Prob.5: The following crucial steps must be shown clearly

1102模模模組組組06-12班班班 微微微積積積分分分3 期期期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. Suppose we are given a function f(x, y) whose second order partial derivatives are continuous. Fix two points P= (1, −1) and Q = (1.2, −1.1) on the xy-plane. It is known that :

ˆ ⟨−3, 2, 2⟩ is a normal vector to the surface z = f(x, y) at (1, −1, f(P)),

ˆ f(x, y) attains an extreme value at Q.

Answer the following questions.

(a) (2%) Find ∂f

∂x(1, −1) and ∂f

∂y(1, −1).

Solution:

Since(f^x(1, −1), f^y(1, −1), −1) is normal to the surface z = f(x, y) at (1, −1, f(P)),

we have (f^x(1, −1), f^y(1, −1), −1) = λ(−3, 2, 2) for some λ (or (f^x(1, −1), f^y(1, −1), −1) // (−3, 2, 2)). (1%) So λ= −1/2, f^x(1, −1) = 3/2 and f^y(1, −1) = −1. (1%)

(b) (4%) Use the linearization of f(x, y) at P = (1, −1) to estimate the value of f(1.2, −1.1) − f(1, −1).

Solution:

The linearization of f at (1, −1) is

L(x, y) = f(1, −1) + f^x(1, −1)(x − 1) + f^y(1, −1)(y + 1) = f(1, −1) +3

2(x − 1) − (y + 1).(2%) Then

f(1.2, −1.1) − f(1, −1) ≈ L(1.2, −1.1) − f(1, −1) =3

2(1.2 − 1) − (−1.1 + 1) = 0.4.(2%)

(c) (1%) (Circle the best answer.) f(Q) is a (i) maximum value (ii) minimum value

Solution:

Answer is (i). (1%)

Since f(Q) is an extreme value and f(Q) > f(P) by (b), we have f(Q) is a maximum value.

(d) (1%) (Circle the best answer.) If f_xy(Q) ≠ 0, then f^xx(Q) is

(i) positive (ii) non-negative (iii) zero (iv) non-positive (v) negative

Solution:

Answer is (v). (1%)

Since f(Q) is a maximum value, we have f^xx(Q) ≤ 0. If f^xx(Q) = 0, we have D(Q) = −[f^xy(Q)]²< 0 which implies that Q is a saddle point of f . It contradicts to (c) and we have f_xx(Q) is negative.

2. Consider the function

f(x, y) =⎧⎪⎪⎪

⎨⎪⎪⎪⎩ xy²

x²+ y² if(x, y) ≠ (0, 0) 0 if(x, y) = (0, 0) .

(a) (4%) Show that f(x, y) is continuous at (0, 0).

(b) (6%) Find lim

(x,y)→(0,0)

f_x(x, y). Is f^x(x, y) continuous at (0, 0)? Explain.

(c) (5%) Let u = ⟨a, b⟩ be a unit vector. Use the definition of directional derivatives to find D^uf(0, 0). (Express your answer in terms of a and b.)

(d) (3%) Using (c), explain why f(x, y) is not differentiable at (0, 0).

Solution:

(a) Sol 1:

∣f(x, y)∣ = ∣x^2y+y²²x∣ =x^2y+y²²∣x∣ ≤ ∣x∣ ≤√ x²+ y².

Hence ∣f(x, y)∣ → 0 as (x, y) approaches (0,0). (3 pts for showing that lim

(x,y)→(0,0)

f(x, y) = 0) Since lim

(x,y)→(0,0)f(x, y) = 0 = f(0, 0), we conclude that f is continuous at (0,0).

(1 pt for showing that f(x, y) is continuous at (0,0).) Sol 2:

By polar coordinates,

∣f(r cos θ, r sin θ)∣ = ∣r³cos θ sin²θ

r² ∣ = ∣r∣∣ cos θ∣ sin²θ≤ ∣r∣ → 0 as r → 0.

Hence lim

(x,y)→(0,0)

f(x, y) = lim_r→0f(r cos θ, r sin θ) = 0 (3 pts for showing that lim

(x,y)→(0,0)

f(x, y) = 0) Since lim

(x,y)→(0,0)f(x, y) = 0 = f(0, 0), we conclude that f is continuous at (0,0).

(1 pt for showing that f(x, y) is continuous at (0,0).) (b) For(x, y) ≠ (0, 0), f^x=^y2(x(x^2+y2+y^2)−xy2)^22⋅2x= (x^y42−x+y^22y)²² (2 pts).

fx(0, 0) = lim_h→0f (h,0)−f (0.0)

h = lim_h→0⁰⁻⁰h = 0 (1 pt) fx(0, y) =^yy⁴⁴ = 1 for all y ≠ 0.

Hence fx(x, y) → 1 ≠ f^x(0, 0) as (x, y) approaches (0,0) along the y-axis.

This shows that fx(x, y) is not continuous at (0,0).

(Or we can also show that fx(x, y) → 0 as (x, y) approaches (0,0) along the x-axis. Hence lim

(x,y)→(0,0)fx(x, y) does not exist. fx(x, y) is not continuous at (0,0).)

(3 pts for showing that fx(x, y) is not continuous at (0,0).) (c)

D_uf(0, 0) = lim_h→0f(ah, bh) − f(0, 0)

h (2 pts for definition of D_uf(0, 0))

= lim

h→0 h³ab²

h² − 0

h = lim

h→0ab²= ab² (3 pts for the answer)

(d) If f(x, y) is differentiable at (0,0), the D^uf(0, 0) = af^x(0, 0) + bf^y(0, 0). (2 pts) However Duf(0, 0) = ab²≠ f^x(0, 0)a + f^y(0, 0)b.

Therefore we know that f us not differentiable at (0,0). (1 pt)

Page 2 of 8

3. Consider the function I(x, y) = ∫_1−y^x(t²+ 3t) ⋅ e^t2dt.

(a) (6%) Find all the critical points of I(x, y).

(b) (6%) Classify the critical points of I(x, y) as local maxima, local minima or saddle points.

Solution:

(a) The partial derivatives are

Ix= (x²+ 3x)e^x2,(2 points) I_y= [(1 − y)²+ 3(1 − y)]e^(1−y)2.(2 points)

The critical points will be the solutions of Ix= 0 and I^y= 0. Namely, x²+ 3x = 0 and (1 − y)²+ 3(1 − y) = 0 which yield x= 0, −3 and y = 1, 4. So the critical points are (0, 1), (0, 4), (−3, 1), and (−3, 4).(2 points) (b) The critical points we found in (a) all have gradient zero, so we use Second Derivatives Test to classify them.

We compute Ixx, Ixy, and Iyy.

Ixx= (2x + 3)e^x2+ (x²+ 3x)e^x22x,

Iyy= [2(y − 1) − 3]e^(1−y)2+ [(1 − y)²+ 3(1 − y)]e^(1−y)22(y − 1).

Both Ixy and Iyx are zero. So

D= I^xxI_yy− Ixy² = [2x + 3 + (x²+ 3x)2x]e^x2e^(1−y)2(2y − 5 + [(1 − y)²+ 3(1 − y)]2(y − 1)).(2 points) D(0, 1) = −9 < 0, so a saddle point. D(0, 4) = 9e⁹> 0, and I^xx(0, 4) = 3 > 0, so a local min. D(−3, 1) = 9e⁹> 0, and Ixx(−3, 1) = −3e⁹< 0, so a local max. Finally, D(−3, 4) = −9e¹⁸< 0, a saddle point. In summary, (0, 1) and (−3, 4) are saddle points, (0, 4) is a local min, and (−3, 1) is a local max. (1 point for each critical point.)

4. (12%) By the method of Lagrange multipliers, find the absolute maximum and minimum values of f(x, y, z) = x²− 2y²− 2z²+ 4xz

on the unit sphere x²+ y²+ z²= 1.

Solution:

Let

f(x, y, z) = x²− 2y²− 2z²+ 4xz, g(x, y, z) = x²+ y²+ z²− 1.

By the method of Lagrange multiplier, we have the set of equations

2x+ 4z = 2λx (1)

−4y = 2λy (2)

4x− 4z = 2λz (3)

along with g(x, y, z) = 0. Equation (2) gives either y = 0 or λ = −2. If λ = −2, the system becomes a set of linear equations

3x+ 2z = 0 x= 0

from which we get(x, y, z) = ±(0, 1, 0). If y = 0, we can eliminate λ by x × (3) − z × (1) and obtain 2x²− 3xz − 2z²= (2x + z)(x − 2z) = 0,

hence z=^x2,−2x. If z = ^x2, we have g(x, 0,^x2) = ⁵4x²−1 = 0, from which we get (x, y, z) = ± (^√2₅, 0,^√1

5). If z = −2x, we have g(x, 0, −2x) = 5x²− 1 = 0, from which we get (x, y, z) = ± (^√1₅, 0,−^√2₅). The values of f at those six critical points are

f(±(0, 1, 0)) = −2, f(±( 2

√5, 0, 1

√5)) = 2, f(±( 1

√5, 0,− 2

√5)) = −3, respectively, so the maximum is 2 and the minimum is−3.

Marking scheme

ˆ Setting up the equation of the Lagrange multiplier correctly (4%).

– Incorrect equations with a very minor mistakes (an incorrect sign etc.) (3%).

ˆ 1% for finding each of the six solutions correctly (6% in total).

– 1% for being aware of the existence of the two cases λ= −2 and y = 0 (with incomplete calculation) – 1% for finiding the fact x= 2z, −z/2 (or the factorization (x − 2z)(2x + z) = 0) for y = 0 case.

ˆ Finding correct maximal value (1%), minimal value (1%)

Page 4 of 8

5. Depicted in the Figure, E is an ‘apple-shaped’ solid that, in spherical coordinates, occupies the region E = {(ρ, ϕ, θ) ∶ 0 ≤ ρ ≤ 1 − cos ϕ, 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π}.

It is known that E has a constant density ρ(x, y, z) = 2.

(a) (6%) Find the mass of E.

(b) (8%) Let(0, 0, z) be the center of mass of E. Find z.

Figure. The apple-shaped solid E

Solution:

Prob.5: The following crucial steps must be shown clearly

(a) Total mass:

M = ∫ ∫ ∫^E2 dV = 2 ∫^02π∫^0π∫^{01−cos ϕ}ρ²sin ϕdρdϕdθ ...2%

= 4π ∫^{0π(1−cos ϕ)3 3}sin ϕ dϕ ...2%

either= ^4π3 ∫⁻¹¹(1 − x)³dx =^4π3 [^(1−x)4 ⁴]⁻¹1 = ^16π3

or= ^4π3 [^{(1−cos ϕ)}4 ⁴]^π0 = ^16π3 ...2%

(b) The z-component of center of mass:

¯

z= M¹ ∫ ∫ ∫^E2z dV =M² ∫^02π∫^0π∫^{01−cos ϕ}ρ³sin ϕ cos ϕ dρdϕdθ ....3%

=M^π ∫^0π(1 − cos ϕ)⁴sin ϕ cos ϕ dϕ ...2%

either= M^π ∫⁻¹¹ x(1 − x)⁴dx =16³[^(1−x)5 ⁵ − ^(1−x)6 ⁶]⁻¹1 = ⁻⁴5

or= M^π[^{(1−cos ϕ)}5 ⁵ − ^{(1−cos ϕ)}6 ⁶]^π0 = 16³

−128

30 = ⁻⁴5 ...3%

6. (a) (9%) Find the volume of the solid that is below the parabolid z= 9 − x²− y² and above the region enclosed by the lemniscate r²= cos(2θ) on the xy-plane (See Figure).

(b) (9%) Evaluate the triple integral

∭_R(36 − x²− 4y²− 9z²) dV where R= {(x, y, z) ∈ R³ ∶ x²+ 4y²+ 9z²≤ 36}.

Figure. The lemniscate r²= cos(2θ)

Solution:

(a) In cylindrical coordinates, the volume is

∫

π/4

−π/4∫

√ cos(2θ)

0 ∫

9−r² 0

r dz dr dθ+ ∫_3π/4^5π/4∫

√ cos(2θ)

0 ∫

9−r² 0

r dz dr dθ By symmetry, the two integrals are the same value.

= 2 ∫_−π/4^π/4∫

√ cos(2θ)

0 ∫

9−r² 0

r dz dr dθ= 2 ∫_−π/4^π/4∫

√ cos(2θ) 0

9r− r³ dr dθ

= 2 ∫_−π/4^π/49 cos(2θ)

2 −cos²(2θ)

4 dθ= ∫_−π/4^π/49 cos(2θ) −1+ cos(4θ)

4 dθ= 9 −π 8 (b) Let x= 6u, y = 3v, z = 2w, then the triple integral becomes

∭_u2+v²+w²≤1(36 − 36x²− 36y²− 36z²)∣J∣ dV = 6⁴∭_u2+v²+w²≤1(1 − u²− v²− w²) dV Use spherical coordinates u= ρ sin ϕ cos θ, v = ρ sin ϕ sin θ, w = ρ cos ϕ.

= 6⁴∫

2π

0 ∫

π 0 ∫

1

0 (1 − ρ²)ρ²sin ϕ dρ dϕ dθ= (6⁴)(2π)(2) (1 3−1

5) =3456π 5 Grading: In general, -2% for each big mistake and -1% for each small mistake.

(a) 5% for cylindrical coordinates integral setup, 4% for computation (they get partial credit in computation if their setup is similar to the answer).

(b) 5% for setting up an integral that they can evaluate, 4% for computation (they get partial credit in compu- tation if their setup is similar to the answer).

Note: Both problems can be evaluated using xyz-coordinates.

Page 6 of 8

7. (a) (i) (2%) For each fixed value of y, find ∫

√ 3 1

cos(xy) dx.

(ii) (6%) Use your result in (i) to transform

I= ∫₀^∞e^−y⋅ (sin(√

3y) − sin(y))

y dy

into a double integral and then evaluate I by Fubini’s Theorem.

(b) (10%) Use the change of variables u= xy and v = y to evaluate

∬_R y

1+ x²y²dA where R is the region enclosed by the curves xy= 1, xy =√

3, x= 1 and y = 3.

Solution:

(a) (i) Marking scheme for 7(a)(i).

All or nothing. 1M for sign error.

∫

√ 3 1

cos(xy) dx = [sin(xy)

y ]

√ 3

1

= sin(√

3y) − sin(y) y

(ii)

Marking scheme for 7(a)(ii).

(1M) Use (a) to transform I into a double integral (1M) Use Fubini’s theorem (correctly)

(2M)* Correct antiderivative for dy

(1M) For realising limy→∞e^−y⋅ (bounded function) = 0 (1M) Correct answer

Remark for *. 1M for any candidates who apply IBP twice but yield incorrect anti-derivative.

∫

∞ 0

e^−y⋅ (sin(√

3y) − sin(y))

y dy^(a)= ∫₀^∞∫

√ 3

1 e^−y⋅ cos(xy) dx dy

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M ) Fubini

= ∫

√ 3

1 ∫

∞ 0

e^−y⋅ cos(xy) dy dx

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M )

IBP= ∫

√ 3 1

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎣

e^−y⋅ (x sin (xy) − cos (xy)) x²+ 1

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(2M )

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎦

∞

0

dx

= ∫

√ 3 1

1 1+ x²

´¹¹¹¹¹¹¸¹¹¹¹¹¹¶

(1M )

dx

= π

¯12

(1M )

Marking scheme for 7(b).

(2M) correct Jacobian

(1M+1M) correct u- and v-components of the transformed region (okay to just sketch the region)

given region is transformed as a trapezoidal region {(u, v) ∶ 1 ≤ u ≤√

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶3

(1M )

and u≤ v ≤ 3

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M )

}.

∬_R y

1+ x²y²dA= ∫

√ 3

1 ∫

3 u

v 1+ u² ⋅1

vdv du

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(3M )

= ∫

√ 3 1

3− u 1+ u²du

=

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎣

3 tan⁻¹(u)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M )

−1

2ln(1 + u²)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1M )

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎦

√ 3

1

=π 4 −ln 2

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¶2

(1M )

Page 8 of 8