HP 49g+ - Manual (Page 487)

Page 16-13
L{df/dt} = s ⋅F(s) - f
o
.
Example 1 – The velocity of a moving particle v(t) is defined as v(t) = dr/dt,
where r = r(t) is the position of the particle. Let r
o
= r(0), and R(s) =L{r(t)}, then,
the transform of the velocity can be written as V(s) = L{v(t)}=L{dr/dt}= s ⋅R(s)-r
o
.
• Differentiation theorem for the second derivative. Let f
o
= f(0), and
(df/dt)
o
= df/dt|
t=0
, then L{d 2 f/dt 2 } = s 2 ⋅F(s) - s⋅f
o
– (df/dt)
o
.
Example 2 – As a follow up to Example 1, the acceleration a(t) is defined as
a(t) = d
2 r/dt 2 . If the initial velocity is v
o
= v(0) = dr/dt|
t=0
, then the Laplace
transform of the acceleration can be written as:
A(s) = L{a(t)} = L{d 2 r/dt 2 }= s 2 ⋅R(s) - s⋅r
o
– v
o
.
• Differentiation theorem for the n-th derivative.
Let f
(k)
o
= d k f/dx k |
t = 0
, and f
o
= f(0), then
L{d n f/dt n } = s n ⋅F(s) – s n-1 ⋅f
o
−…– s⋅f (n-2)
o
– f (n-1)
o
.
• Linearity theorem. L{af(t)+bg(t)} = a⋅L{f(t)} + b⋅L{g(t)}.
• Differentiation theorem for the image function. Let F(s) = L{f(t)}, then
d n F/ds n = L{(-t) n ⋅f(t)}.
Example 3 – Let f(t) = e
–at , using the calculator with ‘EXP(-a*X)’ ` LAP, you
get ‘1/(X+a)’, or F(s) = 1/(s+a). The third derivative of this expression can
be calculated by using:
‘X’ ` ‚¿ ‘X’ `‚¿ ‘X’ ` ‚¿ µ
The result is
‘-6/(X^4+4*a*X^3+6*a^2*X^2+4*a^3*X+a^4)’, or
d 3 F/ds 3 = -6/(s 4 +4 ⋅a⋅s 3 +6 ⋅a 2 ⋅s 2 +4 ⋅a 3 ⋅s+a 4 ).