Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 3
To be solved during the tutorial session Thursday/Friday, 4/5 February 2016
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) = ln(x2 − y 2 )
Solution: The domain consists of the points satisfying x2 − y 2 > 0, and it is an open set
p
(b) f (x, y) = (x2 + 2x + y 2 )(−1 − x2 − 2x − 4y 2 + 4y)
Solution: The domain consists of the points satisfying
((x + 1)2 + y 2 − 1)(−(x + 1)2 − 4(y − 1/2)2 + 1) ≥ 0, and it is a closed set
2. Sketch the level curve z = k for the specified values of k
z = x2 + 2x + 4y 2 − 4y ,
k = −2, −1, 2 .
Solution: The level curve equation x2 + 2x + 4y 2 − 4y = k can be written as
1
(x + 1)2 + 4(y − )2 = k + 2 ,
2
and, therefore, the level curves are ellipses with the centre located at (−1, 1/2). For
k = −2 the level curve is just the point (−1, 1/2).
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3. Determine whether the limit exists. If so, find its value
(a)
lim
(x,y)→(0,0)
sin(x2 − y 2 )
p
1 + x2 + y 2 − 1
Solution: The limit does not exist because the limit along y = 0 gives
lim
x→0,y=0
sin(x2 )
x2
sin(x2 − y 2 )
p
= lim √
= lim 1 2 = 2 ,
1 + x2 + y 2 − 1 x→0 1 + x2 − 1 x→0 2 x
(1)
while the limit along x = 0 gives
lim
y→0,x=0
sin(x2 − y 2 )
− sin(y 2 )
−y 2
p
= lim p
= lim 1 2 = −2 .
1 + x2 + y 2 − 1 y→0 1 + y 2 − 1 y→0 2 y
(2)
(b)
lim
arctan(xy) ln(ex
2 +y 2
− 1)
(x,y)→(0,0)
Solution: The limit exists as can be seen by switching to the polar coordinates and is equal
to 0.
(c)
lim
(x,y)→(1,1)
1 − cosh(x2 − y 2 )
x−y
sin(x − y)(1 − exp( x+y
))
Solution: The limit exists as can be seen by switching to the coordinates x−y = t , x+y = u
lim
(x,y)→(1,1)
1 − cosh(x2 − y 2 )
1 − cos(ut)
= lim
x−y
sin(x − y)(1 − exp( x+y )) (t,u)→(0,2) sin(t)(1 − exp( ut ))
1 2
4t
1 − cos(2t)
= lim
= lim 2 t = −4 .
t
t→0 sin(t)(1 − exp( ))
t→0 t(− )
2
2
(3)
4. Show that the hyperboloid of two sheets x2 + y 2 − z 2 = −1 is not a connected surface that
is there are two points of the hyperboloid which cannot be connected by a curve which
lies on the hyperboloid.
p
Solution:
The
equation
of
the
hyperboloid
is
equivalent
to
z
=
1 + x2 + y 2 ≥ 1, z =
p
− 1 + x2 + y 2 ≤ −1. Since there is a gap in z, the points satisfying z ≥ 1 cannot be
connected to the points satisfying z ≤ 1. The hyperboloid is a surface of revolution, and
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it can be obtained from the two branches of the hyperbola x2 − z 2 = −1 by rotating it
about the z-axis.
5. Sketch the graph of the function and identify it
p
(a) z = x2 + 2x + y 2
Solution: It is half of a hyperboloid of one sheet with the axis of symmetry parallel to
the z-axis and through (−1, 0, 0).
(b) z = 2x − x2 − y 2
Solution: It is a paraboloid with the vertex at (1, 0, 1).
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