August 15, 2009
Mathematics of Physics and Engineering II: Homework answers
You are encouraged to disagree with everything that follows.
Homework 1.
Problem 1.
(1)
→
PQ
=
→
OQ

→
OP
=
h
1
,
0
,
2
i  h
1
,
1
,
1
i
=
h
2
,

1
,
1
i
,
→
PR
=
h
0
,

2
,

2
i
,
→
PS
=
h
a,

1
,

2
a
i
.
(2) The vertex of the angle is
P
, so you need
→
PR
·
→
PS
= 0, or 2+4
a
= 0. Therefore,
a
=

1
/
2
.
(3) The area is (1
/
2)

→
PQ
×
→
PR

and
→
PQ
×
→
PR
=
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
i
j
k

2

1
1
0

2

2
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
=
h
4
,

4
,
4
i
= 4
h
1
,

1
,
1
i
.
Consequently, the area is
2
√
3
.
(4) The normal vector to the plane is any vector parallel to
→
PQ
×
→
PR
, for example,
h
1
,

1
,
1
i
.
Taking
P
as the point on the plane, we get the equation (
x

1)

(
y

1) + (
z

1) = 0 or
x

y
+
z
= 1
.
(5) You are welcome to compute the scale triple product using the determinant, but, given the
work you already did, you do not have to compute another determinant. The answer is
4

1

a

(it has to be nonnegative).
(6) You want the coordinates of
S
to satisfy the equation of the plane through
P,Q,R
, that is,
1 +
a

0 + 1

2
a
= 1 or
a
= 1
. You can also see it immediately from the volume formula.
(7) The direction vector for the line is the normal vector to the plane, that is,
h
1
,

1
,
1
i
. Then
the equation of the line is
r
(
t
) =
h
1 +
t,

t,
1 +
t
i
.
At the point of intersection, (1 +
t
)

(

t
) + (1 +
t
) = 1 or
t
=

1
/
3, so the point is
(2
/
3
,
1
/
3
,
2
/
3).
Problem 2.
(1)
r
(1) =
h
0
,
1
,
2
i
, so the coordinates of the point are
(0
,
1
,
2)
.
(2)
v
(
t
) = ˙
r
(
t
) =
h
2
t,
3
t
2
,
2
t
i
.
(3)

v
(
t
)

=
√
4
t
2
+ 9
t
4
+ 4
t
2
=
t
√
8 + 9
t
2
.
(4)
a
(
t
) = ˙
v
(
t
) =
h
2
,
6
t,
2
i
.
(5) The particle is at (0
,
1
,
2) when 1

t
2
= 0 or
t
= 1 (by assumption,
t
≥
0). Therefore, the
equation of the tangent line is
R
(
u
) =
h
0
,
1
,
2
i
+
u
˙
r
(1). Next, ˙
r
(1) =
h
2
,
3
,
2
i
, and so the
equation of the line is
R
(
u
) =
h
2
u,
1 + 3
u,
2

2
u
i
.
(6) You want the coordinates of the particle to satisfy the equation of the plane. Then 1 +
t
2

(1

t
2
) = 2 or
t
= 1 (remember,
t
≥
0) So the point of intersection is
(0
,
1
,
2)
.
(7) According to the formula, the distance is
Z
1
0

v
(
t
)

dt
=
Z
1
0
p
8 + 9
t
2
tdt
= (simply guess antiderivative)
1
27
(8 + 9
t
2
)
3
/
2

t
=1
t
=0
=
17
3
/
2

8
3
/
2
27
.
Homework 2.
Problem 1
.
(1)
∇
f
(
x,y
) =
h
f
x
(
x,y
)
,f
y
(
x,y
)
i
=
h
4
x

y

1
,
2
y

x
+ 1
i
.
(2) The direction vector is
a
=
h
1
,

1
i
. Therefore, the rate is
∇
f
(1
,
1)
·
a

a

=
h
2
,
2
i · h
1
,

1
i
√
2
=

4
/
√
2
.
The rate is negative, so the function is
decreasing
in that direction.
(3) The direction is given by
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Friedlander
 Math, Derivative, Fourier Series

Click to edit the document details