hS0, , g(0 h) 2 g(0), f (x h cos u, y h sin u) 2 f (x, y), lim, ., h, hS0, h, , (6), , Second, by the Chain Rule, (8) of Section 9.4,, g9(t) f1(x t cos u, y t sin u), , d, d, (x t cos u) f2(x t cos u, y t sin u) (x t sin u), dt, dt, , (7), , f1(x t cos u, y t sin u) cos u f2(x t cos u, y t sin u) sin u., Here the subscripts 1 and 2 refer to the partial derivatives of f (x t cos u, y t sin u) with respect, to x t cos u and y t sin u, respectively. When t 0, we note that x t cos u and y t sin u, are simply x and y, and therefore (7) becomes, g (0) fx(x, y) cos u fy(x, y) sin u., , (8), , Comparing (4), (6), and (8) then gives, Du f (x, y) fx(x, y) cos u fy(x, y) sin u, [ fx(x, y) i fy(x, y) j] (cos u i sin u j), f (x, y) u., , Directional Derivative, , EXAMPLE 3, , Find the directional derivative of f (x, y) 2x2y3 6xy at (1, 1) in the direction of a unit vector, whose angle with the positive x-axis is p/6., SOLUTION, , Since, , 0f, 0f, 4xy3 6y and, 6x 2y2 6x, we have, 0x, 0y, , f (x, y) (4xy3 6y) i (6x 2y2 6x) j, , and, , Now, at u p/6, u cos u i sin u j becomes u , Du f (1, 1) =f (1, 1) u (10 i 12 j) a, , f (1, 1) 10 i 12 j., , 1, "3, i j. Therefore,, 2, 2, , "3, 1, i jb 5"3 6., 2, 2, 9.5 Directional Derivative, , |, , 503
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EXAMPLE 4, , Directional Derivative, , Consider the plane that is perpendicular to the xy-plane and passes through the points P(2, 1), and Q(3, 2). What is the slope of the tangent line to the curve of intersection of this plane with, the surface f (x, y) 4x2 y2 at (2, 1, 17) in the direction of Q?, !, SOLUTION We want Du f (2, 1) in the direction given by the vector PQ i j. But since, !, PQ is not a unit vector, we form u (1/ "2) i (1/ "2) j. Now,, f (x, y) 8x i 2y j, , f (2, 1) 16 i 2 j., , and, , Therefore, the required slope is, Du f (2, 1) (16 i 2 j) a, , 1, "2, , i, , 1, "2, , jb 9"2., , Functions of Three Variables For a function w F(x, y, z) the directional derivative, , is defined by, , DuF (x, y, z) lim, , hS0, , F(x h cos a, y h cos b, z h cos g) 2 F(x, y, z), ,, h, , where a, b, and g are the direction angles of the unit vector u measured relative to the positive, x-, y-, and z-axes, respectively.* But in the same manner as before, we can show that, Du F(x, y, z) F(x, y, z) u., , (9), , Notice that since u is a unit vector, it follows from (10) of Section 7.3 that, Du f (x, y) compuf (x, y), , and, , Du F(x, y, z) compuF(x, y, z)., , In addition, (9) reveals that, Dk F(x, y, z) , , EXAMPLE 5, , 0w, ., 0z, , Directional Derivative, , Find the directional derivative of F(x, y, z) xy2 4x 2y z2 at (1, 1, 2) in the direction, of 6 i 2 j 3 k., SOLUTION We have, , 0F, 0F, 0F, y2 8xy,, 2xy 4x 2, and, 2z so that, 0x, 0y, 0z, , F( x, y, z) ( y2 8xy) i (2xy 4x2) j 2z k, F(1, 1, 2) 9 i 6 j 4 k., Since i 6i 2j 3k i 7 then u 67 i 27 j 37 k is a unit vector in the indicated direction., It follows from (9) that, 6, 2, 3, 54, DuF(1, 1, 2) (9 i 6 j 4 k) a i 1 j 1 kb 5 ., 7, 7, 7, 7, , Maximum Value of the Directional Derivative Let f represent a function of either, two or three variables. Since (5) and (9) express the directional derivative as a dot product, we, see from (5) of Theorem 7.3.2 that, Du f i f i i u i cos f i f i cos f,, , ( i u i 1),, , *Note that the numerator in (4) can be written f (x h cos a, y h cos b) – f (x, y), where b (p/2) a., , 504, , |, , CHAPTER 9 Vector Calculus
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where f is the angle between f and u. Because 0 f p, we have 1 cos f 1 and,, consequently, i f i Du f i f i . In other words:, The maximum value of the directional derivative is i f i and it occurs when u, has the same direction as f (when cos f 1)., , (10), , The minimum value of the directional derivative is i f i and it occurs when u, and f have opposite directions (when cos f 1)., , (11), , Max/Min of Directional Derivative, , EXAMPLE 6, , In Example 5 the maximum value of the directional derivative of F of (1, 1, 2) is, i F(1, 1, 2)i "133. The minimum value of DuF(1, 1, 2) is then "133., , Gradient Points in Direction of Most Rapid Increase of f Put yet another way,, , (10) and (11) state:, , The gradient vector f points in the direction in which f increases most rapidly,, whereas f points in the direction of the most rapid decrease of f., , Direction of Steepest Ascent, , EXAMPLE 7, , Each year in Los Angeles there is a bicycle race up to the top of a hill by a road known to be, the steepest in the city. To understand why a bicyclist with a modicum of sanity will zigzag, , road, , u, , ∇f, (a), , (b), , FIGURE 9.5.3 Model of a hill in, Example 7, , up the road, let us suppose the graph of f (x, y) 4 23 "x 2 y 2 , 0 z 4, shown in, FIGURE 9.5.3(a) is a mathematical model of the hill. The gradient of f is, f (x, y) , , 2, y, 2, x, 3, c, i, jd , r,, 2, 2, 2, 3 "x 2 y 2, "x y, "x y 2, , where r x i y j is a vector pointing to the center of the circular base., Thus the steepest ascent up the hill is a straight road whose projection in the xy-plane is a, radius of the circular base. Since Du f compuf, a bicyclist will zigzag, or seek a direction, u other than f, in order to reduce this component., EXAMPLE 8, , Direction to Cool Off Fastest, , The temperature in a rectangular box is approximated by, T(x, y, z) xyz(1 x)(2 y)(3 z),, , 0 x 1, 0 y 2, 0 z 3., , If a mosquito is located at ( 12 , 1, 1), in which direction should it fly to cool off as rapidly as, possible?, SOLUTION The gradient of T is, T(x, y, z) yz(2 y)(3 z)(1 2x)i xz(1 x)(3 z)(2 2y)j xy(1 x)(2 y)(3 2z)k., Therefore, T ( 12 , 1, 1) 14 k. To cool off most rapidly, the mosquito should fly in the direction, of 14 k; that is, it should dive for the floor of the box, where the temperature is T(x, y, 0) 0., , 9.5, , Exercises, , Answers to selected odd-numbered problems begin on page ANS-22., , In Problems 1–4, compute the gradient for the given function., 2, , 1. f (x, y) x 2 x 3y2 y4, , 2. f (x, y) y e22x y, , 3. F(x, y, z) , , 4. F(x, y, z) xy cos yz, , xy2, z3, , In Problems 5–8, find the gradient of the given function at the, indicated point., 5. f (x, y) x 2 4y2; (2, 4), 6. f (x, y) "x 3y 2 y 4; (3, 2), , 9.5 Directional Derivative, , |, , 505
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7. F(x, y, z) x 2z2 sin 4y; (2, p/3, 1), 8. F(x, y, z) ln(x 2 y2 z2); (4, 3, 5), , In Problems 9 and 10, use Definition 9.5.1 to find Du f (x, y), given that u makes the indicated angle with the positive x-axis., 9. f (x, y) x 2 y2; u 30, 10. f (x, y) 3x y2; u 45, In Problems 11–20, find the directional derivative of the given, function at the given point in the indicated direction., 11. f (x, y) 5x3y6; (1, 1), u p/6, 12. f (x, y) 4x xy2 5y; (3, 1), u p/4, y, 13. f (x, y) tan1 ; (2, 2), i 3 j, x, xy, 14. f (x, y) , ; (2, 1), 6 i 8 j, xy, 15. f (x, y) (xy 1)2; (3, 2), in the direction of (5, 3), 16. f (x, y) x2 tan y; (12, p>3), in the direction of the negative x-axis, 17. F(x, y, z) x 2y2 (2z 1)2; (1, 1, 1), 0, 3, 3, x 2 2 y2, 18. F(x, y, z) , ; (2, 4, 1), i 2 j k, z2, 19. F(x, y, z) "x 2y 2y 2z; (2, 2, 1), in the direction of the, , negative z-axis, 20. F(x, y, z) 2x y2 z2; (4, 4, 2), in the direction of the, origin, In Problems 21 and 22, consider the plane through the points P, and Q that is perpendicular to the xy-plane. Find the slope of the, tangent at the indicated point to the curve of intersection of this, plane and the graph of the given function in the direction of Q., 21. f (x, y) (x y)2; P(4, 2), Q(0, 1); (4, 2, 4), 22. f (x, y) x3 5xy y2; P(1, 1), Q(1, 6); (1, 1, 3), In Problems 23–26, find a vector that gives the direction in, which the given function increases most rapidly at the indicated, point. Find the maximum rate., 23. f (x, y) e2x sin y; (0, p/4), 24. f (x, y) xyexy; (5, 5), 25. F(x, y, z) x 2 4xz 2yz2; (1, 2, 1), 26. F(x, y, z) xyz; (3, 1, 5), In Problems 27–30, find a vector that gives the direction in, which the given function decreases most rapidly at the indicated, point. Find the minimum rate., 27. f (x, y) tan(x 2 y2); ( "p>6, "p>6), 28. f (x, y) x 3 y3; (2, 2), 29. F(x, y, z) "xzey; (16, 0, 9), , xy, b; (12, 16, 13), z, 31. Find the directional derivative(s) of f (x, y) x y 2 at, (3, 4) in the direction of a tangent vector to the graph of, 2x 2 y2 9 at (2, 1)., 32. If f (x, y) x2 xy y2 x, find all points where Du f (x, y), 30. F(x, y, z) ln a, , in the direction of u (1/ "2)(i j) is zero., , 33. Suppose f (a, b) 4i 3j. Find a unit vector u so that, , (a) Du f (a, b) 0,, (b) Du f (a, b) is a maximum, and, (c) Du f (a, b) is a minimum., 506, , |, , CHAPTER 9 Vector Calculus, , 34. Suppose Du f (a, b) 6. What is the value of Du f (a, b)?, 35. (a) If f (x, y) x3 3x2y2 y3, find the directional derivative, , of f at a point (x, y) in the direction of u (1/ "10)(3i j)., (b) If F(x, y) Du f (x, y) in part (a), find DuF(x, y)., 36. Consider the gravitational potential, U(x, y) , , Gm, "x 2 y 2, , ,, , where G and m are constants. Show that U increases or, decreases most rapidly along a line through the origin., 37. If f (x, y) x 3 12x y2 10y, find all points at which, i f i 0., 38. Suppose, Du f (a, b) 7,, u, , Dv f (a, b) 3, , 5, 12, 5, 12, i2, j, v , i, j., 13, 13, 13, 13, , Find f (a, b)., , 39. Consider the rectangular plate shown in FIGURE 9.5.4. The tem-, , perature at a point (x, y) on the plate is given by T(x, y) , 5 2x 2 y2. Determine the direction an insect should take,, starting at (4, 2), in order to cool off as rapidly as possible., y, (4, 2), , x, , FIGURE 9.5.4 Insect in Problem 39, 40. In Problem 39, observe that (0, 0) is the coolest point of the, , plate. Find the path the cold-seeking insect, starting at (4, 2),, will take to the origin. If x(t), y(t) is the vector equation of, the path, then use the fact that T(x, y) x (t), y (t). Why, is this? [Hint: Remember separation of variables?], 41. The temperature at a point (x, y) on a rectangular metal plate, is given by T(x, y) 100 2x 2 y2. Find the path a heatseeking particle will take, starting at (3, 4), as it moves in the, direction in which the temperature increases most rapidly., 42. The temperature T at a point (x, y, z) in space is inversely, proportional to the square of the distance from (x, y, z) to, the origin. It is known that T(0, 0, 1) 500. Find the rate of, change of T at (2, 3, 3) in the direction of (3, 1, 1). In which, direction from (2, 3, 3) does the temperature T increase most, rapidly? At (2, 3, 3) what is the maximum rate of change of T?, 43. Find a function f such that, f (3x2 y3 ye xy) i (2y2 3xy2 xe xy) j., 44. Let fx, fy, fxy, fyx be continuous and u and v be unit vectors., , Show that DuDv f DvDu f.
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49. If F(x, y, z) f1(x, y, z) i f2(x, y, z) j f3(x, y, z) k and, , In Problems 45–48, assume that f and g are differentiable, functions of two variables. Prove the given identity., 45. (cf ) c f, , 46. ( f g) f g, , 47. ( fg) f g gf, , 48. =a b 5, , f, g, , = 5i, , g=f 2 f =g, g2, , 9.6, , 0, 0, 0, 1j, 1k ,, 0x, 0y, 0z, , show that, = 3F5 a, , 0f3, 0f3, 0f2, 0f1, 0f2, 0f1, 2, bi1 a, 2, bj1 a, 2, b k., 0y, 0z, 0z, 0x, 0x, 0y, , Tangent Planes and Normal Lines, , INTRODUCTION The notion of the gradient of a function of two or more variables was, introduced in the preceding section as an aid in computing directional derivatives. In this section, we give a geometric interpretation of the gradient vector., Geometric Interpretation of the Gradient—Functions of Two Variables, , Suppose f (x, y) c is the level curve of the differentiable function z f (x, y) that passes through, a specified point P(x0, y0); that is, f (x0, y0) c. If this level curve is parameterized by the, differentiable functions, , curve, f(x, y) = c, , x g(t), y h(t), , such that, , x0 g(t0), y0 h(t0),, , then the derivative of f (g(t), h(t)) c with respect to t is, , r′(t0), , 0f dx, 0f dy, , 0., 0x dt, 0y dt, P(x0, y0), , ∇f (x0, y0), , When we introduce the vectors, f (x, y) , , FIGURE 9.6.1 Gradient is perpendicular, to tangent vector at P, , (1), , 0f, 0f, i, j, 0x, 0y, , and, , r (t) , , dy, dx, i, j,, dt, dt, , (1) becomes f r 0. Specifically, at t t0, we have, f (x0, y0) r (t0) 0., , (2), , Thus, if r (t0) 0, the vector f (x0, y0) is orthogonal to the tangent vector r (t0) at P(x0, y0). We, interpret this to mean:, , ∇f(2, 3), , f is orthogonal to the level curve at P., See FIGURE 9.6.1., , y, (2, 3), , EXAMPLE 1, x, –x2 + y2 = 5, , Gradient at a Point, , Find the level curve of f (x, y) x 2 y2 passing through (2, 3). Graph the gradient at the point., SOLUTION Since f (2, 3) 4 9 5, the level curve is the hyperbola x 2 y2 5. Now,, f (x, y) 2xi 2yj, , FIGURE 9.6.2 Gradient in Example 1, , and, , f (2, 3) 4 i 6 j., , FIGURE 9.6.2 shows the level curve and f (2, 3)., , Geometric Interpretation of the Gradient—Functions of Three Variables, , Proceeding as before, let F(x, y, z) c be the level surface of a differentiable function w F(x, y, z), that passes through P(x0, y0, z0). If the differentiable functions x f (t), y g(t), z h(t) are the, parametric equations of a curve C on the surface for which x0 f (t0), y0 g(t0), z0 h(t0), then, the derivative of F( f (t), g(t), h(t)) 0 implies that, 0F dx, 0F dy, 0F dz, , , 0, 0x dt, 0y dt, 0z dt, 9.6 Tangent Planes and Normal Lines, , |, , 507
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∇F(x0, y0, z0), , C, , a, , or, , dy, 0F, 0F, dx, dz, 0F, i, j, kb a i , j, kb 0., 0x, 0y, 0z, dt, dt, dt, , In particular, at t t0, (3) is, , F(x0, y0, z0) r (t0) 0., , r′(t0), , P(x0, y0, z0), , (3), , (4), , Thus, when r (t0) 0, the vector F(x0, y0, z0) is orthogonal to the tangent vector r (t0). Since this, argument holds for any differentiable curve through P(x0, y0, z0) on the surface, we conclude that:, , surface, F(x, y, z) = c, , FIGURE 9.6.3 Gradient is normal to level, surface at P, , F is normal ( perpendicular) to the level surface at P., See FIGURE 9.6.3., EXAMPLE 2, , z, , Gradient at a Point, , Find the level surface of F(x, y, z) x 2 y2 z2 passing through (1, 1, 1). Graph the, gradient at the point., , ∇F (1, 1, 1), , SOLUTION Since F(1, 1, 1) 3, the level surface passing through (1, 1, 1) is the sphere, x2 y2 z2 3. The gradient of the function is, , (1, 1, 1), y, , F(x, y, z) 2x i 2y j 2z k,, , x2 + y2 + z2 = 3, , and so, at the given point, F(1, 1, 1) 2 i 2 j 2 k. The level surface and F(1, 1, 1) are, illustrated in FIGURE 9.6.4., , x, , FIGURE 9.6.4 Gradient in Example 2, , tangent, plane at, (x0, y0, z0), , z, , ∇F(x0, y0, z0), , (x0, y0, z0), , surface, F(x, y, z) = c, y, , Tangent Plane In the study of differential calculus a basic problem was finding an equation of a tangent line to the graph of a function. In 3-space the analogous problem is finding an, equation of a tangent plane to a surface. We assume again that w F(x, y, z) is a differentiable, function and that a surface is given by F(x, y, z) c., Definition 9.6.1, , Tangent Plane, , Let P(x0, y0, z0) be a point on the graph of F(x, y, z) c, where F is not 0. The tangent plane, at P is that plane through P that is normal to F evaluated at P., Thus, if P(x, y, z) and P(x0, y0, z0) are points on the tangent plane and r and r0 are their respective, position vectors, then a vector equation of the tangent plane is F(x0, y0, z0) (r r0) 0. See, FIGURE 9.6.5. We summarize this last result:, , (x, y, z), , Theorem 9.6.1, , x, , FIGURE 9.6.5 Tangent plane is normal to, gradient at P, , Equation of Tangent Plane, , Let P(x0, y0, z0) be a point on the graph of F(x, y, z) c, where F is not 0. Then an equation, of the tangent plane at P is, Fx(x0, y0, z0)(x x0) Fy(x0, y0, z0)( y y0) Fz(x0, y0, z0)(z z0) 0., , EXAMPLE 3, , (5), , Equation of Tangent Plane, , Find an equation of the tangent plane to the graph of x 2 4y2 z 2 16 at (2, 1, 4)., SOLUTION By defining F(x, y, z) x2 4y2 z2, the given surface is the level surface, F(x, y, z) F(2, 1, 4) 16 passing through (2, 1, 4). Now, Fx(x, y, z) 2x, Fy(x, y, z) 8y,, and Fz(x, y, z) 2z, so that, F(x, y, z) 2x i 8y j 2z k, 508, , |, , CHAPTER 9 Vector Calculus, , and, , F(2, 1, 4) 4 i 8 j 8 k.
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It follows from (5) that an equation of the tangent plane is, 4(x 2) 8( y 1) 8(z 4) 0, , x 2y 2z 8., , or, , Surfaces Given by z ⴝ f (x, y ) For a surface given explicitly by a differentiable, function z f (x, y), we define F(x, y, z) f (x, y) z or F(x, y, z) z f (x, y). Thus, a point, (x0, y0, z0) is on the graph of z f (x, y) if and only if it is also on the level surface F(x, y, z) 0., This follows from F(x0, y0, z0) f (x0, y0) z0 0., Equation of Tangent Plane, , EXAMPLE 4, , Find an equation of the tangent plane to the graph of z 12 x2 12 y2 4 at (1, 1, 5)., z, , SOLUTION Define F(x, y, z) 12 x2 12 y2 z 4 so that the level surface of F passing, through the given point is F(x, y, z) F(1, 1, 5) or F(x, y, z) 0. Now, Fx x, Fy y,, and Fz 1, so that, , 5, , F(x, y, z) x i y j k, , ∇F(1, –1, 5), y, (1, –1, 0), , and, , F(1, 1, 5) i j k., , Thus, from (5) the desired equation is, (x 1) ( y 1) (z 5) 0, , x, , FIGURE 9.6.6 Tangent plane in, Example 4, , or, , x y z 7., , See FIGURE 9.6.6., , Normal Line Let P(x0, y0, z0) be a point on the graph of F(x, y, z) c, where F is, not 0. The line containing P(x0, y0, z0) that is parallel to F(x0, y0, z0) is called the normal, line to the surface at P. As indicated by its name, this line is normal to the tangent plane to, the surface at P., EXAMPLE 5, , Normal Line to a Surface, , Find parametric equations for the normal line to the surface in Example 4 at (1, 1, 5)., SOLUTION A direction vector for the normal line at (1, 1, 5) is F(1, 1, 5) i j k., It follows that parametric equations for the normal line are, x 1 t,, , y 1 t,, , z 5 t., , Expressed as symmetric equations the normal line to a surface F (x, y, z) = c at P(x0, y0, z0), is given by, x 2 x0, y 2 y0, z 2 z0, , , ., Fx(x0, y0, z0), Fy(x0, y0, z0), Fz(x0, y0, z0), In Example 5, you should verify that symmetric equations of the normal line at (1, 1, 5), are, stream, , x215, , 100, 80 P, 60, , y11, z25, 5, ., 21, 21, , REMARKS, , 40, 30, contours, of a hill, , FIGURE 9.6.7 Stream is perpendicular to, contours, , Water flowing down a hill chooses a path in the direction of the greatest change in altitude., FIGURE 9.6.7 shows the contours, or level curves, of a hill. As shown in the figure, a stream starting at point P will take a path that is perpendicular to the contours. After reading Sections 9.5, and 9.6, the student should be able to explain why., , 9.6 Tangent Planes and Normal Lines, , |, , 509
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Exercises, , 9.6, , Answers to selected odd-numbered problems begin on page ANS-22., 26. x 2 2y2 3z2 33; 8x 4y 6z 5, 27. Find points on the surface x 2 4x y2 z2 2z 11, , In Problems 1–12, sketch the level curve or surface passing, through the indicated point. Sketch the gradient at the point., y 2x, ; (1, 3), 1. f (x, y) x 2y; (6, 1), 2. f (x, y) , x, 3. f (x, y) y x 2; (2, 5), 4. f (x, y) x 2 y2; (1, 3), 2, 2, y, x, 5. f (x, y) , ; (2, 3), 4, 9, y2, 6. f (x, y) ; (2, 2), x, 7. f (x, y) (x 1)2 y2; (1, 1), y21, 8. f (x, y) , ; (p/6, 32), sin x, 9. F(x, y, z) y z; (3, 1, 1), 10. F(x, y, z) x 2 y2 z; (1, 1, 3), , at which the tangent plane is horizontal., , 28. Find points on the surface x 2 3y2 4z2 2xy 16 at, , which the tangent plane is parallel to (a) the xz-plane, (b) the, yz-plane, and (c) the xy-plane., , In Problems 29 and 30, show that the second equation is an, equation of the tangent plane to the graph of the first equation at, (x0, y0, z0)., xx0, yy0, zz0, y2, x2, z2, 29. 2 2 2 1; 2 2 2 1, a, b, c, a, b, c, 2, 2, 2, xx, yy, zz, y, x, z, 0, 0, 0, 30. 2 2 2 2 1; 2 2 2 2 1, a, b, c, a, b, c, 31. Show that every tangent plane to the graph of z2 x2 y2, passes through the origin., 32. Show that the sum of the x-, y-, and z-intercepts of every, , 11. F(x, y, z) "x 2 y 2 z 2 ; (3, 4, 0), 12. F(x, y, z) x 2 y2 z; (0, 1, 1), , In Problems 13 and 14, find the points on the given surface at, which the gradient is parallel to the indicated vector., , tangent plane to the graph of "x "y "z "a,, a 0, is the number a., , 13. z x 2 y2; 4 i j 12 k, 14. x 3 y2 z 15; 27 i 8 j k, , In Problems 15–24, find an equation of the tangent plane to the, graph of the given equation at the indicated point., 15. x 2 y2 z2 9; (2, 2, 1), 16. 5x 2 y2 4z2 8; (2, 4, 1), 17. x 2 y2 3z2 5; (6, 2, 3), 18. xy yz zx 7; (1, 3, 5), 19. z 25 x 2 y2; (3, 4, 0), 20. xz 6; (2, 0, 3), 21. z cos(2x y); (p/2, p/4, 1/ "2), 22. x2y3 6z 10; (2, 1, 1), 23. z ln(x 2 y2); (1/ "2, 1/ "2, 0), 24. z 8e2y sin 4x; (p/24, 0, 4), , In Problems 25 and 26, find the points on the given surface at, which the tangent plane is parallel to the indicated plane., 25. x 2 y2 z2 7; 2x 4y 6z 1, , 9.7, , In Problems 33 and 34, find parametric equations for the normal, line at the indicated point. In Problems 35 and 36, find symmetric equations for the normal line., 33. x 2 2y2 z2 4; (1, 1, 1), 34. z 2x 2 4y2; (3, 2, 2), 35. z 4x 2 9y2 1; ( 12 , 13 , 3), 36. x 2 y2 z2 0; (3, 4, 5), 37. Show that every normal line to the graph x 2 y2 z2 a2, passes through the origin., 38. Two surfaces are said to be orthogonal at a point P of intersection if their normal lines at P are orthogonal. Prove that, the surfaces given by F(x, y, z) 0 and G(x, y, z) 0 are, orthogonal at P if and only if FxGx FyGy FzGz 0., In Problems 39 and 40, use the result of Problem 38 to show that, the given surfaces are orthogonal at a point of intersection., 39. x 2 y2 z2 25; x 2 y2 z2 0, 40. x 2 y2 z2 4; z 1/xy2, , Curl and Divergence, , INTRODUCTION In Section 9.1 we introduced the concept of vector function of one variable., In this section we examine vector functions of two and three variables., , Vector Fields Vector functions of two and three variables,, F(x, y) P(x, y) i Q(x, y) j, F(x, y, z) P(x, y, z) i Q(x, y, z) j R(x, y, z) k, 510, , |, , CHAPTER 9 Vector Calculus
