Mathematics 1c: Solutions, Midterm Examination
Due: Monday, May 3, at 10am
1. Do each of the following calculations.
(a) If a particle follows the curve
c
(
t
) =
e
t

1
i

(
t

1)
j
+ sin(
πt
)
k
and ﬂies oﬀ on a tangent at
t
= 1, where is it at
t
= 2?
Solution.
First note that
c
(1) =
i
. We compute that
c
0
(
t
) =
e
t

1
i

j
+
π
cos(
πt
)
k
.
Therefore
c
0
(1) =
i

j

π
k
. Hence the position of the particle at time
t
= 2 is
i
+ (2

1)(
i

j

π
k
) = 2
i

j

π
k
.
(b) Find the equation of the tangent plane to the surface
x
2

e
xy
+
z
2
= 1 at the point
(1
,
0
,
1).
Solution.
The tangent plane consists of vectors based at (1
,
0
,
1) that are perpen
dicular to the gradient of
f
(
x,y,z
) =
x
2

e
xy
+
z
2
. We compute that
∇
f
(
x,y,z
) = (2
x

ye
xy
,

xe
xy
,
2
z
)
.
Evaluating at (1
,
0
,
1), we obtain
∇
f
(1
,
0
,
1) = (2
,

1
,
2)
.
Therefore, the tangent plane at (1
,
0
,
1) is deﬁned by
(2
,

1
,
2)
·
(
x

1
,y,z

1) = 0
,
namely by
z
=

x
+
y
2
+ 2
.
(c) Let
f
(
x,y,z
) = 5 +
xy

zx
be the concentration of chemical X. Find the direction
at (1
,
1
,
1) in which X is
decreasing
the fastest. In which directions is it decreasing
at 30% of its maximum rate? Give your answer in terms of the angle made with
the direction of fastest decrease.
Solution.
The concentration of
X
is increasing the fastest in the direction of the
gradient of
f
, and hence decreasing the fastest in the opposite of this direction.
We compute that
∇
f
(
x,y,z
) = (
y

z,x,

x
)
,
and hence that
∇
f
(1
,
1
,
1) = (0
,
1
,

1)
.
1