Math 4013 Solutions to Homework Problems from Chapter 2

Transcription

1 Math 4013 Solutions to Homework Problems from Chapter 2 Section Sketch the level curves and graphs of the following functions: a f : R 2 R, x, y x y +2 The level curves are just the lines x y +2C yx+2 C b f : R 2 R, x, y xy 1

2 2 The level curves are the curves where xy C y C x 2. a Describe the behavior, as c varies, of the level curve fx, y cfor the function fx, y x 2 +y The equation of the defining the level curves is just or fx, y c x 2 + y 2 +1c x 2 + y 2 1 c. This is the equation of circle about the origin with radius 1 c. Note that c must be less than or equal to one; otherwise the equation of the level curve has no solutions for real numbers x and y. Below we sketch a few level curves. 3. Sketch or describe the level surfaces and a section of the graph of the following function: f : R 3 R ; x, y, z x 2 y 2 z 2. The level surfaces of this function must be the solution sets of equations of the form or fx, y, z c, x 2 y 2 z 2 c,

3 3 Figure 1 or x 2 + y 2 + z 2 c. The last equation is just the equation of a sphere about the origin in R 3 of radius c. We conclude that the level surfaces of f are just sphere s about the origin. The z 0 section of the graph of f will consist of points in R 4 corresponding to the intersection of the graph of f with the plane z 0: S { x, y, z, t R 4 t x 2 y 2 2} z { x, y, z, t R 4 z 0 } { x, y, z, t R 4 z 0, t x 2 2} y 4. Sketch or describe the surface in R 3 corresponding to the equation 4x 2 + y This surface will just be a two dimensional ellipse extende to ± in the z-direction; a squashed cylinder if you will. 5. Sketch or describe the surface in R 3 corresponding to the equation 2 2 x 4 y 4 + z 9. Section Show that the following subset of R 2 is open. { } B x, y R 2 y>0. By definition, a subset B of R 2 is open if for any point r B there exists a disk D ρ r of radius ρ>0 about r that lies entirely within B.

4 4 Let r x, y be an arbitrary point of B. Consider the disk D y 2 r { x R 2 x <ρ/2 }. Clearly, this disk lies completely with B. This this construction works for any point r B, we conclude that for every point of B there is an open disk containing that point lying completely within B. Thus, B is open. 2. Prove that if U and V are neighborhoods of x R n, then so are U V and V U. By definition, a set S is a neighborhood of a point x R n if i x S; ii S is open. Clearly, since U and V are neighborhoods of x, x U and x V and so x U V and x U V. It thus remains to show that both U V and U V are open. We first prove that U V is open. Let y U V. Then either i y U ii y V Note that these two cases are exhaustive but not exclusive. In the first case, since U is open since it is a neighborhood, there must exist an open ball B ρ1 y about y completely contained in U. But then this ball is also completely contained in U V. In the second case, since V is open, there must exist some open ball B ρ2 y completely contained in V and so completely contained in U V. Thus, in either case there must exist an open ball about y that is completely contained in U V. Hence, U V is open. We now show that U V is open. Let y U V. Clearly, y U and y V. Since U and V are open, there must exist open balls Set B ρ1 y B ρ2 y U V ρ min [ρ 1,ρ 2 ]. Then B ρ y B ρ1 y U and B ρ y B ρ2 y V, and so B ρ y U V. There thus exists an open ball about y that is completely contained in U V. Since y is an arbitrary point of U V, we conclude that U V must be an open set. 3. Compute the following its. The text allows the reader to assume that the exponential, sine, and cosine functions are all continuous. By Example 7 on page 108, any polynomial function on R n is also continuous. a x 3 y x,y 0,1 Since x 3 y is a polynomial function on R 2, it is continuous. Hence, its it at a point r R 2 coincides with its value at the point r. Thus x 3 y x 3 y x,y 0,1 0,1 b e x y x,y 0,1 The exponential function fx, y e x and the linear function gx, y yare both continuous for all x, y R 2. By Theorem 4iii so must be fgx, y fx, ygx, y e x y. Since e x y is continuous, its it at any point must coincide with its value there. Thus, x,y 0,1 e x y e x y 0,1 0 1.

5 5 c sin 2 x x 0 x The function in the denominator vanishes at x 0 and so we cannot apply continuity arguments to evaluate the it. In other words, Theorem 4 iv cannot be applied in this case. We can, however, evaluate this it using l Hospital s rule: if f and g are continuous differentiable functions and then f x 0 gx x xo x x o fx x x o gx x x f x o g x. Applying l Hospital s rule to the case at hand we find sin 2 x x 0 x 2 cosx sinx x 0 1 2cosx sinx x 0 2 cosxsinx x0 0 In the last couple of steps we have used the fact that the sine and cosine functions are continuous - so their its were taken by simply evaluated them at the it point. d sin 2 x x 0 2 x This problem is of course similar to the preceding one. The only difference is that we must apply l Hospital s Rule twice before we get an obviously continuous function. sin 2 x 2cosxsinx x 0 2 x x 0 2x 2sin 2 x+2cos 2 x x 0 2 sin x 0 2 x +cos 2 x 1 4. Compute the following its if they exist. a x 2 + y 2 +3 x,y 0,0 This is the it of a polynomial function on R 2. Since polynomial functions are continuous, we can evaluate the it by simply evaluating the polynomial at the it point. Thus, x,y 0,0 x 2 + y 2 +3 x 2 +y ,0 3. b x,y 0,0 xy x 2 + y 2 +2

6 6 Both fx, y xy and gx, y x 2 + y are continuous for all x. Moreover, g0, 0 2 0, so we can first apply Theorem 4 iv to conclude that 1 is continuous at 0,0 and then Theorem g 4 iii to conclude that f xy g x 2 +y 2 is continuous at 0,0. Therefore, the it at 0, 0 must +2 coincide with evaluation at 0,0. Thus, x,y 0,0 xy x 2 + y 2 +2 xy x 2 + y ,0 0. c x,y 0,0 e xy x +1 This is similar to the preceding problem. Since fx, y e xy is continuous at 0,0 and gx, y x+ 1 is continuous and non-vanishing at 0,0, f/g is continuous at 0,0 and so we can evaluate its it at 0,0 by evaluating f/g at 0,0: x,y 0,0 e xy e xy x +1 x +1 0, d cosx 1 x2 2 x,y 0,0 x 4 + y 4 In this case, the it does not exist. To see this, consider the following curve through 0,0: γ : R R 2, t 0, t, and set fx, y cosx 1 x 2 2 x 4 + y 4. If the it of fx, y asx, y approaches 0,0 exists, it must be coincide with 1 f γt t 0 The right hand side, however, is undefined. We conclude that the it t 0 t 4 does not exist. e x,y 0,0 cosx 1 x2 x 4 + y 4 2 x y 2 x,y 0,0 x 2 + y 2 This it does not exist either. To see this, consider the following two curves through the 0,0: γ 1 : R R 2, t t, 0 γ 2 : R R 2, t t, t and set fx, y If the it e is to exist, we must have f γ 1 t t 0 x y2 x 2 + y 2. x y 2 x,y 0,0 x 2 + y 2 t 0 fγ 2 t

7 7 Evaluating the it on the far left hand side we find t 0 2 f γ 1 t t 0 t 0 t , but the it on the far right hand side is t t 2 f γ 2 t t 0 t 0 t 2 + t 2 0. Since these two its do not agree, we can conclude the it does not exist. 5. Show that the map x y 2 x,y 0,0 x 2 + y 2 f : R R, x x2 e x 2 sinx is continuous. We can prove this using the Theorem 4 and the fact that the exponential function, the sine function, and any polynomial function is continuous. Since x 2 and e x are continuous, so is their product px x 2 e x by Theorem 4 iii. Since the constant function hx 2 and sinx are continuous, so is their sum qx 2 sinx, by Theorem 4 ii. Since 2 sinx is nowhere zero, the quotient function 1/qx is continous everywhere, by Theorem 4 iv. Finally since both px and 1/qx are continuous, so is by Theorem 4 iii. px 1 qx x 2 x e 2 sinx Section Find and a fx, y xy. for each of the following functions. y x b fx, y xcosx cosy. 2. Evaluate cosxcosy xsinxcosy x cosx siny [ ] and for the function z log 1+xy at the points 1,2 and 0, xy 1 1+xy y 2 1+xy x 2 1+xy y 21+xy x 21+xy

8 8 Thus, 1,2 1,2 0,0 0, Find the partial derivatives w w and when w xex2 +y 2. w w e x2 +y 2 + x 2xe 2 x2 +y x 2ye 2 x2 +y 4. Show that the function 2xy fx, y x 2 + y 2 2 is differentiable at each point in its domain. Is this function C 1. The natural domain of fx, y isr 2 {0, 0} the natural domain of a rational function in n variables is R n minus the points where the denominator vanishes. Now 2y x 2 + y 2 2 2xy 2 x 2 + y 2 2x x 2 + y 2 4 2x4 y +4x 2 y 3 +2y 5 8x 4 y 8x 2 y 3 x 2 +y 2 4 2x x 2 + y 2 2 2xy 2 x 2 + y 2 2y x 2 + y 2 4 2x5 +4x 3 y 2 +2xy 4 8x 3 y 2 8xy 4 x 2 + y 2 4 Both partials are rational functions with singularities at the origin. They are therefore continuous everywhere except possibly at 0,0. Since the partial derivatives are continuous everywhere within the domain of f, f is C 1 throughout its domain. 5. Find the equation of the plane tangent to the surface z x 2 + y 3 at 3,1,10. The tangent plane to a graph z fx, y at the point x o,y o,fx o,y o consists of points x, y, z R 3 satisfying the following equation z f x o,y o + For the case at hand, xo,yo x x o + xo,yo y y o. and so we must have 2x 3y z x 3 + 3y 1

9 9 or 6x +3y z Compute the matrix of partial derivatives of the following function: f : R 2 R 3, fx, y xe y +cosy,x,x+e y. Df e y xe y siny e y 7. Find the equation of the tangent plane to z x 2 +2y 3 at 1,1,3. We have 2x 6y 2 and so using the formula appearing in the solution to Problem 2.3.5, the equation of the tangent plane to the graph of x 2 +2y 3 at 1,1,3 is z 3+2x 1 + 6y 1 or 2x +6y z5. Section Find σ t and σ 0 for the following path. We have and so σ t dσ t σt e t,cost, sint. dt d e t, dt dt e t, sint, cost d cost, d dt sint σ 0 e 0, sin0, cos0 1,0,1 2. Determine the velocity and acceleration vectors, and the equation of the tangent line for each of the following curves at the specified value of t. a rt 6ti+3t 2 j+t 3 k, t 0 At t 0,wehave dr dt d 2 r 6i +6tj+3t 2 k dt 2 6j +6tk r0 0i +0j+0k0 r 0 6i

10 10 Therefore, the equation of the tangent line at t 0is ltr0 + tr 0 6ti. b σt We have sin3t, cos3t, 2t 3/2,t1 σ t σ t 3 cos3t, 3 sin3t, 3t 1/2 9 sin3t, 9 cos3t, 32 t 1/2. At t 1 we have σ1 sin3, cos3, 2 σ 1 3 cos3, 3 sin3, 3 so the equation of the tangent line at t 1is ltσ1 + tσ 1 sin3 + 3 cos3t, cos3 3 sin3t, 2+3t c σt cos 2 t, 3t t 3,t, t0 We have σ t 2 costsint,3 3t 2,1 σ t 2sin 2 t+2cos 2 t, 6t, 0. At t 0 we have σ0 1,0,0 σ t 0,3,1 so the equation of the tangent line at t 1is ltσ0 + tσ 0 1, 3t, t d σt 0,0,t, t1 We have σ t 0,0,1 σ t 0,0,0. At t 1 we have σ0 0,0,1 σ t 0,0,1 so the equation of the tangent line at t 1is ltσ1 + tσ 1 0, 0, 1+t. 3. Determine the velocity and acceleration vectors, and the equation of the tangent line for each of the following curves at the specified value of t. a rt costi + sintj, t0 dr dt sinti +costj d 2 r dt 2 costi sintj

11 11 At t 0,wehave r0 1i +0j i r 0 0i +1j j Therefore, the equation of the tangent line at t 0is lt r0 + tr 0 i + tj. b σt tsint, tcost, 3t, t0 We have σ t sint+tcost, cost t sint, 3 At t 0 we have σ t 2 cost t sint, 2sint tcost, 0. σ0 0, 1, 0 σ 0 2, 0, 3 so the equation of the tangent line at t 0is lt σ0 + tσ 0 2t, 1, 3t c rt 2ti+ e t j+e t k,t0 We have At t 0 we have σ t 2i+e t j e t k σ t e t j+e t k. σ0 0,e,e 1 σ t 0,e, e 1 so the equation of the tangent line at t 0is ltσ0 + tσ 0 0,e+et, e 1 e 1 t d σt ti+tj+ 2 3 t3/2 k,t9 We have σ t σ t 1,1,t 1/2 0,0, 12 t 1/2. At t 9 we have σ9 9, 9, 18 σ t 1,1,3 so the equation of the tangent line at t 9is ltσ9 + tσ t, 9+t, t. 4. Find the path σ such that σ0 0, 5, 1 and σ t t, e t,t 2.

12 To 12 Let σt σ x t,σ y t,σ z t. We must have σ xt t σ yt e t σ t t2 z Integrating both sides of these equations with respect to t we obtain σ x t 1 2 t2 + A σ y t e t + B σ z t 1 3 t3 + C Here A, B, C are arbitrary constants of integration. A, B, C will be determined by the initial conditions Thus, 0 σ x A 5 σ y B 1 σ z 0 C σt t2,e t 6, 3 t3 +1 A 0 B 6 C 1 5. Suppose a particle follows a path σt e t,e t,cost until it flies off on a tangent at time t 1. Whereisitattimet2. At time t 1, the particle is at position σ1 e, e 1, cos1 with velocity σ 1 e t, e t, sint t1 e, e 1, sin1. The text intends for us to assume that the particle thereafter travels in straight line with velocity σ 1. Thus, rt σ1 + σ 1t 1 e + et 1,e 1 e 1 t 1, cos1 sin1t 1. At time t 2 then r2 2e, 0, cos1 sin1. Section Write out the chain rule for each of the following functions and justify your answer in case using Theorem 11. a h where hx, y fx, ux, y. compute h we regard hx, y as a composition of a function f : R2 R :v, u fv,u aand a map g : R 2 R 2 ;x, y x, ux, y. Then Df Dg v g1 g2 u g1 g2 1 0 u u

13 13 and so h, h Dh Df g DfDg v u + u u x,u, 1 0 u u u u Therefore, equating the first components on the extreme sides of this equation: h + u u. b dh dx where hx fx, ux,vx. In this case, we regard hx as a composition of two functions f : R 3 R, w, u, v fw, u, v and g : R R 3, x x, ux,vx. The general chain rule then gives dh dx D f g Df gx Dg x w u v dg 1 + w dx u + u dg1 ds dg2 dx dg3 dx du + dx dg 2 + dx v v dv dx dg 3 dx In the last step, we have simply replaced w by x, g 1 by x, g 2 by ux, and g 3 by vx in accordance with the definitions of f and g. c h where hx, y, z fux, y, z,vx, y,wx. In this case, we regard hx, y, z as the composition f g of two functions g : R 3 R 3, x, y, z ux, y, z,vx, y,wx and f : R 3 h h h Dh Df R. The general chain rule gives gx,y,z Dg x,y,z u, v, w gx,y,z g 2 g 3 g 2 g 3 g 2 g 3 So h g1 u u u g2 g3 + + v w v v dw + + w dx 2. Verify the first special case of the chain rule for the composition f c in each of the following cases. a fx, y xy, ct e t,cost. If we compute ft fct explicitly, we find f ct xtyt e t cost,

14 14 so df dt et cost e t sint. On the other hand, according to the chain rule df dt Df Df ct Dc, dc 2 dt yt,xt t ct sint cost,e t sint coste t e t sint which, of course agrees with the preceding computation. b fx, y e xy, ct 3t 2,t 3. If we compute ft fct explicitly, we find f ct e xtyt e 3t5, so df e 3t dt 15t4 On the other hand, according to the chain rule df dt Df Df Dc ct t, 5. ct dc 2 dt yte xtyt,xte xtyt 3t 2 t 3 e 3t5,3t 2 e 3t5 3t 2 6t 4 e 3t5 +9t 4 e 3t5 15t 4 e 3t5 which, of course agrees with the preceding computation. 3. Let f : R 3 R be differentiable. Making the substitution x ρ cosθ sinφ, y ρsinθ sinφ, z ρcosφ spherical coordinates into fx, y, z, compute / ρ, / θ, and / φ. Let g : R 3 R 3 be the map sending If we set ρ, θ, φ ρ cosθsinφ, ρsinθ sinφ,ρcosφ f ρ, θ, φ fxρ, θ, φ,yρ, θ, φ,zρ, θ, φ

15 15 then the Chain Rule tells us that φ Df ρ,θ,φ ρ θ xρ,θ,φ,yρ,θ,φ,zρ,θ,φdg ρ ρ θ θ ρ,θ,φ φ φ sinθ sinφ ρ cosθsinφ ρsinθcosφ cosφ 0 ρsinφ Thus, ρ θ φ cosθsinφ+ sinθ sinφ + cosφ ρ sinθ sinφ + ρ cosθsinφ ρ cosθ cosφ + ρ sinθcosφ ρsinφ where, of course, the partial derivatives, ρcosθ sinφ,ρsinθsinφ,ρcosφ. 4. Let fu, v, are all to be evaluated at the point x, y, z tanu 1 2 e v,u 2 v and gx, y e x y,x y. Calculate f g and D f g1,1. Since both f and g are functions from R 2 to R 2, f g is also a function from R 2 to R 2. We have f gx, y tan ux, y 1 e vx,y, ux, y 2 vx, y 2 tan e x y 1 e x y, e x y 2 x y 2 If we try to compute D f g1,1 directly we re going get a mess; we d have to employ the chain rule for ordinary differentiation a number of times without any simplification appearing until we finally evaluate the result at the point 1,1. Using the chain rule for partial derivatives, we can hope for some simplification as soon as we we evaluate Df at the point g1, 1 1, 0 and Dg at the point 1,1. D f g 1,1 Df g1,1 Dg 1 u 2 u 1 v 2 v sec 2 u 1 2u ,1 u,vg1,1 e v 2v g1 g2 u,v1,0 g1 g2 e x y 1,1 e x y 1 1 1,1 Section Show that the directional derivative of fx, y, z z 2 x+y 3 at 1,1,2 in the direction is 2 5. According to Theorem 12 on page 147, Dnf x o,yo,zo f xo,yo,zo n. 1/ 5, 2/ 5, 0

16 16 Thus, D 1 2, 5,0 5 [ ] z 2 x + y 3 1,1,2 z 2, 3y 2, 2zx 1,1,2 4,3, , 2, , 2, Find the equation of the plane tangent to the surface z fx, y at the indicated point. a z x 3 + y 3 6xy, 1,2,-3. This can be done two different ways. The first is to recognize that z x 3 + y 3 6xy is the graph of the function fx, y. According to the discussion in Section 2.3 see pages , the equation of the plane tangent to the graph of f at x o,y o is z fx o,y o + xo,y o x x o + For the case at hand, this would lead to the following equation xo,y o y y o. or z 3 + 3x 2 6y x 1 + 3y 2 6x y 2 1,2 1,2 3 9x 1 + 6y 2 9x +6y 6 9x 6y + z 6. The second method is to rewrite z x 3 + y 3 6xy as x 3 + y 3 6xy z 0 as the equation of a level surface of a function gx, y, z x 3 +y 3 6xy z. According to the definition of page 150, the equation of the tangent plane to a level surface at a point xo,y o,z ois fxo,y o,z o x xo,y y o,z z o0. For the case at hand, we would have 0 3x 2 6y, 3y 2 6x, 1 or 1,2, 3 9,6, 1 x 1, y 2,z+3 9x+9+6y 12 z 3 9x +6y z 6 9x 6y + z 6 which, of course, is identical to the result of the first method. x 1, y 2,z+3

17 17 b z cosxcosy, 0, π/2,0. To find the equation of the tangent plane, we ll use the second method of part a. The surface z cosxcosy coincides with the level surface fx, y, z 0 of the function fx, y, z cosx cosy z. Thus, the equation of the tangent plane to the level surface fx, y, z 0 at the point 0, π/2,0 is π 0 f 0, 2,0 x 0,y π2,z 0 sinx cosy, cosxsiny, 1 x, 0, π 2,0 y π2,z 0, 1, 1 x, y π2,z or y+ π 2 z y + z π Compute the gradient f for each of the following functions. a fx, y, z 1/ x 2 +y 2 +z 2 f,, x [x 2 + y 2 + z 2 ] 3/2, y [x 2 + y 2 + z 2 ]]]] 3/2, z [x 2 + y 2 + z 2 ] 3/2 b fx, y, z xy + yz + xz f,, y + z, x + z, x + y c fx, y, z 1 x 2 +y 2 +z 2. f,, 2x [x 2 + y 2 + z 2 ] 2, 2y [x 2 + y 2 + z 2 ]]]] 2, 2z [x 2 + y 2 + z 2 ] 2 4. For each of the functions in Exercise 6, what is the direction of fastest increase at 1,1,1? The direction of fastest increase of a differentiable function f atapointx o,y o,z o isgivenbythe value of f at that point. Thus, for the three functions in Problem we have, respectively, 1 1,, , 2, 2 2 2,, as the un-normalized directions of fastest increase.

18 18 5. Captain Ralph is in trouble near the sunny side of Mercury. The temperature of the ship s hull when he is at location x, y, z will be given by T x, y, z e x2 2y 2 3z 2 where x, y, z are measured in meters. He is currently at 1,1,1. a In what direction should he proceed in order to decrease his temperature most rapidly? The direction of fastest increase in temperature around the point 1,1,1 is given by the value of the gradient T at that point. The direction of the fastest decrease in temperature will be the direction exactly opposite T 1, 1, 1. Thus, Captain Ralph should head in the direction T 1, 1, 1 2xe x2 2y 2 3z 2, 4ye x2 2y 2 3z 2, 6ze x2 2y 2 2 3z 1,1,1 2e 6, 4e 6, 6e 6 This vector, however, has a non-zero magnitude. To identify the normalized direction vector corresponding to T 1, 1, 1 we simply divide T 1, 1, 1 by its magnitude. This yields n T 1, 1, 1 2e 6 T1, 1, 1, 4e 6, 6e 6 1, 2, 3. 56e 6 14 b If the ship travels at e 8 meters per second, how fast will the temperature be decreasing if he heads in that direction? Consider the trajectory given by the map ct 1,1,1 + e 8 m/s nt. This would correspond to a straight line trajectory passing through the point 1,1,1 at t 0in the direction of n with a speed of e 8 m/s. The temperature the ship would feel as it travelled along this trajectory would be T t Tct and the rate at which the temperature would decrease would be dt dt d dt T ct DT ct Dc t T ct dc dt t. At t 0,wehavect1,1,1 and dc dt e8 m/sn. Thus, dt dt T 1, 1, 1 e 8, 2e8, 3e e 6, 4e 6, 6e 6 e e 2 e 8, 2e8, 3e c Unfortunately, the metal of the hull will crack if cooled at a rate greater than 14e 2 degrees per second. Describe the set of possible directions in which he may proceed to bring the temperature down at no more than that rate. We now consider straight line trajectories through the point 1,1,1 of the form c t 1,1,1 + e 8 m/sn t

19 19 where now n is an arbitrary unit vector indicating the direction at which the spaceship is heading. Along such a path we have Using the fact that we can write dt dt dt 0 d dt T c t t0 T c 0 dc dt 0 T1, 1, 1 e 8 n a b a b cosθ 0 dt T 1, 1, 1 e8 n cosθ 56e 6 e 8 cosθ 2 14e 2 cosθ We now demand the rate at which the spaceship hull cools be no greater than 14e 2 ; 2 dt 14e < 0 dt 2 14e 2 cosθ which leads to the condition that or cosθ > 1 2 θ<120 We conclude that the angle between the direction of fastest increase in temperature and the direction in which the spaceship heads should be no greater than Compute the second partial derivatives 2 f/ 2, 2 f/, 2 f/, 2 f/ 2 for each of the following functions. Verify Theorem 15 in each case. a fx, y 2xy/ x 2 + y 22,x, y 0. x 2y 2 + y 2 2 2xy2 x 2 + y 2 2x x 2 + y 2 4 2y3 6x 2 y x 2 + y 2 3 2x x 2 + y 2 2 2xy2 x 2 + y 2 2y x 2 + y 2 4 2x3 6xy 2 x 2 + y 2 3 6y 2 6x 2 x 2 +y 2 3 2y 3 6x 2 y 3 x 2 +y 22 2y 6x4 6y 4 +36x 2 y 2 x 2 +y 2 4 x 2 + y 2 6

20 20 Note that 6x 2 6y 2 x 2 +y 2 3 2x 3 6xy 2 3 x 2 + y 22 2y x 2 + y 2 6 6x4 6y 4 +36x 2 y 2 x 2 +y 2 4. b fx, y, z e z +1/x+xe y, x 0. 1 x 2 + e y e y 0 xe y e y 0 Note that e z 0 0 e y, Let fx, y { xy x 2 y 2 / x 2 + y 2, x, y 0,0 0, x, y 0 a If x, y 0, calculate / and /. For all points x, y 0,wehave 3x 2 y y 3 x 2 +y 2 x4 y +4x 2 y 3 y 5 x 2 +y 2 2 x 2 + y 2 2 x 3 y xy 3 2x

21 x 3 3xy x 2 2 +y 2 x 3 y xy 3 2y x 2 + y x5 4x 3 y 2 xy 4 x 2 + y 2 2 b Show that 0,0 0,0 0 0,0 x,y 0,0 At the point 0,0 we will have to be more careful, because fx, y is not obviously differentiable there. c Show that 0,0 0,0 2 f h 0 h 0 0,0 1, f h,0 f 0,0 h f 0,h, f 0,0 h h 0 h f 0,0 h 0 h 1. Again, we have to be a little careful evaluating the partial derivatives at the point 0,0. We have 0,0 0,0 h 0 h 0 1 h 0 0,h h 0 h h,0 h 0 h 0 h 1 d What went wrong? Why are the mixed partials not equal? The second partial derivatives exist, but they are not continuous as functions of two variables. Therefore, Theorem 15 can not be applied in this case. 8. A function u fx, y with continuous second partial derivatives satisfying Laplace s equation 2 u 2 h + 2 u 2 0 h 0,0 0,0

22 22 is called a harmonic function. Show that ux, y x 3 3xy 2 is harmonic. u 3x 2 3y 2 2 u 2 6x u 6xy Therefore and so u is harmonic. 2 u 2 6x 2 u u 2 6x 6x0,