(8 points) Find the equation of the tangent line to the path
c
(
t
) = (cos
t,
sin
t, t
2
) at
t
=
π
.
Solution.
The derivative of this path is given by
c
±
(
t
) = (

sin (
t
)
,
cos (
t
)
,
2
t
)
The tangent line to
c
(
t
) at the point
t
=
π
is given by
L
(
t
)
=
c
(
π
) +
t
·
c
±
(
π
)
=
(
cos (
π
)
,
sin (
π
)
, π
2
)
+
t
·
(

sin (
π
)
,
cos (
π
)
,
2
π
)
=
(

1
,
0
, π
2
)
+
t
·
(0
,

1
,
2
π
)
=
(

1
,

t, π
2
+ 2
πt
)
(Or you may use
L
(
t
) =
c
(
π
) + (
t

π
)
·
c
±
(
π
).)
[2]
(8 points) Let
f
:
R
3
→
R
be deﬁned by:
f
(
x, y, z
)
=
3
z
+
e
x
2

y
2
Let
C
be the set of the heads of unit vectors
v
in
R
3
such that
f
increases at 1
/
3 of its maximum
rate of change in the direction
v
starting from (0
,
0
,
1) . Find the equation(s) which determine(s)
the set
C
. (Hint :
C
is a circle in
R
3
.)
Solution.
The gradient of
f
is given by
∇
f
=
(
2
x
exp
(
x
2

y
2
)
,

2
y
exp
(
x
2

y
2
)
,
3
)
∇
f
(0
,
0
,
1) is a vector in the direction of greatest increase for the function
f
at the point (0
,
0
,
1).
So
∇
f
(0
,
0
,
1)
=
(0
,
0
,
3)
=
3
·
(0
,
0
,
1)
Calculate the maximimum rate of change of
f
by evaluating the directional derivative of
f
.
Df
(0
,
0
,
1)
=
∇
f
·
(0
,
0
,
1)
=
(0
,
0
,
3)
·
(0
,
0
,
1)
=
3
1
/
3 of this rate is just 1, so the goal of this problem is to ﬁnd and equation for the unit vectors
v
= (
x, y, z
) such that
Df
v
= 1 at the point (0
,
0
,
1).