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