94ç-ç‚ }Bù ‚25

94ç-ç‚ }Bù ‚25

(14%) 1. Given curve r = 1 + cos θ.

(a) Find the length of the curve.

(b) Find the area of the region that is bounded by the curve.

Solution: (a)

¶Å S = 2 Z π

0

s

r²+ dr dθ

2

dθ

= 2 Z π

0

p(1 + cos θ)²+ (− sin θ)²dθ

= 2 Z π

0

√2 + 2 cos θ dθ

= 4 Z π

0

r 1 + cos θ

2 dθ

= 4 Z π

0

cosθ 2    

dθ

= 4 · 2 · sin θ 2    

π

0

= 8 (b)

2 Z π

0

1 2r²dθ

= Z π

0

(1 + cos θ)²dθ

= Z π

0

(1 + 2 cos θ + 1 2+ 1

2cos 2θ) dθ

= 3

2π (12%) 2. (a) Find lim

(x,y)→(0,0)

2yx³ y²+ x⁶. (b) Find lim

(x,y)→(0,0)

xy px²+ y². Solution:

(a) Let f (x, y) = 2yx³ y²+ x⁶.

x→0limf(x, 0) = lim

x→00 = 0

x→0lim

f(x, x³) = lim

x→0

2x⁶

x⁶+ x⁶ = lim

x→01 = 1 Since 0 6= 1, limit does not exist.

1

(b) Let f (x, y) = xy

px²+ y², x= r cos θ, y = r sin θ.

lim

r→0f(r cos θ, r sin θ) = lim

r→0

r²cos θ sin θ

r = lim

r→0

rcos θ sin θ = 0 So

(x,y)→(0,0)lim

xy

px²+ y² = 0.

(12%) 3. For f (x, y, z) = ln(x²+ y²) + z.

(a) Find ∇f.

(b) Consider the cylinder of radius 5 with axis along the z-axis. Find the normal vector to the cylinder at the point (3, −4, 4).

(c) Find the rate of change of f in the direction normal to the cylinder at the point (3, −4, 4).

Solution:

∇f =< 2x

x²+ y², 2y

x²+ y²,1 > .

The cylinder is x² + y² = 25. So its normal vector is < 2x, 2y, 0 > which equals < 6, −8, 0 > at (3, −4, 4). A unit vector in this direction is

~u=< 3 5,−4

5,0 > . The rate of change of f in this direction is

∇f · ~u =< 6 5,−8

5,1 > · < 3 5,−4

5,0 >= 46 15.

(10%) 4. Find the parametric equation of the tangent line to the curve of intersection of the surfaces x²+ 2y²+ z² = 4 and x²+ y²− z² = 1 at the point (1, 1, 1).

Solution:

f = x²+ 2y²+ z²− 4, g = x² + y²− z² − 1, P = (1, 1, 1).

∇f =< 2x, 4y, 2z >, ∇g =< 2x, 2y, −2z >.

∇f(P ) =< 2, 4, 2 >, ∇g(P ) =< 2, 2, −2 >.

∇f(P ) × ∇g(P ) =      

i j k

2 4 2

2 2 −2      

=< −12, 8, −4 > .

~(Ñ







x= 1 − 12t y = 1 + 8t z = 1 − 4t

C







x= 1 + 3t y= 1 − 2t z = 1 + t

.

(18%) 5. (a) Suppose fx(0, 0) = 2, fy(0, 0) = 3. Suppose v = g(u) satisfies f (u, v²+v) = 0 and g(0) = 0. Find g⁰(0).

(b) Find all points at which the direction of fastest increase of the function f(x, y) = x²+ y²− 2x − 4y is 2i + j.

2

Solution:

(a) Let x = u, y = v² + v.

df du = ∂f

∂x

∂x

∂u + ∂f

∂y

∂y

∂v dv du. f(u, v²+ v) = 0

f_x(u, v²+ v) · 1 + f^y(u, v²+ v) · (2v + 1) · g⁰(u) = 0.

u= 0, v = 0 Hp

f_x(0, 0) + fy(0, 0) · 1 · g⁰(0) = 0.

2 + 3 · g⁰(0) = 0.

g⁰(0) = −2 3 .

(b) ∇f =< 2x − 2, 2y − 4 >= k < 2, 1 >, k > 0.

2x − 2 = 2(2y − 4)

⇒ x − 2y + 3 = 0, x > 1.

(12%) 6. Find the absolute minimum and maximum values of f (x, y) = xy² on the curve x²+ 7xy + y² = 45, x ≥ 0, y ≥ 0.

Solution:

Let g = x²+ 7xy + y²− 45.

∇g =< 2x + 7y, 7x + 2y >

∇f =< y²,2xy >

 y² = λ(2x + 7y) − −(1) 2xy = λ(7x + 2y) − −(2) (1) × (7x + 2y) − (2) × (2x + 7y) y(2y²− 7xy − 4x²) = 0

⇒ y = 0, y = 4x, y = −^x₂. y= −^x2 .¯.

Hp g

y= 0 ⇒ x = 3√ 5.

y= 4x ⇒ x = 1, y = 4.

f(3√

5, 0) = 0 |üM.

f(1, 4) = 16 |×M.

(10%) 7. Find the volume of the solid in the 1st octant bounded by the surface z = 9 − y² and the plane x = 2.

Solution:

Z 2 0

Z 3

0 (9 − y²) dy dx = Z 2

0 9y −y³ 3

3 0

dx= Z 2

0

18 dx = 36.

(12%) 8. Evaluate Z 1

0

Z ^√y y

sin x

x dx dy.

3

Solution:

Z 1 0

Z ^√y y

sin x

x dx dy = Z 1

0

Z x x²

sin x

x dy dx= Z 1

0 (x − x²)sin x x dx

= Z 1

0 sin x − x sin x dx = [− cos x + x cos x − sin x]¹0

= − cos 1 + cos 1 − sin 1 − (−1) = 1 − sin 1 where

Z

xsin x dx = −x cos x − Z

− cos x dx = −x cos x + sin x + C.

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