(b) (6 pts) Find the maximum and minimum value of the curvature

1072微微微甲甲甲及及及模模模組組組班班班期期期中中中考考考補補補考考考解解解答答答和和和評評評分分分標標標準準準 1. (11 pts) Consider a space curve ⃗r(t) = (sin t,√

3 sin t,− cos t + 1).

(a) (5 pts) Find the unit tangent vector ⃗T(t) and the unit normal vector ⃗N(t).

(b) (6 pts) Find the maximum and minimum value of the curvature.

Solution:

(a) ⃗T(t) = ⃗r^′(t)

∣⃗r^′(t)∣, ⃗N(t) = T⃗^′(t)

∣ ⃗T^′(t)∣. (1 pt.)

⃗r^′(t) = (cos t,√

3 cos t, sin t) (0.5 pt.) and ∣⃗r^′(t)∣ =√

1+ 3 cos²t (0.5 pt.) Hence,

(3 pts.)

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎨⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎩

T⃗(t) = 1

∣⃗r^′(t)∣⃗r^′(t) = 1

√1+ 3 cos²t(cos t,√

3 cos t, sin t), T⃗^′(t) = 1

(1 + 3 cos²t)³²(− sin t, −√

3 sin t, 4 cos t),

∣ ⃗T^′(t)∣ = 2 1+ 3 cos²t,

N⃗(t) = 1

2√

1+ 3 cos²t(− sin t, −√

3 sin t, 4 cos t).

(b) κ(t) = ∣⃗r^′(t) × ⃗r^′′(t)∣

∣⃗r^′(t)∣³ (1 pt.)

(2 pts.)

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎨⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎩

⃗r^′× ⃗r^′′(t) = RRRRR RRRRR RRRRR

⃗i ⃗j ⃗k

cos t √

3 cos t sin t

− sin t −√

3 sin t cos t RRRRR RRRRR RRRRR= (√

3,−1, 0),

∣⃗r^′ × ⃗r^′′(t)∣ = 2,

κ(t) = 2

(1 + 3 cos²t)^3/2.

∵ 0 ≤ cos²t≤ 1 ∴ 1 ≤ (1 + 3 cos²t)³² ≤ 8.

Hence 1

4 ≤ κ(t) ≤ 2,

⎧⎪⎪⎪⎨⎪⎪

⎪⎩

κ(t) = 2 when t = π

2 + nπ, for all n ∈ N.

κ(t) = 1

4 when t= nπ, for all n ∈ N.

Therefore, the maximum value of the curvature is 2. The minimum value of the curvature is 1

4.

(3 pts.) Solution 2: κ(t) = ∣ ⃗T^′(t)∣

∣⃗r^′(t)∣ (1 pt.), then κ(t) = 2

(1 + 3 cos²t)^3/2 (2 pts.).

Page 1 of 10

2. (12 pts) Let f(x, y) =⎧⎪⎪⎪

⎨⎪⎪⎪⎩

sin(x²y)

x⁴+ y² , if (x, y) ≠ (0, 0) 0, if (x, y) = (0, 0) .

(a) (6 pts) Compute f_x and f_y for all (x, y) including (0, 0).

(b) (6 pts) Is f(x, y) continuous at (0, 0)? Is f(x, y) differentiable at (0, 0)? Justify your answers.

Solution:

(a) For (x, y) ≠ (0, 0),

f_x(x, y) = 2xy cos(x²y)

x⁴+ y² −4x³sin(x²y)

(x⁴+ y²)² (2 pts.) fy(x, y) = x²cos(x²y)

x⁴+ y² −2y sin(x²y)

(x⁴+ y²)² (2 pts.) f_x(0, 0) = lim

x→0

f(x, 0) − f(0, 0)

x = lim

x→0

0− 0

x = 0 (1 pt.) fy(0, 0) = lim

y→0

f(0, y) − f(0, 0)

y = lim

y→0

0− 0

y = 0 (1 pt.)

(b) f(x, 0) = sin 0

x⁴ = 0 for all x ≠ 0.

Hence f(x, y) → 0 as (x, y) → (0, 0) along the x-axis.

However, on the curve y= x², f(x, x²) = sin(x⁴) x⁴+ x⁴ = 1

2

sin(x⁴) x⁴ → 1

2 as x→ 0.

Hence, f(x, y) → 1

2 as(x, y) → (0, 0) along the curve y = x².

Because f(x, y) approaches different limits as (x, y) approaches (0, 0) along different paths,

(x,y)→(0,0)lim f(x, y) doesn’t exist.

Then f(x, y) is not continuous at (0, 0).

(4 pts.) Because differentibility implies continuity, we conclude that f is not differentiable at(0, 0).

(2 pts.)

3. (12 pts) Let f(x, y) = xg(y

x), where g is a differentiable function with g(1) = −1, g^′(1) = 2.

(a) (4 pts) Use linear approximation to estimate the value of f(2.01, 1.98).

(b) (4 pts) Suppose that at (x, y) = (2, 2), g(y

x) decreases most rapidly in the direction ⃗u, where

∣⃗u∣ = 1. Find D⃗uf(2, 2).

(c) (4 pts) If near the point (2, 2, −2), the surface z = f(x, y) defines x implicitly as a function of y and z, x= h(y, z). Find ∂x

∂y and ∂x

∂z when(y, z) = (2, −2).

Solution:

(a) The linear approximation of f(x, y) at 2, 2 is

f(2, 2) + fx(2, 2)(x − 2) + fy(2, 2)(y − 2) (1 pt.) f(2, 2) = 2g (2

2) = 2 ⋅ g(1) = −2 f_x(x, y) = g (y

x) − x ⋅ y x²g^′(y

x) = g (y x) − y

xg^′(y x) f_x(2, 2) = g(1) − g^′(1) = −3

fy(x, y) = g^′(y

x) , fy(2, 2) = g^′(1) = 2

(2 pts.) Hence

f(2.01, 1.98) ≈ f(2, 2) + fx(2, 2)(2.01 − 2) + fy(2, 2)(1.98 − 2)

= −2 − 3 × 0.01 + 2 × (−0.02) = −2.07

(1 pt.) (b) Let h(x, y) = g (y

x), ⃗u = − ⃗∇h

∣⃗∇h∣(2, 2)

⃗∇h(x,y) = (− y x²g^′(y

x) ,1 xg^′(y

x)) // (−y, x) Hence ⃗u = − ⃗∇h(2,2)

∣⃗∇h(2, 2)∣ = ( 1

√2,− 1

√2).

(2 pts.) Because y

x is differentiable at (x, y) = (2, 2) and g is differentiable, we know that f(x, y) = x⋅ g (y

x) is differentiable at (x, y) = (2, 2). Therefore, D_⃗uf(2, 2)(1 pt.)

= ⃗∇f(2,2) ⋅ ⃗u = (−3,2) ⋅ ( 1

√2,− 1

√2)(1 pt.)

= − 5

√2.

(c) z = x ⋅ g (y

x) ⇔ F (x, y, z) = 0 where F (x, y, z) = x ⋅ g (y x) − z.

Near the point (2, 2, −2), the level surface F (x, y, z) = 0 defines x implicitly as a function of y and z.

And ∂x

∂y = −F_y F_x, ∂x

∂z = −F_z

F_x. (2 pts.)

Page 3 of 10

At (2, 2, −2),

F_x(2, 2, −2) = fx(2, 2) = −3 F_y(2, 2, −2) = fy(2, 2) = 2 F_z(2, 2, −2) = −1

Hence

∂x

∂y = 2 3, ∂x

∂z = −1

3. (2 pts.)

4. (12 pts) Suppose that (√ 2,√

2) is a critical point of f(x, y) = x³+ αx²y+ y³+ βy, where α, β are constants.

(a) (2 pts) Find values of α and β.

(b) (10 pts) Find and classify all critical points of f(x, y).

Solution:

(a) (√ 2,√

2) is a critical point of f(x, y) (f is differentiable), then f_x(√

2,√

2) = 0, fy(√ 2,√

2) = 0. (1 pt.)

⎧⎪⎪⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎩ f_x(√

2,√

2) = 3x²+ 2αxy∣

(x,y)=(√ 2,√

2)

= 6 + 4α = 0

f_y(√ 2,√

2) = αx²+ 3y²+ β∣

(x,y)=(√ 2,√

2)

= 6 + 2α + β = 0 Hence

α= −3

2, β= −3. (1 pt.) (b) f(x, y) = x³−3

2x²y+ y³− 3y.

Solve

⎧⎪⎪⎨⎪⎪

⎩

f_x(x, y) = 3x²− 3xy = 0 f_y(x, y) = −3

2x²+ 3y²− 3 = 0 ⇒ { x(x − y) = 0 x²− 2y²+ 2 = 0 Hence

(x, y) = (0, ±1) or (x, y) = (√ 2,√

2), (−√ 2,−√

2). (2 pts.) D(x, y) = ∣ f_xx f_xy

f_xy f_yy ∣ = ∣ 6x− 3y −3x

−3x 6y ∣ D(0, 1) = ∣ −3 0

0 6 ∣ <0 ⇒ (0, 1) is a saddle point (2 pts.) D(0, −1) = ∣ 3 0

0 −6 ∣ <0 ⇒ (0, −1) is a saddle point (2 pts.) D(√

2,√

2) = ∣ 3√

2 −3√ 2

−3√

2 6√

2 ∣ = 18 > 0 and fxx(√ 2,√

2) = 3√ 2> 0

⇒ f(√ 2,√

2) is a local minimum value. (2 pts.) D(−√

2,−√

2) = ∣ −3√

2 3√ 2 3√

2 −6√

2 ∣ = 18 > 0 and fxx(−√ 2,−√

2) = −3√ 2< 0

⇒ f(−√ 2,−√

2) is a local maximum value. (2 pts.)

Page 5 of 10

5. (15 pts) (a) (8 pts) Find the shortest distance between the point (0, 0, 1) and the surface y² = x²+ 2z²+ 1.

(b) (7 pts) Let curve C be the intersection of the surface y²= x²+2z²+1 and the sphere x²+y²+z²= 2. Find the points on the curve C which are respectively the closest to and the farthest from the point (0, 0, 1).

Solution:

(a) The square of the distance between (0, 0, 1) and (x, y, z) is f(x, y, z) = x²+ y²+ (z − 1)². Under the constraint g(x, y, z) = x²− y²+ 2z²+ 1 = 0, we want to find the minimum value of f(x, y, z).

By Lagrange’s multiplier method, we solve the equations:

{ ⃗∇f = λ⃗∇g

g(x, y, z) = 0 ⇒

⎧⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎩

f_x= λgx⇒ 2x = λ(2x) − − −(1) f_y = λgy⇒ 2y = λ(−2y) − − −(2) f_z = λgz ⇒ 2(z − 1) = λ(4z) − − −(3) x²− y²+ 2z²+ 1 = 0 − − −(4)

(3 pts.) (1) ⇒ (1 − λ)x = 0 ⇒ x = 0 or λ = 1.

If x= 0 (2) ⇒ (1 + λ)y = 0 ⇒ y = 0 or λ = −1.

Case 1: y= 0 i.e. x = y = 0, (4) cannot be satisfied ⇒ no solution.

Case 2: λ= −1, (3) ⇒ z =1

3, (4) ⇒ y²= 11 9 , y= ±

√11

3 ⇒ (x, y, z) = (0, ±

√11 3 ,1

3).

If λ= 1 (2) ⇒ y = 0, (4) cannot be satisfied ⇒ no solution.

Hence the extreme value of f(x, y, z) may occur at (0,

√11 3 ,1

3) or (0, −

√11 3 ,1

3) (4 pts.) f(0, ±

√11 3 ,1

3) = 5

3 and this should be the minimum value (∵ the surface g(x, y, z) = 0 is unbounded ∴ f(x, y, z) has no upper bound on g = 0.)

Ans: the distance between (0, 0, 1) and surface g(x, y, z) = 0 is

√5 3.

(b) We want to find the extreme values of f(x, y, z) = x² + y² + (z − 1)² under constraints g₁(x, y, z) = x²− y²+ 2z²+ 1 = 0 and g2(x, y, z) = x²+ y²+ z²= 2.

By Lagrange multiplier method, we solve the equations:

⎧⎪⎪⎪⎨⎪⎪

⎪⎩

⃗∇f = λ⃗∇g1+ µ⃗∇g2

g₁(x, y, z) = 0

g₂(x, y, z) = 0 ⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎪⎩

f_x= λg1x+ µg2x

f_y = λg1y+ µg2y

f_z = λg1z+ µg2z

x²− y²+ 2z²+ 1 = 0 x²+ y²+ z²= 2

⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎪⎩

2x= λ(2x) + µ(2x) 2y= λ(−2y) + µ(2y) 2(z − 1) = λ(4z) + µ(2z) x²− y²+ 2z²+ 1 = 0 x²+ y²+ z² = 2

⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎪⎩

(1 − λ − µ)x = 0 − − −(1) (1 + λ − µ)y = 0 − − −(2) (1 − 2λ − µ)z = 1 − − −(3) x²− y²+ 2z²+ 1 = 0 − − −(4) x²+ y²+ z²= 2 − − −(5)

(3 pts.) (1) ⇒ x = 0 or 1 = λ + µ.

If x= 0, (4) and (5) ⇒ z²= 1

3, y² = 5

3⇒ solutions are (x, y, z) =⎛

⎝0,

√5 3, 1

√3

⎞

⎠,⎛

⎝0,−

√5 3, 1

√3

⎞

⎠,

⎛

⎝0,

√5 3,− 1

√3

⎞

⎠, or ⎛

⎝0,−

√5 3,− 1

√3

⎞

⎠. If 1= λ + µ, (2) ⇒ 1 + λ − µ = 0 or y = 0.

Case 1: 1+ λ − µ = 0 ⇒ λ = 0, µ = 1. (3) is not satisfied ⇒ no solutions.

Case 2: y= 0, (4) cannot be satisfied ⇒ no solution.

Hence the extreme values may occur at(x, y, z) =⎛

⎝0,

√5 3, 1

√3

⎞

⎠,⎛

⎝0,−

√5 3, 1

√3

⎞

⎠,⎛

⎝0,

√5 3,− 1

√3

⎞

⎠, or ⎛

⎝0,−

√5 3,− 1

√3

⎞

⎠.

f⎛

⎝0,±

√5 3, 1

√3

⎞

⎠= 3 − 2

√3, f⎛

⎝0,±

√5 3,− 1

√3

⎞

⎠= 3 + 2

√3.

(3 pts.) Ans: ⎛

⎝0,

√5 3, 1

√3

⎞

⎠ and ⎛

⎝0,−

√5 3, 1

√3

⎞

⎠ are closest to the point(0, 0, 1).

⎛

⎝0,

√5 3,− 1

√3

⎞

⎠ and ⎛

⎝0,−

√5 3,− 1

√3

⎞

⎠ are farthest from the point (0, 0, 1).

(1 pt.)

Page 7 of 10

6. (12 pts) Evaluate the double integral (a) (5 pts) ∫

2 0 ∫

4−x² 0

x e^2y

4− y dy dx.

(b) (7 pts) ∫ ∫_Rx²dA,, where R is the region bounded by the ellipse (x − y)²+ 2y² = 1.

Solution:

(a)

∫

2 0 ∫

4−x² 0

xe^2y 4− ydydx

= ∫_y=0⁴ ∫

√4−y x=0

xe^2y

4− ydxdy (2pts.)

= ∫_y=0⁴ e^2y 4− y(x²

2 ) ∣

√4−y

0

dy (1pt.)

= ∫_y=0⁴ 1

2e^2ydy (1pt.)

= 1 4e^2y∣

4

0

= 1

4(e⁸− 1) (1pt.)

(b)

∫ ∫_Rx²dA, R∶ (x − y)²+ 2y²≤ 1.

Let

u = x − y

v = √

2y ⇒

x = u + y = u + v

√2

y = v

√2

(2pts.)

We have

R^′∶ u²+ v² ≤ 1 & ∫ ∫_R′(u + v

√2)

2

∣J∣dudv (1pt.)

where ∣J∣ = ∂(x, y)

∂(u, v) =RRRRR RRRRR RRRRR RR

1 1

√2

0 1

√2 RRRRR RRRRR RRRRR RR

= 1

√2.

Now let

u = r cos θ v = r sin θ We find

R^′′∶ r²≤ 1 (2pts.) and

∫ ∫_R′′ 1

√2(r²cos²θ+ 2

√2r²cos θ sin θ+1

2r²sin²θ) rdrdθ

= ∫_θ=0^2π∫

1 r=0

√1 2r³[1

2(1 + cos θ) + 1

√2sin(2θ) +1

4(1 − cos(2θ))drdθ]

= 3π 8√

2 (2pts.)

7. (14 pts) (a) (7 pts) Find the volume of the wedge cut out of the cylinder x²+ y² = 1, z ≥ 0 by the plane z= −y.

(b) (7 pts) Find the volume of the region cut out of the sphere x²+ y²+ z² = 9 by the cylinder x²+ y²= 3y.

Solution:

(a)

A(x) = 1 2y² =1

2(1 − x²) (3pts.)

V =

1

∫

−1

A(x)dx

= ∫₋₁¹1

2(1 − x²)dx (2pts.)

= 1

2(x − x³ 3 ) ∣

1

−1

= 2

3 (2pts.)

(b) The cylinder

x²+ y²= 3y written in polar coordinate is

r² = 3r sin θ ⇒ r = 2 sin θ (1pt.) The sphere x²+ y²+ z²= 9 written in cylindrical coordinate is

r²+ z²= 9. (1pt.) The integral over cylindrical region is

V_c = ∫_θ=0^π ∫_r=0^{2 sin θ}∫

√9−r²

z=0 rdzdrdθ (1pt.)

= ∫_θ=0^π ∫

3 sin θ r=0 r√

9− r²drdθ (1pt.)

= ∫_θ=0^π −1

3⋅ 9^3/2[(1 − sin²θ)^3/2− 1] dθ

= ∫_θ=0^π −9(cos³θ− 1)dθ (1pt.)

= ∫_θ=0^π −9 [(1 − sin²θ) cos θ − 1] dθ

= 9π. (1pt.)

Thus the volume of the region of the sphere cut by the cylinder is V = 4

3π⋅ 3³− 2 ⋅ 9π = 18π. (1pt.).

Page 9 of 10

8. (12 pts) Find the center of mass of a thin plate occupying the smaller region cut out of the ellipse x²+ 4y²= 12 by the parabola x = 4y² with the density ρ= 5x.

Solution:

The intersection points of x²+ 4y²= 12 and x = 4y² are

x²+ x = 12 ⇒ x = −4(×) or x = 3. (2pts.) The mass over region R is

m = ∫ ∫_RρdA

= ∫

√3/4

y=0 ∫ ^(12−4y

2)^1/2

x=4y² 5xdxdy (1pt.)

= 2 ∫

√3/4 0

5

2(12 − 4y²− 16y⁴)dy (1pt.)

= 5 (12y −4

3y³−16y⁵ 5 ) ∣

√3/4

= 23√ 0

3 (1pt.)

¯

x= ∫ ∫^RxρdA

m = My

m where M_y = 2 ∫

√3/4

0 ∫ ^(12−4y

2)^1/2 4y²

5x²dxdy

= 2 ∫

√3/4 0

5

3[(12 − 4y²)^3/2− (4y²)³] dy (3pts.)

= 15π +765 28

√3. (2pts.)

and ¯y= 0. (2 pts.)