1. No category

Math 203
Solution
week 3
Exc 11.3 no 5) the cylinders z = x2 and z = 4y 2 intersect in two curves, one of which passes through the
point (2,-1,4). Find a parametrization of that curve using t = y as parameter.
solution:
√
√
if t = y, then z = 4y 2 = 4t2 and x = z = 4t2 = ±2t. now since it passes the point (2,-1,4),
by substituting t = −1 we get r = −2ti + tj + 4t2 k
Exc 11.3 no 6) The plane x + y + z = 1 intersect the cylinder z = x2 in a parabola. parametrize the parabola
using t = x as parameter.
solution:
if t = x, then z = t2 and y = 1 − x − z = 1 − t − t2 , then we get r = ti + (1 − t − t2 )j + t2 k
Exc 11.3 no 7) parametrize the given curve of intersection between x2 + y 2 = 9 and z = x + y.
solution:
if we let x = 3cost and y = 3sint, then we get z = 3cost + 3sint. hence, r = 3costi + 3sintj +
(3cost + 3sint)k
Exc 11.3 no 10) parametrize the given curve of intersection between yz + x = 1 and xz − x = 1.
solution:
if we let x = t, then we get z =
ti +
t(1−t)
1+t j
+
1+x
x
=
1+t
t
and y =
1−x
z
=
1−t
z
=
t(1−t)
1+t .
hence r =
1+t
t k
Exc 11.3 no 11) the plane z = 1 + x intersect the cone z 2 = x2 + y 2 in a parabola. try to parametrize using
(a)t = x, (b)t = y, (c)t = z. which of this choices for t leads to a single parametrization that
represent the whole parabola? what is that parametrization? what happen to the other two
choices?
solution:
p
√
√
(a) if x = t, then z √
= 1 + x = 1 + t, y = ± z 2 − x2 = ± (1 + t)2 + t2 = ± 1 + 2t + 2t2 .
hence, r = ti ± 1 + 2t + 2t2 j + (1 + t)k
2
(b) if y = t, then x = z − 1,so z 2 = (z − 1)2 + y 2 = z 2 − 2z + 1 + t2 .then we have z = 1+t
2
2
2
2
and x = t 2−1 hence, r = t 2−1 i + tj + 1+t
k
2
p
√
√
(c) if z = t, then x = z −√1 = t − 1, y = ± z 2 − x2 = ± (t2 + (t − 1)2 = ± 1 − 2t + 2t2 .
hence, r = (t − 1)i ± 1 − 2t + 2t2 j + tk
1
Exc 11.3 no 17) find the length of the conical helix given by the parametrization r = (tcost, tsint, t), f or0 ≤
t ≤ 2π. why is the curve called a conical helix?
solution:
√
υ(t) = R|v| = |(cost −
tsint,
sint
+
tcost,
1)|
=
2 + t2 .
√
√
√
2π
2
so s = 0 υ(t) = π 2 + 4π + ln( 2π + 1 + 2π 2 )
Exc 11.3 no 20) find the length of the piecewise smooth curve r = t3 i + t2 j.f or − 1 ≤ t ≤ 2.
solution:
Exc 11.3 no 23) re-parametrize the given curve in the same orientation in terms of the arc length measured
from the point where t = 0. r = Ati + Btj + Ctk, (A2 + B 2 + C 2 > 0)
solution:
√
v = (A, B, C) then υ = A2 + B 2 + C 2 so the arclength from the point where t = 0 to
arbitrary t is
p
p
s = s(t) = intt0 A2 + B 2 + C 2 = t A2 + B 2 + C 2
so t =
√
s
.
A2 +B 2 +C 2
hence we have r =
Asi+Bsj+Csk
√
A2 +B 2 +C 2
Exc 11.3 no 24) re-parametrize the given curve in the √
same orientation in terms of the arc length measured
t
from the point where t = 0. r = e i + 2tj − e−t k
solution:
2
ˆ
Exc 11.4 no 1) find the unit tangent vector T(t)
for the curve r = ti − 2t2 j + 3t3 k.
solution:
v = i − 4tj + 9t2 k
|v| =
√
ˆ
T(t)
=
1 + 16t2 + 81t4
2
√i−4tj+9t k
1+16t2 +81t4
ˆ
Exc 11.4 no 2) find the unit tangent vector T(t)
for the curve r = asinωti + acosωtk.
solution:
v = (aωcosωt, 0, −aωsinωt)
|v| = |aω|
ˆ
T(t)
=
aωcosωti−aωsinωtk
|aω|
ˆ
Exc 11.4 no 3) find the unit tangent vector T(t)
for the curve r = costsinti + sin2 tj + costk.
solution:
v = (cos2t, sin2t, −sint)
|v| =
√
ˆ
T(t)
=
1 + sin2 t
cos2ti+sin2tj−sintk
√
1+sin2 t
ˆ
Exc 11.4 no 4) find the unit tangent vector T(t)
for the curve r = acosti + bsintj + tk.
solution:
3
v = −asinti + bcostj + k
|v| =
√
ˆ
T(t)
=
a2 sin2 t + b2 cos2 t + 1
√−asinti+bcostj+k
a2 sin2 t+b2 cos2 t+1
Exc 11.4 no 5) show that if κ = 0 for all s, then the curve r = r(s) is a straight line.
solution:
since κ is the rate of turning of the unit tangent, then since its value is zero everywhere, the
tangent line is not changing direction, which implies that it is a straight line.
√
Exc 12.1 no 2) specify the domain of the function f (x, y) = xy.
solution:
the domain is the first and third quadrant since xy ≥ 0
p
Exc 12.1 no 5) specify the domain of the function f (x, y) = 4x2 + 9y 2 − 36.
solution:
The domain consists of all points (x, y) satisfying 4x2 + 9y 2 ≥ 36
Exc 12.1 no 6) specify the domain of the function f (x, y) = √
1
.
x2 −y 2
solution:
The domain consists of all points (x, y) satisfying |x| > |y|
Exc 12.1 no 7) specify the domain of the function f (x, y) = ln(1 + xy).
solution:
The domain consists of all points (x, y) satisfying xy > −1
Exc 12.1 no 14) graph the function f (x, y) = 4 − x2 − y 2 , (x2 + y 2 ≤ 4, x ≥ 0, y ≥ 0).
solution:
the graph is
4
Exc 12.1 no 15) graph the function f (x, y) =
p
x2 + y 2
solution:
the graph is
Exc 12.1 no 16) graph the function f (x, y) = 4 − x2
solution:
the graph is
5
Exc 12.1 no 29) a plane containing y-axis ,sloping uphill in the x direction.
Exc 12.1 no 30) a cylinder parallel to the x-axis, rising from height zero first steeply and then more and more
slowly as y increases.
Exc 12.1 no 31) a right-circular cone with base in xy-plane and vertex at height 5 on the z-axis.
Exc 12.1 no 32) a cylinder(or parabolic) with axis in the yz-plane, sloping upwards in the direction of increasing y.
6