11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR

Transcription

1 SECTION 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 633 wit speed v o along te same line from te opposite direction toward te source, ten te frequenc of te sound eard b te observer is were c is te speed of sound, about 33 ms. (Tis is te Doppler effect.) Suppose tat, at a particular moment, ou are in a train traveling at 34 ms and accelerating at 1. ms. A train is approacing ou from te opposite direction on te oter track at 40 ms, accelerating at 1.4 ms, and sounds its wistle, wic as a frequenc of 460 Hz. At tat instant, wat is te perceived frequenc tat ou ear and ow fast is it canging? Assume tat all te given functions are differentiable. 37. If z f,, were r cos and r sin, (a) find zr and and (b) sow tat z z r z 1 r z 38. If u f,, were e s cos t and e s sin t, sow tat u u u e s s u t 39. z z If z f, sow tat If z f,, were s t and s t, sow tat z z z z s t Assume tat all te given functions ave continuous second-order partial derivatives. 41. Sow tat an function of te form f o c vo c v s f s z f at t at z is a solution of te wave equation [Hint: Let u at, v at.] 4. If u f,, were e s cos t and e s sin t, sow tat es u u u s u t 43. If z f,, were r s, rs, find zr s. (Compare wit Eample 7.) 44. If z f,, were r cos, r sin, find (a) zr, (b), and (c) zr. z 45. If z f,, were r cos, r sin, sow tat z z z r 1 z r 46. Suppose z f,, were ts, t and s, t. (a) Sow tat z t z t z t t z t z (b) Find a similar formula for zs t. 47. Suppose tat te equation F,, z 0 implicitl defines eac of te tree variables,, and z as functions of te oter two: z f,, t, z,, z. If F is differentiable and,, and are all nonzero, sow tat F t F z t a z z z F z t 1 z 1 r z r 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR Recall tat if z f,, ten te partial derivatives and are defined as 1 f 0, 0 lim l 0 f 0, 0 lim l 0 f 0, 0 f 0, 0 f 0, 0 f 0, 0 and represent te rates of cange of z in te - and -directions, tat is, in te directions of te unit vectors i and j. f f

2 634 CHAPTER 11 PARTIAL DERIVATIVES u sin (, ) cos 0 Suppose tat we now wis to find te rate of cange of z at 0, 0 in te direction of an arbitrar unit vector u a, b. (See Figure 1.) To do tis we consider te surface S wit equation z f, (te grap of f ) and we let z 0 f 0, 0. Ten te point P 0, 0, z 0 lies on S. Te vertical plane tat passes troug P in te direction of u intersects S in a curve C. (See Figure.) Te slope of te tangent line T to C at te point P is te rate of cange of z in te direction of u. z FIGURE 1 A unit vector u=ka, bl=kcos, sin l T P(,, z ) Q(,,z) Visual 11.6A animates Figure b rotating u and terefore T. FIGURE If Q,, z is anoter point on C and P, Q are te projections of P, Q on te -plane, ten te vector PBQ is parallel to u and so PBQ u a, b for some scalar. Terefore, 0 a, 0 b, so 0 a, 0 b, and z z z 0 f 0 a, 0 b f 0, 0 If we take te limit as l 0, we obtain te rate of cange of z (wit respect to distance) in te direction of u, wic is called te directional derivative of f in te direction of u. DEFINITION Te directional derivative of f at 0, 0 in te direction of a unit vector u a, b is D u f 0, 0 lim l 0 f 0 a, 0 b f 0, 0 if tis limit eists.

3 SECTION 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 635 B comparing Definition wit Equations (1), we see tat if u i 1, 0, ten D i f f and if u j 0, 1, ten D j f f. In oter words, te partial derivatives of f wit respect to and are just special cases of te directional derivative. Wen we compute te directional derivative of a function defined b a formula, we generall use te following teorem. 3 THEOREM If f is a differentiable function of and, ten f as a directional derivative in te direction of an unit vector u a, b and D u f, f, a f, b PROOF If we define a function t of te single variable b t f 0 a, 0 b ten b te definition of a derivative we ave 4 t t0 t0 lim lim l 0 l 0 D u f 0, 0 f 0 a, 0 b f 0, 0 On te oter and, we can write t f,, were 0 a, 0 b, so te Cain Rule (Teorem 11.5.) gives t f d f d If we now put 0, ten 0, 0, and d d f, a f, b 5 t0 f 0, 0 a f 0, 0 b Te directional derivative D u f 1, in Eample 1 represents te rate of cange of z in te direction of u. Tis is te slope of te tangent line to te curve of intersection of te surface z and te vertical plane troug 1,, 0 in te direction of u sown in Figure 3. z 0 FIGURE 3 (1,, 0) π 6 u Comparing Equations 4 and 5, we see tat If te unit vector u makes an angle wit te positive -ais (as in Figure 1), ten we can write u cos, sin and te formula in Teorem 3 becomes 6 EXAMPLE 1 Find te directional derivative D if f, u f, and u is te unit vector given b angle. Wat is D u f 1,? SOLUTION Formula 6 gives D 3 3 s u f, f, cos 6 f, sin 6 1 [3 s3 3 (8 3s3 )] Terefore D u f 0, 0 f 0, 0 a f 0, 0 b D u f, f, cos f, sin 6 D u f 1, 1 [3s (8 3s3 )] 13 3s3

4 636 CHAPTER 11 PARTIAL DERIVATIVES THE GRADIENT VECTOR Notice from Teorem 3 tat te directional derivative can be written as te dot product of two vectors: 7 D u f, f, a f, b f,, f, a, b f,, f, u Te first vector in tis dot product occurs not onl in computing directional derivatives but in man oter contets as well. So we give it a special name (te gradient of f ) and a special notation (grad f or f, wic is read del f ). 8 DEFINITION If f is a function of two variables and, ten te gradient of f is te vector function f defined b f, f,, f, f f i j EXAMPLE If f, sin e, ten and f, f, f cos e, e f 0, 1, 0 Wit tis notation for te gradient vector, we can rewrite te epression (7) for te directional derivative as 9 D u f, f, u Tis epresses te directional derivative in te direction of u as te scalar projection of te gradient vector onto u. V EXAMPLE 3 Find te directional derivative of te function f, 3 4 at te point, 1 in te direction of te vector v i 5j. SOLUTION We first compute te gradient vector at, 1: f, 3 i 3 4j f, 1 4 i 8 j Note tat v is not a unit vector, but since s9, te unit vector in te direction of v is u v v v s9 i 5 s9 j

5 SECTION 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 637 Te gradient vector f, 1 in Eample 3 is sown in Figure 4 wit initial point, 1. Also sown is te vector v tat gives te direction of te directional derivative. Bot of tese vectors are superimposed on a contour plot of te grap of f. Terefore, b Equation 9, we ave D u f, 1 f, 1 u 4 i 8 j s9 i 5 s9 j s9 3 s9 FUNCTIONS OF THREE VARIABLES ±f(, _1) (, _1) v For functions of tree variables we can define directional derivatives in a similar manner. Again D u f,, z can be interpreted as te rate of cange of te function in te direction of a unit vector u. 10 DEFINITION Te directional derivative of f at 0, 0, z 0 in te direction of a unit vector u a, b, c is FIGURE 4 D u f 0, 0, z 0 lim l 0 if tis limit eists. f 0 a, 0 b, z 0 c f 0, 0, z 0 If we use vector notation, ten we can write bot definitions ( and 10) of te directional derivative in te compact form 11 D u f 0 lim l 0 f 0 u f 0 were 0 0, 0 if n and 0 0, 0, z 0 if n 3. Tis is reasonable since te vector equation of te line troug 0 in te direction of te vector u is given b 0 tu (Equation ) and so f 0 u represents te value of f at a point on tis line. If f,, z is differentiable and u a, b, c, ten te same metod tat was used to prove Teorem 3 can be used to sow tat 1 D u f,, z f,, za f,, zb f z,, zc For a function f of tree variables, te gradient vector, denoted b f or grad f, is f,, z f,, z, f,, z, f z,, z or, for sort, 13 f f, f, f z f f f i j z k

6 638 CHAPTER 11 PARTIAL DERIVATIVES Ten, just as wit functions of two variables, Formula 1 for te directional derivative can be rewritten as 14 D u f,, z f,, z u V EXAMPLE 4 If f,, z sin z, (a) find te gradient of f and (b) find te directional derivative of f at 1, 3, 0 in te direction of v i j k. SOLUTION (a) Te gradient of f is (b) At 1, 3, 0 we ave f 1, 3, 0 0, 0, 3. Te unit vector in te direction of v i j k is u 1 s6 i s6 j 1 s6 k Terefore, Equation 14 gives f,, z f,, z, f,, z, f z,, z sin z, z cos z, cos z D u f 1, 3, 0 f 1, 3, 0 u 3k 1 s6 i s6 j 1 s6 k 3 s6 1 3 MAXIMIZING THE DIRECTIONAL DERIVATIVE Suppose we ave a function f of two or tree variables and we consider all possible directional derivatives of f at a given point. Tese give te rates of cange of f in all possible directions. We can ten ask te questions: In wic of tese directions does f cange fastest and wat is te maimum rate of cange? Te answers are provided b te following teorem. Visual 11.6B provides visual confirmation of Teorem THEOREM Suppose f is a differentiable function of two or tree variables. Te maimum value of te directional derivative D is f u f and it occurs wen u as te same direction as te gradient vector f. PROOF From Equation 9 or 14 we ave 0 0 D u f f u f u cos f cos were is te angle between f and u. Te maimum value of cos is 1 and tis occurs wen. Terefore, te maimum value of D is f u f and it occurs wen, tat is, wen u as te same direction as f.

7 SECTION 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 639 Q 1 ±f(, 0) 0 1 P 3 FIGURE 5 At, 0 te function in Eample 5 increases fastest in te direction of te gradient vector f, 0 1,. Notice from Figure 5 tat tis vector appears to be perpendicular to te level curve troug, 0. Figure 6 sows te grap of f and te gradient vector z FIGURE 6 EXAMPLE 5 (a) If f, e, find te rate of cange of f at te point P, 0 in te direction from P to Q( 1, ). (b) In wat direction does f ave te maimum rate of cange? Wat is tis maimum rate of cange? SOLUTION (a) We first compute te gradient vector: Te unit vector in te direction of PQ l 1.5, is u 3 5, 4 5, so te rate of cange of f in te direction from P to Q is (b) According to Teorem 15, f increases fastest in te direction of te gradient vector f, 0 1,. Te maimum rate of cange is EXAMPLE 6 Suppose tat te temperature at a point,, z in space is given b T,, z 801 3z, were T is measured in degrees Celsius and,, z in meters. In wic direction does te temperature increase fastest at te point 1, 1,? Wat is te maimum rate of increase? SOLUTION Te gradient of T is f, f, f e, e f, 0 1, D u f, 0 f, 0 u 1, 3 5, 4 5 1( 3 5 ) ( 4 5) 1 f, 0 1, s5 T T z i i T j T z k z j z i j 3z k At te point 1, 1, te gradient vector is T1, 1, i j 6 k 5 8i j 6 k B Teorem 15 te temperature increases fastest in te direction of te gradient vector T1, 1, 5 8i j 6 k or, equivalentl, in te direction of i j 6 k or te unit vector i j 6 ks41. Te maimum rate of increase is te lengt of te gradient vector: T1, 1, 5 8 i j 6 k 480z 1 3z k 5s41 8 Terefore, te maimum rate of increase of temperature is 5s418 4Cm.

8 640 CHAPTER 11 PARTIAL DERIVATIVES TANGENT PLANES TO LEVEL SURFACES Suppose S is a surface wit equation F,, z k, tat is, it is a level surface of a function F of tree variables, and let P 0, 0, z 0 be a point on S. Let C be an curve tat lies on te surface S and passes troug te point P. Recall from Section 10.7 tat te curve C is described b a continuous vector function rt t, t, zt. Let t 0 be te parameter value corresponding to P; tat is, rt 0 0, 0, z 0. Since C lies on S, an point t, t, zt must satisf te equation of S, tat is, 16 F(t, t, zt) k If,, and z are differentiable functions of t and F is also differentiable, ten we can use te Cain Rule to differentiate bot sides of Equation 16 as follows: 17 F d dt F d dt F z dz dt 0 But, since F F, F, F z and rt t, t, zt, Equation 17 can be written in terms of a dot product as F rt 0 In particular, wen t t 0 we ave rt 0 0, 0, z 0, so 18 F 0, 0, z 0 rt 0 0 z ±F(,, z ) tangent plane Equation 18 sas tat te gradient vector at P, F 0, 0, z 0, is perpendicular to te tangent vector rt 0 to an curve C on S tat passes troug P. (See Figure 7.) If F 0, 0, z 0 0, it is terefore natural to define te tangent plane to te level surface F,, z k at P 0, 0, z 0 as te plane tat passes troug P and as normal vector F 0, 0, z 0. Using te standard equation of a plane (Equation ), we can write te equation of tis tangent plane as 19 F 0, 0, z 0 0 F 0, 0, z 0 0 F z 0, 0, z 0 z z 0 0 FIGURE 7 Te normal line to S at P is te line passing troug P and perpendicular to te tangent plane. Te direction of te normal line is terefore given b te gradient vector F 0, 0, z 0 and so, b Equation , its smmetric equations are 0 0 F 0, 0, z 0 0 F 0, 0, z 0 z z 0 F z 0, 0, z 0 In te special case in wic te equation of a surface S is of te form z f, (tat is, S is te grap of a function f of two variables), we can rewrite te equation as F,, z f, z 0

9 SECTION 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 641 and regard S as a level surface (wit k 0) of F. Ten so Equation 19 becomes F 0, 0, z 0 f 0, 0 F 0, 0, z 0 f 0, 0 F z 0, 0, z 0 1 f 0, 0 0 f 0, 0 0 z z 0 0 wic is equivalent to Equation Tus our new, more general, definition of a tangent plane is consistent wit te definition tat was given for te special case of Section V EXAMPLE 7 Find te equations of te tangent plane and normal line at te point, 1, 3 to te ellipsoid 4 z 9 3 SOLUTION Te ellipsoid is te level surface (wit k 3) of te function F,, z 4 z 9 Figure 8 sows te ellipsoid, tangent plane, and normal line in Eample 7. 4 Terefore, we ave F,, z F, 1, 3 1 F,, z F, 1, 3 F z,, z z 9 F z, 1, z FIGURE 8 Ten Equation 19 gives te equation of te tangent plane at, 1, 3 as z 3 0 wic simplifies to 3 6 z B Equation 0, smmetric equations of te normal line are 1 1 z 3 3 SIGNIFICANCE OF THE GRADIENT VECTOR We now summarize te was in wic te gradient vector is significant. We first consider a function f of tree variables and a point P 0, 0, z 0 in its domain. On te one and, we know from Teorem 15 tat te gradient vector f 0, 0, z 0 gives te direction of fastest increase of f. On te oter and, we know tat f 0, 0, z 0 is ortog-

10 64 CHAPTER 11 PARTIAL DERIVATIVES onal to te level surface S of f troug P. (Refer to Figure 7.) Tese two properties are quite compatible intuitivel because as we move awa from P on te level surface S, te value of f does not cange at all. So it seems reasonable tat if we move in te perpendicular direction, we get te maimum increase. In like manner we consider a function f of two variables and a point P 0, 0 in its domain. Again te gradient vector f 0, 0 gives te direction of fastest increase of f. Also, b considerations similar to our discussion of tangent planes, it can be sown tat f 0, 0 is perpendicular to te level curve f, k tat passes troug P. Again tis is intuitivel plausible because te values of f remain constant as we move along te curve. (See Figure 9.) ±f(, ) P(, ) level curve f(, )=k 0 curve of steepest ascent FIGURE 9 FIGURE 10 If we consider a topograpical map of a ill and let f, represent te eigt above sea level at a point wit coordinates,, ten a curve of steepest ascent can be drawn as in Figure 10 b making it perpendicular to all of te contour lines. Tis penomenon can also be noticed in Figure 11 in Section 11.1, were Lonesome Creek follows a curve of steepest descent EXERCISES 1 Find te directional derivative of f at te given point in te direction indicated b te angle. 1. f, s5 4, 4, 1,. f, sin,, 0, 3 6 (a) Find te gradient of f. (b) Evaluate te gradient at te point P. (c) Find te rate of cange of f at P in te direction of te vector u. 3. f, 5 4 3, P1,, 4. f, ln, P1, 3, 5. f,, z e z, P3, 0,, 3 6. f,, z s z, P1, 3, 1, 6 u 5 13, u 5, 3 5 u 3, 3, 1 3 u 7, 3 7, Find te directional derivative of te function at te given point in te direction of te vector v. 7. f, 1 s, 3, 4, v 4, 3 8. f, ln,, 1, v 1, 9. ts, t s e t,, 0, v i j 10. f,, z z, 4, 1, 1, v 1,, t,, z 3z 3, 1, 1,, v j k 1. Use te figure to estimate D u f,. (, ) u ±f(, ) 0

11 SECTION 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR Find te directional derivative of f, s at P, 8 in te direction of Q5, Find te directional derivative of f,, z z at P, 1, 3 in te direction of te origin Find te maimum rate of cange of f at te given point and te direction in wic it occurs. 15. f,,, f p, q qe p pe q, 17. f,, z ln z 3, 0, 0 1,, f,, z tan 3z, 5, 1, (a) Sow tat a differentiable function f decreases most rapidl at in te direction opposite to te gradient vector, tat is, in te direction of f. (b) Use te result of part (a) to find te direction in wic te function f, 4 3 decreases fastest at te point, Find te directions in wic te directional derivative of f, sin at te point (1, 0) as te value Find all points at wic te direction of fastest cange of te function f, 4 is i j.. Near a buo, te dept of a lake at te point wit coordinates, is z , were,, and z are measured in meters. A fiserman in a small boat starts at te point 80, 60 and moves toward te buo, wic is located at 0, 0. Is te water under te boat getting deeper or sallower wen e departs? Eplain. 3. Te temperature T in a metal ball is inversel proportional to te distance from te center of te ball, wic we take to be te origin. Te temperature at te point 1,, is 10. (a) Find te rate of cange of T at 1,, in te direction toward te point, 1, 3. (b) Sow tat at an point in te ball te direction of greatest increase in temperature is given b a vector tat points toward te origin. 4. Te temperature at a point,, z is given b T,, z 00e 3 9z were T is measured in C and,, z in meters. (a) Find te rate of cange of temperature at te point P, 1, in te direction toward te point 3, 3, 3. (b) In wic direction does te temperature increase fastest at P? (c) Find te maimum rate of increase at P. 5. Suppose tat over a certain region of space te electrical potential V is given b V,, z 5 3 z. (a) Find te rate of cange of te potential at P3, 4, 5 in te direction of te vector v i j k. (b) In wic direction does V cange most rapidl at P? (c) Wat is te maimum rate of cange at P? 6. Suppose ou are climbing a ill wose sape is given b te equation z , were,, and z are measured in meters, and ou are standing at a point wit coordinates 60, 40, 966. Te positive -ais points east and te positive -ais points nort. (a) If ou walk due sout, will ou start to ascend or descend? At wat rate? (b) If ou walk nortwest, will ou start to ascend or descend? At wat rate? (c) In wic direction is te slope largest? Wat is te rate of ascent in tat direction? At wat angle above te orizontal does te pat in tat direction begin? 7. Let f be a function of two variables tat as continuous partial derivatives and consider te points A1, 3, B3, 3, C1, 7, and D6, 15. Te directional derivative of at in te direction of te vector AB l f A is 3 and te directional derivative at A in te direction of AC l is 6. Find te directional derivative of f at A in te direction of te vector AD l. 8. For te given contour map draw te curves of steepest ascent starting at P and at Q. Q 9. Sow tat te operation of taking te gradient of a function as te given propert. Assume tat u and v are differentiable functions of and and a, b are constants. (a) au bv a u b v (b) uv u v v u (c) u v 30. Sketc te gradient vector f 4, 6 for te function f wose level curves are sown. Eplain ow ou cose te direction and lengt of tis vector. 6 4 v u u v 0 P v _5 _3 _ (d) (4, 6) 40 u n nu n1 u

12 644 CHAPTER 11 PARTIAL DERIVATIVES Find equations of (a) te tangent plane and (b) te normal line to te given surface at te specified point. 31. z z, 3. z 4 arctanz, 1, 1, z 1 e cos z, 1, 0, z ln z, 0, 0, 1 ; Use a computer to grap te surface, te tangent plane, and te normal line on te same screen. Coose te domain carefull so tat ou avoid etraneous vertical planes. Coose te viewpoint so tat ou get a good view of all tree objects. 35. z z 3, 36. z 6, 1,, 3 1, 1, 1, 1, If f, 4, find te gradient vector f, 1 and use it to find te tangent line to te level curve f, 8 at te point, 1. Sketc te level curve, te tangent line, and te gradient vector. 38. If t,, find te gradient vector t3, 1 and use it to find te tangent line to te level curve t, at te point 3, 1. Sketc te level curve, te tangent line, and te gradient vector. 39. Sow tat te equation of te tangent plane to te ellipsoid a b z c 1 at te point 0, 0, z 0 can be written as 0 0 zz0 a b c Find te points on te ellipsoid 3z 1 were te tangent plane is parallel to te plane 3 3z Find te points on te perboloid z 1 were te normal line is parallel to te line tat joins te points 3, 1, 0 and 5, 3, Sow tat te ellipsoid 3 z 9 and te spere z 8 6 8z 4 0 are tangent to eac oter at te point 1, 1,. (Tis means tat te ave a common tangent plane at te point.) 43. Sow tat te sum of te -, -, and z-intercepts of an tangent plane to te surface s s sz sc is a constant. 44. Sow tat ever normal line to te spere z r passes troug te center of te spere. 45. Find parametric equations for te tangent line to te curve of intersection of te paraboloid z and te ellipsoid 4 z 9 at te point 1, 1,. 46. (a) Te plane z 3 intersects te clinder 5 in an ellipse. Find parametric equations for te tangent line to tis ellipse at te point 1,, 1. ; (b) Grap te clinder, te plane, and te tangent line on te same screen. 47. (a) Two surfaces are called ortogonal at a point of intersection if teir normal lines are perpendicular at tat point. Sow tat surfaces wit equations F,, z 0 and G,, z 0 are ortogonal at a point P were F 0 and G 0 if and onl if F G F G F zg z 0 at P (b) Use part (a) to sow tat te surfaces z and z r are ortogonal at ever point of intersection. Can ou see w tis is true witout using calculus? 48. (a) Sow tat te function f, s 3 is continuous and te partial derivatives f and f eist at te origin but te directional derivatives in all oter directions do not eist. ; (b) Grap f near te origin and comment on ow te grap confirms part (a). 49. Suppose tat te directional derivatives of f, are known at a given point in two nonparallel directions given b unit vectors u and v. Is it possible to find f at tis point? If so, ow would ou do it? 50. Sow tat if z f, is differentiable at 0 0, 0, ten lim l 0 f f 0 f [Hint: Use Definition directl.] MAXIMUM AND MINIMUM VALUES As we saw in Capter 4, one of te main uses of ordinar derivatives is in finding maimum and minimum values. In tis section we see ow to use partial derivatives to locate maima and minima of functions of two variables. In particular, in Eample 5 we will see ow to maimize te volume of a bo witout a lid if we ave a fied amount of cardboard to work wit.