University of Toronto Scarborough
Department of Computer & Mathematical Sciences
MAT B41H
2013/2014
Solutions #5
1. A normal vector to the plane 4
x
−
8
y
−
z
= 3 is (4
,
−
8
,
−
1)
and a normal vector to the tangent plane to the graph
of
f
(
x,y
) =
x
2
+
y
2
−
1 is (2
x,
2
y,
−
1).
For the planes
to be parallel we need (2
x,
2
y,
−
1) =
k
(4
,
−
8
,
−
1), for
some
k
. From the third component we see that
k
= 1, so
we have
x
= 2 and
y
=
−
4.
The tangent plane passes
through (2
,
−
4
,f
(2
,
−
4)) = (2
,
−
4
,
19) and has equation
4
x
−
8
y
−
z
= 21.
2
x
4
y
19
z
2.
(a) We have
f
(0
,
0) = 0, so we can compute the partial derivatives from the defini
tion.
∂f
∂x
(0
,
0) = lim
h
→
0
f
(
h,
0)
−
f
(0
,
0)
h
= lim
h
→
0
0
−
0
h
= 0
∂f
∂y
(0
,
0) = lim
h
→
0
f
(0
,h
)
−
f
(0
,
0)
h
= lim
h
→
0
0
−
0
h
= 0
(b) Recall that for a function
f
to be differentiable at a point
a
, we need
lim
x
→
a
bardbl
f
(
x
)
−
f
(
a
)
−
D f
(
a
)
(
x
−
a
)
bardbl
bardbl
x
−
a
bardbl
= 0.
Here
we
need
to
evaluate
lim
(
x,y
)
→
(0
,
0)

x
1
/
3
y
1
/
3
−
0
−
(0
,
0)
·
(
x,y
)

bardbl
(
x,y
)
bardbl
=
lim
(
x,y
)
→
(0
,
0)

x
1
/
3
y
1
/
3

radicalbig
x
2
+
y
2
. If we approach along the line
y
=
x
, with
x >
0, this limit
reduces to lim
x
→
0
+
x
2
/
3
√
2
x
2
= lim
x
→
0
+
x
2
/
3
x
√
2
= lim
x
→
0
+
1
x
1
/
3
√
2
=
∞
. Since this limit is not
0,
f
is not differentiable at (0
,
0).
3. The rate of change in depth is the directional derivative.
(a) The depth will increase most rapidly in the direction of the gradient; i.e., in
direction
∇
D
(1
,
−
2). Now
∇
D
= (
−
6
xy
2
,
−
6
x
2
y
) so the rubber duck swims in
direction
∇
D
(1
,
−
2) = (
−
24
,
12).
(b) The depth will stay constant if the duck stays on the level set which passes through
(1
,
−
2). Since the gradient is orthogonal to level sets it can proceed in direction
(1
,
2) or direction (
−
1
,
−
2) since
±
1 (1
,
2)
·
∇
D
(1
,
−
2) =
±
1 (1
,
2)
·
(
−
24
,
12) = 0.
MATB41H
Solutions # 5
page
2
4.
(a) Let
x
=
u
2
−
v
2
and
y
=
v
2
−
u
2
. Then
g
(
u,v
) =
f
(
x,y
) and the Chain Rule
can be applied to give
∂g
∂u
=
∂f
∂x
∂x
∂u
+
∂f
∂y
∂y
∂u
=
∂f
∂x
(
2
u
)
+
∂f
∂y
(
−
2
u
)
and
∂g
∂v
=
∂f
∂x
∂x
∂v
+
∂f
∂y
∂y