1082 Calculus A 01-10, Mode 01-02 Makeup Midterm

1082 Calculus A 01-10, Mode 01-02 Makeup Midterm

1. (15 pts) Consider the function

f(x, y) =⎧⎪⎪⎪

⎨⎪⎪⎪⎩

sin(xy)

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

(a) (3 pts) Is f(x, y) continuous at (0, 0)?

(b) (4 pts) Find fx(0, 0), fy(0, 0), and fx(x, y), fy(x, y) for (x, y) ≠ (0, 0).

(c) (2 pts) Write down the linear approximation of f(x, y) at (0, 0), L(x, y).

(d) (4 pts) Compute lim

(x,y)→(0,0)

f(x, y) − L(x, y)√

x²+ y² if it exists.

(e) (2 pts) Is f(x, y) differentiable at (0, 0)?

Solution:

(a) We consider polar coordinate, let x= r cos θ and y = r sin θ, then

f(x, y) = f(r, θ) =⎧⎪⎪⎪

⎨⎪⎪⎪⎩

sin(r²cos θ sin θ√ )

r² ,if r≠ 0,

0 ,if r= 0.

By definition of continuous, we can consider the countinuity of f(x, y) at (x, y) = (0, 0) by taking r→ 0⁺. lim

r→0⁺f(r, θ) = lim

r→0⁺

sin(r²cos θ sin θ)

r = lim

r→0⁺

sin(r²cos θ sin θ)

r²cos θ sin θ ⋅ r cos θ sin θ.

Since both lim

r→0⁺

sin(r²cos θ sin θ)

r²cos θ sin θ and lim

r→0⁺r cos θ sin θ are exist with lim

r→0⁺

sin(r²cos θ sin θ) r²cos θ sin θ = 1 and lim

r→0⁺r cos θ sin θ= 0, then

rlim→0⁺

sin(r²cos θ sin θ)

r²cos θ sin θ ⋅ r cos θ sin θ = lim

r→0⁺

sin(r²cos θ sin θ) r²cos θ sin θ ⋅ lim

r→0⁺r cos θ sin θ= 1 ⋅ 0 = 0.

That is lim

(x,y)→(0,0)f(x, y) = f(0, 0), so we have f(x, y) is continuous at (0, 0).

(b) By definition of partial derivative at (0, 0).

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎨⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎩

f_x(0, 0) = lim

h→0

f(h, 0) − f(0, 0)

h = lim

h→0 sin√(h⋅0)

h² − 0

h = 0

f_y(0, 0) = lim

k→0

f(0, k) − f(0, 0)

k = lim

k→0 sin√(0⋅k)

k² − 0

k = 0

and we can derivative for(x, y) ≠ (0, 0) since f(0, 0) is differentiable except (x, y) = (0, 0).

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎨⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎩

f_x(x, y) = cos(xy) ⋅ y ⋅√

x²+ y²− sin(xy) ⋅¹₂^√_x^2x_2+y2 (√

x²+ y²)² = y(x²+ y²) cos(xy) − x sin(xy) (x²+ y²)³²

fy(x, y) = cos(xy) ⋅ x ⋅√

x²+ y²− sin(xy) ⋅¹₂^√_x^2y_2+y2 (√

x²+ y²)² = x(x²+ y²) cos(xy) − y sin(xy) (x²+ y²)³²

(c) Linear approximation of f(x, y) at (0, 0) is

L(x, y) = f(0, 0) + fx(0, 0)x + fy(0, 0)y

= 0 + 0 + 0 = 0

Page 1 of 9

(d)

(x,y)→(0,0)lim

f(x, y) − L(x, y)√

x²+ y² = lim

(x,y)→(0,0)

sin(xy) x²+ y². Consider two straight line x= y and x = −y then

⎧⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎩

(x,y)→(0,0)lim

sin(xy) x²+ y² = lim

x→0

sin(x²) 2x² = 1

2, if x= y

(x,y)→(0,0)lim

sin(xy) x²+ y² = lim

x→0

sin(−x²) 2x² = −1

2, if x= −y (d)

(x,y)→(0,0)lim

f(x, y) − L(x, y)√

x²+ y² = lim

(x,y)→(0,0)

sin(xy) x²+ y². That is the limit lim

(x,y)→(0,0)

f(x, y) − L(x, y)√

x²+ y² does not exist.

(e) No, f(x, y) is not differentiable at (0, 0) since lim

(x,y)→(0,0)

f(x, y) − L(x, y)√

x²+ y² does not exist.

Page 2 of 9

2. (15 pts) Consider the level surface S defined by

S = {(x, y, z) ∈ R³ ∶ z⁵− xz⁴+ yz³= 1}.

This level surface defines z= z(x, y) implicitly as a differentiable function of x and y.

(a) (4 pts) Find an equation of the tangent plane at the point P(0, 0, z(0, 0).) (b) (4 pts) Use linear approximation at P to estimate the value z(0.02, −0.03).

(c) (5 pts) Find ∂²z

∂x∂y ∣

(x,y)=(0,0)

.

(d) (2 pts) Find the directional derivative D_uz(0, 0) along the direction ⃗u parallel to the vector (−3, 4).

Solution:

(a) Let F(x, y, z) = z⁵− xz⁴+ yz³ − 1, then S = {(x, y, z) ∈ R³ ∶ F (x, y, z) = 0}, then the normal vector of tangent plane at each point (x, y, z) is ∇F (x, y, z) = (−z⁴, z³, 5z⁴− 4xz³+ 3yz²), that is the normal vector at (0, 0, z(0, 0)) = (0, 0, 1) is (−1, 1, 5). Then the tangent plane at the point P(0, 0, 1) is −(x − 0) + (y − 0) + 5(z − 1) = 0.

(b) Linear approximate at P is L(x, y) = z(0, 0) + zx(0, 0)(x − 0) + zy(0, 0)(y − 0) = z(0, 0) − F_x(0, 0, 0)

F_z(0, 0, 0)(x − 0) −F_y(0, 0, 0)

F_z(0, 0, 0)(y − 0) ⇒ z(0.02, −0.03) = 1 +1

5⋅ 0.02 −1

5⋅ (−0.03) = 1.01

Actually, by tangent plane at P(0, 0, 1), −(0.02 − 0) + (−0.03 − 0) + 5(z(0.02, −0.03) − 1) = 0 ⇒ 5(z(0.02, −0.03) − 1) = 0.05 ⇒ z(0.02, −0.03) = 1 +0.05

5 = 1.01.

(c) Since ∂z

∂y = −F_y(x, y, z)

F_z(x, y, z) = − z³

5z⁴− 4xz³+ 3yz² = −z 5z²− 4xz + 3y

⇒ ∂²z

∂x∂y∣

(x,y)=(0,0) = ∂

∂x( −z

5z²− 4xz + 3y)∣

(x,y)=(0,0)

= −zx(5z²− 4xz + 3y) + z(10zzx− 4(z + xzx)) (5z²− 4xz + 3y)² ∣

(x,y)=(0,0)

= −¹₅⋅ 5 + 1 ⋅ (10 ⋅ 1 ⋅¹₅ − 4)

5² = −1 + (−2)

25 = − 3 25 (d) D_⃗uz(0, 0) = ∇z(0, 0) ⋅ ⃗u = (1

5,−1

5) ⋅ ± (−3 5 ,4

5) = ± 7 25

Page 3 of 9

3. (14 pts) Let f(x, y) = y(x + 1)².

(a) (6 pts) Find and classify critical point(s) of f(x, y).

(b) (8 pts) Find the extreme values of f(x, y) on the region R = {(x, y)∣x²− 1 ≤ y ≤ 3}.

Solution:

(a) ∇f(x, y) = (2y(x + 1), (x + 1)²) = (0, 0) iff (x, y) = (−1, y)

⇒⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

fxx= 2y

fxy = 2(x + 1) = fyx

fyy= 0

⇒ D∣(x,y)=(−1,y)= fxxf_yy− fxy² ∣(x,y)=(−1,y)= 0 ∀y ∈ R.

If y > 0, f(x, y) ≥ 0 for all x ≠ −1,then (−1, y) is local minimum. With same argument for y< 0, (−1, y) is local maximun. The last one is (−1, 0) is a saddle point.

(b) R= {(x, y)∣x²− 1 ≤ y ≤ 3}.

First, we have the local minimum in the region equal to 0 when (x, y) = (−1, y) and y > 0.

Second, we consider the boundary x²− 1 ≤ y ≤ 3 ⇒ x²− 1 = 3 with −2 ≤ x ≤ 2 and y = 3 then we have the local maximum when (x, y) = (2, 3) is equal to 27.

The last, y= x²−1 ⇒ f(x, y) = f(x, x²−1) = (x²−1)(x+1)²= g(x) with −2 ≤ x ≤ 2, then g^′(x) = 2x(x+1)²+2(x²−1)(x+1) = 2(x+1)(x²+x+x²−1) = 2(x+1)(2x²+x−1) = 2(x+1)²(2x−1) = 0 for x= −1, 1

2 ⇒ (x, y) = (1 2,−3

4) has local minimum is equal to −27 16.

Page 4 of 9

4. (10 pts) Let C be the hyperbola formed by the intersection of the cone 4x²= y²+3z² and the plane x+ y = 6. Find the maximum and the minimum distance between the origin and the point on C (if exist) by the method of Lagrange multipliers.

Solution:

That is to find the extreme of d(x, y, z) = x²+y²+z² with the restriction x+y = 6 and 4x² = y²+3z² By Lagrange Multipliers

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

2x= λ + µ8x 2y= λ + µ(−2y) 2z= 0 + µ(−6z)

⇒⎧⎪⎪⎪

⎨⎪⎪⎪⎩

λ= (2 − 8µ)x = (2 + 2µ)y µ= −1

3 or z = 0 ⇒⎧⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎩

λ=14 3 x= 4

3y or

λ= (2 − 8µ)x = (2 + 2µ)y, z = 0 (i) If λ= 14

3 x= 4

3y, x+ y = 6 and 4x²= y²+ 3z²

⇒ 3 14λ+3

4λ= 6 ⇒ 27

28λ= 6 ⇒ λ = 56

9 ⇒ (x, y) = (4 3,14

3 ) ⇒ (x, y, z) not exists in R³. (ii) If λ= (2 − 8µ)x = (2 + 2µ)y, z = 0, x + y = 6 and 4x² = y²+ 3z²

⇒ 4x² = y² ⇒ 2x = ±y and x + y = 6 ⇒ (x, y) = (2, 4) or (−6, 12) Since there is no maximun distance between origin and the point on hyperbola C, we have the minimun at (2, 4, 0) is equal to √

20.

Page 5 of 9

5. (12 pts)

(a) Compute ∫

1 0 ∫

cos⁻¹y 0

sin x(1 + sin²x)¹³dxdy.

(b) Let f(x) = ∫_x¹e^−t2dt. Find the average value of f on the interval[0, 1].

Solution:

(a)

∫

1 0 ∫

cos⁻¹y 0

sin x(1 + sin²x)¹³dxdy = ∫

π 2

0 ∫

cos x 0

sin x(1 + sin²x)¹³dydx= ∫

π 2

0

sin x(1 + sin²x)¹³ cos xdx

= 1

2(1 + sin²x)⁴³ ⋅3 4∣

π 2

0

=3⋅ 2⁴³ 8 −3

8

(b) Average value of f = ∫

1 0 f dx

1− 0 = ∫₀¹∫

1 x

e^−t2dtdx

⇒ ∫₀¹∫

1 x

e^−t2dtdx= ∫₀¹∫

t 0

e^−t2dxdt= ∫₀¹te^−t2dt= −1 2 e^−t2∣¹

0

=1

2(1 − e⁻¹)

Page 6 of 9

6. (12 pts) Find the center of mass of a lamina

D= {(x, y) ∈ R²∣(x − 2)²

4 + y² ≤ 1 and x ≤ 2}

whose density function at any point is proportional to the square of its distance from the line x= 2.

Solution:

Assume the coordinate of center of mass is (¯x, ¯y), then obvious ¯y = 0.

⇒ ¯x ⋅ ∬

D

C(x²+ y²)dA = ∬

D

Cx(x²+ y²)dA ⇒ ¯x = ∬^Dx(x²+ y²)dA

∬D(x²+ y²)dA

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎨⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎩

∬

D

(2 − x)²dA= ∫₀²∫

√ 1−^(x−2)2₄

−

√

1−^(x−2)2₄ (2 − x)²dydx= ∫₀²2

√

1−(x − 2)²

4 (2 − x)²dx

∬

D

x(2 − x)²dA= ∫₀²∫

√ 1−^(x−2)2₄

−

√

1−^(x−2)2₄ x(2 − x)²dydx= ∫₀²2

√

1−(x − 2)²

4 x(2 − x)²dx

Let t= x− 2

2 ⇒ dx = 2dt

⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎨⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎩

∫

2 0 2

√

1−(x − 2)²

4 (2 − x)²dx= ∫₋₁⁰2√

1− t²4t²⋅ 2dt

∫

2 0

2

√

1−(x − 2)²

4 x(2 − x)²dx= ∫₋₁⁰2√

1− t²⋅ (2t + 2) ⋅ 4t²⋅ 2dt Let t= cos θ ⇒ dt = − sin θdθ

⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎪⎩

− ∫₋₁⁰2√

1− t²4t²⋅ 2dt = ∫π^π 2

16 sin θ cos²θ sin θdθ= 16 ∫π^π 2

sin²θ cos²θdθ

− ∫₋₁⁰2√

1− t²⋅ (2t + 2) ⋅ 4t²⋅ 2dt = ∫π^π 2

32 sin θ cos²θ sin θ(cos θ + 1)dθ = 32 ∫π^π 2

sin²θ cos³θ+ sin²θ cos²θdθ

⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎪⎩

16∫

π

π 2

sin²θ cos²θdθ= 2 ∫π^π 2

sin²2θd2θ= 2 ∫π^π 2

cos²2θd2θ= 2 ∫π^π 2

sin²2θ+ cos²2θ

2 d2θ= 2π − π = π

32∫

π

π 2

sin²θ cos³θdθ= 32 ∫π^π 2

sin²θ(1 − sin²θ) cos θdθ = 32 (1

3sin³θ−1

5sin⁵θ)^π

π 2

= −64 15

⇒ ∬

D

(2 − x)²dA= π and ∬

D

x(2 − x)²dA= −64 15+ 2π

∴¯x = 2π−⁶⁴₁₅

π ⇒ center of mass : (¯x, ¯y) = (2 − 64 15π, 0)

Page 7 of 9

7. (12 pts) Use spherical coordinates to evaluate ∫

√3

0 ∫

√3−y²

−√ 3−y² ∫

√4−x²−y² 1

√ 1

x²+ y²+ z²dzdxdy.

Solution:

By sperical coordinates, let ⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

x= ρ cos θ sin ϕ y= ρ sin θ sin ϕ z= ρ cos ϕ

, where 0≤ ρ ≤ 2 0 ≤ ϕ ≤ π and 0 ≤ θ ≤ 2π.

⇒ ∫

√3

0 ∫

√3−y²

−√ 3−y² ∫

√4−x²−y² 1

√ 1

x²+ y²+ z²dzdxdy = ∫₀^π∫

π 3

0 ∫

2 sec ϕ

ρ²sin ϕ

ρ dρdϕdθ=π 2 ∫

π 3

0 (4 − sec²ϕ) sin ϕdϕ

= π 2 ∫

π 3

0

4 sin ϕ− sec ϕ tan ϕdϕ = π

2(−4 cos ϕ − sec ϕ)∣

π 3

0

= π 2

Page 8 of 9

8. (10 pts) Compute∬_Rln(x²y+x)dA, where R is the region bounded by curves xy = 1, xy = 3, x = 1 and x= e.

Solution:

By change variables, let ⎧⎪⎪

⎨⎪⎪⎩

u= xy, 1 ≤ u ≤ 3

v= x, 1 ≤ v ≤ e ⇒ ∣J∣ = ∣∂(x, y)

∂(u, v)∣ = ∣ 0 1 1 v

−u v²

∣ = 1 v

⇒ ∬_Rln(x²y+ x)dA = ∫₁^e∫

3 1

ln(uv + v)∣J∣dudv = ∫₁^e∫

3 1

ln[(u + 1)v]1 vdudv

= ∫₁^e∫₁³ln(u + 1) + ln v

v dudv

= ∫₁^e[(u + 1) ln(u + 1) − (u + 1)]³1

1

v + (3 − 1)ln v v dv

= ln e[(3 + 1) ln(3 + 1) − (3 + 1) − 2 ln 2 + 2] + (3 − 1) 2 (ln e)²

= 8 ln 2 − 4 − 2ln2 + 2 + 1 = 6 ln 2 − 1

Page 9 of 9