This means that the discriminant of this equation ought to be equal to
zero, i.e. we ought have 4(36
−
H
) = 0. Thus the special value of
H
is
H
= 36. In this case, the differential equation becomes
dP
dt
= 12
P
−
P
2
−
36 =
−
(
P
−
6)
2
.
The unique equilibrium is at
P
= 6.
If
H >
36 the equation has no equilibria, and moreover we have that
dP
dt
= 12
P
−
P
2
−
H
=
−
(
P
−
6)
2
+ (36
−
H
)
is always negative. This shows that if we harvest the trout at rate higher
than the critical harvesting rate
H
= 36, then the population will
steadily decrease and we will
eventually empty the pond completely
.
In contrast, if we harvest at a rate smaller than the critical harvesting
rate
H
= 36 and if we start with enough trout in the pond, then we
will always have enough fish to harvest .
Solution of problem 1.9:
(a) From the parametric equation
vector
r
(
t
) = (1 +
2
t
)
hatwide
ı
+ (1 + 3
t
)
hatwide
+ (1 + 4
t
)
hatwide
k
we can extract a direction vector for the
line. It is the vector
−→
v
whose components are given by the coefficients
of the parameter
t
in the parametric equation. Thus
−→
v
= 2
hatwide
ı
+3
hatwide
+4
hatwide
k
.
Also, from the equation 2
x
+ 3
y
−
4
z
= 9 of the plane we can extract a
normal vector
−→
n
to the plane. It is the vector whose components are
the coefficients of the equation of the plane. Thus
−→
n
= 2
hatwide
ı
+ 3
hatwide
−
4
hatwide
k
.
15
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The line and the plane will be perpendicular when
−→
v
is parallel to
−→
n
, that is when
−→
v
is proportional to
−→
n
. But if
−→
v
=
c
·
−→
n
for some
constant
c
, then we will have 2 = 2
c
, 3 = 3
c
, and 4 =
−
4
c
.
From
he first equation we get
c
= 1 but from the last equation we have
c
=
−
1. This is a contradiction. Therefore the line and the plane are
not perpendicular, and so (a) is
False
.
(b) The equation
x
2
=
z
2
depends only on two variables so it describes
a cylinder with a base in the
xz
plane. Hence
x
2
=
z
2
can not be an
ellipsoid and (b) is
False
.
(c) If the length of
vectora
×
vector
b
is zero, then the vector
vectora
×
vector
b
must be the zero
vector. This can happen either when one of
vectora
or
vector
b
is the zero vector,
or when
vectora
is parallel to
vector
b
. For instance if
−→
a
is any vector and
−→
b
=
−→
a
we will have
vectora
×
vector
b
=
vectora
×
vectora
=
−→
0 . Hence (c) is
False
.
Solution of problem 1.10:
(a)
The unit normal vector is given by
−→
N
=
d
−→
T
dt
vextendsingle
vextendsingle
vextendsingle
d
−→
T
dt
vextendsingle
vextendsingle
vextendsingle
where
−→
T
is the unit tangent vector. To compute
−→
T
we compute the
velocity vector
d
−→
r /dt
=
(
2 cos(2
t
)
,
−
2 sin(2
t
)
,
1
)
and normalize
−→
T
=
1

d
−→
r /dt

d
−→
r /dt
=
1
radicalbig
4 cos
2
(2
t
) + 4 sin
2
(2
t
) + 1
(
2 cos(2
t
)
,
−
2 sin(2
t
)
,
1
)
=
(bigg
2
3
cos(2
t
)
,
−
2
3
sin(2
t
)
,
1
3
)bigg
.
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 Fall '07
 Temkin
 Math, Calculus, Quadratic equation, Parametric equation

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