Math 53: Worksheet 5
September 26
1. Sketch the following surfaces.
(a) y 2 + 4 = x2 + 4z 2 .
This equation represents a hyperboloid of one sheet wrapped around the y-axis.
2
1
0
-1
2.6904
-2
0.53808
Z axis
1.61424
-0.53808
-1.61424
-2.6904
5.38516
3.23459
1.08402
5
xis
3
-1.06656
Xa
1
Y axis
-1
-3.21713
-3
-5.3677
-5
(b) y 2 + 2z 2 = 6.
This equation represents a cylinder
wrapped around the
√
√ x-axis. Note that the
cylinder will have a radius of 6 in the y direction and 3 in the z direction.
1.5
1
0.5
0
-0.5
1.73065
-1
-1.5
0.346129
-0.346129
-1.03839
-1.73065
-5
-2.44155
-3
-1.46334
-0.485132
Y axis
1
0.493075
3
1.47128
5
2.44949
1
-1
is
X ax
Z axis
1.03839
(c) x2 − y 2 − z 2 = 4.
We have x2 = 4 + y 2 + z 2 . This represents a hyperboloid of two sheets about the
x-axis.
4
2
0
-2
5.65227
-4
3.39136
Z axis
1.13045
-1.13045
-3.39136
-5.65227
-5.63851
-3.37944
-6
-3.6
-1.12036
Y axi
s
-1.2
1.13871
1.2
3.39778
X axis
3.6
5.65685
6
2. Find a vector function that represents the curve of intersection of the cylinder x2 +y 2 =
9 and the plane x + 2y + z = 3.
Note that the cylinder can be parametrized as x = 3 ∗ cos(t), y = sin(t), where
0 ≤ t < 2π, with z ∈ R. Plugging these in the equation of the plane gives
z = 3 − x − 2y = 3 − 3 cos(t) − 6 sin(t).
The curve of intersection is therefore given by
r(t) =< 3 cos(t), 3 sin(t), 3 − 3 cos(t) − 6 sin(t) > , 0 ≤ t < 2π.
3. Find the point of intersection of the tangent lines to the curve r(t) =< sin(πt), 2 sin(πt), cos(πt) >
at the points where t = 0 and t = 0.5.
We have
r0 (t) =< π cos(πt), 2π cos(πt), −π sin(πt) > .
At t = 0 and t = 0.5, we get v := r0 (0) =< π, 2π, 0 > and w := r0 (0.5) =< 0, 0, −π >.
Also let a := r(0) =< 0, 0, 1 > and b := r(0.5) =< 1, 2, 0 >. The two tangent lines
have equations
r1 (s) = a + sv =< πs, 2πs, 1 >
r1 (u) = b + uw =< 1, 2, −πu >
where s, u ∈ R. At the point of intersection, we require
πs = 1, 2πs = 2, 1 = −πu
so that s =
1
π
and u = − π1 . The point of intersection is therefore
r1 (1/π) = r2 (−1/π) = < 1, 2, 1 > .
2
4. Show that the curve with parametric equations x = sin(t), y = cos(t), z = sin2 (t) is
the curve of intersection of the surfaces z = x2 and x2 + y 2 = 1. Use this fact to help
sketch the curve.
Note that the parametric equations of the curve obey both z(t) = x(t)2 and x(t)2 +
y(t)2 = 1 for all values of t.
1
0.9
0.8
0.7
z-axis
0.6
0.5
0.4
0.3
0.2
0.1
0
1
0.5
1
0
0.5
0
-0.5
-1
y-axis
-0.5
-1
x-axis
5. Draw contour maps of the following functions showing several level curves.
(a) f (x, y) = y sec(x).
Observe that f (x, y) = k ⇒ y sec(x) = k ⇒ y = k cos(x). The level curves of
f are therefore just cosine waves of differing amplitudes. Note that the points
where the level curves meet are not actually part of the domain of f .
3
2
y
1
0
-1
-2
-3
0
1
2
3
4
x
(b) f (x, y) =
x
.
x2 +y 2
3
5
6
7
1 2
Let k 6= 0 first. Observe that f (x, y) = k ⇒ k1 x = x2 + y 2 ⇒ (x − 2k
) + y 2 = 4k12 .
The level curves of f are therefore just circles centered at different points on the
x-axis and passing through the origin. Note again that the origin is not actually
part of the domain of f . For k = 0, we simply get f (x, y) = 0 ⇒ x = 0, i.e, the
y-axis.
0.8
0.6
0.4
0.2
y
0
-0.2
-0.4
-0.6
-0.8
-1
-0.5
0
0.5
1
x
(c) f (x, y) = sin(xy).
Let −1 ≤ k ≤ 1. Observe that f (x, y) = k ⇒ sin(xy) = k ⇒ xy = arcsin(k) ⇒
y = arcsin(k)
. The level curves of f therefore are the inverse proportionality curves
x
between x and y.
3
2
y
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
x
6. Find the limit, if it exists, or show that the limit does not exist.
(a) lim(x,y)→(0,0)
5x2 y
.
x2 +y 2
We show that the limit is zero. Switching to polar coordinates gives
5x2 y
5r3 cos2 (θ) sin(θ)
=
lim
= lim 5r cos2 (θ) sin(θ).
r→0
r→0
(x,y)→(0,0) x2 + y 2
r2
lim
4
Since 5 cos2 (θ) sin(θ) is bounded, we conclude that the limit is zero.
(b) lim(x,y)→(0,0)
sin(xy 2 )
.
x2 +y 4
Along the y-axis, we have x = 0 so we get limy→0
sin(0)
y4
= 0. Along the curve
4)
limy→0 sin(y
2y 4
= 21 . Since we get different values along
x = y 2 , however, we get
different paths, we conclude that the required limit does not exist.
(c) lim(x,y)→(0,0)
2x
.
x2 +x+y 2
Along the y-axis, we have x = 0 so we get limy→0 y02 = 0. Along the x-axis, we
have y = 0 so we get limx→0 x22x+x = 2. Since we get different values along different
paths, we conclude that the required limit does not exist.
(d) lim(x,y)→(0,0) √
x2 +y 2
x2 +y 2 +1−1
.
Note that rationalization yields
p
p
x2 + y 2 + 1 + 1
(x2 + y 2 )( x2 + y 2 + 1 + 1)
x2 + y 2
×p
= lim
lim p
(x,y)→(0,0)
x2 + y 2 + 1 − 1
x2 + y 2 + 1 − 1
x2 + y 2 + 1 + 1 (x,y)→(0,0)
p
whence we get lim(x,y)→(0,0)
x2 + y 2 + 1 + 1. Since this function is continuous at
√
(0, 0), the limit is 0 + 0 + 1 + 1 = 2.
2
(e) lim(x,y)→(0,0)
2
e−x −y −1
.
x2 +y 2
Switching to polar coordinates gives
2
2
2
2
e−r − 1
−2re−r
e−x −y − 1
2
lim
=
lim
=
lim
= lim −e−r = −1
2
2
2
r→0
r→0
r→0
(x,y)→(0,0)
x +y
r
2r
where we used L’Hopital’s rule.
5