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
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View Full DocumentTherefore the concentration of
X
is decreasing fastest in the direction
∇
f
(1
,
1
,
1) = (0
,

1
,
1)
.
Let
n
be a unit vector, and let
θ
be the angle between
n
and
∇
f
(1
,
1
,
1). Then
the directional derivative of
f
in the direction
n
is given by
∇
f
(1
,
1
,
1)
·
n
=
k∇
f
(1
,
1
,
1)
kk
n
k
cos(
π

θ
). This is 30% of its most negative value when cos(
π

θ
) =

0
.
3, which is equivalent to cos(
θ
) = 0
.
3. This means that
θ
= cos

1
(0
.
3).
2. Answer each of the following questions.
(a) Let
f
(
r,s
) be a (smooth) function of
r
and
s
and, let
r
=
x
+ 2
y
and
s
=
x

2
y
.
Deﬁne the function
h
by
h
(
x,y
) =
f
(
x
+ 2
y,x

2
y
). Calculate
∂
2
h
∂x∂y
in terms of the partial derivatives of
f
.
Solution.
Let
r
(
x,y
) =
x
+ 2
y
and
s
(
x,y
) =
x

2
y
. Then we have that
∂r
∂x
= 1
,
∂r
∂y
= 2
,
∂s
∂x
= 1
,
∂s
∂y
=

2
.
Applying the chain rule to
h
(
x,y
) =
f
(
r
(
x,y
)
,s
(
x,y
)), we ﬁnd that
∂
2
h
∂x∂y
=
∂
∂x
±
∂h
∂y
²
=
∂
∂x
±
∂f
∂r
∂r
∂y
+
∂f
∂s
∂s
∂y
²
=
∂
∂x
±
2
∂f
∂r

2
∂f
∂s
²
= 2
∂
∂x
±
∂f
∂r
²

2
∂
∂x
±
∂f
∂s
²
= 2
±
∂
2
f
∂r
2
∂r
∂x
+
∂
2
f
∂s∂r
∂s
∂x
²

2
±
∂
2
f
∂r∂s
∂r
∂x
+
∂
2
f
∂s
2
∂s
∂x
²
= 2
±
∂
2
f
∂r
2
+
∂
2
f
∂s∂r
²

2
±
∂
2
f
∂r∂s
+
∂
2
f
∂s
2
²
.
Because
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 Spring '08
 Ramakrishnan
 Math, Chain Rule, Optimization, Global Minimum

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