Question

i need to learn derivatives of transform...can someone give me a demo?

for example
\[\text{Find} \; \mathcal L \lbrace t^2 \sin kt \rbrace\]

Answer 1

I'm learning from this question :)

Answer 2

good for you! :D

Answer 3

waiting for the answer .. sami , please :)

Answer 4

btw...please dont post a full solution...i learn better when it's step by step :DD

Answer 5

yes let me try.
\[\Large L[t^nf(t)]=(-1)^n \frac{d^n}{ds^n}[F(s)]\]
ok

Answer 6

what does that mean?

Answer 7

this means first find the Laplace of f(t) here
f(t)=sin(kt)
what is its Laplace.?

Answer 8

L(sin kt)=k/(s^2+k^2)
\[L(t \sin kt)=--d/ds(k/s ^{2}+k ^{2})\]

Answer 9

uhh let's do this slowly... what does this mean \[\mathcal L \lbrace t^n f(t) \rbrace\]

Answer 10

t^n is t^2 and f(t) is sinkt?

Answer 11

\[L( t sinkt)=-d/ds(k/s ^{2}+k ^{2})\]

Answer 12

its means Laplace when a function f(t) is multiplied by t^n here n=2 in your post .
so first you need Laplace of f(t)=sin(kt)

Answer 13

...so im right?

Answer 14

yes you are right

Answer 15

nice...

Answer 16

so \[\large (-1)^n \frac{d^n}{ds^n} [F(s)]\]
is this a formula?

Answer 17

yes it is formula .
you need first of all the Laplace of sin(kt )
once you find the Laplace of sin(kt), then take its nth derivative with respect to s (here n=2 so second derivative)
then multiply the result of derivative with (-1)^n, in this case (-1)^2 =1

Answer 18

that is the EXACT formula?

Answer 19

yes it is !

Answer 20

my notes says something like \[f'(s) = \int _0^\infty (-t) e^{-st} f(t) dt\]
are those the same?

btw...im not too sure about this formula...my notes are a little vague

Answer 21

your notes are talking about transform of derivative .it is correct.
while the question you have posted is different .it needs the formula that i have provided.

Answer 22

uhmm okay... F(s) = L{f(t)} corret?

Answer 23

correct*

Answer 24

yes

Answer 25

uhmm wait...when do i use the formula you gave and when do i use the formula in my notes?

Answer 26

the formula i gave you always use it when you see a function is multiplied by t,t^2,t^3....
while the formula in your notes apply that whenever you encounter derivative . f',f'',f'''
and you need their Laplace.

Answer 27

@lgbasallote i just checked the formula you have mentioned is used during the proof of the formula i have provided !!!!!

Answer 28

but you would not use that one. its end result is what i have mentioned. stick to that one.

Answer 29

uhh so in layman's terms...?

Answer 30

the formula you gave is applicable when i have L{t^n f(t)}?

Answer 31

always.

Answer 32

what do you mean always?

Answer 33

always applicable when you have L{t^n f(t)}

Answer 34

oh. it's only applicable when it's like that?

Answer 35

*making notes*

Answer 36

or it applies in some other specific forms too?

Answer 37

yes it is applied to that form .and
it has applications when you will need inverse Laplace .

Answer 38

in layman's terms...?

Answer 39

in Layman's terms
whenever you encounter multiplication of any standard function by t and powers of t this formula will be the only option there to save you.
for example
L(te^t)
L(t^3sin(t)
L(t^4e^at)
L(t^2cosh(t))
L(t^3e^(at)sin(t))
the only formula is the above one this is the only only and only way you will be able to get their Laplace.

Answer 40

yes...but as far as getting laplace is concerned L{t^n f(t)} is the only time the formula you gave is applicable?

Answer 41

for now yes only when L{t^n f(t)}.
But
you will see this as you go through your course.

Answer 42

well anyway....on to the solution

Answer 43

ok
find Laplace of sin(kt) first.
i hope you can do this ?

Answer 44

n = 2 and f(t) = sin kt

Answer 45

yes correct.

Answer 46

\[F(s) = \mathcal L \lbrace \sin kt \rbrace \implies \frac{k}{s^2 + k^2}\]

Answer 47

yes it is correct
now differentiate this two times (because n=2) with respect to s.

Answer 48

so \[\mathcal L \lbrace t^2 \sin kt \rbrace = (-1)^2 \frac{d^2}{ds^2} (\frac{k}{s^2 + k^2})\]
??

Answer 49

yes

Answer 50

is the first derivative \[\frac{-k(2s+k^2)}{(s^2+k^2)^2}\]

Answer 51

use Quotient rule .
in the numerator it should be just -2s because k is constant.

Answer 52

oh lol yeah..

Answer 53

wait...shouldnt it be -2sk?

Answer 54

yes it should be .

Answer 55

is the second derivative \[\frac{-2k(s^2 + k^2)^2 + 8s^2k (s^2 + k^2)}{(s^2 + k^2)^4}\]??

Answer 56

ok wait let me calculate as well !

Answer 57

okay...

Answer 58

i think you made a mistake i am getting
\[\Large \frac{2k(3s^2-k^2)}{(s^2+k^2)^3}\]
check it again.

Answer 59

but how :O

Answer 60

now i have \[\huge \frac{-2k (-3s^2 + k^2)}{(s^2 + k^2)^3}\]
i think they're same right?

Answer 61

seems you just simplified farther than i did

Answer 62

ok
after first derivative we have
\[\Large \frac{-2ks}{(s^2+k^2)^2}\]
i am always comfortable with product rule :P
so we can write it as
\[\Large -2ks(s^2+k^2)^{-2}\]
now by product rule
\[ \frac{d}{ds}(-2ks(s^2+k^2)^{-2})=-2k((s^2+k^2)^{-2}-2ks(-2)(s^2+k^2)^{-3}*2s\]
can you simply it now

Answer 63

so this is the answer for \[\huge \mathcal L \lbrace t^2 \sin kt \rbrace\]
since (-1)^2 = 1

Answer 64

i got it already lol

Answer 65

ok

Answer 66

it was just a difference in simplification

Answer 67

so is this the final answer?

Answer 68

yes it is the answer because multiplication by (-1)^2=1 will not make any difference

it is final answer !!!!!!!! after 1 and half hrs :P

Answer 69

wow cool..my book agrees :D

Answer 70

and s>0 right?

Answer 71

yes