Math 243
Instructor: Longfei Li
Name:
Quiz III
You have 30 min to work individually. Calculator, notes and textbook are NOT allowed.
1. Evaluate the limit or show the limit does not exit. [10 points]
2x2 y
(a)
lim
(x,y)→(0,0) x2 + 2y 2
2xy
(b)
lim
2
x
+ 2y 2
(x,y)→(0,0)
Solution:
(a)
0≤
2x2 y x2 ≤
1
⇒
0
≤
2
≤ |2y|
2
2
2
x + 2y
x + 2y
Since
|2y| = 0
lim
(x,y)→(0,0)
by Squeeze Theorem, we have
2x2 y
=0
(x,y)→(0,0) x2 + 2y 2
lim
(b) Let (x, y) → (0, 0) along y = kx, then
f (x, y) =
x2
2kx2
2k
2k
=
→
as (x, kx) → (0, 0)
+ 2k 2 x2
1 + 2k 2
1 + 2k 2
So for different k values we get different limits. Thus,
2. Find all the second partial derivatives of w =
Solution:
wu = √
So
√
wuu =
2xy
DNE
+ 2y 2
u2 + v 2 . [15 points]
u
v
, wv = √
2
2
+v
u + v2
2
√ u
u2 +v 2
v2
u2 + v 2 −
wuv = wvu =
wvv =
(x,y)→(0,0) x2
u2
u2 +
√
√
lim
− √uuv
2 +v 2
u2 + v 2
=
=−
2
√ v
u2 +v 2
v2
u2 + v 2 −
u2 +
1
=
v2
3
(u2 + v 2 ) 2
uv
(u2
3
+ v2) 2
u2
3
(u2 + v 2 ) 2
√
3. Let f (x, y, z) = x2 y + x 1 + z. [20 points]
(a) Find the directional derivative of f at the point (1, 2, 3) in the direction v = 2i + j − 2k
(b) Find the maximum rate of change of f at the point (1, 2, 3) and direction it occurs.
(c) Find equation for the tangent plane to the surface given by f (x, y, z) = 4 at the point (1, 2, 3).
(d) Find parametric equations of the normal line to the surface given by f (x, y, z) = 4 at the
point (1, 2, 3).
Solution:
∇f =< 2xy +
(a) |v| =
v:
√
x
1
1 + z, x2 , √
>⇒ ∇f (1, 2, 3) =< 6, 1, >
4
2 1+z
√
9 = 3 is not a unit vector. We have to find the unit vector u in the same direction as
u=
v
2 1 2
=< , , − >
|v|
3 3 3
So the directional derivative in the direction v at the point (1,2,3) is
Du = ∇f (1, 2, 3) · u =< 6, 1,
1
2 1 2
25
> · < , , − >=
4
3 3 3
6
(b) the maximum rate of change of f at (1, 2, 3) occurs in the direction
a = ∇f (1, 2, 3) =< 6, 1,
1
>
4
and the maximum rate is
1
Da f (1, 2, 3) = |∇f (1, 2, 3)| = | < 6, 1, > | =
4
√
593
4
(c) We know the normal vector at (1, 2, 3) of f (x, y, z) = 4 is ∇f (1, 2, 3) =< 6, 1, 14 >, so the
tangent plane is
1
6(x − 1) + (y − 2) + (z − 3) = 0
4
(d) The parametric equations of the normal line are
1
x = 1 + 6t, y = 2 + t, z = 3 + t
4
2
4. If z = x4 + x2 y, x = s + 2t − u and y = stu2 , find
[15 points]
∂z ∂z
∂z
,
and
when s = 1, t = 0, u = −1.
∂s ∂t
∂u
Solution:
∂z
∂z ∂x ∂z ∂y
=
+
= (4x3 + 2xy)(1) + x2 tu2
∂s
∂x ∂s ∂y ∂s
∂z ∂x ∂z ∂y
∂z
=
+
= (4x3 + 2xy)(2) + x2 su2
∂t
∂x ∂t
∂y ∂t
∂z
∂z ∂x ∂z ∂y
=
+
= (4x3 + 2xy)(−1) + x2 (2stu)
∂u
∂x ∂u ∂y ∂u
when s = 1, t = 0, u = −1, we have x = 2, y = 0. So
∂z
= 32
∂s
∂z
= 32(2) + 4 = 68
∂t
∂z
= −32
∂u
3