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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