Final Exam 1516 - Calculus 1 | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố HCM

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Môn: Calculus 1 (MA001IU)

Trường: Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh

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Profile Mai Nguyệt
INTERNATIONAL UNIVERSITY-VNUHCM
FINAL EXAMINATION
Semester 3, Academic Year 2015-2016
Duration: 120 minutes
SUBJECT:
Calculus 2
Chair of Department of Mathematics Lecturers:
Signature: Signature:
Full name: Assoc.Prof. Nguyen Dinh Full names: Assoc.Prof. Mai Duc Thanh
• Each student is allowed a maximum of two double-sided sheets of reference
material (of size A4 or similar) and a scientific calculator. All other documents
and electronic devices are forbidden.
• Each question carries 20 marks.
Question 1.
a) Find the first partial derivatives of the function f (x, y) = .e
4x−y
2
b) Find the directional derivative
D
u
f(x, y) of the function f(x, y) = e
4x−y
2
at the point
(1
, 2) in the direction of the vector u =< 1/
√
2 1, − /
√
2 >.
Question 2. Find the local maximum and minimum values and saddles point(s) of the function
f
(x, y .) = e
x
(x
2
− y
2
)
Question 3. a) Evaluate the double integral
I
=
ZZ
D
2 1
y dA, D = {(x, y) | 0 ≤ x ≤ , x .
2
≤ y ≤
√
x}
b) Find the volume of the solid under the surface z = 1 + 2xy and above the region in the
xy
-plane bounded by y = 1 −x and y = 1 − x
2
.
Question 4. a) Evaluate the triple integral
ZZZ
E
x dV, E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, .0 ≤ z ≤ x + 2y}
b) Evaluate the line integral
Z
C
(2
x
2
− y) dx + 2y dy
where
C is the arc of the curve y = x
2
from (0, 0) to (2, 4).
Question 5. Let a vector field F(x, y, z) = xyi + yj + zk be given.
a) Evaluate the line integral
R
C
F · dr, where C has the parametric equations =x = t, y
t
2
, z = e
t
, 0 ≤ t ≤ 1.
b) Evaluate the surface integral
RR
S
F · dS, where S is the part of the paraboloid =z
16
− 3x
2
− 3y
2
that lies above the rectangle 0 ≤ ≤x 2, 0 ≤ y ≤ 1.
1
SOLUTIONS OF FINAL EXAM
Subject: CALCULUS 2
Question 1.
a) Find the first partial derivatives of the function f (x, y) = .e
4x−y
2
f f
x
(x, y) = 4e
4x−y
2
y
(x, y) = −2ye
4x−y
2
.
b) Find the directional derivative of the function
f(x, y) = e
4x−y
2
at the point (1 2) in the,
direction of the vector
u = (1/
√
2, −1/
√
2).
D
u
f f(x, y) = ∇ (x, y) · u =< 4, −4 > · < 1/
√
2, −1/
√
2 >= 4 2
√
Question 2. Find the local maximum and minimum values and saddles point(s) of the function
f
(x, y .) = e
x
(x
2
− y
2
)
Partial derivatives
f
x
(x, y x, y) = e
x
(x
2
− y
2
+ 2x) f
y
( ) = −2ye
x
Critical points:
f f
x
(x, y) = e
x
(x
2
− y
2
+ 2x) = 0
y
(x, y) = −2ye
x
= 0
or
y = 0, x(x + 2) = 0
so that x = 0 or x = −2 and y = 0. So, there are two critical points (0, 0) and (−2, 0).
Second partial derivatives:
f
xx
(x, y) = e
x
(x
2
− y
2
+ 4x + 2), f , f
xy
(x, y) = −2ye
x
yy
(x, y .) = −2e
x
Consider
D
= f f f
xx
(x, y)
yy
(x, y) − (
xy
(x, y .))
2
At M
1
(0, 0), it holds
D
= f f
xx
(0, 0)f
yy
(0, 0) − (
xy
(0, 0))
2
= 2(−2) − 0 = −4 < 0,
so (0,0) is a saddle point.
At M
2
(−2, 0) it holds
D
= f f
xx
(−2, 0)f
yy
(−2, 0) − (
xy
(−2, 0))
2
= e
−4
(−6)(−2) − 0 = 12e
−4
> 0
and
f
xx
(−2, 0) = −6e
−4
< 0. So f(−2, 0) = 4e
−2
is a local maximum value.
Question 3. a)
Z
1
0
Z
√
x
x
2
2
y dy dx =
Z
1
0
y
2
y
=
√
x
y
=x
2
dx
=
Z
1
0
(
x − x
4
)dx = (x
2
/2 − x
5
/5)
1
0
= 1 5 = 3/ /2 − 1 /10.
2
b) The volume is given by
V
=
ZZ
D
(1 + 2
xy) dA, D = {(x, y)|0 ≤ x ≤ 1, 1 − x ≤ y ≤ 1 − x
2
},
Therefore
V
=
Z
1
0
Z
1−x
2
1−x
(1 + 2xy)dydx
=
Z
1
0
( )
y + xy
2
y
=1−x
2
y
=1−x
dx
=
Z
1
0
[1
− x
2
+ x(1 (1− x
2
)
2
− −x) −x(1 − x)
2
]dx
=
Z
1
0
[1
− x
2
+ x(1 (1− 2x
2
+ x
4
) − −x) − x(1 − 2x + x
2
)]
=
Z
1
0
( )
x
5
− 3x
3
+ x
2
+ x dx = (x
5
/5 − 3x
4
/4 + 3 +x
3
/ x
2
/2)
1
0
= 17/60 = 0.2833.
Question 4. a) Evaluate the triple integral
ZZZ
E
x dV, E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, .0 ≤ z ≤ x + 2y}
ZZZ
E
xdV
=
Z
1
0
Z
x
0
Z
0
x+2y
xdzdydx
=
Z
1
0
Z
x
0
x
(x + 2y)dydx =
Z
1
0
Z
x
0
( )
x
2
+ 2xy dydx
=
Z
1
0
(
x
2
y + xy
2
)
x
0
dx =
Z
1
0
2
x
3
dx
=
x
4
/2
1
0
= 1 2/
b) Evaluate the line integral
Z
C
(2
x
2
− y) dx + 2y dy
where
C is the arc of the curve y = x
2
from (0, 0) to (2, 4).
We have
y
0
= 2x, 0 ≤ x ≤ 2. Thus,
Z
C
(2
x
2
− y) dx + 2y dy =
Z
2
0
[(2
x x
2
−
2
) + 2x
2
2x] dx
=
Z
2
0
(4 3)
x
3
+ x
2
) dx = (x
4
+ x
3
/
2
0
= 2
3
(2 + 1/3) = 56 3 = 18/ .67.
Question 5. Let a vector field F(x, y, z) = xyi + +yj zk be given.
a) Evaluate the line integral
R
C
F · dr, where C has the parametric equations =x = t, y
t
2
, z = e
t
, 0 ≤ t ≤ 1.
3
Vector equation of
C: r(t) =< t, t
2
, e
t
>, 0 ≤ t ≤ 1. We have r
0
(t) =< 1, 2t, e
t
>, and
F
(r(t)) =< t > .
3
, t , e
2 t
Thus
Z
C
F
· dr =
Z
1
0
F
(r r(t)) ·
0
(t)dt
=
Z
1
0
(
t
3
+ 2t
3
+ e
2t
)dt =
Z
1
0
(3
t
3
+ e
2t
)dt
= (3
t
4
/4 + e
2t
/2)
1
0
= 3/4 + e
2
/2 − 1/2 = e
2
/ /2 + 1 4 = 3.9445
b) Evaluate the surface integral
RR
S
F · dS, where S is the part of the paraboloid =z
16
− 3x
2
− 3y
2
that lies above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.
Set F(x, y, z) =< P, Q, R >, where P = xy, Q = y, R = z, and S is given by z = g(x, y) =
16
− 3x
2
− 3y
2
. Applying the formula
ZZ
S
F
· dS =
ZZ
D
(−P g
x
− Qg
y
+ R) dA
=
ZZ
D
[
−xy(−6 6 3x) − y(− y) + (16 − x
2
− 3y
2
)] dA
=
ZZ
D
(6
x
2
y + 3 + 16)y
2
− 3x
2
dA
=
Z
2
0
Z
1
0
(6
x
2
y + 3 + 16)y
2
− 3x
2
dydx
=
Z
2
0
(3
x
2
y
2
+ y
3
− 3x
2
y + 16y)
1
0
)dx
=
Z
2
0
(3
x
2
+ 1 + 16)− 3x
2
dx =
Z
2
0
17dx = 34
4
| 1/4
| | Tải xuống

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INTERNATIONAL UNIVERSITY-VNUHCM FINAL EXAMINATION
Semester 3, Academic Year 2015-2016 Duration: 120 minutes SUBJECT: Calculus 2
Chair of Department of Mathematics Lecturers: Signature: Signature:
Full name: Assoc.Prof. Nguyen Dinh
Full names: Assoc.Prof. Mai Duc Thanh
• Each student is allowed a maximum of two double-sided sheets of reference
material (of size A4 or similar) and a scientific calculator. All other documents
and electronic devices are forbidden.
• Each question carries 20 marks.
Question 1. a) Find the first partial derivatives of the function f (x, y) = e4x−y2.
b) Find the directional derivative Duf (x, y) of the function f (x, y) = e4x−y2 at the point √ √
(1, 2) in the direction of the vector u =< 1/ 2, −1/ 2 >.
Question 2. Find the local maximum and minimum values and saddles point(s) of the function f (x, y) = ex(x2 − y2).
Question 3. a) Evaluate the double integral Z Z √ I = 2y dA,
D = {(x, y) | 0 ≤ x ≤ 1, x2 ≤ y ≤ x}. D
b) Find the volume of the solid under the surface z = 1 + 2xy and above the region in the
xy-plane bounded by y = 1 − x and y = 1 − x2.
Question 4. a) Evaluate the triple integral Z Z Z x dV,
E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ x + 2y}. E b) Evaluate the line integral Z (2x2 − y) dx + 2y dy C
where C is the arc of the curve y = x2 from (0, 0) to (2, 4).
Question 5. Let a vector field F(x, y, z) = xyi + yj + zk be given.
a) Evaluate the line integral R F · dr, where C has the parametric equations x = t, y = C t2, z = et, 0 ≤ t ≤ 1.
b) Evaluate the surface integral RR F · dS, where S is the part of the paraboloid z = S
16 − 3x2 − 3y2 that lies above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1. 1 SOLUTIONS OF FINAL EXAM Subject: CALCULUS 2
Question 1. a) Find the first partial derivatives of the function f (x, y) = e4x−y2. fx(x, y) = 4e4x−y2 fy(x, y) = −2ye4x−y2.
b) Find the directional derivative of the function f (x, y) = e4x−y2 at the point (1, 2) in the √ √
direction of the vector u = (1/ 2, −1/ 2). √ √ √
Duf (x, y) = ∇f(x, y) · u =< 4, −4 > · < 1/ 2, −1/ 2 >= 4 2
Question 2. Find the local maximum and minimum values and saddles point(s) of the function f (x, y) = ex(x2 − y2). Partial derivatives
fx(x, y) = ex(x2 − y2 + 2x) fy(x, y) = −2yex Critical points:
fx(x, y) = ex(x2 − y2 + 2x) = 0 fy(x, y) = −2yex = 0 or y = 0, x(x + 2) = 0
so that x = 0 or x = −2 and y = 0. So, there are two critical points (0, 0) and (−2, 0). Second partial derivatives:
fxx(x, y) = ex(x2 − y2 + 4x + 2), fxy(x, y) = −2yex, fyy(x, y) = −2ex. Consider
D = fxx(x, y)fyy(x, y) − (fxy(x, y))2. At M1(0, 0), it holds
D = fxx(0, 0)fyy(0, 0) − (fxy(0, 0))2 = 2(−2) − 0 = −4 < 0, so (0,0) is a saddle point. At M2(−2, 0) it holds
D = fxx(−2, 0)fyy(−2, 0) − (fxy(−2, 0))2 = e−4(−6)(−2) − 0 = 12e−4 > 0
and fxx(−2, 0) = −6e−4 < 0. So f(−2, 0) = 4e−2 is a local maximum value. Question 3. a) √ √ Z 1 Z x Z 1 y= x 2y dy dx = y2  dx  0 x2 0 y=x2 Z 1 1 =
(x − x4)dx = (x2/2 − x5/5) = 1/2 − 1/5 = 3/10.  0 0 2 b) The volume is given by Z Z V = (1 + 2xy) dA,
D = {(x, y)|0 ≤ x ≤ 1, 1 − x ≤ y ≤ 1 − x2}, D Therefore Z 1 Z 1−x2 V = (1 + 2xy)dydx 0 1−x Z 1 y=1−x2 = (y + xy2) dx  0 y=1−x Z 1 =
[1 − x2 + x(1 − x2)2 − (1 − x) − x(1 − x)2]dx 0 Z 1 =
[1 − x2 + x(1 − 2x2 + x4) − (1 − x) − x(1 − 2x + x2)] 0 Z 1 1 =
(x5 − 3x3 + x2 + x)dx = (x5/5 − 3x4/4 + x3/3 + x2/2) = 17/60 = 0.2833.  0 0
Question 4. a) Evaluate the triple integral Z Z Z x dV,
E = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ x + 2y}. E Z Z Z Z 1 Z x Z xdV = 0x+2yxdzdydx E 0 0 Z 1 Z x Z 1 Z x = x(x + 2y)dydx = (x2 + 2xy)dydx 0 0 0 0 Z 1 Z 1  x = (x2y + xy2) dx = 2x3dx  0 0 0  1 = x4/2  = 1/2 0 b) Evaluate the line integral Z (2x2 − y) dx + 2y dy C
where C is the arc of the curve y = x2 from (0, 0) to (2, 4).
We have y0 = 2x, 0 ≤ x ≤ 2. Thus, Z Z 2 (2x2 − y) dx + 2y dy = [(2x2 − x2) + 2x22x] dx C 0 Z 2  2 =
(4x3 + x2) dx = (x4 + x3/3) = 23(2 + 1/3) = 56/3 = 18.67.  0 0
Question 5. Let a vector field F(x, y, z) = xyi + yj + zk be given.
a) Evaluate the line integral R F · dr, where C has the parametric equations x = t, y = C t2, z = et, 0 ≤ t ≤ 1. 3
Vector equation of C: r(t) =< t, t2, et >, 0 ≤ t ≤ 1. We have r0(t) =< 1, 2t, et >, and
F(r(t)) =< t3, t2, et > . Thus Z Z 1 F · dr = F(r(t)) · r0(t)dt C 0 Z 1 Z 1 = (t3 + 2t3 + e2t)dt = (3t3 + e2t)dt 0 0 1
= (3t4/4 + e2t/2) = 3/4 + e2/2 − 1/2 = e2/2 + 1/4 = 3.9445 0
b) Evaluate the surface integral RR F · dS, where S is the part of the paraboloid z = S
16 − 3x2 − 3y2 that lies above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.
Set F(x, y, z) =< P, Q, R >, where P = xy, Q = y, R = z, and S is given by z = g(x, y) =
16 − 3x2 − 3y2. Applying the formula Z Z Z Z F · dS = (−P gx − Qgy + R) dA S D Z Z =
[−xy(−6x) − y(−6y) + (16 − 3x2 − 3y2)] dA D Z Z = (6x2y + 3y2 − 3x2 + 16) dA D Z 2 Z 1 = (6x2y + 3y2 − 3x2 + 16)dydx 0 0 Z 2 1 =
(3x2y2 + y3 − 3x2y + 16y) )dx  0 0 Z 2 Z 2 = (3x2 + 1 − 3x2 + 16)dx = 17dx = 34 0 0 4