Math 203 Hw 4
Spring 2015
Chapter 1
Math 203 Hw 4
1.1
12.2
* Evaluate the indicated limit or explain why it does not exist for 1, 3, 5, 6.
1) lim(x,y)→(2,−1) xy + x2
= 2(−1) + 22 = 2
3) lim(x,y)→(0,0)
x2 +y 2
y
It does not exist. If (x, y) → (0, 0) along x = 0, then
y = x2 , then
x2 +y 2
y
5) lim(x,y)→(1,π)
x2 +y 2
y
= y → 0. If (x, y) → (0, 0) along
= 1 + x2 → 1.
cos(xy)
1−x−cos y
=
2
cos π
= −1
1 − 1 − cos π
2
(y−1)
6) lim(x,y)→(0,1) xx2 +(y−1)
2
2
(y−1)2 2
2
We have 0 ≤ xx2 +(y−1)
2 ≤ x , and x → 0 as (x, y) → (0, 1). So the limit is 0.
15) What is the domain of f (x, y) = xx−y
2 −y 2 ? Does f (x, y) have a limit as (x, y) → (1, 1)?
Can the domain of f be extended so that the resulting function is continuous at (1, 1)? Can the
domain be extended so that the resulting function is continuous everywhere in the xy-plane?
f (x, y) =
x−y
x−y
=
2
2
x −y
(x − y)(x + y)
1
Since f (x, y) = x+y
at all points of the line x = y and so is defined at some points in any
neighborhood of (1, 1), it approaches 1/(1 + 1) as (x, y) → (1, 1). If we define f (1, 1) = 1/2,
1
then f becomes continuous at (1, 1). Similarly, f (x, y) can be defined to be 2x
at any point on
1
2
Math 203 Hw 4
the line x = y except the origin and becomes continuous at such points. However there is no
1
way to define f (x, y) so that f becomes continuous at y = −x, since |f (x, y)| = |x+y|
→ ∞ as
y → −x.
19) What condition must the constants a, b, and c satisfy to guarantee that
lim(x,y)→(0,0)
xy
ax2 +bxy+xy 2
exist? Prove your answer.
Suppose (x, y) → (0, 0) along y = kx. Then
xy
k
=
2
+ bxy + xy
a + bk + ck 2
thus f (x, y) has a different constant values along different rays from the origin unless a = c = 0
and b 6= 0.
f (x, y) =
ax2
20) Can be the function f (x, y) =
becomes continuous there? If so, how?
sin x sin3 y
1−cos(x2 +y 2 )
be defined at (0, 0) in such a way that it
f (x, y) has no limit at (0, 0) so it cannot be defined at (0, 0) to become continuous. f (x, 0) =
0, so if the limit exists it must be 0. But
f (x, x) =
1.2
sin4 x
sin4 x
1
=
→
as x → 0
2
1 − cos(2x2 )
2
2 sin x2
12.3
* Find all the first partial derivatives of the function specified, and evaluate then at the given
point for 1, 3, 4, 5, 7.
1) f (x, y) = x − y + 2, (3, 2)
fx (x, y) = 1 = fx (3, 2)
fy (x, y) = −1 = fy (3, 2)
,
3) f (x, y, z) = x3 y 4 z 5 , (0, −1, −1)
4 g(x, y, z) =
fx (x, y, z) = 3x2 y 4 z 5
,
fx (0, −1, −1) = 0
fy (x, y, z) = 4x3 y 3 z 5
,
fy (0, −1, −1) = 0
fz (x, y, z) = 5x3 y 4 z 4
,
fz (0, −1, −1) = 0
,
gx (1, 1, 1) =
xz
, (1, 1, 1)
y+z
gx (x, y, z) =
z
y+z
1
2
1.2 12.3
3
gy (x, y, z) =
−xz
(y + z)2
gz (x, y, z) =
5) z = tan−1
x
y
gy (1, 1, 1) = −
,
xy
(y + z)2
,
gz (1, 1, 1) =
1
4
1
4
, (−1, 1)
∂z
y
=− 2
∂x
x + y2
,
x2
∂z
= 2
∂y
x + y2
√
7) f (x, y) = sin(x y),
∂z 1
=−
∂x (−1,1)
2
1
∂z =−
∂y (−1,1)
2
,
π
,4
3
√
√
y cos(x y)
,
fx
x
√
fy (x, y) = √ cos(x y)
2 y
,
fy
fx (x, y) =
π
3
, 4 = −1
π
,4 = −
3
24
π
* Find equations of the tangent plane and normal line to the graph of the given function at
the point with specified values of x and y for 13, 15, 17.
13) f (x, y) = x2 − y 2 , (−2, 1)
fx (x, y) = 2x
fx (−2, 1) = −4
,
fy (x, y) = −2y
fy (−2, 1) = −2
,
Tangent plane: z = 3 − 4(x + 2) − 2(y − 1)
Normal line:
15) f (x, y) = cos
x
y
y−1
z−3
x+2
=
=
−4
−2
−1
, (π, 4)
1
fx (x, y) = − sin
y
x
y
,
1
fx (π, 4) = − √
4 2
x
fy (x, y) = 2 sin
y
x
y
,
fy (π, 4) =
π
√
16 2
4
Math 203 Hw 4
1
Tangent plane: z = √
2
1
π
1 − (x − π) + (y − 4)
4
16
√
√
1
16 2
(y − 4) = −z + √
Normal line: − 4 2(x − π) =
π
2
17) f (x, y) =
x
, (1, 2)
x2 +y 2
fx (x, y) =
fy (x, y) = −
y 2 − x2
(x2 + y 2 )2
,
fx (1, 2) =
2xy
+ y 2 )2
,
fy (1, 2) = −
(x2
Tangent plane: z =
Normal line:
3
25
4
25
1
3
4
+ (x − 1) − (y − 2)
5 25
25
y−2
5z − 1
x−1
=
=
3
−4
−125
23) Find the coordinates of all points on the surface with equation z = x4 − 4xy 3 + 6y 2 − 2
where the surface has a horizontal tangent plane.
∂z
= 4x3 − 4y 3 = 4(x − y)(x2 + xy + y 2 )
∂x
∂z
= −12xy 2 + 12y = 12y(1 − xy)
∂y
The tangent plane will be horizontal at points where both first partials are zero. Thus we
require x = y and either y = 0 or xy = 1. So the points are (0, 0), (1, 1), (−1, −1).
* Show that the given function satisfies the given partial differential equation for 25, 28 and
31.
∂z
25) z = xey , x ∂x
=
∂z
∂y
∂z
= ey
∂x
,
∂z
∂z
∂z
= xey =⇒ x
= xey =
∂y
∂x
∂y
28) w = x2 + yz, x ∂w
+ y ∂w
+ z ∂w
= 2w
∂x
∂y
∂z
∂w
= 2x ,
∂x
∂w
=z
∂y
,
∂w
∂w
∂w
∂w
= y =⇒ x
+y
+z
= 2(x2 + yz) = 2w
∂z
∂x
∂y
∂z
∂z
∂z
31) z = f (x2 − y 2 ), where f is any differentiable function of one variable, y ∂x
+ x ∂y
=0
1.2 12.3
5
∂z
= 2xf 0 (x2 − y 2 )
∂x
1.3
,
∂z
∂z
∂z
= −2yf 0 (x2 − y 2 ) =⇒ y
+x
=0
∂y
∂x
∂y
12.3
* Find all the second derivatives of the given function for 1, 2, 3, 5.
1) z = x2 (1 + y 2 )
∂z
= 2x(1 + y 2 )
∂x
∂ 2z
= 2(1 + y 2 )
∂x2
,
∂z
= 2x2 y
∂z
,
∂ 2z
= 2x2
∂y 2
∂ 2z
∂z
=
= 4xy
∂x∂y
∂y∂x
,
2) f (x, y) = x2 + y 2
fx (x, y) = 2x
fxx (x, y) = 2
,
,
fyy = 2
fy (x, y) = 2y
,
fxy (x, y) = f( yx)(x, y) = 0
3) w = x3 y 3 z 3
∂w
= 3x2 y 3 z 3
∂x
,
∂w
= 3x3 y 2 z 3
∂y
,
∂w
= 3x3 y 3 z 2
∂z
∂ 2w
= 6xy 3 z 3
∂x2
,
∂ 2w
= 6x3 yz 3
∂y 2
,
∂ 2w
= 6x3 y 3 z
∂z 2
∂ 2w
∂ 2w
=
= 9x2 y 2 z 3
∂x∂y
∂y∂x
∂ 2w
∂ 2w
=
= 9x2 y 3 z 2
∂x∂z
∂z∂x
,
,
∂ 2w
∂ 2w
=
= 9x3 y 2 z 2
∂y∂z
∂z∂y
5) z = xey − yex
∂z
= ey − yex
∂x
∂ 2z
= −yex
∂x2
,
∂ 2z
= xey
∂y 2
∂z
= xey − ex
∂y
,
,
∂ 2z
∂ 2z
=
= ey − ex
∂x∂y
∂y∂x
* Show that the functions are harmonic in the plane regions indicated for 8, 9, 10.
6
Math 203 Hw 4
8) f (x, y) = A(x2 − y 2 ) + Bxy in the whole plane (A and B are constants.)
fx = 2Ax + By
fxx = 2A
fy = −2Ay + Bx
,
fyy = −2A =⇒ fxx + fyy = 2A − 2A = 0 so f is harmonic.
,
9) f (x, y) = 3x2 y − y 3 in the whole plane (Can you think of another polynomial of degree
3 in x and y that is also harmonic?)
fx (x, y) = 6xy
fxx (x, y) = 6y
,
fy (x, y) = 3x2 − 3y 2
,
fyy = −6y =⇒ fxx + fyy = 6y − 6y = 0 so f is harmonic.
Also g(x, y) = x3 − 3xy 2 is harmonic.
10) f (x, y) =
x
x2 +y 2
everywhere except at the origin.
fx (x, y) =
fxx (x, y) =
y 2 − x2
(x2 + y 2 )2
,
fy (x, y) = −
2x3 − 6xy 2
(x2 + y 2 )3
,
fyy (x, y) =
=⇒ fxx (x, y) + fyy (x, y) =
(x2
2xy
+ y 2 )2
−2x3 + 6xy 2
(x2 + y 2 )3
2x3 − 6xy 2 −2x3 + 6xy 2
+
= 0 so f is harmonic.
(x2 + y 2 )3
(x2 + y 2 )3
15) Let the functions u(x, y) and v(x, y) have continuous second partial derivatives and
satisfy the Cauchy-Riemann equations
∂u
∂v
=
and
∂x
∂y
Show that u and v are both harmonic. Solution:
∂v
∂u
=−
∂x
∂y
∂ 2u
∂ 2v
∂ 2v
∂ 2u
∂ 2u ∂ 2u
=
=
=
−
=⇒
+
= 0 so u is harmonic.
∂x2
∂x∂y
∂y∂x
∂y 2
∂x2 ∂y 2
The proof for v is similar.
1
x2
17) Show that the function u(x, t) = t− 2 e− 4t satisfies the partial differential equation
∂u
∂ 2u
=
∂t
∂x2
1.3 12.3
7
This equation is called the one-dimensional heat equation because it models heat diffusion
in an insulated rod (with u(x, t) representing the temperature at position x at time t) and other
similar phenomena. Solution:
∂u
=
∂t
1 − 3 1 − 5 2 − x2
2
2
− t + t x e 4t
2
4
∂ 2u
=⇒
=
∂x2
,
3
x2
∂u
1
= − xt− 2 e− 4t
∂x
2
∂u
1 − 3 1 − 5 2 − x2
− t 2 + t 2 x e 4t =
2
4
∂t
18) Show that the function u(x, y, t) = t−1 e−
equation
x2 +y 2
4t
satisfies the two-dimensional heat
∂u
∂ 2u ∂ 2u
=
+
∂t
∂x2 ∂y 2
Solution:
∂u
=
∂t
1 − 3 x2 + y 2 − x2 +y2
− t 2+
e 4t
2
4t3
1 − x2 +y2 x2 − x2 +y2
∂ 2u
= − 2 e 4t + 3 e 4t
∂x2
2t
4t
,
,
x2 +y 2
∂u
x
= − 2 e− 4t
∂x
2t
x2 +y 2
∂u
y
= − 2 e− 4t
∂y
2t
,
1 − x2 +y2 y 2 − x2 +y2
∂ 2u ∂ 2u
∂ 2u
∂u
4t
4t
= − 2e
=
+
+ 3e
=⇒
∂x2
2t
4t
∂t
∂x2 ∂y 2
21) Verify that u(x, y) = x4 − 3x2 y 2 is biharmonic.
∂ 2u
∂
3
2
=
4x
−
6xy
= 12x2 − 6y 2
∂x2
∂x
∂ 4u
∂
=
(24x) = 24
4
∂x
∂x
=⇒
,
∂ 2u
∂
2
=
−6x
y
= −6x2
∂y 2
∂y
,
∂ 4u
∂
=
(−12x) = −12
2
2
∂x ∂y
∂x
∂ 4u
=0
∂y 4
,
∂ 4u
∂ 4u
∂ 4u
=
+2
=
+
== 24 + 2(−12) + 0 = 0 so u is biharmonic.
∂x4
∂x2 ∂y 2
∂y 4
22) Show that if u(x, y) is harmonic, then v(x, y) = xu(x, y) and w = yu(x, y) are biharmonic.
We have
∂2u
∂x2
+
∂2u
∂y 2
∂ 2v
∂
=
2
∂x
∂x
= 0.
∂u
u+x
∂x
∂u
∂ 2u
=2
+x 2
∂x
∂x
,
∂ 2v
∂
=
2
∂y
∂y
∂u
x
∂y
=x
∂ 2u
∂y 2
8
Math 203 Hw 4
∂ 2v ∂ 2v
∂u
=⇒
+x
+
=
2
∂x2 ∂y 2
∂x
Since u is harmonic, so is
Thus
∂2v
∂x2
+
∂2v
∂y 2
∂ 2u ∂ 2u
+
∂x2 ∂y 2
=2
∂u
∂x
∂u
:
∂x
∂2
∂2
+
∂x2 ∂y 2
∂u
∂
=
∂x
∂x
∂ 2u ∂ 2u
+
∂x2 ∂y 2
=
∂
(0) = 0
∂x
is harmonic, so v is biharmonic. The proof for w is similar.