1072ᵬᵬᵬ555IIIʟʟʟὃὃὃYYYȤȤȤᑖᑖᑖ 1. (11 pts) Railway tracks cannot have large curvature at any point, otherwise the train might derail. Let r(t)=20t, a(t− t

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1. (11 pts) Railway tracks cannot have large curvature at any point, otherwise the train might derail.

Let r(t) = ⟨20t, a(t − t²), 5⟩, a ≠ 0 be the vector function describing a track starting at (0, 0, 5) and ending at (20, 0, 5).

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

(b) (6 pts) If the curvature needs to be smaller than 0.001, what is the largest possible value of a?

Solution:

(a) Solution 1: Book method.

By definition

T(t) = r^′(t)

∣r^′(t)∣

Here

r^′(t) = ⟨20, a(1 − 2t), 0⟩, ∣r^′(t)∣ =√

400+ a²(1 − 2t)² Therefore

T(t) = ⟨ 20

√400+ a²(1 − 2t)²,√ a(1 − 2t)

400+ a²(1 − 2t)², 0⟩ Next is to find N(t). By definition

N(t) = T^′(t)

∣T^′(t)∣

Here

T^′(t) = ⟨ 40a²(1 − 2t)

(400 + a²(1 − 2t)²)^3/2,√ −2a

400+ a²(1 − 2t)² + 2a³(1 − 2t)²

(400 + a²(1 − 2t)²)^3/2, 0⟩

T^′(t) = ⟨ 40a²(1 − 2t)

(400 + a²(1 − 2t)²)^3/2, −800a

(400 + a²(1 − 2t)²)^3/2, 0⟩ = 40a

(400 + a²(1 − 2t)²)^3/2⟨a(1 − 2t), −20, 0⟩

∣T^′(t)∣ = 40∣a∣

400+ a²(1 − 2t)²

Solution 2: TNB method.

Find the first and second derivative of the vector function r^′(t) = ⟨20, a(1 − 2t), 0⟩

r^′′(t) = ⟨0, −2a, 0⟩

The cross product r^′(t) × r^′′(t) is in the same direction as B(t).

r^′(t) × r^′′(t) = ⟨0, 0, −40a⟩

and we know N(t) is in the direction of B(t) × T(t) when a ≠ 0

⟨0, 0, −40a⟩ × ⟨20, a(1 − 2t), 0⟩ = ⟨40a²(1 − 2t), −800a, 0⟩

Make the vectors unit length

T(t) = ⟨ 20

√400+ a²(1 − 2t)²,√ a(1 − 2t)

400+ a²(1 − 2t)², 0⟩ N(t) = ⟨√ a(1 − 2t)

400+ a²(1 − 2t)² (a

∣a∣) ,√ −20

400+ a²(1 − 2t)² (a

∣a∣) , 0⟩

Other possible solutions: Use geometry to find N(t). Since the space curve is on a plane we can use the slope of T(t) to find the slope of N(t). Orthogonal projection of r^′′(t) to r^′(t) can also find N(t).

(For all methods) Grading scheme: 2 points for T (t) and 2 points for N (t) in a > 0 case (for each vector 1 point for direction 1 point for unit length). 1 point for correctly identifying the cases.

For example: 2 points if the answers are not unit vectors and only for a > 0.

(b) Solution 1: Use answer in (a) and definition.

By definition of the curvature κ(t)

κ(t) = ∣T^′(t)∣

∣r^′(t)∣

Use answer in (a) to get

κ(t) = 40∣a∣

(400 + a²(1 − 2t)²)^3/2 The curvature is largest when t= 1

2 (when the denominator is minimized). So we can set κ(1

2) < 0.001.

40∣a∣

8000 < 0.001, ∣a∣ < 0.2.

The largest possible value of a is 0.2.

Solution 2: Use formula in book.

κ(t) = ∣r^′(t) × r^′′(t)∣

∣r^′(t)∣³ κ(t) = 40∣a∣

(400 + a²(1 − 2t)²)^3/2 And the rest of the problem is the same.

(For all methods) Grading scheme: 1 point for a formula for κ (t), 3 points for finding the correct expression (absolute value is 1 point), 2 points for analyzing and getting the answer (1 point for realizing t = 1

2 , 1 point solving inequality).

2. (12 pts) Let f(x, y) =⎧⎪⎪⎪

⎨⎪⎪⎪⎩

sin xy²

x²+ y⁴, 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= (cos xy²

x²+ y⁴) (y²(x²+ y⁴) − xy²⋅ 2x

(x²+ y⁴)² ) = (cos xy²

x²+ y⁴) y⁶− x²y²

(x²+ y⁴)² (2 points) fy = (cos xy²

x²+ y⁴) (2xy(x²+ y⁴) − xy²(4y³)

(x²+ y⁴)² ) = (cos xy²

x²+ y⁴)2x³y− 2xy⁵

(x²+ y⁴)² (2 points) f_x(0, 0) = lim

x→0

f(x, 0) − f(0, 0)

x = lim

x→0

0

x = 0 (1 point)

f_y(0, 0) = lim

y→0

f(0, y) − f(0, 0)

y = 0 (1 point)

(b) If f(x, y) is continuous at (0, 0), by definition any path (x(t), y(t)) → (0, 0). We have limt→0f(x(t), y(t)) = f(0, 0)

along x= y², lim

y→0f(y², y) = lim

y→0sin y⁴

y⁴+ y⁴ = sin1

2 (2 points)

sin1

2 ≠ f(0, 0). ∴ f(x, y) is not continuous at (0, 0) (1 point) or

Take two paths with different limits (2 points)

different limits ⇒ lim

(x,y)→(0,0)f(x, y) does not exist ∴ f is not continuous at (0, 0) (1 point) If f(x, y) is differentiable at (0, 0), it means that f(x, y) = f(0, 0) + fx(0, 0)x + fy(0, 0)y +

₁x+ 2y lim

(x,y)→(0,0)₁= 0, lim

(x,y)→(0,0)₂= 0 (the definition of differentiability 1 point)

∴ lim

(x,y)→(0,0)f(x, y) = f(0, 0) ⇒ f(x, y) is continuous at (0, 0) (1 point) Because f(x, y) is not continuous at (0, 0), it implies that f(x, y) is not differentiable at

(0, 0) (1 point)

Remark: If give wrong reasons or not state the reason for not differentiability, clearly only get one point.

3. (12 pts) Consider the level surface z²+ z tan⁻¹ y x =π

4 + 1.

(a) (5 pts) Find the tangent plane for the level surface at (1, 1, 1).

(b) (4 pts) The level surface defines z implicitly as a function of x and y, z = f(x, y). Compute the directional derivative of f(x, y) at (1, 1) in the direction of −3⃗i+ 4⃗j.

(c) (3 pts) Use linear approximation to estimate the value of f(0.98, 1.04).

Solution:

(a) Define F(x, y, z) = z²+ z tan⁻¹ y x

∇F =⎛

⎝z −_x^y2

1+ (^y_x)², z

1 x

1+ (^y_x)², 2z+ tan⁻¹ y x

⎞

⎠= ( −yz

x²+ y², xz

x²+ y², 2z+ tan⁻¹ y

x) (2 points)

∇F (1, 1, 1) = (−1 2,1

2, 2+ tan⁻¹1) = (−1 2,1

2, 2+π

4) (1 point)

tangent plane (−1

2)(x − 1) +1

2(y − 1) + (2 +π

4)(z − 1) = 0

⇒ −x 2+y

2+ (2 +π

4)z − (2 +π 4) = 0

⇒ x − y − (4 + π

2)z + (4 +π

2) = 0 (2 points)

The problem (b) and (c) need to specify at (1, 1, 1). Please see remark below.

(b) F(x, y, z(x, y)) = π

4 + 1 ⇒ ∂F

∂x +∂F

∂z

∂z

∂x = 0

∵ ∂F

∂z∣

(1,1,1)≠ 0

∴ ∂z

∂x∣

(1,1,1)= −^∂F_∂F^∂x

∂z

∣(1,1,1)= − −¹₂

(2 +^π₄) = 1

4+^π₂ (1 point)

Similarly, ∂z

∂y∣

(1,1,1)= −

∂F

∂y

∂F

∂z

∣(1,1,1)= − 1

4+^π₂ (1 point)

u= (−3 5,4

5), Duf =< ∇f, u >= −3

5(4 +^π₂)− 4

5(4 +^π₂) = −7

5(4 +^π₂) (2 points)

(c)

f(0.98, 1.04) ≈ 1 + 1

4+^π₂(0.98 − 1) − 1

4+^π₂(1.04 − 1) (2 points)

≈ 1 −0.02+ 0.04

4+^π₂ = 1 − 0.06

4+^π₂ (1 point)

Remark:

for x= 1, y = 1, z²+ z tan⁻¹1=π 4 + 1 That is, z²+π

4z =π

4 + 1 ⇒ z = 1 or z = −1 − π For z= 1, as discussed above. 4

For z= −1 − π

4, ∇F (1, 1, −1 −π

4) = (1+^π₄

2 ,−1 −^π₄

2 ,−2 −π 4)

4. (12 pts) Find and classify all critical points of f(x, y) = 4x³+ 2xy²+2

3y³+ 6x².

Reminder: each critical point must be shown to be either a local maximal point, a local minimal point, or neither of the above.

Solution:

Step 1: Compute the gradient (partial derivatives) 3pts.

We have f_x= 12x²+ 12x + 2y², f_y = 4xy + 2y² = 2y(2x + y).

Partial credits: If the student only get 1 of them correct, 2pts.

Step 2: Find the critical points. 3pts

Set f_x = fy = 0, we can find the critical points. From fy = 0, either y = 0 or y = −2x. If y = 0, from f_x= 12x²+ 12x − 2y² = 0, we have (0, 0) or (−1, 0). If y = −2x, from fx= 12x²+ 12x + 2y²= 12x²+ 12x + 8x² = 0, we have (0, 0) or (−3

5,6 5).

Partial credits for this part:

ˆ Only set fx= fy = 0 without fining any correct critical points. 1pts

ˆ If the student find some correct critical points but not all the critical points or there is some incorrect critical points. 2pts

Step 3: Second derivative test. 6pts

Now, consider the second derivative test. f_xx = 24x + 12, fxy = fyx = 4y, fyy = 4x + 4y. D = 48(2x + 1)(y + x) − 16y²= 16(−y²+ (6x + 3)(y + x)).

ˆ At (−1, 0), we have D = 48 > 0, fxx= −12 < 0, it is a local maximum.

ˆ At (−3 5,6

5), we have D = 16(−(6 5)²−3

5⋅3

5) < 0, it is neither a local maximum nor a local minimum.

ˆ At (0, 0), D = 0, the second derivative test fails. However, if we approach (0, 0) from y axis, f(0, y) = 2

3y³. f(0, y) > 0 for positive y, f(0, y) < 0 for negative y. Therefore, (0, 0) is neither a local maximum nor a local minimum.

Partial credits for this part:

ˆ 2pts for each correct pair of critical point and the classification. it is okay to write “Saddle”

in the case of “Neither a local max nor a local min”.

ˆ For (0, 0),

– If the answer only says “Second derivative fails” 1pts.

– If the answer is “Neither a local max nor a local min” or “Saddle” 2pts.

ˆ If there is no correct pair of critical point and the classification, but the student knows how to use second derivative test, the student can obtain at most 3pts for this part.

5. (15 pts) Consider the part of an elliptic paraboloid defined by z= x² 16+y²

8, z≤ 6. Find the points on the surface segment which are respectively the farthest from and the closest to the point(0, 0, 8).

Solution:

Step 1 Compute the interior candidate for extremal value. 7pts

Let f(x, y, z) = x²+ y² + (z − 8)². Let g(x, y, z) = x² 16 + y²

8 − z = 0 be the constraint. In the interior of the surface, the extremal point happens when ∇f = (2x, 2y, 2(z − 8)) is parallel to

∇g = (x 8,y

4,−1). From the x, y components, we have xy 2 = xy

4 . Either x= 0 or y = 0.

ˆ If (x, y) = (0, 0), we have z = 0 and f(0, 0, 0) = 64.

ˆ If x = 0, y ≠ 0, we have z = 8 − 4 = 4, y²= 32, y = ±4√

2, f(x, y, z) = 32 + 16 = 48.

ˆ If y = 0, x ≠ 0, we have z = 0. We have x = 0 again.

Partial credits for this part:

ˆ Setting up the function f and the constraint g = 0. 1pt

ˆ Correct answer for ∇f, ∇g. worth 1pt each.

ˆ Setting up the equation ∇f = λ∇g. 1pt.

ˆ Solving ∇f = λ∇g together with the constraint. 3pts. Each solution worth 1pt.

ˆ It is possible to solve this problem without using Lagrange multiplier, partial credits may be given to incomplete answers.

Step 2 Compute the boundary candidate for extremal value. 6pts

Let h(x, y, z) = z − 6. At the intersection of g and h, we consider it as a optimization problem with 2 constraint. From ∇h = (0, 0, 1), use Lagrange multiplier, we have

∇f = λ∇g + µ∇h.

We obtain the equation (2x, 2y, 2(z − 8)) = (λx 8, λy

4,−λ + µ) together with 2 constraints g = 0, h= 0. From h = 0, we have z = 6. (2x, 2y, −4) = (λx

8, λy

4,−λ + µ).

Use the same argument as above, we have either x= 0 or y = 0.

ˆ If (x, y) = (0, 0), we have z = 0. This contradicts z = 6

ˆ If x = 0, z = 6, we have y²= 48, y = ±4√

3, f(x, y, z) = 48 + 4 = 52.

ˆ If y = 0, z = 6, we have x²= 96, x = ±4√

6, f(x, y, z) = 96 + 4 = 100.

Partial credits for this part:

ˆ Setting up the constraint h = 0 or other ways to describe the boundary. 1pt

ˆ Setting up the equation ∇f = λ∇g + µ∇h. 2pts.

ˆ Solving ∇f = λ∇g + µ∇h together with the constraint. 3pts.

ˆ It is possible to solve this problem without using Lagrange multiplier(Eg. paramatrize the boundary ellipse), partial credits may be given to incomplete answers.

Step 3 Conclusion. 2pts

ˆ The longest distance happens at (±√

96, 0, 6) and the distance is 10.(1pt)

ˆ The shortest distance happens at (0, ±4√

2, 4) and the distance is 4√

3.(1pt)

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

8 0 ∫

2

x¹³ 1

1+ y⁴ dy dx.

(b) (7 pts) Evaluate the double integral ∬

R

(x²+ y²) dA, where R is the region bounded by the ellipse 9x²+ y²= 36.

Solution:

(a)

I = ∫₀⁸∫

2 x¹³

1

1+ y⁴ dy dx= ∫_y=0² ∫

y³ x=0

1

1+ y⁴ dx dy (到此步驟得 3分)

= ∫_y=0² y³

1+ y⁴ dy (到此步驟得 4分)

= [1

4ln∣1 + y⁴∣]²

0

(到此步驟得 5分)

=1 4ln 17.

(b)

I = ∫ ∫_9x2+y²=36(x²+ y²) dA (let x = 2u, y = 6v) (到此步驟得 2分)

= ∫ ∫_u2+v²=1[(2u)²+ (6v)²] ∣∂(x, y)

∂(u, v)∣ du dv (let u = r cos θ, v = r sin θ)

(到此步驟得 4分)

= ∫_θ=0^2π∫

1

r=0[(2r cos θ)²+ (6r sin θ)²] 12 r dr dθ (到此步驟得 6分)

= 120π. (到此步驟得 7分)

7. (14 pts) (a) (7 pts) Find the volume of the region R, where R is bounded by the surfaces z = 0, x= 0, x = y², and z= 1 − y².

x

y z

(b) (7 pts) LetR be the part of the solid ball x²+y²+z² ≤ 1 which is above the cone z =

√x²+ y² 3 . Evaluate ∭_R(x²+ y²)dV.

Solution:

(a) Notice that R, as seen from the given figure, can be expressed as the region under the graph of the function f(x, y) = 1 − y² over the plane region D (in the xy-plane) given by

D = {−1 ≤ y ≤ 1 , 0 ≤ x ≤ y²} .

(suitable description of the region R: 3 points) Therefore, the volume in question is

V = ∬_D(1 − y²) dA (integral representing V : 1 point)

= ∫₋₁¹∫ ^y

2

0 (1 − y²) dx dy (Fubini’s Theorem: 1 point)

= ∫₋₁¹y²(1 − y²) dy (computation: 1 point)

= 2 ∫₀¹(y²− y⁴) dy = 2(1 3−1

5) = 4

15 . (answer: 1 point)

(b) In spherical coordinated, x²+ y²= (ρ sin φ cos θ)²+ (ρ sin φ sin θ)²= ρ²sin²φ. It implies that ρ cos φ=

√x²+ y²

3 = ρ

√3sin φ⇒√

3 cos φ= sin φ ⇒ φ = π 3. Thus R = {(ρ, φ, θ) ∶ 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π

3, 0≤ θ ≤ 2π} (2 points).

∭_R(x²+ y²)dV = ∭_Rρ⁴sin³φdρdφdθ

= ∫₀^2π∫

π/3

0 ∫

1 0

ρ⁴sin³φdρdφdθ (2 points)

= 2π ∫

π 3

0 [1

5ρ⁵]¹0sin³φdφ

= 2π 5 ∫

π 3

0

sin³φdφ (1 points)

= 2π 5 ∫

π 3

0 (1 − cos²θ)(−d cos θ)

= 2π 5 [1

3cos³θ− cos θ]^π/30 (1 points)

= π

12 (1 points)

8. (12 pts) A lamina with constant density ρ occupies the region

R ∶= {(x, y) ∣ 1 ≤ x⁵³y≤ 9 , 1 ≤ xy ≤ 4, x > 0} . Find the coordinates(¯x, ¯y) of its center of mass.

Solution:

The coordinates of the center of mass are given by

¯ x= ρ

m ∬R

x dA and y¯= ρ m ∬R

y dA , where m∶= ρ ∬

R

dA . The integrals are computed via a change of variables

s∶= x⁵³y and t∶= xy ( ⇐⇒ x=(s t)

3 2

and y= t⁵² s³² ) with the corresponding Jacobian determinant given by

∂(x, y)

∂(s, t) =(∂(s, t)

∂(x, y))

−1

=RRRRR RRRRR RR 5

3x²³y x⁵³

y x

RRRRR RRRRR RR

−1

=(2

3x⁵³y)⁻¹ = 3 2s .

(proper strategy (e.g. change of variables) for the computation: 3 points) As the region R is mapped to the rectangle {(s, t) ∣ 1 ≤ s ≤ 9 , 1 ≤ t ≤ 4} , the coordinates of the center of mass can now be computed as

m= ρ ∬

R

dA= ρ ∫₁⁹∫

4 1 ∣ 3

2s∣ dt ds = 3ρ

2 (4 − 1) ln ∣9

1∣ = 9ρ ln 3 ,

¯ x= ρ

m ∬^Rx dA= ρ m ∫

9 1 ∫

4 1 (s

t)

3 2∣ 3

2s∣ dt ds = 3ρ 2m ∫

4 1

1

t³² dt⋅ ∫₁⁹s¹² ds

= 3ρ 2m( 2

1¹² − 2 4¹²) ⋅2

3(9³² − 1³²)

= 26 9 ln 3 ,

¯ y= ρ

m ∬R

y dA= ρ m ∫

9 1 ∫

4 1

t⁵² s³² ∣ 3

2s∣ dt ds = 3ρ 2m ∫

4 1

t⁵² dt⋅ ∫₁⁹ 1 s⁵² ds

= 3ρ 2m ⋅2

7(4⁷² − 1⁷²) ⋅2 3( 1

1³² − 1 9³²)

=2²⋅ 13 ⋅ 127 3⁵⋅ 7 ln 3 .

(correct execution of the strategy (e.g. correct use of the change-of-variable formula): 3 points) (computations for m, m¯x and m¯y : 2+2+2 points)