(t, t2), a ÑL<ÝÉ¡b, ½- ?´A ? (i) lim t→0f (ϕ(t

92^{çB$ø`çü5}(^-ç‚)

Ch11.Ch14 ^¡b. ^”è™

1. [a] _#8ƒb f (x, y) =

( 2x²y

x⁴+y² J (x, y) 6= (0, 0) 0 ^J (x, y) = (0, 0)

I ϕ(t) = (t, at), ψ(t) = (t, t²), a ^ÑL<ÝÉ¡b, ^½-

?´A ? (i) lim

t→0f (ϕ(t)) = lim

t→0f (ψ(t)), (ii) f ^Ê (0, 0) ^©/. sol.

(i)

limt→0f (ϕ(t)) = lim

t→0

2t²·at

t⁴+a²t² = lim

t→0 2at

a²+t² = 0 limt→0f (ψ(t)) = lim

t→0 2t²·t²

t⁴+t⁴ = lim

t→01 = 1 limt→0f (ϕ(t)) 6= lim

t→0f (ψ(t))

(ii) ^J f ^Ê (0, 0) ^©/, ^† lim

(x,y)→(0,0

f (x, y) = f (0, 0),

Oâ¤, ^ú‡H ϕ, ψ

limt→0f (ϕ(t)) = 0 = f (0, 0) = lim

t→0f (ψ(t)) = 1 =⇒

0 = 1, ^8e
] f ^Ê (0, 0) ^.©/

[b] _#8ƒb f (x, y) =

( |x|

y²e⁻

|x|

y2 J y 6= 0

0 ^J y = 0

½”Ì lim

(x,y)→(0,0)

f (x, y) ^æÊ´?

sol.

f (x, y) =

( |x|

y²e⁻

|x|

y2 J y 6= 0

0 ^J y = 0

5? ϕ(t) = (t², t), lim

t→0f (ϕ(t)) = lim

t→0f (t², t) = limt→0

|t²| t² e⁻

|t2|

t2 = lim

t→0e⁻¹ = e⁻¹.

5? ψ(t) = (t³, t), f (ψ(t)) = lim

t→0

|t³| t² e⁻

|t3|

t2 ,

Y t > 0, t 6= 0 ^C t < 0, lim

t→0f (ψ(t)) = lim

t→0±te^t = 0.

B¤, lim

t→0f (ϕ(t)) 6= lim

t→0f (ψ(t)), ^] lim

(x,y)→(0,0)

f (x, y)^.

æÊ

2. ^{ø”õÊÆ£ì(} (circular helix) −→r (t) ^«,

−

→r (t) = (a cos ωt, a sin ωt, bωt) [a] ^½¤õ5§0 (speed) ^u´Ñb?

sol.

velocity vector= ^{d−→r (t)}_dt = (−aω sin ωt, aω cos ωt, bω)

|−→r | = √

a²ω²sin²ωt + a²ω²cos²ωt + b²ω² speed= |−→r | = √

a² + b²ω, constant.

[b] ^{°¤”õ5§²¾} (velocity vector) ^D z ^W5>i sol.−

→z ^5ÀP²¾ −→

k = (0, 0, 1), −→r ^D −→

k ^íHiÑ θ,

† cos θ = ^{−→r ·}

−

→k

|−→r ||−→

k | = ^bω

(a²+b²)¹²ω

= ^b

(a²+b²)¹²

,

θ = cos⁻¹



b (a²+b²)¹²



, ^ø_b

[c] ^‚à (b) ^C¦( −→r (t) ^,íù_Ôyõ, ^j-½

æ: ^²_¾ƒb(^à−→r (t))?´À‰bõbMƒbf : [a, b] → R ⁵ Mean Value Theorem ?(^cì f ^x_5FbíGß4

”) sol.

J −→r (t) = (a cos ωt, a sin ωt, bωt)^ÑW, ^¦ t₁ = 0, t₂ =

2π ω , ^†

−

→r (t₁) = (a, 0, 0), −→r (t₂) = (a, 0, 2πb), ^⤠−→r (t₂) −

−

→r (t₁) = (0, 0, 2πb), ^{¤²¾5j²uòk} x, y ^Þ (^²

,, ^Jb > 0)

J −→r ¥_²¾ƒbé°À‰bõbMƒb5 Mean Value Theorem, ^† ξ _k t₁^D t₂ ^5ÈU) −→r (t₂) − −→r (t₁) =

(t₂−t₁)−→r ⁰(ξ), _¹(0, 0, 2πb) = ^2π_ω (−aω sin ωξ, aω cos ωξ, bω),

ku sin ωξ = 0, cos ωξ = 0, ^¤Ñ.ª?59
] (c) ^5

u´ìí

[d] ^°Æ£ì( (^ÊLSøõ) ^í0 sol.

”õ5 unit velocity vector= −→

T = _|−^−→→^vv |, ^Js^ÑÆ£(5 parameter,

−

→T = d−→r dt /ds

dt = d−→r ds

= d

ds(a cos ωt, a sin ωt, bωt)

= dt ds · d

dt(−aω sin ωt, aω cos ωt, bω)

b²ì^dt

ds, 1 = (dt

ds)² · (a²ω²sin²ωt + a²ω²cos²ωt + b²ω²)

= (dt

ds)² · (a² + b²)ω²

dt

ds = ^{√ 1}

a²+b²ω, ^â¤−→

T = ^{√ 1}

a²+b²(−a sin ωt, a cos ωt, bω) d−→

T

ds = 1

√a² + b^{2 d}_ds(−a sin ωt, a cos ωt, bω)

= 1

√a² + b^{2 dt}_dsdt^d (−a sin ωt, a cos ωt, bω)

0K =   

d−→ T ds

 = ^{√ 1}

a²+b²

 ^dt

ds

·√

a²ω²cos²ωt + a²ω²sin²ωt

B¤Æ£ì($ÊLSøõ50Ñ ^a

a²+b². 3. Let r(t) = (e^t − t, 2√

6e^2t,√

3t), −1 ≤ t ≤ 1, [a] Find the length of the curve;

sol.

r⁰(t) = (e^t − 1,√

6e^2t,√ 3),

||r⁰(t)|| = ((e^t − 1)² + (√

6e^2t)² + (√

3)²)¹² = (e^2t + 4e^t + 4)¹² = e^t + 2.

L = R1

−1||r⁰(t)||dt = R 1

−1(e^y + 2)dt = e − e⁻¹ + 4.

[b] Find the curvature of r(t) at t = 0;

sol.

r⁰⁰(t) = (e^t, ¹₂√

6e^2t, 0), r⁰(0) = (0,√

6,√

3), r⁰⁰(0) = (1,

√6

2 , 0), r⁰(0)×r⁰⁰(0) = (⁻³

√2 2 ,√

3, −√ 6).

So k(0) = ^||r0(0)×r_||r0(0)||^00(0)||₃ = ¹₂.

4. Let r(t) = (t², − sin t + t cos t, cos t + t sin t), t > 0, find an equation of the osculating plane of the curve r(t) at the point (π², −π, −1).

sol.

r⁰(t) = (2t, −t sin t, t cos t),

||r⁰(t)|| = ((2t)²+ (−t sin t)²+ (t cos t)²)¹² = √

5|t| =

√5t (by t > 0).

T(t) = _||r^r0₀^(t)_(t)|| = ^√1

5(2, − sin t, cos t), T⁰(t) = ^√1

5(0, − cos t, − sin t), ||T⁰(t)|| = ^√1

5, N(t) = _||T^T00^(t)(t)|| = (0, − cos t, − sin t),

B(t) = T(t) × N(t) = ^√1

5(1, 2 sin t, −2 cos t).

So the osculating plane of the curve r(t) at the point (π², −π, −1) has normal vector B(π) = ^{√ 1}

5(1,0,2), so

an equation is 1(x − π²) + 0(y + π) + 2(z + 1) = 0 or x + 2z = π² − 2.

5. Let f (x, y, z) =

( xy+yz³

x²+z⁶ if (x, y, z) 6= (0, 0, 0)

0 if (x, y, z) = (0, 0, 0) , de- termine the set of points at which f is continuous.

sol.

The function f (x, y, z) is continuous for (x, y, z) 6=

(0, 0, 0) since it is equal a rational function there.

For (x, y, z) → (0, 0, 0) along curve r(t) = (kt³, kt³, t), we see that f (x, y, z) = f (kt³, kt³, t) = ^(kt3)2+kt3t3

(kt³)²+t⁶ =

(k²+k) k²+1 .

So f (x, y, z) → ^(k2+k)

k²+1 as (x, y, z) → (0, 0, 0) along r(t) = (kt³, kt³, t).

Therefore, different path leads to different limit val- ues, hence the limit of f (x, y, z) at (0, 0, 0) dose not exist, so f (x, y, z) is not continuous at (0, 0, 0).

Ch15 _Rûb

6. ¹⁶₃ (x³+y²)+xyz²+8z = 0. ^°ÊÞ, (−1, −1, 0)^øõ

5 z_x, z_y, z_xx, z_xy ^{£¬¤õ5~Þ}
(^C_É°z_x, z_xx, z_xy^£

~Þ) sol.

z_x:

16x² + yz² + 2xyzz_x + 8z_x = 0 16 + 8z_x = 0,∴ z^x = −2

z_y:

16

3 2y + xz² + 2xyzz_y + 8z_y = 0

32

3 = 8z_y ∴ zy = ⁴₃

~Þ:

p(x + 1) + q(y + 1) + (−1) · z = 0

−2(x + 1) + ⁴₃(y + 1) − z = 0 z_xx :

32x + 2yzz_x + 2yzz_x + 2xyz_x² + 2xyzz_xx+ 8z_xx = 0 32x + 2xyz_x² + 8z_xx = 0

−32 + 2(4) + 8z_xx = 0

−24 + 8z_xx = 0

∴ z^xx = 3 z_xy:

z² + 2yzz_y + 2xzz_x + 2xyz_yz_x + 2xyzz_xy + 8_xy = 0 2z_yz_x + 8z_xy = 0

∴ z^xy = ^−z₄^xzy = −¹₄(−2)(⁴₃) = ²₃.

7. ^° z_x, z_y, ^~Þ, ^1°(−₁₀⁹ , −¹¹₁₀) ^FÊ5z⁵_,lM

sol.

z = 0 + −2(0.1) + ⁴₃(−0.1)

= −0.2 − ₃₀⁴

= −₁₀² − ₃₀⁴ = −¹⁰₃₀ = −¹₃.

8. Find the absolute maximum and minimum values of

f (x, y) = 4x + 6y − x² − y² in x² + y² ≤ 1.

sol.

At a critical point, f_x = 4 − 2x = 0, f_y = 6 − 2y = 0 =⇒ x = 2, y = 3. So there are no critical points in x²+y² < 1. Need only look at boundary x²+y² = 1.

method 1

Use x = cos θ, y = sin θ g(θ) = 4 cos θ + 6 sin θ − 1

g⁰(θ) = −4 sin θ + 6 cos θ = 0 if tan θ = ³₂

=⇒ sec²θ = 1 + tan²θ = 1 + ⁹₄ = ¹³₄

=⇒ cos²θ = ₁₃⁴ , sin²θ = 1 − cos²θ = ₁₃⁹ . So 2 possibilities, (^√2

13, ^√3

13),(−^√2

13, −^√3

13). The first gives

√26

13 − 1 = 2√

13 − 1 ←− maximum.

The second gives −2√

13 − 1 ←− minimum.

method 2.

Lagrange multipliers:

4 − 2x = 2λx (1)

6 − 2y = 2λy (2)

x² + y² = 1 (3)

(1),(2)=⇒ 2(1 + λ)x = 4, 2(1 + λ)y = 6 =⇒

3(1 + λ)x = 2(1 + λ)y =⇒ x = ²₃y. Then (3)=⇒

4

9y² + y² = 1 =⇒ y² = ₁₃⁹ =⇒ y = ±^√3

13. So get (^√2

13, ^√3

13) or (−^√2

13, −^√3

13). Finish as in method 1.

9. Find the maximum value of f (x, y, z) = xy + xz + yz subject to the constraints x ≥ 0, y ≥ 0, z ≥ 0 and x + y + z = 1.

sol.

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 = 1 must hold for some λ. Then

y + z = x + z =⇒ x = y x + z = x + y =⇒ y = z So x = y = z = ¹₃ =⇒ f (x, y, z) = ³₉ = ¹₃.

The other possibility is that the maximum occurs where one of x, y, z is 0.

Suppose x = 0. Then y + z = 1 and f (0, y, z) = yz = y(1 − y). The maximum of this in 0 ≤ y ≤ 1 is ¹₄. Similarly, if y = 0 or z = 0 we get a maximum of ¹₄. Hence the maximum value must be ¹₃.

10. The plane 2x + y + z = 10 intersects the paraboloid z = x² + y² in an ellipse. Find the highest point on the ellipse.

sol.

We want to maximize f (x, y, z) = z subject to x² + y² − z = 0, 2x + y + z − 10 = 0.

By the method of Lagrange multipliers, we get

0 = λ2x + 2µ (1)

0 = λ2y + µ (2)

1 = −λ + µ (3)

x² + y² − z = 0 (4)

2x + y + z − 10 = 0 (5) (1),(2)=⇒ λx = 2λy =⇒ λ = 0 or x = 2y.

If λ = 0, then (1) =⇒ µ = 0 which contradicts (3).

So x = 2y. Then (4)=⇒ z = x² + y² = 5y².

(5)=⇒ 4y + y + 5y²− 10 = 0 =⇒ y²+ y − 2 = 0 =⇒

(y + 2)(y − 1) = 0 =⇒ y = −2, 1.

y = −2 =⇒ (x, y, z) = (−4, −2, 20) ←− highest point.

y = 1 =⇒ (x, y, z) = (2, 1, 5) ←− lowest point.

11. Show that the ellipsoid 3x² + 2y² + z² = 9 and the sphere x² + y² + z² − 8x − 6y − 8z + 24 = 0 are tangent to each other at he point (1, 1, 2).

sol.

Let f (x, y) = 3x² + 2y² + z² − 9 and g(x) = 3x² + 2y²+ z²− 9. Then f (1, 1, 2) = g(1, 1, 2) = 0 and the ellipsoid and the sphere are the level surfaces of f and g. Thus ∇f (1, 1, 2) and ∇g(1, 1, 2) are orthogonal to the ellipsoid and the sphere (1, 1, 2).

∇f (x, y, z) = h6x, 4y, 2zi and ∇g(x, y, z) = h2x −

8, 2y − 6, 2z − 8i. Hence ∇f (1, 1, 2) = h6, 4, 4i is parallel ∇g(1, 1, 2) = h−6, −4, −4i. This implies that the ellipsoid 3x² + 2y² + z² = 9 and the sphere x²+ y²+ z² − 8x − 6y − 8z + 24 = 0 are tengent to each other at he point (1, 1, 2).

12. If f (x, y) = 0 define y as a function of x, show that d²y

dx² = f_xxf_y² − 2f_xyf_xf_y + f_yyf_x² f_y³

sol.

Differentiate f (x, y) = 0 w.r.t x. We have f_x + f_ydy

dx = 0 (1)

Hence if f_y 6= 0, ^dy_dx = ^−f_f ^x

y . Differentiate (1) w.r.t x.

We have

f_xx + f_xydy

dx + f_yxdy

dx + f_yy(dy

dx)² + f_yd²y dx² = 0 We assume that f_xy and f_yx are both continuous, and hence f_xy = f_yx. Also, from the equation (1), we have

f_xx + f_xy−f_x

f_y + f_yx−f_x

f_y + f_yy(−f_x

f_y )² + f_yd²y dx² = 0 Thus

d²y

dx² = f_xxf_y² − 2f_xyf_xf_y + f_yyf_x² f_y³

13. Let f (x, y) = x²(x²+ y²)⁻¹²e^sin(x2y) if (x, y) 6= (0, 0) and f (0, 0) = 0.

(1) Let −→u = hcos θ, sin θi and find D^−→_uf (0, 0).

(2) Prove that f is continuous at (0, 0).

sol.

D^−→_uf (0, 0) = lim

h→0

f (h cos θ, h sin θ) − f (0, 0) h

= lim

h→0

h(cos θ)²e^{sin(h3cos2θ sin θ)}

h = (cos θ)²

|x²(x² + y²)⁻¹²e^sin(x2y)| ≤ |x|e¹. Since lim

(x,y)→(0,0)

|x| = 0, it follows that lim

(x,y)→(0,0)

f (x, y) = 0 = f (0, 0).

Ch16 ^{½ }}

14. D = {(x, y)|x²+y² ≤ 4}, ^t°RR

D

p4 − x² − y²dA.

sol.

I u = 4 − r²

Ÿ =

Z 2π 0

Z 2 0

p4 − r²rdrdθ

= 1 2

Z 2π 0

Z 4 0

√ududθ

= 1 2

Z ^π₂

0

2 3u³²

4

0

dθ

= 8 3

Z 2π 0

dθ = 16 3 π 15. ^°R1

0

R1

x cos(y²)dydx.

sol.

‰² }ßå,

Ÿ = Z 1

0

Z 4 0

cos y²dxdy

= Z 1

0

y cos y²dy

= sin y² 2

1

0

= sin 1 2 16. ^°R1

0

R1 0

√ xy

x²+y²+1dydx.

sol.

I u = x² + y² + 1

Ÿ = Z 1

0

1 2x

Z 2+x² 1+x²

√du udx

= Z 1

0

x 2

√

2 + x²dx − Z 1

0

x 2

√

1 + x²dx

= 1

3(2 + x²)³²    

1

0

− 1

3(1 + x²)³²    

1

0

= 1 3[3√

3 − 2√

2 − 2√

2 + 1]

= 1 3[3√

3 − 4√

2 + 1]

17. ^t° R1 0

R1

√y ye^x2

x³ dxdy sol.

x = √

y =⇒ y = x²

Ÿ = Z 1

0

Z x² 0

ye^x2

x³ dydx

= Z 1

0

x⁴e^x2 2x³ dx

= 1 2

Z 1 0

xe^x2dx

= 1 4e^x2

1

0

= 1

4(e − 1) 18. (a) ^t° RR

D 1

(x²+y²)ⁿ²dA, n ^Ñcb. D : {(x, y), a ≤ px² + y² ≤ b}

sol.

Ÿ =

Z 2π 0

Z b a

1

rⁿrdrdθ

= 2π Z b

a

1 rⁿ⁻¹dr (b) ^ç a −→ 0⁺, ^t°n_{MU }í”ÌæÊ}

sol.

(^ÿ )

19. ^t°Þ z = p

x² + y^{2 D6Þ} x² + y² = 2x ^£ xy ^

ÞFˇ5ñ

sol.

(x − 1)² + y² = 1, r = 2 cos θ

Ÿ = Z ^π₂

−^π₂

Z 2 cos θ 0

rrdrdθ

= 2 Z ^π₂

0

Z 2 cos θ 0

r²drdθ

20. ρ(x, y) = xy², D = {(r, θ), 0 ≤ r ≤ a, 0 ≤ θ ≤ ^π₂},

t°D^í”-

sol.

(^ÿ )

21. Consider the iterated integralR8 0

R 2 y¹³

√1 + x⁴dx dy.

Sketch the region of integration and then evaluate it by reversing the order of integration.

sol.

R2

0 dxRx³

0 dy√

1 + x⁴ = R2

0 x³√

1 + x⁴dx = ²₃ · ¹₄(1 + x⁴)³²   

2 0

=

1

6{17³² − 1}

22. Consider the iterated integralR2 0

R 4

y² y sin(x²)dx dy.

Sketch the region of integration and then evaluate it by reversing the order of integration.

sol.

R4 0 dxR

√x

0 dyy sin(x²) = R4 0

x

2 sin(x²)dx = ¹₂¹₂(− cos(x²))|⁴₀ =

1

4(1 − cos 16)

23. By making an appropriate change of variables to evaluate the integral RR

R

dA

1+9x²+9y², where R is the region bounded by the ellipse 9x² + 9y² = 4.

sol.

x = r cos θ, y = r sin θ R2π

0 dθR ²₃

0 dr ^r

1+9r² = 2π^ln(1+9r₁₈ ²⁾   

2 3

0 = ^π₉ ln 5