Exam 1: Solutions
Problem 1. Compute the limit if it exists or otherwise show that
the limit does not exist:
(a)
(x+y)2
2
2 .
(x,y)→(0,0) x +y
(b)
x2 y 2
2 +y 2 .
x
(x,y)→(0,0)
lim
lim
Solution. (a) The limit along the line x = 0 is
y2
= 1.
y→0 y 2
lim
The limit along the line y = x is
4x2
= 2.
lim
x→0 2x2
Since the two partial limits are different, the limit does not exist.
(b) Just checking that two (or three, or one hundred) partial limits
coincide does not necessarily mean the existence of a limit. You need
a proof and computation for the general case. Here are two ways to
solve this problem:
Way 1.
x2 y 2
1
1
= lim
=
= 0.
2
2
−2
−2
(x,y)→(0,0) x + y
(x,y)→(0,0) y
+x
∞
lim
Way 2. We have
0≤
Since
lim
x2 y 2
(x2 + y 2 )2
≤
= x2 + y 2 .
2
2
2
2
x +y
x +y
(x2 +y 2 ) = 0, by the squeeze theorem
(x,y)→(0,0)
x2 y 2
2 +y 2
x
(x,y)→(0,0)
lim
= 0.
Problem 2. (a) Find the two points on the hyperboloid x2 + 4y 2 −
z 2 = 4 where the tangent plane is parallel to the plane 2x + 2y + z = 5.
(b) Calculate the velocity and acceleration vectors for the cycloid
→
−c (t) = (t − sin t, 1 − cos t) at t = π .
4
Solution. (a) The normal vectors of the two planes must be proportional, i.e.,
→
−
−
→
−
→
−
→
−
→
− →
2x i + 8y j − 2z k = c(2 i + 2 j + k )
1
2
for some constant c, so that x = c, y = c/4, z = −c/2. Plugging this
point in the equation of the hyperboloid, we obtain
c2 c2
−
= 4.
4
4
Thus c = ±2 and the two solutions are (2, 1/2, −1) and (−2, −1/2, 1).
c2 +
−c 0 (t) = (1−cos t, sin t), and the accelera(b) The velocity is equal to →
→
−
tion is equal to c 00 (t) = (sin t, cos t). At the point where t = π4 they are
−c 0 (π/4) = (1 − 1/√2, 1/√2) and →
−c 00 (π/4) = (1/√2, 1/√2)
equal to →
respectively.
Problem 3. A bug finds itself in a toxic environment. the toxicity
level is given by T (x, y) = 2x2 − 4y 2 . the bug is at (−1, 2). In what
direction should it move to lower the toxicity the fastest? Find the
unit vector of that direction.
Solution. The gradient is perpendicular to level surfaces of a function, the level increases the fastest in the direction of gradient and
decreases the fastest in the opposite direction. So, the best option for
→
−
→
−
the poor bug is to crawl in the direction of −∇T = −4x i +8y j which
→
−
→
−
is at (−1, 2) equal to −∇T (−1, 2) = 4 i + 16 j . The unit vector of
that direction is
→
−
→
−
−∇T (−1, 2)
4 i + 16 j
1 →
4 →
−
−
→
−
v =
= √
=√ i +√ j.
2
2
k − ∇T (−1, 2)k
4 + 16
17
17
Problem 4. Compute
∂z
∂x
and
∂z
∂y
if
u2 + v 2
, u = e−x−y , v = exy
u2 − v 2
(a) by substitution and direct calculation,
z=
(b) by the chain rule.
Solution. (a) z =
xy). Then
e−2x−2y +e2xy
e−2x−2y −e2xy
=
∂z
1+y
=
,
2
∂x
sinh (x + y + xy)
e−x−y−xy +ex+y+xy
e−x−y−xy −ex+y+xy
= − coth(x + y +
∂z
1+x
=
.
2
∂y
sinh (x + y + xy)
2
2u(u2 −v 2 )−2u(u2 +v 2 )
∂z
4e−x−y+2xy
= − (u24uv
= − (e−2x−2y
,
=
∂u
(u2 −v 2 )2
−v 2 )2
−e2xy )2
2v(u2 −v 2 )+2v(u2 +v 2 )
4u2 v
4e−2x−2y+xy
∂u
−x−y
= (u2 −v2 )2 = (e−2x−2y −e2xy )2 , ∂x = −e
,
(u2 −v 2 )2
(b)
∂z
∂v
∂u
∂y
=
=
3
−e−x−y ,
∂v
∂x
= yexy ,
∂v
∂y
= xexy . Then we have
∂z
∂z ∂u ∂z ∂v
=
+
∂x
∂u ∂x ∂v ∂x
4e−2x−2y+xy
4e−x−y+2xy
−x−y
(−e
)
+
· yexy
= − −2x−2y
(e
− e2xy )2
(e−2x−2y − e2xy )2
2
2
1+y
= . . . = −x−y−xy
.
(1 + y) =
2
x+y+xy
e
−e
sinh (x + y + xy)
Analogously,
∂z
∂z ∂u ∂z ∂v
=
+
∂y
∂u ∂y ∂v ∂y
2
2
1+x
.
= . . . = −x−y−xy
(1
+
x)
=
2
e
− ex+y+xy
sinh (x + y + xy)
Problem 5. Show that the following functions are harmonic, i.e.,
satisfy the Laplace equation ∇2 f = 0:
(a) f (x, y) = arctan xy .
(b) f (x, y) = ln(x2 + y 2 ).
Solution. (a)
1
1+(y/x)2
·
1
x
=
∂f
∂x
=
∂2f
x
,
x2 +y 2 ∂y 2
∇2 f =
1
1+(y/x)2
−
y
x2
∇2 f =
∂2f
∂x2
=
2xy
, ∂f
(x2 +y 2 )2 ∂y
=
= − (x22xy
, so that
+y 2 )2
∂ 2f
∂ 2f
2xy
2xy
+
=
−
= 0.
∂x2
∂y 2
(x2 + y 2 )2 (x2 + y 2 )2
2
2
2
2
∂f
= x22x
, ∂ f = 2(x (x+y2 +y)−(2x)
2 )2
∂x
+y 2 ∂x2
2(x2 +y 2 )−(2y)2
2x2 −2y 2
=
, so that
2
2
2
(x +y )
(x2 +y 2 )2
(b)
y
= − x2 +y
2,
=
2y 2 −2x2
, ∂f
(x2 +y 2 )2 ∂y
=
2
2y
, ∂f
x2 +y 2 ∂x2
=
∂ 2f
2y 2 − 2x2
2x2 − 2y 2
∂ 2f
+
=
+
= 0.
∂x2
∂y 2
(x2 + y 2 )2 (x2 + y 2 )2
Problem 6. (a) Compute the divergence and curl of the vector field
→
−
→
−
→
−
→
−
F (x, y, z) = (x + y)3 i + sin(xy) j + cos(xyz) k at the point (2, 0, 1).
→
−
→
−
→
−
→
−
(b) Show that the vector field V (x, y, z) = 2x i − 3y j + 4z k is
not the curl of any vector field.
4
→
−
Solution. (a) div F =
∂
(x
∂x
+ y)3 +
∂
(sin(xy))
∂y
∂
(cos(xyz))
∂z
=
→
−
3(x+y) +x cos(xy)−xy sin(xyz). Then div F (2, 0, 1) = 3·4+2 cos 0−
0 = 12 + 2 = 14.
→
→
−
→
−
−i
j
k
→
−
∂
∂
∂
curl F = ∂x
∂y
∂z
(x + y)3 sin(xy) cos(xyz)
→
−
→
−
→
−
= −xz sin(xyz) i + yz sin(xyz) j + (y cos(xy) − 3(x + y)2 ) k .
→
−
→
−
→
−
→
−
→
−
Hence curl F (2, 0, 1) = 0 i + 0 j − 12 k = −12 k .
→
−
∂
∂
∂
(b) div V = ∂x
(2x) + ∂y
(−3y) + ∂z
(4z) = 2 − 3 + 4 = 3 6= 0, therefore
→
−
V is not a curl (curls are divergence-free!).
2
+