Find the area of the region that 1 ≤ r ≤ 3 + 3 cos θ

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1. (10 %) Find the area of the region that 1 ≤ r ≤ 3 + 3 cos θ.

Sol.

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3 + 3 cos θ ≥ 1 3 cos θ ≥ −2 cos θ ≥ ⁻²₃

∴ θ range from − cos⁻¹(⁻²₃ ) to cos⁻¹(⁻²₃ ), Let θ₁ = cos⁻¹(−2 3 )

∴ area =

Z θ1

−θ1

(3 + 3 cos θ)²− 1

2 dθ

= Z θ1

0

(3 + 3 cos θ)²− 1dθ

= 25

2 θ + 18 sin θ +9 4sin 2θ

θ1

0

= 25

2 θ1+ 18 sin θ1+ 9

4sin(2θ1)

= 5√ 5 + 25

2 cos⁻¹(−2 3 ) here we need to use sin θ₁ =

√5 3

sin(2θ₁) = 2 sin θ₁cos θ₁ = −4 9

√5

2. (10 %) Assume x > 0. Solve the initial value problem x²y⁰+ 3y = ln x with y(1) = 0.

Sol.

y⁰ +³_xy = ^lnx_x2

then v(x) = e^R ³^x = e^{3 ln x} = x³ so

x³y˙⁰ + x³3˙

xy = x ln x we have (y ˙x³)⁰ = x ln xdx (y ˙x³) = R x ln xdx

using integration by parts, so x³y = ¹₂x²ln x −¹₂R xdx = ¹₂x²ln x −^x₄² + C and then, y = _x¹3(¹₂x²ln x −^x₄² + C) = ^{ln x}_2x −_4x¹ +_x^C3

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with initial value y(1) = 0 so C = ¹₄

hence y = _x¹3(¹₂x²ln x −^x₄² + C) = ^{ln x}_2x − _4x¹ + _4x¹3

3. (20 %) Given the ordinary differential equation y⁰ = y²(y²− 1).

(a) Identify the equilibrium values and state those which are stable and which are unstable.

(b) Make a phase line of the differential equation and identify the signs of y⁰ and y⁰⁰ on it.

(c) Sketch several solution curves.

(d) Solve the equation by separation of variables. An equality in y and t is enough.

Sol.

(a) y²(y²− 1) = 0 ⇔ y = −1, 0, 1

y⁰ < 0 on (−1, 1) and y⁰ > 0 on (−∞, −1) and (1, ∞) Hence -1 is stable, and 0,1 are unstable.

(b)

y⁰⁰ = −2yy⁰ + 4y³y⁰ = (4y³− 2y)y⁰ = (4y³− 2y)y²(y²− 1) = 2y³(2y²− 1)(y²− 1)

= 4y³(y − 1

√2)(y + 1

√2)(y − 1)(y + 1) Thus, on (−∞, −1): y⁰ > 0, y⁰⁰ < 0

on (−1, −^√¹

2): y⁰ < 0, y⁰⁰ > 0 on (−^√¹

2, 0): y⁰ < 0, y⁰⁰ < 0 on (0,^√¹

2): y⁰ < 0, y⁰⁰> 0 on (^√¹

2, 1): y⁰ < 0, y⁰⁰< 0 on (1, ∞): y⁰ > 0, y⁰⁰ > 0

(d) y⁰ = y²(y²− 1)

⇒ 1

y²(y²− 1)dy = dt

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⇒ (−1

y² + 1

2(y − 1) + −1

2(y + 1))dy = dt

⇒ Z −1

y² dy + 1 2

Z 1

y − 1dy − 1 2

Z 1

y + 1dy = Z

dt

⇒ 1 y + 1

2ln |y − 1| − 1

2ln |y + 1| = t + C

4. (15 %) At the point P₀(1, 3), a function f (x, y) has a derivative of −√

5 in the direction from P₀(1, 3) toward P (2, 1) and a derivative of√

5 in the direction from P₀(1, 3) toward P (3, 2).

(a) Find the directions in which the function f increases and decreases most rapidly at P₀(1, 3).

Then find the rates of change in these directions. (12%)

(b) Estimate how much the value of f (x, y) will change if the point P (x, y) moves from P₀(1, 3) straightly toward P (0, 0) with 0.05 unit. (3%)

Sol.

(a) Let (∇f (x, y))_P₀ = ∇f (1, 3) = f_x(1, 3)i + f_y(1, 3)j. By the theorem of the directional derivative, we have the formula:

df ds

u,P0

= (∇f )_P₀ · u

Since the unit vector from P₀(1, 3) toward P (2, 1) is ^√¹

5i −^√²

5j and the unit vector from P₀(1, 3) toward P (3, 2) is ^√²

5i −^√¹

5j, we get

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√1

5f_x(1, 3) − ^√²

5f_y(1, 3) = −√ 5

√2

5f_x(1, 3) − ^√¹₅f_y(1, 3) =√ 5

⇒

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f_x(1, 3) = 5 f_y(1, 3) = 5

Thus, the direction in which the function f increases most rapidly at P₀(1, 3) is

∇f (1, 3)

|∇f (1, 3)| = 1

√2i + 1

√2j,

and the rate of change in this direction is |∇f |_P₀ =√

5²+ 5² = 5√ 2.

The direction in which the function f decreases most rapidly at P₀(1, 3) is

− ∇f (1, 3)

|∇f (1, 3)| = − 1

√2i − 1

√2j,

and the rate of change in this direction is −|∇f |_P₀ = −√

5²+ 5² = −5√ 2.

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(b) Since the unit vector from P₀(1, 3) straightly toward P (0, 0) is u = −^√¹

10i −^√³

10j, we get

df = (∇f |_P₀ · u)(ds) = (5i + 5j) ·

− 1

√10i − 3

√10j

(0.05) = − 1

√10.

5. (15 %) Let w = f (x, y), x = r cos θ, and y = r sin θ. Express ∂w

∂r and ∂²w

∂r² in terms of r, θ, and partial derivatives of f (x, y) with respect to x and y.

Sol.

∂w

∂r = ∂f

∂x

∂r +∂f

∂y

∂r = f_xcos θ + f_ysin θ

∂²w

∂²r = ∂

∂r(f_xcos θ + f_ysin θ) = ( ∂

∂r(f_x) cos θ + ∂

∂r(f_y) sin θ)

= (fxxsin θ + fxycos θ) sin θ + (fyxcos θ + fyysin θ) sin θ

= f_xxcos²θ + (f_xy+ f_yx) sin θ cos θ + f_yycos²θ (or = fxxcos²θ + 2fxysin θ cos θ + fyycos²θ)

6. (10 %) Assume f (x, y) = 3xy²− x³−6

5y⁵. Find all the local maximum, local minimum and saddle points of f . Please still determine the types of critical points even if the second derivative test is inconclusive.

Sol.

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f_x = 3y²− 3x² = 0 f_y = 6xy − 6y⁴ = 0

⇒

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x = ±y

x = y³ or y = 0

⇒ (x, y) = (0, 0), (−1, −1), (1, 1)

Second derivative test:

fxx = −6x, fyy = 6x − 24y³, fxy = 6y = fyx

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H_f =

f_xx f_xy f_yx f_yy

= f_xxf_yy− f_xy² = −36(x²− 4xy³ + y²)

(0, 0) : H_f(0, 0) = 0 ⇒ inconclusive,

consider f (x, 0) = −x³ ⇒ f (x, 0) > 0 when x < 0, f (x, 0) < 0 when x > 0

⇒ (0, 0) is a saddle point.

(−1, −1) : H_f(−1, −1) = 72 > 0 and f_xx(−1, −1) = 6 > 0

⇒ (−1, −1) is a local minimum point.

(1, 1) : Hf(1, 1) = 72 > 0 and fxx(1, 1) = −6 < 0

⇒ (1, 1) is a local maximum point.

7. (20 %) Suppose Γ is the intersection curve of x + y + 2z = 2 and z² = 2x²+ 2y². (a) Find the tangent line of Γ at (1

2,−1 2 , 1).

(b) Find all the points on Γ that lie closest and farthest from the origin.

Sol.

(a) Let f (x, y, z) = x + y + 2z − 2, g(x, y, z) = 2x²+ 2y²− z², P = (¹₂, −¹₂, 1)

∇f = (1, 1, 2), ∇f |_P = (1, 1, 2)

∇g = (4x, 4y, −2z), ∇g|_P = (2, −2, −2) The direction of tangent vector of Γ is

∇f |_P × ∇g|_P = (1, 1, 2) × (2, −2, −2) = (2, 6, −4) The tangent line of Γ at (¹₂,⁻¹₂ , 1) is

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x = ¹₂ + 2t y = −¹₂ + 6t z = 1 − 4t

t ∈ R

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(b)

d(x, y, z) = x²+ y²+ z², ∇d = λ∇f + s∇g and f = 0, g = 0 ⇒

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2x = λ + 4sx 2y = λ + 4sy 2z = 2λ − 2sz f = 0

g = 0

⇒

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(2 − 4s)x = λ (2 − 4s)y = λ (2 + 2s)z = λ 2x²+ 2y²− z² = 0 x + y + 2z − 2 = 0

Case1. (2 − 4s) = 0 ⇒ λ = 0 ⇒ z = 0 ⇒

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2x²+ 2y² = 0 x + y = 2

⇒ x = y = 0 and x + y = 2, no solution

Case2. (2 − 4s) 6= 0 ⇒ x = y ⇒

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4x² = z² 2x + 2z = 2

⇒ z = ±2x and 2x + 2z = 2 If z = 2x ⇒ (x, y, z) = (¹₃,¹₃,²₃)

If z = −2x ⇒ (x, y, z) = (−1, −1, 2) At (x, y, z) = (¹₃,¹₃,²₃) ⇒ d = ²₃ At (x, y, z) = (−1, −1, 2) ⇒ d = 6

So (¹₃,¹₃,²₃) is the closest point from (0, 0, 0) So (−1, −1, 2) is the farthest point from (0, 0, 0)