788 SECTION 16.1 CHAPTER 16 SECTION f =(6x y) i +(1 x) j 2. f =(2Ax + By)i +(Bx +2Cy)j. 3. f = e xy [(xy +1)i + x 2 j]

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1 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 788 SECTION 6. CHAPTER 6 SECTION 6.. f =6x y) i + x) j. f =Ax + By)i +Bx +Cy)j. f = e xy [xy +)i + x j] 4. f = x + y ) [y x +xy)i +y x xy)j] 5. f = [ y sinx +)+4x y cosx +) ] i +4xy sinx +)j 6. f = x x + y i + y x + y j 7. f =e x y + e y x ) i + e x y e y x ) j =e x y + e y x )i j) 8. f = AD BC [ yi xj ] Cx + Dy) 9. f =z +xy) i +x +yz) j +y +zx) k 0. f = x x + y + z i + y x + y + z j + z x + y + z k. f = e z xy i + x j x y k). [ xyz f = x + y + z [ + ] [ + yz lnx + y + z) i + xyz + xy lnx + y + z) x + y + z ] k xyz + xz lnx + y + z) x + y + z. f = e x+y cos z + ) i +e x+y cos z + ) j ze x+y sin z + ) k 4. f = e yz /x yz x 4 i + z x j + yz ) x k 5. f = 6. f = [ y cosxy)+ ] i +xcosxy) j + x z k ) xy z 4 i + x x ) z z j y z +xz k 7. f =4x y) i +8y x) j; at, ), f = i +8j 8. f = x y) yi +xj), f, ) = i + j ] j 9. f = x x + y i + y x + y j; at, ), f = 4 5 i + 5 j

2 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 0. f = tan y/x) xy ) ) x π x + y i + x + y j, f, ) = 4 ) i + j. f = sin xy + xy cos xy) i + x cos xy j; at,π/), f = i SECTION f = e x +y ) [y x y)i +x xy )j], f, ) = e i j). f = e x sin z +y) i +e x cos z +y) j + e x cos z +y) k; at 0,π/4,π/4), f = i +j + k) 4. f = cos πzi cos πzj πx y) sin πzk, f, 0, ) = πk 5. f = i y y + z j z y + z k; at,, 4), f = i + 5 j 4 5 k 6. f = sinxyz )yz i + xz j +xyzk), f π, 4 ), = 7. a) f0, ) = 4 i b) f 4 π, 6 π) = c) f,e)= e) i j 8. a) f,, ) = 8 i + j 7 c) f,e,π/6) = i + π e j + k 9. For the function fx, y) =x xy + y, we have + ) i + 4 i + π j π ) k ) + j 8 k b) f,, ) = 5 8 i + 9 j + 8 k fx + h) fx) =fx + h,y+ h ) fx, y) =x + h ) x + h )y + h )+y + h ) [ x xy + y ] = [6x y) i + x) j] h i + h j)+h h h = [6x y) i + x) j] h +h h h The remainder gh) =h h h =h i h j) h i + h j), and gh) h = h i h j h cos θ h i h j h Since h i h j 0 as h 0 it follows that f =6x y) i + x) j 0. fx + h) fx) =[x +y) i +x +y) j] [h i + h j]+ h +h h + h ; gh) = h +h h + h is oh).

3 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 790 SECTION 6.. For the function fx, y, z) =x y + y z + z x, we have fx + h) fx) =fx + h,y+ h,z+ h ) fx, y, z) =x + h ) y + h )+y + h ) z + h )+z + h ) x + h ) x y + y z + z x ) = xy + z ) h + yz + x ) h + xz + y ) h +xh + yh + h h ) h + yh + zh + h h ) h +zh + xh + h h ) h = [ xy + z ) i + yz + x ) j + xz + y ) k ] h + gh) h, where gh) =xh + yh + h h ) i +yh + zh + h h ) j +zh + xh + h h ) k Since gh) h 0 as h 0 it follows that f = xy + z ) i + yz + x ) j + xz + y ) k [ ]. fx + h) fx) = xy +h x + h y) i +x j + zz + h ) k h i + h j + h k)+h ; gh) =h h is oh) and f =4xy i =x j + z k.. f = Fx, y) =xy i + +x ) j x =xy fx, y) =x y + gy) for some function g. Now, y = x + g y) =+x g y) = gy) =y + C, C a constant. Thus, fx, y) =x y + y + C 4. f =xy + x)i +x + y)j = f x =xy + x = fx, y) =x y + x + gy) Now, f y = x + g y) =x + y = g y) =y = gy) = y + C Thus, fx, y) =x y + x + y + C 5. f = Fx, y) =x + sin y) i +xcos y y) j x = x + sin y fx, y) = x + x sin y + gy) for some function g. Now, y = x cos y + g y) =x cos y y g y) = y gy) = y + C, C a constant. Thus, fx, y) = x + x sin y y + C. 6. f = yzi +xz +yz)j +xy + y )k = f x = yz = fx, y, z) =xyz + gy, z). f y = xz + g y = xz +yz = g y =yz = gy, z) =y z + hz) = fx, y, z) =xyz + y z + hz). f x = xy + y + h z) =xy + y = h z) =0 = hz) =C. Thus, fx, y, z) =xyz + y z + C. 7. With r =x + y + z ) / we have r x = x r, a) r y = y r, r z = z r. ln r) = x ln r) i + y ln r) j + ln r)k z = r r x i + r r y j + r r z k = x r i + y r j + z r k = r r

4 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION b) sin r) = x sin r) i + y sin r) j + sin r)k z = cos r r r r i + cos r j + cos r x y z k = cos r) x r i + cos r)y r j + cos r)z r k cos r ) = r r ) e c) e r r = r [ same method as in a) and b) ] r 8. With r n =x + y + z ) n/ we have r n x = n x + y + z ) n/) x) =nx + y + z ) n )/ x = nr n x. Similarly Therefore r n y = nrn y and r n z = nrn z. r n = nr n xi + nr n yj + nr n zk = nr n xi + yj + zk) =nr n r 9. a) f =x i +y j = 0 = x = y =0; f = 0 at 0, 0). b) c) f has an absolute minimum at 0, 0) 40. a) f = xi + yj) =0 at 0, 0) 4 x y b) c) f has a maximum at 0, 0)

5 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 79 SECTION a) Let c = c i + c j + c k. First, we take h = hi. Since c h is oh), c h 0 = lim h 0 h = lim c h h 0 h = c. Similarly, c = 0 and c =0. b) y z) h =[fx + h) fx) z h]+[y h fx + h)+fx)]=oh)+oh) =oh), so that, by part a), y z = lim gh) = lim h gh) ) ) = lim h 0 h 0 h h lim h 0 h 0 ) gh) =0)0) =0). h 4. a) In Section 5.6 we showed that f was not continuous at 0, 0). It is therefore not differentiable at 0, 0). b) For x, y) 0, 0), x = y y 4 = y x = yy x ) x + y. As x, y) tends to 0, 0) along the positive y-axis, ) tends to. SECTION 6.. f =xi +6yj, f, ) = i +6j, u = i j), f u, ) = f, ) u =. f = [ + cosx + y)]i + cosx + y)j, f0, 0) = i + j, u = 5 i + j), f u0, 0) = f0, 0) u = 5. f =e y ye x ) i +xe y e x ) j, f, 0) = i + e)j, u = i +4j), 5 f u, 0) = f, 0) u = 7 4e) 5 4. f = x y) yi +xj), f, 0) = j, u = i j), f u, 0) = f, 0) u = a b)y b a)x a b 5. f = i + j, f, ) = x + y) x + y) 4 i j), u = i j), f u, ) = f, ) u = 4 a b) d c 6. f = [d c)yi +c d)xj], f, ) = cx + dy) c + d) i j), u = ci dj), c + d f u, d c ) = f, ) u = c + d) c + d 7. f = x x + y i + y x j, f0, ) = j, u = 8 i + j), + y 65 f u0, ) = f0, ) u = 65

6 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 8. f =xyi +x + sec y)j, f, π 4 )= π i +j, u = i j) 5 f u, π ) = f, π ) u = π ) SECTION f =y + z)i +x + z)j +y + x)k, f,, ) = j, u = 6 6i +j + k), f u,, ) = f,, ) u = 6 0. f =z +xy)i +x +yz)j +y +zx)k, f, 0, ) = i + j +k, u = j k) 0 0 f u, 0, ) = f, 0, ) u = 0. f = x + y + z ) i +yj +z k ), f,, ) = 6i j +k), u = i + j), f u,, ) = f,, ) u =. f =Ax + Byz)i +Bxz +Cy)j + Bxyk, f,, )=A + B)i +B +4C)j +Bk u = Ai + Bj + Ck); A + B + C f. f = tan y + z) i + x +y + z) j + u = i + j k), f u, 0, ) = f, 0, ) u = π 4 = u,, ) = f,, ) u = A + B +AB +6BC A + B + C x +y + z) k, f, 0, ) = π 4 i + j + k, π 4. f =y cos z πyz cos πx +6zx)i +xy cos z z sin πx)j + xy sin z 4yz sin πx +x )k f0,,π)=π )i; u = i j +k), f u0,,π)= f0,,π) u = π ). 5. f = x x + y i + y x + y j, u = xi yj), x + y f ux, y) = f u = x + y 6. f = e xy [ y + xy y )i +x )y + xy )j ], f0, ) = j u = i +j), f u0, ) = f0, ) u = f =Ax +By) i +Bx +Cy) j, fa, b) =aa +bb)i +ab +bc) j a) u = i + j), f u a, b) = fa, b) u = [ab A)+bC B)] b) u = i j), f u a, b) = fa, b) u = [aa B)+bB C)] 8. f = z x i z y j +ln x y ) k, f,, ) = i j u = i + j k); f u,, ) = f,, ) u = 0

7 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 794 SECTION f = e y z i +xyj xzk), f,, ) = i +4j +4k, r t) =i sin t ) j e t k, at,, ) t =, r ) = i k, u = 5 5i k), f u,, ) = f,, ) u = f =xi + zj + yk, f,, ) = i +j k Direction: r ) = i +j k, u = i +j k), f u,, ) = f,, ) u =. f =x +yz) i + xz z ) j +xy yz) k, f,, ) = 6 i k The vectors v = ± i + j k) are direction vectors for the given line; u = ± are corresponding unit vectors; f u,, ) = f,, ) ±u) =± 8 4 ) [ i + j k] 4. f = e x cos πyzi πz sin πyzj πy sin πyzk), f0,, )= π j π k The vectors v = ± i +j +5k) are direction vectors for the line; u = ± are corresponding unit vectors; f u0,, )= f0,, π ) ±u) = 8 ) [ i +j +5k] 8. f =y e x i +ye x j, f0, ) = i +j, f = f, f = i + j) f increases most rapidly in the direction u = i + j); the rate of change is. f decreases most rapidly in the direction v = i + j); the rate of change is. 4. f = [ + cosx +y)]i + cosx +y)j, f0, 0) = i +j Fastest increase in direction u = i + j), rate of change f0, 0) = Fastest decrease in direction v = i + j), rate of change 5. f = x x + y + z i + y x + y + z j + f,, ) = 6 i j + k), f = z x + y + z k, f increases most rapidly in the direction u = i j + k); the rate of change is. 6 f decreases most rapidly in the direction v = i j + k); the rate of change is f =xze y + z )i + x ze y j +x e y +xz)k, f, ln, ) = i +4j +6k Fastest increase in direction u = 6i +j +k), rate of change f, ln, ) =4 7 Fastest decrease in direction v = 6i +j +k), 7 rate of change 4 7. f = f x 0 ) i. If f x 0 ) 0, the gradient points in the direction in which f increases: to the right if f x 0 ) > 0, to the left if f x 0 ) < 0.

8 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION ; the vector c = y x 0,y 0 )i x x 0,y 0 )j is perpendicular to the gradient fx 0,y 0 ) and points along the level curve of f at x 0,y 0 ). fh, 0) f0, 0) h 9. a) lim = lim h 0 h h 0 h 0. a) h = lim h 0 h does not exist b) no; by Theorem 6..5 f cannot be differentiable at 0, 0) gx + h)oh) h [gx + h) gx)] fx) h b) h by Schwarz s inequality = gx + h) oh) h gx)0) = 0. λx, y) = 8 xi 6yj a) λ, ) = 8 λ, ) i =6j, u = [gx + h) gx)] fx) h h = gx + h) gx) fx) 0 8 λ, ) = i 6j, λ u, ) = λ, ) u = b) u = i, λ u, ) = λ, ) u = 8 i j) i = 8 c) u = i + j), λ u, ) = λ, ) u = 6 i j) [ ] i + j) = 6. I = 4xi yj. We want the curve rt) =xt)i + yt)j which begins at, ) and has tangent vector r t) in the direction I. We can satisfy these conditions by setting These equations imply that x t) = 4xt), x0) = ; y t) = yt), y0) =. xt) = e 4t, yt) =e t. Eliminating the parameter, we get x = y ; the particle will follow the parabolic path x = y toward the origin.. a) The projection of the path onto the xy-plane is the curve C : rt) = xt)i + yt)j which begins at, ) and at each point has its tangent vector in the direction of f. Since we have the initial-value problems f =xi +6yj, x t) = xt), x0) = and y t) = 6yt), y0) =. From Theorem 7.6. we find that xt) =e t and yt) =e 6t. Eliminating the parameter t, we find that C is the curve y = x from, ) to 0, 0).

9 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 796 SECTION 6. b) Here so that x t) = xt), x0) = and y t) = 6yt), y0) = xt) =e t and yt) = e 6t. Eliminating the parameter t, we find that the projection of the path onto the xy-plane is the curve y = x from, ) to 0, 0). 4. z = fx, y) = x y ; f = xi yj, so we choose the projection rt) =xt)i + yt)j of the path onto the xy-plane such that x t) =xt), y t) = yt) a) With initial point,, ), we get xt) = et,yt) =e t, or y = x from, ), in the direction of decreasing x. b) With initial point, 0, ), we get xt) =et,yt) =0, or the x-axis from, 0), in the direction of increasing x. 5. The projection of the path onto the xy-plane is the curve C : rt) = xt)i + yt)j which begins at a, b) and at each point has its tangent vector in the direction of f = a xi +b yj ). We can satisfy these conditions by setting so that x t) = a xt), x0) = a and y t) = b yt), y0) = b xt) =ae a t and yt) =be bt. Since [ x ] b = a ) b [ e a t y ] a =, b C is the curve b) a x b =a) b y a from a, b) to 0, 0). 6. The particle must go in he direction T = e y cos xi e y sin xj, so we set x t) = e yt) cos xt), y t) = e yt) sin xt). Dividing, we have y t) sin xt) x = t) cos xt), or dy = tan x. dx 0, 0), we get y =ln sec x, in the direction of decreasing x since x 0) < 0). 7. We want the curve C : rt) = xt)i + yt)j which begins at π/4, 0) and at each point has its tangent vector in the direction of From T = e y sin x i e y cos x j. x t) = e y sin x and y t) = e y cos x With initial point

10 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION we obtain so that dy dx = y t) x t) = cot x y =ln sin x + C. Since y = 0 when x = π/4, we get C =ln and y =ln sin x. As T π/4, 0) = i j, the curve y =ln sin x is followed in the direction of decreasing x. 8. z = x)i + 6y)j, so the projection of the path onto the xy-plane satisfies x t) = xt), y dy t) = 6yt), or dx = 6y. With initial point 0, 0), this gives the curve x y =x ) +, in the direction of increasing x. 9. a) b) f +h, + h) ) f, 4) + h) ++h) 6 lim = lim h 0 h h 0 h [ ] 4h + h = lim h 0 4 = lim h h) =6 h ) ) h +8 h +8 f, 4+h f, 4) +4+h) lim = lim h 0 h h 0 h = lim h 0 6 h +h ++4+h 6 h = lim h 0 6 h +4) =4 c) u = 7 7 i +4j), f, 4) = i + j; f u, 4) = f, 4) u = d) The limits computed in a) and b) are not directional derivatives. In a) and b) we have, in essence, computed f, 4) r 0 taking r 0 = i +4j in a) and r 0 = 4i + j in b). is r 0 a unit vector. GMm 40. f = xi + yj + zk) = GMm x + y + z ) / r r 4. a) u = cos θ i + sin θ j, fx, y) = x i + y j; f ux, y) = f u = x i + ) y j b) f = x +y y ) i +x xy) j, f, ) = i +j cos θ i + sin θ j) = x f u, ) = cosπ/) + sinπ/) = cos θ + y sin θ In neither case

11 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 798 SECTION f ux, y) = x cos 5π 4 + y sin 5π ) ) 4 =xey +x e y = xe y + x) f u, ln ) = e ln +)= 4 4. fg)= fg) x i + fg) j + fg) k = f g y z x + g ) x 44. ) f = g x = f g x i + g y j + g z k = f g + g f ) f i + g y ) f j + g z i + f g y + g y ) + g ) f k g ) j + f g z + g z ) x i + y j + z k ) k = x g f g x g i + y g f g y g j + z g f g z g k = gx) fx) fx) gx) g x) 45. f n = n x i + n y j + n n n n k = nf i + nf j + nf z x y z k = nf n f SECTION 6.. fb) =f, ) = ; fa) =f0, ) = 0; fb) fa) = f = x y ) i x j; b a = i +j and f b a) =x y x The line segment joining a and b is parametrized by Thus, we need to solve the equation x = t, y =+t, 0 t t +t) t =, which is the same as t 4t +=0, 0 t The solutions are: t =,t=. Thus, c =, 5 ) satisfies the equation. Note that the endpoint b also satisfies the equation.. f =4zi yj +4x +z)k, fa) =f0,, ) = 0, fb) =f,, ) = b a = i +j + k, so we want x, y, z) such that f b a) =4z 4y +4x +z =6z 4y +4x = fb) fa) = Parametrizing the line segment from a to b by xt) = t, yt)= + t, zt) = + t, we get t =, or c =,, ). a) fx, y, z) =a x + a y + a z + C b) fx, y, z) =gx, y, z)+a x + a y + a z + C 4. Using the mean-value theorem 6.., there exists c such that fc) b a) =0

12 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 5. a) U is not connected b) i) gx) =fx) ii) gx) = fx) 6. By the mean-value theorem fx ) fx )= fc) x x ) for some point c on the line segment x x. Since Ω is convex, c is in Ω. Thus SECTION fx ) fx ) = fc) x x ) fc) x x M x x. by Schwarz s inequality 7. f =xyi + x j; frt)) r t) = i + e t j ) e t i e t j)=e t 8. f = i j; frt)) r t) =i j) ai ab sin atj) =a + b sin at) 9. f = x +y x ) i + frt)) r t) = 0. f = x + y 4xi +y j) y sin t +y x j, frt)) = ) + cos t i + cos t + cos t j ) 4 sin t cos t cos t i sin t j) = + cos = t sin t + cos t i + cos t + cos t j frt)) r t) = e 4t + t 4et i +t / j) e t i + t / j)= 8e4t + e 4t + t. f =e y ye x ) i +xe y + e x ) j; frt)) = t t ln t) i + frt)) r t) =. f = t t ln t) i + x + y xi + yj + zk) + z frt)) r t) =. f = yi +x z)j yk; t t ln t + t ) ) j t i +[+lnt] j +e 4t sin ti + cos tj + et k) cos ti sin tj +e t k)= 4e4t +e 4t frt)) r t) = t i + t t ) j t k ) i +tj +t k ) =t 5t 4 sin t + cos t t t ln t + ) j t ) ) t = t t +lnt + [ln t] + t 4. f =x i +y j frt)) r t) =a cos ωti +b sin ωtj) ωa sin ωti + ωb cos ωtj + bωk) =ωb a ) sin ωt cos ωt 5. f =xi +yj + k; frt)) r t) =a cos ωt i +b sin ωt j + k) aω sin ωt i + bω cos ωt j + bωk) =ω b a ) sin ωt cos ωt + bω

13 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 800 SECTION f = y cosx + z)i +ysinx + z)j + y cosx + z)k frt)) r t) = [cos t cost + t )i + cos t sint + t )j + cos t cost + t )k] i sin tj +t k) = cos t[+t ) cos t cost + t ) sin t sint + t )] 7. du dt = dx x dt + dy =x y) sin t)+4y x)cos t) y dt = cos t sin t + sin t cos t = sin t cos t 8. du dt = x dx dt + y dy dt = + = y x ) x t + ) y t ) + t ) t +t ) t ) =t +4+ t du dt = dx x dt + dy y dt =e x sin y + e y cos x) ) +e x cos y + e y sin x) ) = e t/ sin t + cos t) + e t cos t + sin t) du dt = x dx dt + y dy =4x y) sin t)+y x) cos t dt = sin tsin t 4 cos t) + cos t sin t cos t) du dt = dx x dt + dy y dt =ex sin y)t)+e x cos y)π) = e t [t sinπt)+π cosπt)] du dt = x dx dt + y dy dt + z dz = t e t t + + et +ln t t + dt = z x t + z y ) e t + t) du dt = dx x dt + dy y dt + dz z dt =y + z)t)+x + z) t)+y + x)t ) y ) t +ln e t + t) x = t)t)+t t + ) t)+tt ) = 4t +6t 4t 4. du dt = x dx dt + y dy dt + z dz = sin πy + πz sin πx)t + πx cos πy ) cos πx t) dt =t [ sin[π t)] + π t ) sinπt ) ] πt cos[π t)]+tcosπt ) 5. V = πr h, dv dt = V r dr dt + V h dh dt = πrh ) dr dt + πr ) dh dt.

14 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 At the given instant, dv dt = π80)) + 88 π96) ) = π. The volume is increasing at the rate of 88 π in. / sec. 6. v = πr h, dr dt =, dv dt = v r dr dt + v h dh dt =πrhdr dh + πr dt dt dh dv =, r =, h =8 = dt dt = 49π : SECTION decreasing at the rate of 49π cm /sec. 7. A = xy sin θ; da dt = A dx x dt + A dy y dt + A [ dθ θ dt = y sin θ) dx ] dy dθ +xsin θ) +xy cos θ). dt dt dt At the given instant da dt = [ sin ) 0.5)+.5sin ) 0.5).5) cos ) 0.)] = 0.87 ft /s = 4.4 in /s dz dt =xdx dt + y dy dt. = dz dt =xdx dt + y But x + y = = x dx dt x ) dx = x dx y dt dt = 5. +y dy dt s = x x s + y =x y)cos t)+ x)t cos s) y s =scos t t sin s cos t st cos s cos t t = x x t + y =x y) s sin t)+ x)sin s) y t = s cos t sin t + st sin s sin t s cos t sin s s = x x s + y y s = [cosx y) sinx + y)]t +[ cosx y) sinx + y)]s dy =0 = dt = x dx y dt z is increasing 5 centimeters per second =t s) cosst s + t ) t +s) sinst + s t ) t = x x t + y y t = [cosx y) sinx + y)]s +[ cosx y) sinx + y)] t) =s +t) cosst s + t ) s t) sinst + s t ). s = x x s + y y s =xtan y)st)+ x sec y ) ) =4s t tan s + t ) + s 4 t sec s + t ) t = x x t + y y t =xtan y) s ) + x sec y ) t) =s 4 t tan s + t ) +s 4 t sec s + t )

15 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 80 SECTION 6... s = x x s + y y s + z z s = z y sec xy tan xyt)+z x sec xy tan xy +zsec xyst) = sec[sts t )] s 4 t s t ) tan[sts t )]+s t tan[sts t )]+4s t ) t = z y sec xy tan xys)+z x sec xy tan xy t)+zsec xys ) = sec[sts t )] s 5 t s t ) tan[sts t )] 4s 5 t 4 tan[sts t )]+s 4 t ) s = x x s + y y s + z z s =x y)cos t)+ x) cos t s))+ztcos s) =scos t sin t s) cos t + s cos t cos t s)+t sin s cos s t = x x t + y y t + z z t =x y) s sin t)+ x)cos t s))+zsin s) = s cos t sin t + s sin t s) sin t s cos t cos t s)+t sin s 4. s = x x s + y y s + z z s = e yz s + xz e yz 0+xyze yz s = s et s +t ) +4st s + t ) lnst)e t s +t ) t = x x t + y y t + z z t = e yz t + xz e yz t +xyze yz t = t et s +t ) + t s + t )s +7t ) lnst)e t s +t ) d dt [frt))] = [ frt)) r ] t) r r t) t) = f ut) rt)) r t) where ut) = r t) r t) x [fr)] = d r [fr)] dr x = f r) r x = f r) x r ; similarly y [fr)] = f r) y r and z [fr)] = f r) z r. Therefore fr) =f r) x r i + f r) y r j + f r) z r k = f r) r r. 7. a) cos r) r r b) r cos r + sin r) r r 8. a) r ln r)=+lnr) r r b) e r )= re r r r = e r r

16 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION a) r cos r sin r) r r b) sin r r cos r sin r ) r r 40. a) b) du dt = dx ds x ds dt + y dy ds ds dt 4. a) b) r = x w x w r + x t ) t + y w r y w r + y t To obtain / s, replace each r by s. ) t + z w r z w r + z t ) t. r 4. a) b) r = x x r + z z r + w w r, v = y y v + w w v

17 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 804 SECTION du dt = dx x dt + dy y dt d u dt = d x x dt + dx dt and the result follows. s = x x s + y y s u s = x [ u x dx dt + u y x x s + x s ] dy + d y dt y dt + dy [ ] u dx dt x y dt + u dy y dt u x x s + u y x ) y s + y y s + y ) u x s x y s + u y y s 45. a) b) = u x ) x + u s x y x y s s + u y r = x x r + y y r = cos θ + x y sin θ y s ) + x x s + y y s θ = x x θ + y y θ = r sin θ)+r cos θ) x y ) ) = cos θ + ) cos θ sin θ + sin θ, r x x y y ) ) r = sin θ ) cos θ sin θ + cos θ, θ x x y y ) + ) ) r r = cos θ + sin θ ) ) + sin θ + cos θ ) = θ x y ) + x ) y 46. a) By Exercise 45 a) w r = w w w cos θ + sin θ, x y w Solve these equations simultaneously for and x b) To obtain the first pair of equations set w = r; to obtain the second pair of equations set w = θ. θ = w w r sin θ + r cos θ. x y w y. c) θ is not independent of x; r = x + y gives r x = x x + y = r cos θ = cos θ r 47. Solve the equations in Exercise 45 a) for x x = r cos θ r θ Then and y : sin θ, y = r sin θ + r θ cos θ u = x i + y j = r cos θ i + sin θ j)+ sin θ i + cos θ j) r θ 48. ur, θ) =r = u =re r

18 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 49. ux, y) =x xy + y = r r cos θ sin θ = r 50. u = r e r + r θ = x x θ + y r = r sin θ), θ e θ = r sin θ)e r r cos θ e θ y = r sin θ + r cos θ θ x y u = sin θ r θ x r sin θ u x x r + u y x sin θ) θ = r cos θ SECTION ) y + cos θ ) r y + r cos θ u x x y r + u y y r = sin θ ) u + cos θ + r sin θ cos θ x y y u x + rcos θ sin θ) u x y 5. From Exercise 45 a), u r = u x cos θ + u y x sin θ cos θ + u y sin θ ) u θ = u x r sin θ u y x r sin θ cos θ + u y r cos θ r cos θ + x y sin θ. The term in parentheses is r. Now divide the second equation by r and add the two equations. The result follows. 5. ux, y) =x xy + y 4 4, dy dx = x = y y x 4y x = =x y, x y x y x y = x +4y 5. Set u = xe y + ye x x y. Then x = ey + ye x 4xy, y = xey + e x x dy dx = / x / y = ey + ye x 4xy xe y + e x x. 54. ux, y) =x / + y /, = dy dx = x y y / = x) x = x /, y = y / 55. Set u = x cos xy + y cos x. Then = cos xy xy sin xy y sin x, x y = x sin xy + cos x dy dx = / x cos xy xy sin xy y sin x = / y x. sin xy cos x

19 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 806 SECTION Set ux, y, z) =z 4 + x z + y + xy. Then x =xz + y, y =y + x, z =4z +x z z x = x z = xz + y 4z +x z, z y = y z = y + x 4z +x z 57. Set u = cos xyz +ln x + y + z ). Then x = yz sin xyz + x x + y + z, z = xy sin xyz + z x + y + z. y z x = / x / z z y = / y / z = xz sin xyz + y x + y + z, and x yz x + y + z ) sin xyz = z xy x + y + z ) sin xyz, y xz x + y + z ) sin xyz = z xy x + y + z ) sin xyz. 58. a) Use du dt = du dt i + du dt j and apply the chain rule to u,u. b) i) du dt = tex cos yi + e x sin yj)+π e x sin yi + e x cos yj) = te t / cos πti + sin πtj)+πe t / sin πti + cos πtj) ii) ut) =e t / cos πti + e t / sin πtj du / sin πt + te dt = πet t / cos πt)i +πe t / cos πt + te t / sin πt)j 59. s = x x s + y y s, t = x x t + y y t 60. du dt = dx x dt + dy y dt + dz z dt where x = x i + x j + x k, y = y i + y j + y k, z = z i + z j + z k. SECTION 6.4. Set fx, y) =x + xy + y. Then, f =x + y)i +x +y)j, f, ) = i j. normal vector i + j; tangent vector i j tangent line x + y + = 0; normal line x y =0

20 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION Set fx, y) =y x) x, f = y x +)i +y x)j, f, 4) = 6i +4j normal vector i + j; tangent vector i + j tangent line x y + = 0; normal line x +y 6 = 0. Set fx, y) = x + y ) 9 x y ). Then, f =[4xx + y ) 8x]i + [ 4y x + y ) +8y ] j, ) f, = 6 i +0j. normal vector i 5 j; tangent vector 5i + j tangent line x 5y + = 0; normal line 5x + y 6 =0 4. Set fx, y) =x + y, f =x i +y j, f, ) = i +j normal vector i +4j; tangent vector 4i j tangent line x +4y 9 = 0; normal line 4x y =0 5. Set fx, y) =xy x + y +5x. Then, f =y 4x +5)i +xy +)j, f4, ) = 7i +7j. normal vector 7i 7j; tangent vector 7i + 7j tangent line 7x 7y + 6 = 0; normal line 7x +7y 8=0 6. Set fx, y) =x 5 + y 5 x. f =5x 4 6x )i +5y 4 j, f, ) = i +5j normal vector i 5j; tangent vector 5i + j tangent line x 5y + 4 = 0; normal line 5x + y 6=0 7. Set fx, y) =x x y x + y. Then, f =6x xy ) i + x y +)j, f, ) = 5i +5j. normal vector i j; tangent vector i + j tangent line x y = 0; normal line x + y +=0 8. Set fx, y) =x + y +x. f =x +)i +yj, f, )=5i +6j normal vector 5i +6j; tangent vector 6i 5j tangent line 5x +6y = 0; normal line 6x 5y +=0 9. Set fx, y) =x y + a y. By 5.4.4) m = / x / y = xy x + a. At 0,a) the slope is Set fx, y, z) =x + y ) z. f =4xx + y )i +4yx + y )j k, f,, 4) = 8i +8j k Tangent plane: 8x +8y z = 0 Normal: x =+8t, y =+8t, z =4 t

21 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 808 SECTION 6.4. Set fx, y, z) =x + y xyz. Then, f =x yz) i +y xz) j xyk, f,, ) = 6i + 5 j 6k; tangent plane at,, ) : 6x ) + 5 y ) 6 z ) =0, which reduces to 4x 5y +4z =0. Normal: x =+4t, y = 5t, z = +4t. Set fx, y, z) =xy +z. f = y i +xyj +4zk, f,, ) = 4i +4j +8k Tangent plane: x + y +z 7=0 Normal: x =+t, y =+t, z =+t. Set z = gx, y) =axy. Then, g = ayi + axj, g tangent plane at, a, ) : z =x ) + a y a, ) = i + aj. ) a, which reduces to x + ay z =0 Normal: x =+t, y = a + at, z = t 4. Set fx, y, z) = x + y + z. f = x i + y j + z k, f, 4, ) = i + 4 j + k Tangent plane: x + y +z 8=0 Normal: x =+t, y =4+t, z =+t 5. Set z = gx, y) = sin x + sin y + sin x + y). Then, g = [cos x + cos x + y)] i + [cos y + cos x + y)] j, g0, 0) = i +j; tangent plane at 0, 0, 0): z 0=x 0)+y 0), x +y z =0. Normal: x =t, y =t, z = t 6. Set fx, y, z) =x + xy + y 6x + z. f =x + y 6)i +x +y)j k, f4,, 0) = k Tangent plane: z = 0 Normal: x =4, y =, z = 0 + t 7. Set fx, y, z) =b c x a c y a b z. Then, f x 0,y 0,z 0 )=b c x 0 i a c y 0 j a b z 0 k; tangent plane at x 0,y 0,z 0 ): b c x 0 x x 0 ) a c y 0 y y 0 ) a b z 0 z z 0 )=0, which can be rewritten as follows: b c x 0 x a c y 0 y a b z 0 z = b c x 0 a c y 0 a b z 0 = f x 0,y 0,z 0 )=a b c. Normal: x = x 0 +b c x 0 t, y = y 0 a c y 0 t, z = z 0 a b z 0 t

22 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION Set fx, y, z) = sinx cos y) z. f = cos y cosx cos y)i x sin y cosx cos y)j k, f, π, 0) = j k Tangent plane: y + z = π Normal: x =, y = π + t, z = t 9. Set z = gx, y) =xy + a x + b y. g = y a x ) i + x b y ) j, g = 0 = y = a x and x = b y. Thus, y = a b 6 y 4, y = b 6 a, y = b /a, x = b y = a /b and g a /b, b /a ) =ab. The tangent plane is horizontal at a /b, b /a, ab ). 0. z = gx, y) =4x +y x + xy y. g =4 x + y)i ++x y)j g = 0 = 4 x + y =0, +x y =0 = x = 0,y= 8 The tangent plane is horizontal at 0, 8, 8 ).. Set z = gx, y) =xy. Then, g = yi + xj. The tangent plane is horizontal at 0, 0, 0). g = 0 = x = y =0.. z = gx, y) =x + y x y xy. g =x y)i +y x)j g =0 = x y =0=y x =0 = x =,y= The tangent plane is horizontal at,, ).. Set z = gx, y) =x +xy y 5x +y. Then, g =4x +y 5) i +x y +)j. g = 0 = 4x +y 5=0=x y + = x =, y = 6. The tangent plane is horizontal at, 6, ). 4. a) Set fx, y, z) =xy z. f = yi + xj k, f,, ) = i + j k upper unit normal = i j + k) b) Set fx, y, z) = x y z. f = x i + j k, f,, 0) = i + j k y lower unit normal: = i + j k) 5. x x 0 / x)x 0,y 0,z 0 ) = y y 0 / y)x 0,y 0,z 0 ) = z z 0 / z)x 0,y 0,z 0 )

23 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 80 SECTION All the tangent planes pass through the origin. To see this, write the equation of the surface as xfx/y) z =0. The tangent plane at x 0,y 0,z 0 ) has equation [ ) )] [ )] x0 x x 0 ) f x0 x0 x0 + f y y 0 ) y 0 y 0 y 0 y f x0 z z 0 )=0. 0 y 0 The plane passes through the origin: x 0 y 0 f x0 y 0 ) x 0 f x0 y 0 ) + x 0 y 0 f x0 y 0 ) + z 0 = z 0 x 0 f x0 y 0 ) =0. 7. Since the tangent planes meet at right angles, the normals F and G meet at right angles: F x G x + F y G y + F z G z =0. 8. The sum of the intercepts is a. To see this, note that the equation of the tangent plane at x 0,y 0,z 0 ) can be written x x 0 + y y 0 + z z 0 =0. x0 y0 z0 Setting y = z = 0 we have x x 0 x0 = y 0 + z 0. Therefore the x-intercept is given by x = x 0 + x 0 y 0 + z 0 )=x 0 + x 0 a x 0 )= x 0 a. Similarly the y-intercept is y 0 a and the z-intercept is z0 a. The sum of the intercepts is x 0 + y 0 + z 0 ) a = a a = a. 9. The tangent plane at an arbitrary point x 0,y 0,z 0 ) has equation y 0 z 0 x x 0 )+x 0 z 0 y y 0 )+x 0 y 0 z z 0 )=0, which simplifies to x y 0 z 0 x + x 0 z 0 y + x 0 y 0 z =x 0 y 0 z 0 and thus to + y + z =. x 0 y 0 z 0 The volume of the pyramid is V = Bh = [ ] x0 )y 0 ) z 0 )= 9 x 0y 0 z 0 = 9 a. 0. The equation of the tangent plane at x 0,y 0,z 0 ) can be written x 0 / x x 0 )+y 0 / y y 0 )+z 0 / z z 0 )=0 Setting y = z =0, we get the x-intercept x = x 0 + x / 0 y / 0 + z / 0 )=x 0 + x / 0 a / x / 0 ) = x = x / 0 a / Similarly, the y-intercept is y / 0 a / and the z-intercept is z / 0 a /. The sum of the squares of the intercepts is x 0 / + y 0 / + z 0 / )a 4/ = a / a 4/ = a.

24 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION The point,, ) is the tip of r). Since r t) =i t j 4tk, we have r ) = i j 4k. Now set fx, y, z) =x + y +z 5. The function has gradient xi +yj +6zk. At the point,, ), The angle θ between r ) and the gradient gives cos θ = i j 4k) 9 f = i +j 6k). i +j 6k) 7 = 9 7 = Therefore θ =.04 radians. The angle between the curve and the plane is π θ = = 0.58 radians.. The curve passes through the point,, ) at t =, and its tangent vector is r ) = i +4j +k. For the ellipsoid, set fx, y, z) =x +y +z. f =xi +4yj +6zk, f,, ) = 6i +8j +6k, which is parallel to r ).. Set fx, y, z) =x y +x + z. Then, f =xy +)i +x yj +z k, f,, )=6i +8j +k. The plane tangent to fx, y, z) = 6 at,, ) has equation 6x )+8y ) + z )=0, or x +4y +6z =. Next, set gx, y, z) =x + y z. Then, g =6xi +yj k, g,, ) = i +j k. The plane tangent to gx, y, z) =9at,, ) is x )+y ) z )=0, or 6x + y z =. 4. Sphere: fx, y, z) =x + y + z 8x 8y 6z +4, f =x 8)i +y 8)j +z 6)k f,, ) = 4i 6j 4k Ellipsoid: gx, y, z) =x +y +z, g =xi +6yj +4zk g,, )=4i +6j +4k Since their normal vectors are parallel, the surfaces are tangent. 5. A normal vector to the sphere at,, ) is xi +y 4) j +z )k =i j +k. A normal vector to the paraboloid at,, ) is 6xi +4yj k =6i +4j k.

25 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 8 SECTION 6.4 Since i j +k) 6i +4j k) =0, the surfaces intersect at right angles. 6. Surface A: Set fx, y, z) =xy az, f = yi + xj azk Surface B: Set gx, y, z) =x + y + z, Surface C: Set hx, y, z) =z +x cz +y ). g =xi +yj +zk h =4xi 4cyj + c)zk Where surface A and surface B intersect, f g =4xy az )=0 Where surface A and surface C intersect, f h = 4 c)xy az )=0 Where surface B and surface C intersect, g h = 4[x cy + c)z ]=0 7. a) x +4y + 6 = 0 since plane p is vertical. b) y = 4 x +6)= 4 [4t )+6]= t z = x +y +=4t ) + t) +=4t 6t +6 rt) =4t )i tj + 4t 6t +6)k c) From part b) the tip of r) is,, ). r ) = 4i j +70j as d to write We take Rs) =i j +k)+s4i j +70k). d) Set gx, y) =x +y +. Then, g =xi +6yj and g, ) = 4i 8j. An equation for the plane tangent to z = gx, y) at,, ) is z = 4x ) 8y + ) which reduces to 4x 8y z =9. e) Substituting t for x in the equations for p and p, we obtain From the first equation and then from the second equation Thus, ) rt) =ti 4 t + )j + 5 t ) k. t +4y + 6 = 0 and 4t 8y z =9. y = 4 t +) z =4t 8 [ 4 t +)] 9 = 5 t. Lines l and l are the same. To see this, consider how l and l are formed; to assure yourself, replace t in ) by4s + to obtain Rs) found in part c).

26 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 8. a) normal vector: 5 i j; normal line: x =+ 5 t, y = b) tangent line: x =+ 4 5 c) y t, y =+ 5 t SECTION t x 9. a) normal vector: i +j +4k; normal line: x =+t, y =+t, z =+4t b) tangent plane: x )+y )+4z ) = 0 or x + y +z 7=0 c) 40. a) b) c) f = 0 at ±, 0), - 0 0, ± ) /

27 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 84 SECTION a) b) c) f = 4x 4x ) i 4y 4y ) j; f =0: 4x 4x =0 = x =0, ±; 4y 4y =0 = y =0, ± f = 0 at 0, 0), ±, 0), 0, ±), ±, ±) 4. a) b) c) f = 0 at 0, 0), ± /, ± / ) SECTION 6.5. f = x) i y j = 0 only at, 0). The difference f + h, k) f, 0) = [ + h) + h) k ] = h k 0 for all small h and k; there is a local maximum of at, 0).. f = x) i ++y) j = 0 only at, ). The difference f + h, +k) f, ) = [ + h)+ +k) + h) + +k) +5] 5= h + k does not keep a constant sign for small h and k;, ) is a saddle point.

28 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION f =x + y +)i +x +y) j = 0 only at, ). The difference f +h, +k) f, ) =[ +h) + +h) + k)++k) + +h)+] ) = h + hk + k is nonnegative for all small h and k. To see this, note that h + hk + k h h k + k = h k ) 0; there is a local minimum of at, ). 4. f =x ) i + j is never 0 ; there are no stationary points and no local extreme values. 5. f =x + y 6) i +x +y) j = 0 only at 4, ). f xx =, f xy =, f yy =. At 4, ), D=> 0 and A => 0 so we have a local min; the value is f =x +y +)i +x +6y + 0) j = 0 only at, ). f x =, f y x =, f y =6; D = 6 > 0, A= = local min; the value is f =x 6y) i + y 6x ) j = 0 at, ) and 0, 0). f xx =6x, f xy = 6, f yy =6y, D =6xy 6. At, ), D= 08 > 0 and A => 0 so we have a local min; the value is 8. At 0, 0), D= 6 < 0 so we have a saddle point. 8. f =6x + y +5)i +x y 5) j = 0 at f x =6, f y x =, 5 ), 5. f y = ; D =6 ) < 0; 5/, 5/) is a saddle point. 9. f =x 6y +6)i +y 6x +)j = 0 at 5, 7 ) and, ). f xx =6x, f xy = 6, f yy =, D =x 6. At 5, 7 7 ),D=4> 0 and A =0> 0 so we have a local min; the value is 4. At, ),D= 4 < 0 so we have a saddle point. 0. f =x y ) i + x +4y +5)j = 0 at, ). f x =, the value is 4. f y x =, f y =4; D = 4 ) > 0, A= = local minimum;. f = sin y i + x cos y j = 0 at 0,nπ) for all integral n. f xx =0, f xy = cos y, f yy = x sin y. Since D = cos nπ = < 0, each stationary point is a saddle point.

29 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 86 SECTION 6.5. f = sin y i ++x cos y) j = 0 at, nπ) and, n +)π) for all integral n. f x =0, f = cos y, y x so they are all saddle points f y = x sin y; D =0 x sin y) cos y<0 at the above points. f =xy ++y ) i + x +xy + ) j = 0 at, ) and, ). f xx =y, f xy =x +y, f yy =x, D =4xy 4x + y). At both, ) and, ) we have saddle points since D = 4 < f = y + y ) x i + x y ) j = x + y x x i x + y y xy j is never 0; no stationary points, no local extreme values. 5. f =y x ) i + x 8y ) j = 0 only at, 4). f xx =x, f xy =, f yy =6y, D =x y. At, 4),D=> 0 and A =6> 0 so we have a local min; the value is f =x y) i + x y) j = 0 only at 0, 0) f x =, f y x =, f y = ; D = ) ) < 0; 0, 0) is a saddle point. 7. f =y x ) i + x y ) j = 0 only at, ). f xx =x, f xy =, f yy =y, D =4x y. At, ), D=> 0 and A => 0 so we have a local min; the value is. 8. f =xy y ) i +x xy +)j = 0 at, ),, ) f x =y, f =x y), y x, ) and, ) are saddle points. f y = x; D = 4xy 4x y) < 0 at the above points; 9. f = x y ) x + y +) i + 4xy x + y j = 0 at, 0) and, 0). +) f xx = 4x +xy +x x + y +), f xy = 4y +4y x y x + y +), f yy = 4x +4x xy x + y +). f, 0) = ; f, 0) = point A B C D result, 0) 0 loc. min., 0) 0 loc. max.

30 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 0. f = ln xy + ) i + x j = 0 at, /) x y SECTION f x = x + x, f y x = y, f y = x y At, /), f x =, f y x =, f = 0 and D = 9 < 0 = saddle point. y. f = 4x 4x ) i +y j = 0 at 0, 0),, 0), and, 0). f xx =x 4, f xy =0, f yy =. f±, 0) =. point A B C D result 0, 0) saddle, 0) loc. min., 0) loc. min.. f =xe x y + x + y ) i +ye x y x y ) j = 0 at 0, 0), 0, ), 0, ) A = f y x =xex x +x +xy )+e x y +6x +y ) B = f y x = y x +x +xy )+e yex x y 4xy) C = f y = y yex y yx y )+e x y x 6y ) At 0, 0), AC B = )) = 4 > 0,A>0 local minimum; the value is 0. At 0, ±), AC B =4e ) 4e )= 8e > 0, saddle points. f = cos x sin y i + sin x cos y j = 0 at π, π), π, π), π, π), π, π), π, π). f xx = sin x sin y, f xy = cos x cos y, f yy = sin x sin y point A B C D result π, π) 0 loc. max. π, π) 0 loc. min. π, π) 0 0 saddle π, π) 0 loc. min. π, π) 0 loc. max. f π, π) = f π, π) =; f π, π) = f π, π) = 4. f = sin x cosh y i + cos x sinh y j = 0 at π, 0), 0, 0), π, 0). f xx = cos x cosh y, f xy = sin x sinh y, f yy = cos x cosh y D = cos x cosh y sin x sinh y<0; π, 0), 0, 0), π, 0) are saddle points.

31 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 88 SECTION a) f =x + ky) i +y + kx) j and f0, 0) = 0 independent of the value of k. b) f xx =, f xy = k, f yy =, D =4 k. Thus, D<0 for k > and 0, 0) is a saddle point c) D =4 k > 0 for k <. Since A = f xx => 0, 0, 0) is a local minimum. d) The test is inconclusive when D =4 k = 0 i.e., for k = ±. If k = ±, fx, y) =x ± y) and 0, 0) is a minimum.) 6. a) f =x + ky) i +kx +8y) j = 0 at 0, 0). b) f x =, f y x = k, f y = 8; we want 6 k < 0, or k > 4 c) We want 6 k > 0, or k < 4 d) k = ±4. If k = ±4, fx, y) =x ± y) and 0, 0) is a minimum.) 7. Let P x, y, z) be a point in the plane. We want to find the minimum of fx, y, z) = x + y + z. However, it is sufficient to minimize the square of the distance: F x, y, z) =x + y + z. It is clear that F has a minimum value, but no maximum value. Since P lies in the plane, x y +z =6 which implies y =x +z 6 = x + z 8). Thus, we want to find the minimum value of Now, The gradient is 0 when F x, z) =x +4x + z 8) + z F =[x +8x + z 8)] i + [8x + z 8)+z] k x +8x + z 8) = 0 and 8x + z 8)+z =0 The only solution to this pair of equations is: x = z = 9, from which it follows that y = 6 9. The point in the plane that is closest to the origin is P 9, 6 9, ) 9. The distance from the origin to the plane is: F P )= 6. Check using.6.5): dp, 0) = ) + = We want to minimize x +) +y ) +z 4) on the plane. Since z = 6 x +y, we need to minimize fx, y) =x +) +y ) + 0 x +y) ; f = x 6y +6) i x +0y) j = 0 at 4, 6) Closest point 4, 6, ), distance= 4)) + 6) +4 ) = 9 [ 9. fx, y) =x ) +y ) + z =x ) +y ) + x +y since z = ] x +y f = [x )+x] i + [y ) + 4y] j = 0 = x =,y=. f xx =4> 0, f xy =0, f yy =6, D =4> 0. Thus, f has a local minimum at /, /). The shortest distance from,, 0) to the cone is f, ) = 6 4

32 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION V =8xyz, x + y + z = a = V x, y) =8xy a x y, x > 0, y>0, x + y <a V = 8ya x y ) 8x y a x y i + 8xa x y ) 8xy a x y j = 0 at a/,a/ ) dimensions: a a a, maximum volume: 8 9 a. a) b) c) f =4y 4x ) i +4x 4y ) j = 0 at 0, 0),, ),, ). f xx = x, f xy =4, f yy = y, D = 44x y 6 point A B C D result 0, 0) saddle, ) 4 8 loc. max., ) 4 8 loc. max. f, ) = f, ) =. a) b) c) f =4x 4x) i +yj = 0 at 0, 0),, 0),, 0). f xx =x 4, f x,y =0, f yy =, D =4x 8 point A B C D result 0, 0) saddle, 0) loc. min., 0) loc. min. f, 0) = f, 0) = 0

33 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 80 SECTION 6.5. a) 4 b) f, ) = is a local max.; f has a saddle at 0, 0). 4. a) b) f0, 0) = 0 is a local min.; f0, ) = f0, )=e are local maxima; f has a saddle at ±, 0). 5. a) b) f, 0) = is a local min.; f, 0) = is a loc. max.

34 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION a) b) fπ/,π/) = fπ/, 5π/) = f5π/,π/) = f5π/, 5π/) = are local maxima; π/, π/), π/,π/), π/, π/), π/, 5π/), 5π/, π/) are saddle points of f; f7π/6, 7π/6) = fπ/6, π/6) = are local minima. SECTION 6.6. f =4x 4) i +y ) j = 0 at, ) in D; f, ) = Next we consider the boundary of D. We y 4 parametrize each side of the triangle: C : r t) =t i, t [0, ], C : r t) =i + t j, t [0, 4], C : r t) =t i +tj, t [0, ], Now, x f t) =fr t)) = t ), t [0, ]; critical number: t =, f t) =fr t)) = t ) +, t [0, 4 ]; critical number: t =, f t) =fr t)) = 6t 8t +, t [0, ]; critical number: t =. Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 0) = f 0) = f0, 0) = ; f ) = f, 0)=0; f ) = f 0) = f, 0) = ; f ) = f, )=; f 4) = f ) = f, 4) = 0; f /) = f/, 4/) =. f takes on its absolute maximum of 0 at, 4) and its absolute minimum of at, ).. f = i +j 0; no stationary points in D; Next we consider the boundary of D. We parametrize each side of the triangle: C : r t) =t i, t [0, 4], C : r t) =t i + t +6) j, t [0, 4], C : r t) =t j, t [0, 6],

35 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 8 SECTION 6.6 and evaluate f: f t) =fr t)) = t, t [0, 4 ]; no critical numbers, f t) =fr t)) = 6t +4, t [0, 4 ]; no critical numbers, f t) =fr t)) = + t, t [0, 6 ]; no critical numbers. Evaluating these functions at the endpoints of their domains, we find that: f 0) = f 0) = f0, 0) = ; f 4) = f 4) = f4, 0) = 0; f 0) = f 6) = f0, 6) = 4; f takes on its absolute maximum of 4 at 0, 6) and its absolute minimum of 0 at 4, 0).. f =x + y 6) i +x +y) j = 0 at 4, ) in D; f4, ) = Next we consider the boundary of D. We parametrize each side of the rectangle: C : r t) = t j, t [0, ] C : r t) =t i j, t [0, 5] C : r t) =5i tj, t [0, ] C 4 : r 4 t) =t i, t [0, 5] Now, f t) =fr t)) = t, t [0, ]; no critical numbers y x f t) =fr t)) = t 9t +8, t [0, 5 ]; critical number: t = 9 f t) =fr t)) = t 5t 6, t [0, ]; critical number: t = 5 f 4 t) =fr 4 t)) = t 6t, t [0, 5 ]; critical number: t = Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 0) = f 4 0) = f0, 0) = ; f ) = f 0) = f0, ) = 8; f 9/) = f9/, ) = 49 4 ; f 5) = f ) = f5, ) = ; f 5/) = f5, 5/) = 49 4 ; f 0) = f 4 5) = f5, 0) = 6. f 4 ) = f, 0) = 0 f takes on its absolute maximum of 8 at 0, ) and its absolute minimum of at 4, ). 4. f =x +y) i +x +6y) j = 0 at 0, 0) in D; f0, 0) = 0 Next we consider the boundary of D. We parametrize each side of the square: C : r t) =t i j, t [, ], C : r t) =i + t j, t [, ], C : r t) =t i +j, t [, ], C 4 : r 4 t) = i + t j, t [, ],

36 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION and evaluate f: f t) =fr t)) = t 4t +, t [, ]; no critical numbers, f t) =fr t)) = 4 + 4t +t, t [, ]; critical number: t =, f t) =fr t)) = t +4t +, t [, ]; no critical numbers, f 4 t) =fr 4 t)) = 4 4t +t, t [, ]; critical number: t =. Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f ) = f 4 ) = f, ) = 4; f ) = f ) = f, ) = 8; f /) = f, /) = 8 ; f ) = f ) = f, ) = 4; f ) = f 4 ) = f, ) = 8; f 4 /) = f, /) = 8. f takes on its absolute maximum of 4 at, ) and, ) and its absolute minimum of 0 at 0, 0). Note that x +xy +y =x + y) +y. The results follow immediately from this observation. 5. f =x +y) i +y +x) j = 0 at 0, 0) in D; f0, 0) = Next we consider the boundary of D. We parametrize the circle by: C : rt) = cos t i + sin t j, t [0, π ] The values of f on the boundary are given by the function F t) =frt)) = 6 + sin t cos t, t [0, π ] F t) = cos t sin t : F t) =0 = cos t = ± sin t = t = 4 π, 4 π, 5 4 π, 7 4 π Evaluating F at the endpoints and critical numbers, we have: F 0) = F π) =f, 0)=6; F 4 π) = F 5 4 π) ) = f, F 4 π) = f, ) = F 7 4 π) ) = f, =0 f takes on its absolute maximum of at, ) and at, ) ; minimum of 0 at, ) and at, ). = f, ) = ; 6. f = yi +x )j = 0 at, 0), which is not in the interior of D. The boundary is f takes on its absolute rt) = cos t i + sin t j. frt)) = sin t cos t ) = 9 sin tcos t ), t [0, π]. df dt = 9 cos t cos t ); df dt =0 = cos t =, which yields the points A, 0), B, ),C, ): fa) =0, fb) = 7 4 min, fc) = 7 4 max 7. f =x )i +y ) j = 0 only at, ) in D. As the sum of two squares, fx, y) 0. Thus, f, ) = 0 is a minimum. To examine the behavior of f on the boundary of D, we note that f represents the square of the distance between x, y) and, ). Thus, f is maximal at the point of the boundary furthest from, ). This is the point, ) ; the maximum value of f is f, ) =6+4.

37 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 84 SECTION f =y +)i +x ) j = 0 at, ) which is not in the interior of D. Next we consider the boundary of D. C : r t) =t j + t j, t [, ], C : r t) =t i +4j, t [, ], and evaluate f: We parametrize the boundary by: f t) =fr t)) = t t + t +, t [, ]; no critical numbers, f t) =fr t)) = 5t, t [, ]; no critical numbers. Evaluating these functions at the endpoints of their domains, we find that: f ) = f ) = f, 4) = ; f ) = f ) = f, 4) = 9. f takes on its absolute maximum of 9 at, 4) and its absolute minimum of at, 4). 9. f = x y x + y +) i + 4xy x + y j = 0 at, 0) and, 0) in D; f, 0) =, f, 0) =. +) Next we consider the boundary of D. We parametrize each side of the squre: Now, C : r t) = i + t j, t [, ] C : r t) =t i +j, t [, ] C : r t) =i + t j, t [, ] C 4 : r 4 t) =t i, t [, ] f t) =fr t)) = 4 t, +5 t [, ]; critical number: t =0 f t) =fr t)) = t t, +5 t [, ]; no critical numbers f t) =fr t)) = 4 t, +5 t [, ]; critical number: t =0 f 4 t) =fr 4 t)) = t t, +5 t [, ]; no critical numbers Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f ) = f 4 ) = f, ) = 4 9 ; f 0) = f, 0) = 4 5 ; f ) = f ) = f, ) = 4 9 ; f 4 ) = f ) = f, ) = 4 9 ; f 0) = f, 0) = 4 5 ; f ) = f ) = f, ) = 4 9. f takes on its absolute maximum of at, 0) and its absolute minimum of at, 0). 0. f = x y x + y +) i + 4xy x + y j = 0 at, 0) in D; f, 0) =. +) Next we consider the boundary of D. We parametrize each side of the triangle: C : r t) =t i t j, t [0, ] C : r t) =i + t j, t [, ] C : r t) =t i + t j, t [0, ],

38 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 and evaluate f: SECTION f t) =fr t)) = t t +, t [0, ]; critical number: t =/, f t) =fr t)) = 4 t, +5 t [, ]; critical number: t =0 f t) =fr t)) = t t +, t [0, ]; critical number: t =/. Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f 0) = f 0) = f0, 0) = 0; f / ) = f/, / ) = / ; f ) = f ) = f, ) = 4 9 ; f 0) = f, 0) = 4 5 ; f ) = f ) = f, ) = 4 9 ; f / ) = f/, / ) = /. f takes on its absolute maximum of 0 at 0, 0) and its absolute minimum of at, 0).. f =4 4x) cos y i 4x x ) sin y j = 0 at, 0) in D: f, 0) = Next we consider the boundary of D. We parametrize each side of the rectangle: C : r t) =t j, t [ 4 π, 4 π ] C : r t) =t i 4π j, t [0, ] C : r t) =i + t j, t [ 4 π, 4 π ] C 4 : r 4 t) =t i + 4π j, t [0, ] Now, f t) =fr t)) = 0; f t) =fr t)) = 4t t ), t [0, ]; critical number: t =; f t) =fr t)) = 0; f 4 t) =fr 4 t)) = 4t t ), t [0, ]; critical number: t =; f at the vertices of the rectangle has the value 0; f ) = f 4 ) = f, 4 π) = f, 4 π) =. f takes on its absolute maximum of at, 0) and its absolute minimum of 0 along the lines x =0 and x =.. f =x )i +yj = 0 at, 0) which is not in the interior of D. Boundary: On y = x, f =x ) + x 4 df, dx =x )+4x =0atx = =, ) On y =4x, f =x ) +6x, df dx =x )+x =0atx = 7 = 7, 7 ). So the maximum and minimum occur at one or more of the following points: 0, 0), 4, 6),, ), 7, 7 ). Evaluating f at these points, we find that f, ) = 5 is the absolute minimum of f; f4, 6) = 57 is the absolute maximum of f.

39 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 86 SECTION 6.6. f =x y) i + x y ) j = 0 at, ) in D; f, ) = Next we consider the boundary of D. We parametrize each side of the triangle: C : r t) = i + t j, t [, ], C : r t) =t i + t j, t [, ], C : r t) =t i +j, t [, ], - - x - y and evaluate f: f t) =fr t)) = 8+6t t, t [, ]; critical numbers: t = ±, f t) =fr t)) = t, t [, ]; critical number: t =0, f t) =fr t)) = t 6t 8, t [, ]; critical numbers: t = ±. Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f ) = f ) = f, ) = ; f ) = f, ) = 8 4 =.66; f ) = f, ) = 8+4 =.4; f ) = f ) = f, ) = 4; f 0) = f0, 0)=0; f ) = f ) = f, ) = ; f ) = f, ) = 8+4 ; f ) = f, ) = 8 4 f takes on its absolute maximum of at, ) and its absolute minimum of 8 4 at, ) and, ). 4. f =x 4)i +yj = 0 at 4, 0) which is not in the interior of D. Next we examine f on the boundary of D: C : r t) =ti +4tj, t [0,, ], Note that Next and C : r t) =ti + t j, t [0, ]. f t) =f r t)) = 7t 8t +6, f t) =f r t)) = t 4) + t 6. f t) =4t 8=0 = t =4/7 and gives x =4/7, y =6/7 f t) =6t 5 +t 8=0 = t = and gives x =, y =. The extreme values of f can be culled from the following list: f0, 0) = 6, f, 8)=68, f 4 7, ) 6 7 = 56 7, f, ) = 0. We see that f, ) = 0 is the absolute minimum and f, 8) = 68 is the absolute maximum.

40 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 SECTION xy 5. f = x + y +) i + y x x + y j = 0 at 0, ) and 0, ) in D; +) f0, ) =, f0, ) = Next we consider the boundary of D. We parametrize the circle by: C : rt) = cos t i + sin t j, t [0, π ] The values of f on the boundary are given by the function F t) =frt)) = 4 5 sin t, t [0, π ] F t) = 4 5 cos t : F t) =0 = cos t =0 = t = π, π. Evaluating F at the endpoints and critical numbers, we have: F 0) = F π) =f, 0) = 0; F π) = f0, ) = 4 5 ; F π) = f0, ) = 4 5 f takes on its absolute maximum of at 0, ) and its absolute minimum of at0, ). 6. f =x +)i +8y ) j = 0 at, 4) in D; f, ) 4 = Next we consider the boundary of D. We parametrize the ellipse by: C : rt) = cos t i + sin t j, t [0, π ] The values of f on the boundary are given by the function F t) =frt)) = 4 cos t + 4 sin t + cos t sin t =4+cost sin t, t [0, π ] F t) = sin t cos t : F t) =0 = cos t = sin t = t = 4 π, or 7 4 π Evaluating F at the endpoints and critical numbers, we have: F 0) = F π) =f, 0) = 6; F 4 π) = f, ) / =4 ; F 7 4 π) ) = f, / =4+ f takes on its absolute maximum of 4 + at, / ) ; f takes on its absolute minimum of at, 4). 7. f =x y)i x y) j = 0 at each point of the line segment y = x from 0, 0) to 4, 4). Since fx, x) = 0 and fx, y) 0,f takes on its minimum of 0 at each of these points. Next we consider the boundary of D. We parametrize each side of the triangle: C : r t) =tj, t [0, ] C : r t) =ti, t [0, 6] C : r t) =ti + t) j, t [0, 6]

41 P: PBU/OVY P: PBU/OVY QC: PBU/OVY T: PBU JWDD07-6 JWDD07-Salas-v December 7, 006 6:6 88 SECTION 6.6 and observe from fr t)) = t, t [0, ] fr t)) = t, t [0, 6] fr t)) = t ), t [0, 6] that f takes on its maximum of 44 at the point 0, ). 8. f = xi yj) 0 in D. Note that fx, y) is the reciprocal of the distance of x, y) x + y / ) from the origin. The point of D closest to the origin draw a figure) is, ). Therefore f, ) = / is the maximum value of f. The point of D furthest from the origin is, 4). Therefore f, 4) = /5 is the least value taken on by f. 9. Using the hint, we want to find the maximum value of fx, y) =8xy x y xy in the triangular region. The gradient of f is: The gradient is 0 when D = 8y xy y ) i + 8x x xy ) j 8y xy y = 0 and 8x x xy =0 The solution set of this pair of equations is: 0, 0), 8, 0), 0, 8), 6, 6). It is easy to verify that f is a maximum when x = y =6. The three numbers that satisfy x + y + z =8 and maximize the product xyz are: x = 6,y = 6,z = fy, z) =0yz y z yz, f = 0z yz z )j + 60yz y z yz )k = 0 at ) 5, 5 ) 5 other points are not in the interior); f, 5 = On the line y + z =0,fy, z) = 0 so the maximum of xyz occurs at x = y = 5, z =5. fx, y) =xy x y), 0 x, 0 y x. [ dom f) is the triangle with vertices 0, 0),, 0), 0, ).] f =y xy y )i + x xy x ) j = 0 = x = y =0,x=,y=0,x=0,y=,x=y =. Note that [ 0, 0 ] is not an interior point of the domain of f.) f xx = y, f xy = x y, f yy = x. At, ), D = > 0 and A<0 so we have a local max; the value is /7. Since fx, y) = 0 at each point on the boundary of the domain, the local max of /7 is also the absolute max.