Module MA1132 (Frolov), Advanced Calculus
Homework Sheet 3
Each set of homework questions is worth 100 marks
Due: at the beginning of the tutorial session Thursday/Friday, 11/12 February 2016
Name:
1. Sketch the domain of f (you may use Mathematica). Use solid lines for portions of the
boundary included in the domain and dashed lines for portions not included. Determine
whether the domain is an open set, closed set, or neither.
√
(a) f (x, y) = √x−y
x+y
Solution: The domain consists of the points satisfying x − y ≥ 0 , x + y > 0, and it is neither
an open set nor a closed set
10
5
0
-5
-10
-10
-5
0
5
10
√
(b) f (x, y) = ln(x+y)
x−y
Solution: The domain consists of the points satisfying x − y > 0 , x + y > 0, and it is an
open set
10
5
0
-5
-10
-10
(c) f (x, y) =
q
-5
0
5
10
x2 +y 2 −1
2−x2 +2x−4y 2 +4y
Solution: The domain consists of the points satisfying x2 +y 2 −1 ≥ 0 , 2−x2 +2x−4y 2 +4y > 0
and x2 + y 2 − 1 ≤ 0 , 2 − x2 + 2x − 4y 2 + 4y < 0, and it is neither an open set nor a
closed set
1
2. Sketch the level curve z = k for the specified values of k
z = 4x2 + 4x + y 2 − 2y ,
k = −2, −1, 2 .
Solution: The level curve equation 4x2 + 4x + y 2 − 2y = k can be written as
1
4(x + )2 + (y − 1)2 = k + 2 ,
2
and, therefore, the level curves are ellipses with the centre located at (−1/2, 1). For
k = −2 the level curve is just the point (−1/2, 1).
3
2
1
-1.5
-1.0
0.5
-0.5
-1
3. Determine whether the limit exists. If so, find its value
(a)
−√
lim
(x,y)→(0,0)
1
e x2 +y2
ex2 +y2 − 1
Solution: The limit exist as can be seen by switching to the polar coordinates
−√
lim
(x,y)→(0,0)
1
1
1
e− r
e− r
e x2 +y2
= lim+ r2
= lim+ 2 = lim t2 e−t = 0 .
2 +y 2
x
t→+∞
e
− 1 r→0 e − 1 r→0 r
2
(1)
(b)
1 − cosh(x + y)
2x
)
sin(x2 − y 2 ) ln( x−y
lim
(x,y)→(1,−1)
Solution: The limit exist as can be seen by switching to the coordinates x + y = t , x − y = u
lim
(x,y)→(1,−1)
1 − cosh(x + y)
1 − cosh(t)
1 − cosh(t)
=
lim
=
lim
t+u
2x
sin(x2 − y 2 ) ln( x−y ) (t,u)→(0,2) sin(tu) ln( u ) t→0 sin(2t) ln(1 + 2t )
− 12 t2
1
= lim
=− .
t
t→0 2t
2
2
(2)
(c)
lim
(x,y)→(0,0)
3 + cos(2x) − 4 cosh(y)
p
1 − 4 1 + x2 + y 2
Solution: The limit exist as can be seen by switching to the polar coordinates
3 + cos(2x) − 4 cosh(y)
3 + cos(2r cos φ) − 4 cosh(r sin φ)
p
√
=
lim
r→0
(x,y)→(0,0)
1 − 4 1 + r2
1 − 4 1 + x2 + y 2
3 + 1 − 12 (2r)2 cos2 φ − 4(1 + 12 r2 sin2 φ) + O(r4 )
= 8.
= lim
r→0
− 14 r2 + O(r4 )
lim
(3)
4. A cone is the union of a set of half-lines that start at a common apex point and go
through
a base which can be any parametric curve. Show that the graph of z =
p
2
x − 4x + 4y 2 + 8y + 8 is a cone. Where is its apex? What can be chosen as its base?
Sketch the base.
Solution: We write x2 − 4x + 4y 2 + 8y + 8 = (x − 2)2 + 4(y + 1)2 . Then the apex is obviously
at the
p point A = (2, −1, 0). Let (x0 , y0 , z0 ) be any point but the apex of the surface
z = (x − 2)2 + 4(y + 1)2 . We need to show that the half-line that start at the apex and
go through the point belongs to the surface. The half-line has the parametric equations
p
x = 2+t(x0 −2) , y = −1+t(y0 +1) , z = tz0 , t ≥ 0 , z0 = (x0 − 2)2 + 4(y0 + 1)2 ,
(4)
where t = 0 corresponds to the apex,pand t = 1 corresponds to the point (x0 , y0 , z0 ).
Substituting
these equations into z = (x − 2)2 + 4(y + 1)2 one finds that the equation
p
z =
(x − 2)2 + 4(y + 1)2 is satisfied for any t ≥ 0 which means that this half-line
belongs to the surface. As a base of the cone one can choose the intersection of the cone
with the plane z = 1 which gives the curve (x − 2)2 + 4(y + 1)2 = 1. This curve is an
ellipse.
-0.6
-0.8
-1.0
-1.2
-1.4
1.0
1.5
2.0
3
2.5
3.0
5. Show that the hyperboloid of one sheet x2 + y 2 − z 2 = 1 is a connected surface, that is any
two points of the hyperboloid can be connected by a curve which lies on the hyperboloid.
Sketch the hyperboloid.
Solution: Let (x0 , y0 , z0 ) and (x1 , y1 , z1 ) be any two points on the hyperboloid. It is clear that
if (a, b, c) is a point on the hyperboloid then all points satisfying x2 + y 2 = a2 + b2 , z = c
are also on the hyperboloid. These points obviously form a circle. Thus, the points
(x0 , y0 , z0 ) and (x1 , y1 , z1 ) can be connected by arcs of their circles to the points (x00 , 0, z0 )
2
02
2
and (x01 , 0, z1 ) in the xz-plane where x00 > 0 and x01 > 0, and x02
0 −z0 = 1, x1 −z1 =√1. The
equation x2 − z 2 = 1 , x > 0 gives obviously one branch of the hyperbola x = 1 + z 2
and all points on this branch are connected to each other. The
√ hyperboloid is a surface of
revolution, and it can be obtained from the hyperbola x = 1 + z 2 by rotating it about
the z-axis.
6. Sketch the graph of the function and identify it
p
(a) z = − 2x − x2 − y 2
Solution: It is a semi-sphere centred at (1, 0, 0).
(b) z =
p
2y − y 2
4
Solution: It is half of a cylinder of radius 1 with the axis of symmetry parallel to the
y-axis and through (1, 0, 0).
5