Calculus II for Engineering
Spring, 2010
QUIZ 1 { SOLUTION
Score: 10/10
Date: Tuesday, February 16, 2010
1. (5 points) Sketch the plane curve dened by the given parametric equations and nd a corresponding x{y equation for the curve: x = cos t and y = 3 sin t
1.
Answer.
cos t = x;
sin t =
y+1
=)
3
x2 +
(y + 1)2
= cos2 t + sin2 t = 1;
32
i:e:;
p
x2 +
(y + 1)2
= 1;
32
which is an ellipse passing through (0; 2), (0; 4) and (2 2=3; 0).
2. (5 points) Find parametric equations describing the circle of radius 4 centered at ( 2; 1), drawn
counterclockwise.
Answer 1. The x{y equation for the given circle is
(x + 2)2 + (y
Letting x + 2 = 4 cos and y
1)2 = 42 :
1 = 4 sin , we have
(x +2)2 +(y 1)2 = 42 cos2 +42 sin2 = 42 cos2 + sin2 = 42
=)
(x +2)2 +(y 1)2 = 42 :
Therefore, the parametric equations is
x = 4 cos 2;
y
= 4 sin + 1;
0 2:
Here, since the orientation is given as the counterclockwise direction, so we choose the interval
0 2 .
Answer 2. For the given circle, we set up
x = a + b cos ;
y
= c + d sin ;
where a, b, c and d are constants to be determined.
cos t =
x
a
b
;
sin t =
y
c
d
=)
=)
(x
a )2
+
(y
c)2
b2
d2
d2 (x a)2 + b2 (y
= cos2 + sin2 = 1
c)2
= b2 d2 :
Since it should represent the circle with radius 4 and centered at ( 2; 1), we should get a = 2
and c = 1 and b = d = 4. Therefore, we deduce
x=
2 + 4 cos ;
y
= 1 + 4 sin ;
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0 2:
Calculus II for Engineering
Spring, 2010
QUIZ 2 { SOLUTION
Score: 10/10
Date: Tuesday, February 23, 2010
1. (Total 5 points)
(1.1) (21=2 points) Find all polar coordinates (r; ) for the rectangular point (x; y ) = (3; 6).
Answer.
The formula implies
p
p
= x2 + y 2 = 32 + 62 = 45 =) r = 45 = 3 5
6
y
tan = = = 2 =) = tan 1 2 = 63:4349 = 1:10715 radian:
x
3
r2
Since the point (x; y ) = (3; 6) is in the rst quadrant, thus we deduce the corresponding
polar coordinates,
p
(r; ) = 3 5; 1:10715 radian
and (r; ) =
All polar coordinates are
p
(r; ) = 3 5; 1:10715 + 2n radian
and (r; ) =
p
3 5; 1:10715 + radian :
p
3 5; 1:10715 + + 2n radian ;
where n is any integer.
(1.2) (21=2 points) Find the rectangular coordinates (x; y ) for the polar point (r; ) = (3; 4).
Answer.
The formula implies
x = r cos = 3 cos 4;
and
y
= r sin = 3 sin 4:
Thus, the corresponding rectangular coordinates is (x; y ) = (3 cos 4; 3 sin 4).
2. (5 points) Find a polar equation corresponding to the rectangular equation
x2 + ( y
1)2 = 1.
(Simplify your answer as much as you can.)
Answer.
Expanding the given equation gives
1 = x2 + ( y
1)2 = x2 + y 2
2y + 1
=)
0 = x2 + y 2
2y:
By the formula, the equation becomes
0 = (r cos )2 + (r sin )2
= r2 cos2 + sin2 Since r 6= 0, thus we have r
2r sin = r2 cos2 + r2 sin2 2r sin = r2
2r sin = r (r
2r sin 2 sin ) :
2 sin = 0, i.e., r = 2 sin with 0 2 .
Page 1 of 1
Spring, 2010
Calculus II for Engineering
QUIZ 3 – SOLUTION
Score: 10/10
ID No: Solution
1/2 (5 points) Find 2a + 3b, kak and ka − bk for a = 4i + j and b = ⟨ 1, −2 ⟩. Here, i and j are the
standard basis vectors of the V2 space.
A NSWER. We observe a = 4i + j = ⟨ 4, 1 ⟩ and b = i − 2j = ⟨ 1, −2 ⟩. One can use either the form of
standard basis vectors or the component form.
2a + 3b = 2 (4i + j ) + 3 (i − 2j ) = 11i − 4j = ⟨ 11, −4 ⟩
kak = k ⟨ 4, 1 ⟩ k =
p
42 + 12 =
p
17
a − b = ⟨ 4, 1 ⟩ − ⟨ 1, −2 ⟩ = ⟨ 3, 3 ⟩ = 3 ⟨ 1, 1 ⟩
ka − bk = k3 ⟨ 1, 1 ⟩ k = 3k ⟨ 1, 1 ⟩ k = 3
p
p
12 + 12 = 3 2.
ä
2/2 (5 points) The thrust of an airplane’s engines produces a speed of 300 mph in still air. The wind
velocity is given by ⟨ 50, 0 ⟩. In what direction should the airplane head to fly due north?
A NSWER. Let v = ⟨ x, y ⟩ be the direction of the plane and w = ⟨ 50, 0 ⟩ represent the wind velocity.
We want v + w = ⟨ 0, c ⟩, where c > 0 so that the plane is traveling due north. We have
⟨ 0, c ⟩ = v + w = ⟨ x, y ⟩ + ⟨ 50, 0 ⟩ = ⟨ x + 50, y ⟩ ,
i.e.,
x + 50 = 0,
and
y = c.
That is, x = −50 and y = c and so v = ⟨ −50, c ⟩. We have kv k = 300, which implies
300 = kv k = k ⟨ −50, c ⟩ k =
i.e.,
(−50)2 + c2 = 3002 ,
p
(−50)2 + c2 ,
c2 = 3002 − (−50)2 = 87500,
p
c = 50 35,
where we take the positive square root to have the plane moving north ( c > 0). Therefore, we
conclude that the plane should fly in the direction:
p
p
v = ⟨ −50, 50 35 ⟩ = 50 ⟨ −1, 35 ⟩ .
Page 1 of 1
ä
Calculus II for Engineering
Spring, 2010
QUIZ 4 { SOLUTION
Score: 10/10
ID No: Solution
!
1/2 (5 points) Find the displacement vectors
!
and QR and determine whether the points
P (2; 3; 1), Q(0; 4; 2) and R(4; 1; 4) are colinear (i.e., on the same line).
Answer.
!
PQ = h 0
2; 4
3; 2
PQ
1 i = h 2; 1; 1 i ;
We observe that there does not
exist
! h
PQ =
!=h4
PR
2; 1
3; 4
1 i = h 2; 2; 3 i :
a scalar s such that
!
2; 1; 1 i = s h 2; 2; 3 i = sP R;
i.e., any scalar s does not satisfy all the equations at the same time:
2 = 2s;
!
1 = 2s;
1 = 3s:
!
It implies that the vectors, P Q and P R, are not parallel and thus the points are not colinear.
Remark:
One may use the dot product:
! ! h 2; 1; 1 i h 2; 2; 3 i
3
p
=p
6= 1:
cos = ! ! = p
4+1+1 4+4+9
112
kP QkkP Rk
!
!
So the angle between two vectors P Q and P R is neither 0 nor . It means that those two
PQ PR
vectors are not parallel and thus the points are not colinear.
2/2 (5 points) A car makes a turn on a banked road. When the road is banked at 30 , the vector
parallel to the road is h cos 30 ; sin 30 i. If the car has weight 2500 pounds, nd the component
of the weight vector along the road vector.
Answer.
The vector b = h cos 30 ; sin 30 i represents the direction of the banked road. The
vector deduced by the weight of the car is w = h 0; 2500 i. The component of the weight
vector in the direction of the bank is
Compb w =
wb
kbk =
2500 sin 30 = 1250 lbs
toward the inside of the curve.
Page 1 of 1
Calculus II for Engineering
Spring, 2010
QUIZ 5 { SOLUTION
ID No: Solution
1/2
Score:
(5 points) Use the cross product to determine the angle between the vectors
b = h 0; 0; 2
Answer.
i, assuming that 0 2 .
The cross product of
a b = h 4;
4; 0
a
i;
and
b
a
=
10/10
h 2 ; 2 ; 1 i,
is
p
ka bk = 4 2;
and
kak = 3;
kbk = 2:
We recall the formula:
p
p
4 2
2 2
k
a bk
ka bk = kak kbk sin =) sin = kak kbk = 3(2) = 3
p!
1 2 2
= sin
1:23096 radian 70:5288:
3
2/2
(5 points) Use the parallelepiped volume formula to determine whether the vectors
h 1;
i
3; 1 ,
Answer.
b = h 2;
1; 0
i and c = h 0;
The cross product of
a
and
b
5; 1
is
i are coplanar.
a b = h 1; 2; 5
a
=
i.
The volume formula implies
j b)j = jh 0;
V = c (a
5; 1
i h 1; 2; 5 ij = 5:
Since the parallelepiped has the volume 5, we conclude those three vectors are not coplanar.
Page 1 of 1
Calculus II for Engineering
Spring, 2010
QUIZ 6 { SOLUTION
1/2
; ;
(5 points) Find parametric equations for the line through the point (2 1 0) and perpendicular to both
i+j
and
Answer.
j + k,
where i,
j
and
are standard basis vectors of
k
V3 .
The line is parallel to the vector,
(i + j )
(j + k ) = i j + i k + j j + j k = k
j+i=h
1
;
; i:
1 1
Thus, its parametric equations are
x = 2 + t;
2/2
(5 points) Find an equation of the plane with
Answer.
Suppose
x + y + z
=
y = 1 t;
z = t:
x{intercept a, y{intercept b, and z {intercept c.
is the equation of the plane having the intercepts given in the
problem. Then,
a = ; b = ; c = )
=
= ; = ; = :
a
b
c
So the equation of the plane becomes
x+ y+ z =
a
b
c
) a1 x + 1b y + 1c z = 1
=
) bcx + acy + abz = abc:
=
Calculus II for Engineering
Spring, 2010
QUIZ 7 { SOLUTION
1/2 (5 points) Find the partial derivative fxy =
Answer.
fx
@y@ x
of f (x; y ) = y 8
= 16xy + ln y 2 = 16xy + 2 ln y;
2/2 (5 points) Use the Chain
z
2
@ f
Rule
to nd the partial derivative
= ey sin x ;
(
x s; t
fxy
zs
=
) = 3s cos t;
8x2 y + x ln
= 16x +
@z
@s
2
y
y
2
+ sin y .
:
, where
(
) = 4st2 :
y s; t
Answer.
zx
= ey sin x y cos x = yz cos x;
xs
= 3 cos t;
zs
= zx xs + zy ys = yz cos x(3 cos t) + z sin x(4t2 ) = z 3y cos x cos t + 4t2 sin x
ys
zy
= ey sin x sin x = z sin x;
= 4t2 ;
= e4st
2 sin(3s cos t)
12st2 cos(3s cos t) cos t + 4t2 sin(3s cos t)
2 sin(3s cos t)
= 4t2 e4st
(3s cos(3s cos t) cos t + sin(3s cos t)) :
Calculus II for Engineering
Spring, 2010
Quiz 8 – Solution
1. (5 points) Find the directional derivative of f (x, y, z) =
√
xy sin(z) in the direction of the
vector v = ⟨ 2, 3, −π ⟩.
Answer.
f
fx =
,
2x
√
fz = xy cos(z) ,
f
fy = ,
2y
⟨
∇f = ⟨ fx , fy , fz ⟩ =
f f √
, , xy cos(z)
2x 2y
⟩
.
Let u be the unit vector in the direction of v. Then,
v
1
=√
⟨ 2, 3, −π ⟩ ,
∥v∥
13 + π 2
(
)
1
f
3f
√
Du f = ∇f · u = √
+
− π xy cos(z)
x 2y
13 + π 2
(
)
√
xy
sin(z) 3 sin(z)
=√
+
− π cos(z) .
x
2y
13 + π 2
u=
2. (5 points) At the point (1, −1, 4), find the direction of the greatest rate of decrease in function
f (x, y, z) = x2 y + z 3 .
Answer.
fy = x2 , fz = 3z 2 ,
⟨
⟩
∇f = ⟨ fx , fy , fz ⟩ = 2xy, x2 , 3z 2 ,
fx = 2xy,
∇f (1, −1, 4) = ⟨ −2, 1, 48 ⟩ .
√
The greatest rate of decrease is −∥∇f (1, −1, 4)∥ = − 2309 and it occurs in the direction,
−
∇f (1, −1, 4)
⟨ −2, 1, 48 ⟩
=− √
.
∥∇f (1, −1, 4)∥
2309
Calculus II for Engineering
Quiz 9 – Solution
Spring, 2010
Quiz 9 – Solution
1. (5 points) Let Q be the solid bounded by the three coordinate planes and the plane 2x + 3y + 5z = 30. Set up,
but do not compute, the iterated integral representing the volume of the solid Q.
Answer. The plane passes through the points (15, 0, 0), (0, 10, 0) and (0, 0, 6). Putting three points in the
three–dimensional space, we can sketch the solid Q. We take the projection R of Q onto the xy–plane as the
base of the solid Q, i.e.,
R = { (x, y) | Bounded by 2x + 3y = 30 and x = 0 and y = 0 }
{
}
2x
= (x, y) 0 ≤ x ≤ 15, 0 ≤ y ≤ 10 −
(Vertical Cut)
3
From the figure or since R is the base of Q, so the z–component of a point in Q should be the height of the
2x 3y
solid Q. Inside Q, z moves from z = 0 (R on the xy–plane) to the surface z = 6 −
− . Therefore, the
5
5
volume V of the solid Q is
)
)
¨ (
ˆ x=15 ˆ y=10− 2x (
3
2x 3y
2x 3y
V =
6−
−
− 0 dA =
6−
−
dy dx = 150
5
5
5
5
R
x=0
y=0
¨
2. (5 points) Evaluate the given integral by changing to polar coordinates:
(x + y) dA, where R is the region
R
2
that lies to the left of the y–axis between the circles x2 + y 2 = 1 and x2 + y = 4.
Answer. The region R is represented by
{
}
R = (x, y) | 1 ≤ x2 + y 2 ≤ 4, x ≤ 0 .
Using the polar coordinates, the region R can be expressed by
{
}
π
3π
R = (r, θ) ≤ θ ≤
, 1≤r≤2 .
2
2
Using this representation with the polar coordinates, the integral becomes
¨
ˆ
(x + y) dA =
R
θ= 3π
2
ˆ
θ= π2
[ˆ
θ= 3π
2
=
θ= π2
ˆ
r=2
(r cos(θ) + r sin(θ))r dr dθ =
r=1
] [ˆ
r=2
r=1
Page 1
θ= π2
ˆ
r=2
r2 (cos(θ) + sin(θ)) dr dθ
r=1
]θ= 3π2 [ r3 ]r=2
14
.
=
−
r dr = sin(θ) − cos(θ)
3 r=1
3
θ= π2
]
2
(cos(θ) + sin(θ)) dθ
θ= 3π
2
[