Part I: (5 points for each problem) Multiple Choice - Single Answer

Calculus II Quiz 3 June 4, 2009

Name: Student ID number:

Part I: (5 points for each problem)

Multiple Choice - Single Answer ( 選擇題- 單選題).

1. B Determine the set of points where the function is continuous, f (x, y) =√ x²− y²

(A) (B) (C)

2. C Determine the function that has the contour plot below.

(a) f (x, y) = x²+ y² (b) f (x, y) = e^x2+y2,

(c) f (x, y) = xy (d) f (x, y) = x + y

Fill-In Problems( 填充)

3. (5 pts) Suppose that g(t) = f (x(t), y(t)), where f is a differentiable function of x and y and where x = x(t) and y = y(t) both have first-order derivatives.

Given that

x⁰(1) = 4, y⁰(1) = 5 x(1) = 2, y(1) = 3, f_x(4, 5) = 7, f_y(4, 5) = 8, f_x(2, 3) = 7, f_y(2, 3) = 8, f (2, 3) = 5, f (4, 5) = 6.

g⁰(1) = 68

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Part II: Problem-Solving Problems ( 計算題 Show all work)

4. (15 pts) Let f (x, y) = e^xsin y with x(t) = t² and y(t) = e^t. Find the derivative of g(t) = f (x(t), y(t)) with respect to t.

g⁰(t) = f_x(x(t), y(t))x⁰(t) + f_y(x(t), y(t))y⁰(t)

= e^xsin y· 2t + e^xcos y· e^t

= 2te^t2sin (e^t) + e^te^t2cos (e^t)

5. (15 pts) Given f (x, y) = −2x²+ y³− 3y, find all critical points and determine their types (local maximum, local minimum or saddle).

f_x =−4x = 0 ⇒ x = 0, f_y = 3y²− 3 = 0 ⇒ y = ±1, Critical Points: (0,±1)

f_xx =−4, fyy = 6y, f_xy = f_yx= 0

At (0, 1), D = (−4)(6) − 0 = −24 < 0 ⇒ Saddle point.

At (0,−1), D = (−4)(−6) − 0 = 24 > 0, fxx(0,−1) = −4 < 0 ⇒ local maximum.

6. (20 pts) Let f (x, y) = x²⁰⁰⁹+ y² and ~u =< ^√₂³,¹₂ >.

(a) Compute ∇f(x, y) and the directional derivative of f at (0, 2) in the di- rection of ~u.

∇f(x, y) =< 2009x²⁰⁰⁸, 2y > .∇f(0, 2) =< 0, 4 >

|~u| =

√√3 2

2

+ 1 2

2

= 1

D_~u(0, 2) =∇f(0, 2) · ~u = 0 ·

√3

2 + 4·1 2 = 2.

(b) Find the direction of maximum and minimum change of the given function at the point (0, 2).

Maximum change in the direction of ∇f(0, 2) =< 0, 4 > ⇒ < 0, 1 >

Minimum change in the direction of−∇f(0, 2) =< 0, 4 > ⇒ < 0, −1 >

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7. (15 pts) Find the equation of the tangent plane to the surface at the given point.

z = f (x, y) = 2x³+ y² at (1, 2, 6)

f_x(1, 2) = 6(1)² = 6, f_y(1, ) = 2(2) = 4 Equation of the tangent plane is:

z = 6 + 6(x− 1) + 4(y − 2)

8. (a) (10 pts) Use the Method of Lagrange Multiplier to find the absolute maximum and minimum values of the function f (x, y) = x²+ y²− 6y + 3 subject to the constraint x² + y² = 16.

Let g(x, y) = x²+ y²− 16 = 0.

{ ∇f = λ∇g

g(x, y) = 0 ⇒





2x = λ(2x)

2y− 6 = λ(2y) x²+ y²− 16 = 0 ⇒





x(1− λ) = 0 2y(1− λ) = 6 x²+ y²− 16 = 0 x(1− λ) = 0 ⇒ x = 0 or λ = 1. Since plug λ = 1 into 2y(1 − λ) = 6 ⇒ contradiction, x = 0.

Plug x = 0 into x²+ y²− 16 = 0 ⇒ y = ±4.

Since f (0, 4) = −5 and f(0, −4) = 43, f attains absolute maximum at (0,−4) and absolute minimum at (0, 4).

(b) (10 pts) Find the absolute maximum and minimum values of the function f (x, y) = x²+ y²− 6y + 3 subject to the constraint x²+ y² ≤ 16.

i. Critical points in the region:

∇f = 0 ⇒

{ 2x = 0

2y− 6 = 0 ⇒ (0, 3).

Since 0²+ 3² = 9 < 16, (0, 3) is in the region.

ii. Candidates on the boundary:

By 8a, candidates on the boundary are (0, 4) and (0,−4).

iii. Since f (0, 3) =−6, f(0, 4) = −5 and f(0, −4) = 43, f attains absolute maximum at (0,−4) and absolute minimum at (0, 3).

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