Find the vectors T, N, B, the curvature κ and the torsion τ of the curve C at t= 0

1092 !D01-05í ®M3 ãTŒU–

1. (10 pts) Let the curve C be given by r(t) = ²₃(2 + t)³²i+²₃(2 − t)³²j+ at k, a ≠ 0, t ∈ (−2, 2). Find the vectors T, N, B, the curvature κ and the torsion τ of the curve C at t= 0.

Solution:

r^′(t) = ⟨(2 + t)^1/2,−(2 − t)^1/2, a⟩ r^′′(t) = 1

2⟨(2 + t)^−1/2,(2 − t)^−1/2, 0⟩ r^′′′(t) = 1

4⟨−(2 + t)^−3/2,(2 − t)^−3/2, 0⟩

∣r^′(t)∣ = (4 + a²)^1/2 r^′(t) × r^′′(t) = 1

2⟨−a(2 − t)^−1/2, a(2 + t)^−1/2, 4(4 − t²)^−1/2⟩

∣r^′(t) × r^′′(t)∣ =√

4+ a²/√

4− t², (r^′(t) × r^′′(t)) ⋅ r^′′′(t) = a

2(4 − t²)^−3/2

T(t) = r^′(t)

∣r^′(t)∣ =

√ 1

4+ a² ⟨(2 + t)^1/2,−(2 − t)^1/2, a⟩ N(t) = T^′

∣T^′∣ = 1

2⟨(2 − t)^1/2,(2 + t)^1/2, 0⟩ B(t) = T × N(t) = 1

2√

4+ a² ⟨−a(2 + t)^1/2, a(2 − t)^1/2, 4⟩ κ(t) = ∣T^′(t)∣

∣r^′(t)∣ = ∣r^′× r^′′∣

∣r^′∣³ = 1 (4 + a²)√

4− t² τ(t) = (r^′× r^′′) ⋅ r^′′′

∣r^′× r^′′∣² = a 2

1 (4 + a²)√

4− t² At t= 0 (Ï 2 )

T= 1

√4+ a² ⟨√ 2,−√

2, a⟩ N= 1

2⟨2, 2, 0⟩

B= 1

2√

4+ a² ⟨−a√ 2, a√

2, 4⟩

κ= 1

2(4 + a²)

τ = a

4(4 + a²)

Page 1 of 9

2. (12 pts) Consider the following function on R²:

f(x, y) = { ^x

2y

x⁴+(sin y)² if (x, y) ≠ (0, 0);

0 if (x, y) = (0, 0).

(a) (6 pts) Is f(x, y) continuous at (x, y) = (0, 0)? Justify your answer.

(b) (4 pts) Do the partial derivatives f_x(0, 0) and f^y(0, 0) exist? (Compute them if you think they exist; otherwise, prove that they do not exist.)

(c) (2 pts) Is f differentiable at (0, 0)? Justify your answer.

Solution:

(a) Let (x, y) tend to (0, 0) along special paths. One may gain 6 points if the students find such a path that along which the limit of f is different from 0= f(0, 0). For example, one may consider (x, mx²): if x ≠ 0,

f(x, mx) = x²⋅ mx²

x⁴+ sin(mx)² = m

1+ (^{m sin(mx}mx^{2 2)})² → m

1+ m² ≠ 0 = f(0, 0) if m ≠ 0 as x → 0.

Therefore f is not continuous at (0, 0). One may also consider for example the curve y = sin⁻¹(mx²). ~ï‘&ªðuP I¼f(0, 0) = 0„ûU °À/¤ï c1ó2

(b) Compute by definition that f_x(0, 0) = 0 = f^y(0, 0). N—PœGcº—4&G

—0

(c) Since f is not continuous at(0, 0), it is not differentiable at (0, 0). One may also argue by definition. 1E—2&G—0

3. (10 pts) Let g(x, y, z) be a function defined on R³ with continuous partial derivatives. Suppose that

∣∇g(2, 1, 3)∣²= 24 and gz(2, 1, 3) > 0.

Moreover, the trajectories of the two curves

r₁(s) = ⟨2s, s², 1+ 2s⟩ and r2(t) = ⟨2e^t, cos t, 3+ t + 5t²⟩ lie on the level surface g(x, y, z) = 0 completely.

(a) (5 pts) Find the vector∇g(2, 1, 3).

(b) (5 pts) Suppose that f(x, y, z) is a function defined on R³ with continuous partial derivatives such that

f(2, 1, 3) ⩾ f(x, y, z) for every point (x, y, z) on the level surface g(x, y, z) = 0.

If f(2, 1, 3) = 5, ∣∇f(2, 1, 3)∣²= 6 and f^y(2, 1, 3) > 0, estimate the value of f(2.01, 0.9, 3.02) by the linear approximation of f at (2, 1, 3).

Solution:

(a) Note that r₁(1) = (2, 1, 3) = r²(0). Thus

(2, 2, 2) = r^′1(1) ⊥ ∇g(2, 1, 3) ⊥ r^′2(0) = (2, 0, 1),

0áiÏý—cºï—1 and hence ∇g(2, 1, 3) is parallel to (2, 2, 2) × (2, 0, 1) = (2, 2, −4). ú∇g(2, 1, 3)sL¼iÏMï—2M—cº

—1 By (i), we see that ∇g(2, 1, 3) = (−2, −2, 4). (0ö(i)†zš∇g(2, 1, 3)„¹

ï—1

(b) We need to find ∇f(2, 1, 3) for the linear approximation. By (a) and by the Lagrange multiplier method we see that ∇f(2, 1, 3) = λ∇g(2, 1, 3) for some λ ∈ R.ª0(

Lagrange—1 By (b) we see that λ= ⁻¹2 and∇f(2, 1, 3) = (1, 1, −2).zš∇f(2, 1, 3) = (1, 1, −2)„1cºï—2 Therefore

f(2.01, 0.9, 3.02) ≈ f(2, 1, 3) + ∇f(2, 1, 3) ⋅ (0.01, −0.1, 0.02) = 4.87.

Ú'<фbcº—1—cº—1

Page 3 of 9

4. (10 pts) Let f(x, y) = ^xy(x+y)_e^x+y be defined on the first quadrant D ∶ x > 0 and y > 0 (without boundary). Find all critical points of f in D and classify them (as local maximum points, local minimum points, or saddle points). Please provide details of calculation.

Solution:

The first derivatives of f are

∂f

∂x = e^−x−y(2xy + y²− x²y− xy²) ,

∂f

∂y = e^−x−y(2xy + x²− x²y− xy²) . (2 points) There is only one critical point of f in D, which is

(x, y) = (3 2,3

2) . (2 points) The second partial derivatives of f are

∂²f

∂x² = e^−x−y(2y − 4xy − 2y²+ x²y+ xy²) ,

∂²f

∂x∂y = e^−x−y(2x + 2y − x²− 4xy − y²+ x²y+ xy²) ,

∂²f

∂y² = e^−x−y(2x − 4xy − 2x²+ x²y+ xy²) . (3 points) Since

det∇²f(3 2,3

2) = e⁻⁶∣ −¹⁵⁴ −³4

−³₄ −¹⁵₄ ∣ > 0

and ∂²f

∂x² (3 2,3

2) = −15

4 e⁻³ < 0 (2 points) , the critical point (³2,³₂) is a local maximum point of f (1 point).

5. (12 pts) A pentagon is formed by placing an isosceles triangle on a rectangle. The side lengths are denoted by a, b, and c as shown in the figure.

(a) (3 pts) Write down the area of pentagon in terms of a, b, and c.

(b) (9 pts) Find the maximum area of pentagon if the perimeter is fixed as 2.

Solution:

(a) The height of the upper triangle equals √

a²− c² (2 points). Therefore the area is given by

A= c ⋅√

a²− c²+ 2bc (2 points).

(b) We use the method of Lagrange multiplier. We are looking for the maximum value of A under the conditions a, b, c> 0 and g(a, b, c) = 1 where g(a, b, c) = a+b+c. When A achieves the extremum, we have

√ ac

a²− c² = λ 2c= λ a²− 2c²

√a²− c² + 2b = λ a+ b + c = 1

(4 points).

The first two equations imply 3a² = 4c² or equivalently c = ^√2³a. Plugging into the third equation and replacing λ by 2c=√

3a, we have b= ¹⁺2^√3a. The last equation then reduces to ³⁺²

√ 3

2 a= 1 so we obtain a= 2

3+ 2√

3, b= 1+√ 3 3+ 2√

3, c=

√3 3+ 2√

3. In this circumstance,

A= 6+ 3√ 3 (3 + 2√

3)² (4 points).

Page 5 of 9

6. (26 pts) (a) (6 pts) Find the average value of f(x) = ∫_x^ae^−t2dt on the interval [0, a], where a > 0 is a constant.

(b) (10 pts) Compute ∭

E

e^3y−y3dV , where E is the solid bounded by x= 0, y = 0, x = z, y = z, and z = 1.

(c) (10 pts) Compute ∫_−a^a∫

√ a²−x²

0 ∫ ^a+

√

a²−x²−y²

a

√ 1

x²+ y²+ z² dzdydx.

(Hint: Use Spherical coordinates.)

Solution:

(a) The average value of f(x) on [0, a] is ¹_a∫^0af(x) dx. (1 pt for the definition of average value.)

1 a ∫

a

0

f(x) dx = 1 a ∫

a

0 ∫_x^ae^−t2dt dx

= 1 a ∫

a

0 ∫₀^te^−t2dx dt (3 pts for changing the order of integration)

= 1 a ∫

a

0

te^−t2dt= 1 a(−1

2e^−t2) ∣^t=a

t=0 = 1

2a(1 − e^−a2) (2 pts for the final answer) (b) Solution 1: E = {(x, y, z)∣0 ≤ y ≤ 1, y ≤ z ≤ 1, 0 ≤ x ≤ z}

Hence ∭

E

e^3y−y3dV = ∫₀¹∫_y¹∫₀^ze^3y−y3dxdzdy

(5 pts. If students correctly project E onto the yz-plane and write down the correct range of x, they get 2 pts. 3 pts for correct iterated integrals.)

∫₀¹∫_y¹∫₀^ze^3y−y3dxdzdy= ∫₀¹∫_y¹ze^3y−y3dz (1 pt)

= ∫₀¹ 1

2(1 − y²)e^3y−y3dy (1 pt)

let u=3y−y³

ÔÔÔÔÔÔÔ

du=(3−3y²)dy ∫₀²1 6e^udu

The integral is

∫₀¹∫_y¹(1 − x)e^3y−y3dxdy+ ∫₀¹∫₀^y(1 − y)e^3y−y3dxdy (1 pt)

= ∫₀¹(1 − y) −1

2(1 − y²)e^3y−y3dy+ ∫₀¹y(1 − y)e^3y−y3dy (1 pt)

= ∫₀¹1

2(1 − y²)e^3y−y3dy= 1

6(e²− 1) (3 pts) (c) Solution 1: The integral is ∭

E

√ 1

x²+ y²+ z²dV , where E = {(ρ, θ, ϕ)∣0 ≤ θ ≤ π, 0 ≤ ϕ ≤

π

4,_{cos ϕ}^a ≤ ρ ≤ 2a cos ϕ}.

⎛⎜⎜

⎝

1 pt for the range of θ 2 pts for the range of ϕ 2 pts for the range of ρ

⎞⎟⎟

⎠

∭

E

√ 1

x²+ y²+ z² = ∫₀^π∫

π 4

0 ∫ ^{2a cos ϕ}a cos ϕ

1

ρρ²sin ϕ dρdϕdθ (1 pt for Jacobian)

= π ∫

π 4

0

1

2(4a²cos²ϕ− a²

cos²ϕ) sin ϕ dϕ (1 pt)

=π 2a²∫

π 4

0 (4 cos²ϕ− 1

cos²ϕ) sin ϕ dϕ

u=cos ϕ

ÔÔÔÔÔÔ

du=− sin ϕ dϕ

π 2a²∫

√1 2

1 (4u²− 1

u²) (−du) (2 pts for substitution)

=π 2a²[4

3u³+ 1 u] ∣^u=1

u=√¹ 2

=π 2a²[7

3−4 3

√2] (1 pt for final answer.)

Solution 2: Use cylindrical coordinates. The integral is ∭

E

√ 1

x²+ y²+ z²dV , where E = {(r, θ, z)∣0 ≤ θ ≤ π, 0 ≤ r ≤ a, a ≤ z ≤ a +√

a²− r²}.

⎛⎜⎜

⎝

1 pt for the range of θ 1 pt for the range of r 1 pt for the range of z

⎞⎟⎟

⎠

∭

E

√ 1

x²+ y²+ z² = ∫₀^π∫₀^a∫ ^a+

√ a²−r²

a

√ 1

r²+ z²r dzdrdθ (1 pt for Jacobian) Note that ∫ 1

√a²+ t²dt= ln(t +√

a²+ t²) + c. (2 pts)

Page 7 of 9

7. (10 pts) Let D be an xy-plane region bounded by one loop of r² = cos 2θ. Find the area of the part of the upper half sphere z=√

1− x²− y² that is above D.

Solution:

The first derivatives of f(x, y) =√

1− x²− y² are

∂f

∂x =√ −x 1− x²− y²,

∂f

∂y =√ −y

1− x²− y². (2 points) The area of the graph of f above D is given by

∬_D

¿Á ÁÀ(∂f

∂x)²+ (∂f

∂y)²+ 1 dx dy = ∬_D 1

√1− x²− y²dx dy (2 points)

= ∫

π 4

−^π

4 ∫

√ cos(2θ)

0

√ r

1− r² dr dθ (3 points)

= ∫

π 4

−^π

4

(1 −√

1− cos (2θ)) dθ

= 2 ∫

π 4

0 (1 −√

2 sin θ) dθ = 2 (π

4 + 1 −√

2) (3 points) .

8. (10 pts) Let D= {(x, y)∣x > 0, y > 0, y ≤ x²≤ 2y, 3x ≤ y²≤ 4x}. Evaluate ∬

D

xy dA.

Solution:

Method 1:

A(3^1/3, 3^2/3) B(4^1/3, 4^2/3) C(2^4/3, 2^5/3) D(2^2/33^1/3, 2^1/33^2/3)

1

1 22 33 44 55

1 1 2 2 3 3

0 0

a a

A A B B

C C

D D

I= ∫₃1/3^41/3∫ ^x

2

√ 3x

xy dydx+ ∫₄1/3^22/331/3∫

√ 4x

√ 3x

xy dydx+ ∫₂2/3^24/33^1/3∫

√ 4x

x²/2

xy dydx= 7 4 6

TH (⁷4)4  Method 2:

Make a change of variables u= ^x_y², v= ^y_x², which transforms D to E= {(u, v)∣1 ≤ u ≤ 2, 3 ≤ v ≤ 4}.

Observe that uv= xy, ^∂(u,v)∂(x,y) = ∣ ^2xy −^x_y²²

−^yx²² 2y

x

∣ = 4 − 1 = 3, ^∂(x,y)∂(u,v) = ¹3 (4å’Öxc 1 )

I = ∬_E uv

ŠÛŒ„ýx 2 ¯

∣∂(x, y)

∂(u, v)∣dudv= 1

3 ∫₁²u du∫₃⁴v dv

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

2D
P 1 q 2 

= 1 3(u²

2 )²

1(v² 2 )⁴

3= 7

®4

2

ûU H!H„ŠÛ†‚hË%f2 D
P 1 q 2 Jacobian 4

ŠÛŒ„«Mýx 2 TH 2 (—N Å)

Page 9 of 9