Density Function f
Y
(y) =
1
3
1
(1)
√
2π
exp
−
(y + 4)
2
18(1)
2
, −∞ ≤ 0 ≤ ∞
91
4.10. Given the fluid drag equation F
D
= C
D
V
2
, where C
D
is a constant (with dimensions) and
f
V
(v) = 0.1, for the range 10 ≤ v ≤ 20, derive f
F
D
and sketch both density functions.
Solution: For this case there are two roots
v = ±
F
D
/C
D
, and
dv
dF
D
= ±
1
2
√
F
D
C
D
.
The general transformation is given by
f
F
D
(F
D
) =
1
2
√
C
D
F
D
1
f
V
3
F
D
C
D
+ f
V
−
3
F
D
C
D
2
u (F
D
) ,
where u(·) is the unit step function. However, since v has a positive range we must drop the negative root.
Therefore,
f
F
D
(F
D
) =
1
2
√
C
D
F
D
1
10
u (F
D
)
=
1
20
√
C
D
F
D
u (F
D
) .
If C
D
= 2.0,then
f
F
D
(F
D
) =
1
20
√
2F
D
, 200 ≤ F
D
≤ 800.
The density plots are shown in the next two figures.
Density Function f
V
(v) =
1
10
, 10 ≤ v ≤ 20
Density Function f
F
D
=
1
20
√
2F
D
, 200 ≤ F
D
≤ 800
If 10 ≤ v ≤ 20 and −10 ≤ v ≤ −20, then f
V
(e)= 0.05 and we would retain both positive and negative
roots.
92
4.11. Given the fluid drag equation, F
D
= C
D
V
2
, where C
D
is a constant (with dimensions) and V is
standard normal N(0, 1). (Note that F
D
≥ 0.) Derive f
F
D
and sketch both density functions.
Solution:
f
V
(v) =
1
√
2π
exp
−
v
2
2
.
The roots are v = ±
F
D
/C
D
, and here, since V is governed by a normal density, it can be positive or
negative and we retain both roots. Thus,
dv
dF
D
= ±
1
2
√
F
D
C
D
f
F
D
(F
D
) =
1
2
√
C
D
F
D
1
f
V
3
F
D
C
D
+ f
V
−
3
F
D
C
D
2
=
1
√
2πC
D
F
D
exp
−
F
D
2C
D
, F
D
> 0.
The density function f
V
(e) is shown in the figure below.
Density Function f
V
(v) =
1
√
2π
exp
−
v
2
2
, −∞ ≤ 0 ≤ ∞
For C
D
= 2.0, the density function f
F
D
(F
D
) is shown in the figure below.
Density Function f
F
D
(F
D
) =
1
√
2πC
D
F
D
exp
−
F
D
2C
D
, F
D
> 0
93
Note that a better model of the drag force is F
D
= C
D
V |V | where the following derivative is useful:
d(v |v|)
dv
= |v| + v signum(e).
94
4.12. For the function Y = a tan X, a > 0, derive the general relation for f
Y
(y). Then, assume X is
uniformly distributed over [−π, π] and derive f
Y
(y). Sketch the density functions for X and Y.
Solution: The general transformation is found as follows:
x
n
= arctan (y/a) , n = ..., −1, 0, 1, ...
g
(x) =
a
cos
2
x
=
a
2
+ y
2
a
f
Y
(y) =
a
a
2
+ y
2
∞
n=−∞
f
X
(x
n
) .
We have used the geometrical relation implied by the function y = a tan x. That is, tan x = y/a and therefore,
cos x = a/
a
2
+ y
2
.
For the range [−π, π] , x = arctan(y/a), there is one root. Then,
f
Y
(y) =
a
a
2
+ y
2
1
2π
=
a
2π(a
2
+ y
2
)
.
Graphs of the density functions are shown below.
a tan X for a = 1 f
Y
(y) =
a
2π(a
2
+ y
2
)
for a = 1
95
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