1092 ®Y01-03í -ãTŒU– 1. (15 pts) Let f(x, y, z)= sin(xy+ z), and P be the point(0,−2, π 3

1092 ®Y01-03í -ãTŒU–

1. (15 pts) Let f(x, y, z) = sin(xy + z), and P be the point (0, −2,^π₃).

(a) (6 pts) Compute ∇f(x, y, z).

(b) (2 pts) At P , find the direction along which f obtains maximum directional derivative.

(c) (4 pts) Calculate the directional derivative ^∂f_∂u(P), where u is a unit vector making an angle

π

6 with the gradient ∇f(P).

(d) (3 pts) The level surface f(x, y, z) = ^√2³ defines z implicitly as a function of x and y near P . Compute _∂x^∂z at P .

Solution:

(a)

∇f = (y cos(xy + z), x cos(xy + z), cos(xy + z))

= cos(xy + z)(y, x, 1)(each component 2%) (b) ∇f(0, −2,^π3) = ¹2(−2, 0, 1) (each component ²₃%)

(c)

∂f

∂⃗u(P) = ∇f(P) ⋅ u(2%)= ∣∇f(P)∣ ⋅ ∣u∣ ⋅ cosπ

6(1%)=

√5 2

√3 2 =

√15 4 (1%).

(d)

sin(xy + z(x, y)) =

√3

2 ⇒ cos(xy + z(x, y))(y + z^x) = 0(2%)

⇒ cos(π

3)(−2 + z^x) = 0 ⇒ z^x = 2(1%)

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2. (12 pts) Assume that f(x, y, z) and g(x, y, z) have continuous partial derivatives and (1, 2, −1) lies on the level surface f(x, y, z) = 3. Suppose the tangent plane of f(x, y, z) = 3 at (1, 2, −1) is 2x− y + 3z + 3 = 0 and fy(1, 2, −1) = 2.

(a) (4 pts) Find ∇f(1, 2, −1).

(b) (4 pts) Estimate f(1.1, 2.01, −0.98) by the linear approximation of f at (1, 2, −1).

(c) (4 pts) Suppose that when restricted on the surface f(x, y, z) = 3, g(x, y, z) obtains maximum value at (1, 2, −1) and g^x(1, 2, −1) = −2. Find ∇g(1, 2, −1) and the maximum directional derivative of g at the point (1, 2, −1).

Solution:

(a) The tangent plane of f(x, y, z) = 3 at (1, 2, −1) is

f_x(1, 2, −1)(x − 1) + f^y(1, 2, −1)(y − 2) + f^z(1, 2, −1)(z + 1) = 0. (2 pts) Since the tangent plane is 2x− y + 3z + 3 = 0, ∇f(1, 2, −1) (2, −1, 3). (1 pt)

Because f_y(1, 2, −1) = 2, we know that ∇f(1, 2, −1) = −2(2, −1, 3) = (−4, 2, −6). (1 pt).

Ans: ∇f(1, 2, −1) = (−4, 2, −6).

(b) The linear approximation of f at (1, 2, −1) is

L(x, y, z) = f(1, 2, −1) + f^x(1, 2, −1)(x − 1) + f^y(1, 2, −1)(y − 2) + f^z(1, 2, −1)(z + 1) (2 pts for the definition of L(x, y, z).)

f(1.1, 2.01, −0.98) ≈ L(1.1, 2.01, −0.98)

= 3 − 4(0.1) + 2(0.01) − 6(0.02) (1 pt for plugging in correct partial derivatives and x, y, z)

= 2.5 (1 pt for computation)

(c) By the method of Lagrange multiplies, we know that ∇g(1, 2, −1) = λ∇f(1, 2, −1) (1 pt) Hence ∇g(1, 2, −1) = λ(−4, 2, −6).

By g_x(1, 2, −1) = −2, we know that ∇g = ¹2(−4, 2, −6) = (−2, 1, −3) (1 pt).

The maximum directional derivative of g at (1, 2, −1) is ∣∇g(1, 2, −1)∣ =√ 14

(1 pt for knowing the maximum directional derivative is ∣∇g∣. 1 pt for the final answer.)

3. (25 pts) f(x, y) = x²+ xy + y²+ 3x.

(a) (7 pts) Find critical point(s) of f(x, y) and determine whether it is a saddle point or f(x, y) obtains local maximum or local minimum at it.

(b) (15 pts) Find the maximum and minimum value of f(x, y) on the curve x² + y² = 9 by the method of Lagrange multiplies.

(c) (3 pts) Find the maximum value of f(x, y) on the region x²+ y²≤ 9.

Solution:

(a) To find critical points of f(x, y), we solve ⎧⎪⎪

⎨⎪⎪⎩

f_x = 2x + y + 3 = 0 f_y = x + 2y = 0 (1 pt for setting f_x = f^y = 0. 1 pt for correct f^x and f_y.) The solution is (x, y) = (−2, 1). (1 pt)

At (−2, 1), f^xx = 2, f^xy = 1, f^yy= 2 (1 pt) D(−2, 1) = ∣f_xx f_xy

f_xy f_yy∣ (−2, 1) = ∣2 1

1 2∣ =3 (1 pt)

∵ D(−2, 1) = 3 > 0 and fxx(−2, 1) = 2 > 0 (1 pt) We know that f(−2, 1) is a local minimum. (1 pt)

(b) By the method of Lagrange multiplies, to find extreme values of f(x, y) on the constraint g(x, y) = x²+ y²= 9, we solve ⎧⎪⎪

⎨⎪⎪⎩

∇f = λ∇g

g(x, y) = 9 (2 pts)

which is ⎧⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎩

2x+ y + 3 = 2λx (1) x+ 2y = 2λy (2) x²+ y²= 9 (3)

(2 pts) Method 1:

(2) ⇒ x = 2(λ − 1)y

plug in (1) ⇒ (1 − 4(λ − 1)²)y = −3 ⇒ y = 4(λ−1)^{3 2}−1, x= 4(λ−1)^6(λ−1)2−1. (3 pts) plug in (3) ⇒ (4(λ−1)^{9+36(λ−1)2}−1)²² = 9

Let (λ − 1)²= u, we have 1 + 4u = (4u − 1)² ⇒ u = 0 or ³4 i.e. λ= 1 or 1 ±^√2³ (3 pts) when λ= 1, y = −3, x = 0

when λ= 1 ±^√2³, y= ³2, x= ±^3√2³

Critical points are (x, y) = (0, −3), (^3√2³,³₂), and (−^3√2³,³₂). (3 pts) f(0, −3) = 9, f(^3√2³,³₂) = 9 +²⁷4^√3, f(−^3√2³,³₂) = 9 −²⁷4^√3.

f(−3√ 3 2 ,3

2) < f(0, −3) < f(3√ 3 2 ,3

2).

Hence the maximum value is f(^3√2³,³₂) = 9 +²⁷4^√3. The minimum value is f(−^3√2³,³₂) = 9 − ²⁷4^√3. (2 pts) Method 2:

Note that if λ= 0, then (1) and (2) ⇒ (x, y) = (2, −1) but (3) is not satisfied. If y = 0, then (2) implies that x is also 0, but (3) is not satisfied for (x, y) = (0, 0).

Hence we conclude that λ≠ 0 and y ≠ 0.

Therefore both sides of (2) are not zero.

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Then we can divide ⁽¹⁾₍₂₎. (1)(2) ⇒

2x+ y + 3 x+ 2y =

x

y ⇒ 2xy + y²+ 3y = x²+ 2xy

⇒ x² = y²+ 3y plug in (3) ⇒ 2y²+ 3y = 9

⇒ y = 3

2 or − 3.

If y= ³2, x= ±^3√2³. If y= −3, x = 0.

Hence critical points are (x, y) = (0, −3), (^3√2³,³₂) and (−^3√2³,³₂).

(c) Inside the disc x²+ y²< 9, f(x, y, ) has a critical point (−2, 1). f(−2, 1) = −3.

On the boundary x²+ y²= 9, f(x, y) obtains maximum f(^3√2³,³₂) = 9 +^27√4³. because f(^3√2³,³₂) > f(−2, 1) (or ∵ f(−2, 1) is a local minimum) (2 pts)

∴ The maximum value of f on the region x²+ y²≤ 9 is f(^3√2³,³₂) = 9 +²⁷4^√3. (1 pt)

4. (18 pts) (a) (8 pts) Reverse the order of integration and evaluate it. ∫₀⁴∫^√1y

2

√x³+ 3 dx dy.

(b) (10 pts) Compute ∬

Ω

(ln y)⁻¹dA, where Ω is bounded by y= e^x and y= e^√x.

1

Solution:

(a)

∫₀⁴∫^√1_y/2√

x³+ 3 dx dy = ∫₀¹∫ ^4x

2

0

√x³+ 3 dy dx(4%)

= ∫₀¹4x²√

x³+ 3 dx(2%)

= 4 3⋅2

3(x³+ 3)^3/2∣

1

0

(1%)

= 8

9(8 − 3√

3)(1%)

= 64 9 −8

3

√3

(b)

y= e^x ⇒ x = ln y, y= e^√x⇒√

x= ln y ⇒ x = (ln y)². Then

∫ ∫_Ω(ln y)⁻¹dA= ∫₁^e∫_{(ln y)}^{ln y}2(ln y)⁻¹dx dy(4%)

= ∫₁^e(ln y)⁻¹(ln y − ln²y) dy(3%)

= ∫₁^e(1 − ln y) dy(2%)

= (e − 1) − (y ln y − y)∣^e₁(1%)

= e − 2.

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5. (18 pts) (a) (8 pts) Evaluate∬

D

e^−x2−y2dA, where D is the upper disc, x²+ y² ≤ 25 and y ≥ 0.

(b) (10 pts) Calculate the area of the region inside the cardioid r= 1 − sin θ.

-1.5

-1.5 -1-1 -0.5-0.5 0.50.5 11 1.51.5

-2 -2 -1.5 -1.5 -1 -1 -0.5 -0.5 0.5 0.5

0 0

Solution:

(a) Note that

D= {[r, θ] ∶ 0 ≤ r ≤ 5, 0 ≤ θ ≤ π} (2%), we have

∬_De^−(x2+y2)dA= ∫₀^π∫₀⁵e^−r2⋅ r dr dθ (4%)

= ∫₀^π(−e^−r2

2 ) ∣⁵₀ dθ (1%)

= π(1 − e⁻²⁵)/2 (1%).

(b)

Area= ∬_Ω1 dA= ∫₀^2π∫₀^{1−sin θ} r dr dθ (4%)

= ∫₀^2π(1 − sin θ)²

2 dθ (1%)

= ∫₀^2π1− 2 sin θ + sin²θ

2 dθ (1%)

= ∫₀^2π(1

2− sin θ +1− cos 2θ

4 ) dθ (2%)

= 3

4⋅ 2π = 3π

2 (2%).

6. (12 pts) Evaluate ∬

D

e^xydxdy, where D is bounded by curves xy = 10, xy = 20, x²y = 20 and x²y= 40.

Solution:

Method 1: Let u= xy and v = x²y. Then x= v/u and y = u²/v (2%). Further, (x, y) maps Ω= {(u, v) ∶ 10 ≤ u ≤ 20, 20 ≤ v ≤ 40}

to D (2%). Since

J(u, v) = ∣−v/u² 2u/v

1/u −u²/v²∣ = −1/v (2%), we have

∬_De^xydx dy= ∫₁₀²⁰∫₂₀⁴⁰e^u(∣ − v⁻¹∣) du dv (4%)

= ∫₁₀²⁰e^udu⋅ ∫₂₀⁴⁰v⁻¹dv

= (e²⁰− e¹⁰) ln 2 (2%).

Method 2 Let u= xy and v = x²y (2%). Then(u, v) maps D to Ω= {(u, v) ∶ 10 ≤ u ≤ 20, 20 ≤ v ≤ 40} (2%).

Since

J(x, y) = ∣y 2xy

x x² ∣ = −x²y= −v (1%), we have

J(u, v) = J(x, y)⁻¹= −v⁻¹ (1%).

We have,

∬_De^xydx dy= ∫₁₀²⁰∫₂₀⁴⁰e^u(∣ − v⁻¹∣) du dv (4%)

= ∫₁₀²⁰e^udu⋅ ∫₂₀⁴⁰v⁻¹dv

= (e²⁰− e¹⁰) ln 2 (2%).

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