ASSIGNMENT 2 - SOLUTIONS
1. Does the limit
lim
(x,y)→(0,0)
x3 y
x4 + y 4
exist?
Solution. The x-axis (y = 0) has parametric equations (x, y) = (t, 0), with (0, 0)
corresponding to t = 0. Hence
lim
along
(x,y)→(0,0)
f (x, y) = lim f (t, 0) = lim 0 = 0.
t→0
y=0
t→0
The line (y = x) has parametric equations (x, y) = (t, t), with (0, 0) corresponding to
t = 0. Hence
t4
1
lim
f (x, y) = lim f (t, t) = lim 4 = .
t→0
t→0 2t
2
(x,y)→(0,0) along x=y
Since f has two different limits along two different smooth curves, lim (x,y)→(0,0) f (x, y)
does not exist.
2. Find the points (x, y) ∈ R2 where
f (x, y) = cos (x2 + y 2 )
is differentiable. Show that all directional derivatives of f at the point (0, 0) are equal
to zero.
Solutions. The first order partial derivatives
∂f
(x, y) = −2x sin(x2 + y 2 ),
∂x
∂f
(x, y) = −2y sin(x2 + y 2 ),
∂y
are continuous functions on R2 , hence f is differentiable at all points (x, y) ∈ R 2 .
It is clear that ∇f (0, 0) = (fx (0, 0), fy (0, 0)) = (0, 0). Hence for any unit verctor
u = (u1 , u2 ), we have
Du (0, 0) = ∇f (0, 0) · u = 0.
3. Let f be a differentiable function of three variables and define
w(x, y, z) = f (x − y, y − z, z − x).
Use the Chain Rule to show that
∂w ∂w ∂w
+
+
= 0.
∂x
∂y
∂z
Solution. Put
a = a(x, y, z) = x − y,
b = b(x, y, z) = y − z,
c = c(x, y, z) = z − x.
Then
w(x, y, z) = f (a(x, y, z), b(x, y, z), c(x, y, z)).
1
2
ASSIGNMENT 2 - SOLUTIONS
By the Chain Rule
(1)
∂w
∂x
=
=
(2)
∂w
∂y
=
=
∂f ∂a ∂f ∂b
∂f ∂c
+
+
∂a ∂x
∂b ∂x
∂c ∂x
∂f
∂f
∂f
·1+
·0+
· (−1).
∂a
∂b
∂c
∂f ∂a ∂f ∂b ∂f ∂c
+
+
∂a ∂y
∂b ∂y
∂c ∂y
∂f
∂f
∂f
· (−1) +
·1+
· 0.
∂a
∂b
∂c
∂f ∂a ∂f ∂b ∂f ∂c
+
+
∂a ∂z
∂b ∂z
∂c ∂z
∂f
∂f
∂f
=
·0+
· (−1) +
· 1.
∂a
∂b
∂c
Adding up the relations (1)-(2)-(3) we get
∂w ∂w ∂w
∂f
∂f
∂f
∂f
∂f
∂f
+
+
=
−
−
+
−
+
= 0.
∂x
∂y
∂z
∂a
∂c
∂a
∂b
∂b
∂c
(3)
∂w
∂z
=
4. Find the equation of the tangent plane to the hyperboloid
z 2 x2 y 2
−
−
=1
3
9
4
at the point (3, 2, 3).
z 2 x2 y 2
− − . The given hyperboloid is the level surface
3
9
4
F (x, y, z) = 1. Since
2x
y 2z ∇F (x, y, z) = −
,− ,
,
9
2 3
we have
2
∇F (3, 2, 3) = − , − 1, 2 .
3
The equation of the tangent plane to the surface at (3, 2, 3) is
Solution. Let F (x, y, z) =
∇F (3, 2, 3) · (x − 3, y − 2, z − 3) = 0
or
−
2
, − 1, 2 · (x − 3, y − 2, z − 3) = 0,
3
that is,
2
− (x − 3) − (y − 2) + 2 (z − 3) = 0.
3
The equation of the plane is therefore
−2x − 3y + 6z − 6 = 0.