2x 2 y x 4 +y 2 J (x, y) (0, 0) 0 J (x, y) = (0, 0) I ϕ(t) = (t, at), ψ(t) = (t, t 2 ), a ÑL<ÝÉ b, ½-? A? 2t 2 at t 4 +a 2 t 2 = lim

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1 9çB$ø`çü5 (-ç ) Ch.Ch4 b. è. [a] #8ƒb f(x, y) = { x y x 4 +y J (x, y) (, ) J (x, y) = (, ) I ϕ(t) = (t, at), ψ(t) = (t, t ), a ÑL<ÝÉ b, ½-? A? (i) lim f(ϕ(t)) = lim f(ψ(t)), (ii) f Ê (, ) /. t t (i) lim f(ϕ(t)) = lim t t lim f(ψ(t)) = lim t lim t t t at t 4 +a t = lim f(ϕ(t)) lim t f(ψ(t)) t t t t 4 +t 4 = lim t = at a +t = (ii) J f Ê (, ) /, lim f(x, y) = f(, ), (x,y) (, Oâ, ú H ϕ, ψ lim f(ϕ(t)) = = f(, ) = lim f(ψ(t)) = = t =, 8e ] f Ê (, ). / ½ Ì [b] #8ƒb f(x, y) = lim (x,y) (,) { f(x, y) æê? t x e x y y J y J y =

2 { x f(x, y) = e x y y J y J y = 5? ϕ(t) = (t, t), lim f(ϕ(t)) = lim f(t, t) = t t t lim e t t t t = lim t e = e. 5? ψ(t) = (t 3 t, t), f(ψ(t)) = lim 3 e t3 t t t, Y t >, t C t <, lim f(ψ(t)) = lim ±te t =. t t B, lim f(ϕ(t)) lim f(ψ(t)), ] lim f(x, y). t t æê (x,y) (,). ø õêæ ì( (circular helix) r (t) «, r (t) = (a cos ωt, a sin ωt, bωt) [a] ½ õ5 (speed) u Ñb? velocity vector= d r (t) dt = ( aω sin ωt, aω cos ωt, bω) r = a ω sin ωt + a ω cos ωt + b ω speed= r = a + b ω, constant. [b] õ5 ²¾ (velocity vector) D z W5>i z 5ÀP²¾ k = (,, ), r D k íhiñ θ, cos θ = r k r k = bω (a +b ) ω b =, (a +b )

3 θ = cos ( b (a +b ) ), ø_b [c] à (b) C ( r (t),íù_ôyõ, j-½ æ: ²¾ƒb (à r (t))? À bõbmƒb f : [a, b] R 5 Mean Value Theorem?(cì f x5fbígß4 ) J r (t) = (a cos ωt, a sin ωt, bωt)ñw, t =, t = π ω, r (t ) = (a,, ), r (t ) = (a,, πb), â r (t ) r (t ) = (,, πb), ²¾5j²uòk x, y Þ (²,, Jb > ) J r _²¾ƒbé À bõbmƒb5 Mean Value Theorem, ξ k t D t 5ÈU) r (t ) r (t ) = (t t ) r (ξ), ¹ (,, πb) = π ω ( aω sin ωξ, aω cos ωξ, bω), ku sin ωξ =, cos ωξ =, Ñ.ª?59 ] (c) 5 u ìí [d] Æ ì( (ÊLSøõ) í õ5 unit velocity vector= T = v v, JsÑÆ (5 parameter, 3

4 b²ì dt ds, T = d r dt /ds dt = d r ds = d (a cos ωt, a sin ωt, bωt) ds = dt ds d ( aω sin ωt, aω cos ωt, bω) dt = ( dt ds ) (a ω sin ωt + a ω cos ωt + b ω ) = ( dt ds ) (a + b )ω dt ds =, a +b ω â T = a +b( a sin ωt, a cos ωt, bω) d T ds = a + b d ds ( a sin ωt, a cos ωt, bω) = a + b dt d ds dt ( a sin ωt, a cos ωt, bω) K = d T ds = dt a a +b ds ω cos ωt + a ω sin ωt B Æ ì($êlsøõ5ñ a a +b. 3. Let r(t) = (e t t, 6e t, 3t), t, [a] Find the length of the curve; 4

5 r (t) = (e t, 6e, t 3), r (t) = ((e t ) + ( 6e) t + ( 3) ) = (e t + 4e t + 4) = e t +. L = r (t) dt = (ey + )dt = e e + 4. [b] Find the curvature of r(t) at t = ; r (t) = (e t, t 6e, ), r () = (, 6, 3), r () = (, 6, ), r () r () = ( 3, 3, 6). So k() = r () r () = r () Let r(t) = (t, sin t + t cos t, cos t + t sin t), t >, find an equation of the osculating plane of the curve r(t) at the point (π, π, ). r (t) = (t, t sin t, t cos t), r (t) = ((t) +( t sin t) +(t cos t) ) = 5 t = 5t (by t > ). T(t) = r (t) r (t) = 5 (, sin t, cos t), T (t) = 5 (, cos t, sin t), T (t) = 5, N(t) = T (t) T = (, cos t, sin t), (t) B(t) = T(t) N(t) = 5 (, sin t, cos t). So the osculating plane of the curve r(t) at the point (π, π, ) has normal vector B(π) = 5(,,), so 5

6 an equation is (x π ) + (y + π) + (z + ) = or x + z = π. { xy+yz 3 if (x, y, z) (,, ) 5. Let f(x, y, z) = x +z 6 if (x, y, z) = (,, ), determine the set of points at which f is continuous. The function f(x, y, z) is continuous for (x, y, z) (,, ) since it is equal a rational function there. For (x, y, z) (,, ) along curve r(t) = (kt 3, kt 3, t), we see that f(x, y, z) = f(kt 3, kt 3, t) = (kt3 ) +kt 3 t 3 = (kt 3 ) +t 6 (k +k) k +. So f(x, y, z) (k +k) as (x, y, z) (,, ) along k + r(t) = (kt 3, kt 3, t). Therefore, different path leads to different limit values, hence the limit of f(x, y, z) at (,, ) dose not exist, so f(x, y, z) is not continuous at (,, ). Ch5 Rûb (x3 +y )+xyz +8z =. ÊÞ, (,, ) øõ 5 z x, z y, z xx, z xy õ5~þ(cé z x, z xx, z xy ~Þ) z x : 6

7 6x + yz + xyzz x + 8z x = 6 + 8z x =, z x = z y : 6 3 y + xz + xyzz y + 8z y = 3 3 = 8z y z y = 4 3 ~Þ: p(x + ) + q(y + ) + ( ) z = (x + ) (y + ) z = z xx : 3x + yzz x + yzz x + xyzx + xyzz xx + 8z xx = 3x + xyzx + 8z xx = 3 + (4) + 8z xx = 4 + 8z xx = z xx = 3 z xy : z + yzz y + xzz x + xyz y z x + xyzz xy + 8 xy = z y z x + 8z xy = z xy = z xz y 4 = 4 ( )(4 3 ) = z x, z y, ~Þ, ( 9, ) FÊ5z5,lM z = + (.) (.) =. 4 3 = 4 3 = 3 = Find the absolute maximum and minimum values of 7

8 f(x, y) = 4x + 6y x y in x + y. At a critical point, f x = 4 x =, f y = 6 y = = x =, y = 3. So there are no critical points in x +y <. Need only look at boundary x +y =. method Use x = cos θ, y = sin θ g(θ) = 4 cos θ + 6 sin θ g (θ) = 4 sin θ + 6 cos θ = if tan θ = 3 = sec θ = + tan θ = = 3 4 = cos θ = 4 3, sin θ = cos θ = 9 3. So possibilities, ( 3 3, 3 ),( 3, 3 3 ). The first gives 6 3 = 3 maximum. The second gives 3 minimum. method. Lagrange multipliers: 4 x = λx () 6 y = λy () x + y = (3) (),()= ( + λ)x = 4, ( + λ)y = 6 = 3( + λ)x = ( + λ)y = x = 3y. Then (3)= 4 9 y + y = = y = 9 3 = y = ± 3 3. So get ( 3 3, 3 ) or ( 3, 3 3 ). Finish as in method. 8

9 9. Find the maximum value of f(x, y, z) = xy + xz + yz subject to the constraints x, y, z and x + y + z =. If the maximum occurs in the interior of the set, then the equation y + z = λ, x + z = λ, x + y = λ, x + y + z = must hold for some λ. Then y + z = x + z = x = y x + z = x + y = y = z So x = y = z = 3 = f(x, y, z) = 3 9 = 3. The other possibility is that the maximum occurs where one of x, y, z is. Suppose x =. Then y + z = and f(, y, z) = yz = y( y). The maximum of this in y is 4. Similarly, if y = or z = we get a maximum of 4. Hence the maximum value must be 3.. The plane x + y + z = intersects the paraboloid z = x + y in an ellipse. Find the highest point on the ellipse. We want to maximize f(x, y, z) = z subject to x + y z =, x + y + z =. 9

10 By the method of Lagrange multipliers, we get = λx + µ () = λy + µ () = λ + µ (3) x + y z = (4) x + y + z = (5) (),()= λx = λy = λ = or x = y. If λ =, then () = µ = which contradicts (3). So x = y. Then (4)= z = x + y = 5y. (5)= 4y +y +5y = = y +y = = (y + )(y ) = = y =,. y = = (x, y, z) = ( 4,, ) highest point. y = = (x, y, z) = (,, 5) lowest point.. Show that the ellipsoid 3x + y + z = 9 and the sphere x + y + z 8x 6y 8z + 4 = are tangent to each other at he point (,, ). Let f(x, y) = 3x + y + z 9 and g(x) = 3x + y +z 9. Then f(,, ) = g(,, ) = and the ellipsoid and the sphere are the level surfaces of f and g. Thus f(,, ) and g(,, ) are orthogonal to the ellipsoid and the sphere (,, ). f(x, y, z) = 6x, 4y, z and g(x, y, z) = x

11 8, y 6, z 8. Hence f(,, ) = 6, 4, 4 is parallel g(,, ) = 6, 4, 4. This implies that the ellipsoid 3x + y + z = 9 and the sphere x + y + z 8x 6y 8z + 4 = are tengent to each other at he point (,, ).. If f(x, y) = define y as a function of x, show that d y dx = f xxfy f xy f x f y + f yy fx fy 3 Differentiate f(x, y) = w.r.t x. We have dy f x + f y dx = () Hence if f y, dy dx = f x f y. Differentiate () w.r.t x. We have dy dy f xx + f xy dx + f yx dx + f yy( dy d y dx ) + f y dx = We assume that f xy and f yx are both continuous, and hence f xy = f yx. Also, from the equation (), we have f xx + f xy f x Thus f y + f yx f x f y + f yy ( f x ) d y + f y f y dx = d y dx = f xxf y f xy f x f y + f yy f x f 3 y

12 3. Let f(x, y) = x (x + y ) e sin(x y) if (x, y) (, ) and f(, ) =. () Let u = cos θ, sin θ and find D u f(, ). () Prove that f is continuous at (, ). f(h cos θ, h sin θ) f(, ) D u f(, ) = lim h h h(cos θ) e sin(h3 cos θ sin θ) = lim = (cos θ) h h x (x + y ) e sin(xy) x e. Since x =, it follows that lim lim (x,y) (,) = f(, ). f(x, y) = (x,y) (,) Ch6 ½ } 4. D = {(x, y) x +y 4}, t D 4 x y da.

13 I u = 4 r 5. Ÿ = = = = 8 3 x cos(y )dydx. ² }ßå, 6. Ÿ = π = = π 4 π 3 u3 π 4 4 r rdrdθ ududθ 4 dθ dθ = 6 3 π sin y = sin xy x +y + dydx. cos y dxdy y cos y dy 3

14 I u = x + y + Ÿ = x x +x +x du u dx = + x dx = 3 ( + x ) 3 x + x dx 3 ( + x ) 3 = 3 [3 3 + ] = 3 [ ] 7. t y ye x x 3 dxdy 4

15 x = y = y = x 8. (a) t D x + y b} Ÿ = = = (x +y ) n x ye x x 4 e x x dx 3 = 4 ex xe x dx = (e ) 4 x 3 dydx da, n Ñcb. D : {(x, y), a Ÿ = π b = π a b a r nrdrdθ r n dr (b) ç a +, t nmu }í ÌæÊ (ÿ ) 5

16 9. t Þ z = x + y D6Þ x + y = x xy ÞFˇ5ñ (x ) + y =, r = cos θ Ÿ = = π cos θ π π cos θ rrdrdθ r drdθ. ρ(x, y) = xy, D = {(r, θ), r a, θ π }, t Dí - (ÿ ). Consider the iterated integral 8 ( y 3 + x4 dx) dy. Sketch the region of integration and then evaluate it by reversing the order of integration. dx x 3 dy + x 4 = x3 + x 4 dx = 3 6 {73 } ( 4. Consider the iterated integral y y sin(x )dx dy. Sketch the region of integration and then evaluate it by reversing the order of integration. 6 4 ( + x4 ) 3 = )

17 4 4 dx x dyy sin(x ) = 4 ( cos 6) x sin(x )dx = ( cos(x )) 4 = 3. By making an appropriate change of variables to evaluate the integral da R, where R is the +9x +9y region bounded by the ellipse 9x + 9y = 4. x = r cos θ, y = r sin θ π dθ 3 dr r = π ln(+9r ) 3 +9r 8 = π 9 ln 5 7