Question

mx'' = gm - kx'

Come up with a change of variables to solve this IVP, giving an equation for the position of the object at time t.

Answer 1

try V= dx/dt and dV/dt = d^2x/(dt^2)

Answer 2

hmm ... ok

Answer 3

\[\frac{ dV }{dt }+ \frac{ k }{m }V =g\]
\[\huge u = e ^{\int\limits_{ }^{ }\frac{ k }{m }dt}\]
\[\huge u =Ce ^{\frac{ kt }{m } }\]
\[ Ce ^{\frac{ kt }{m } } *(\frac{ dV }{ dt } +\frac{ k }{m } V =g)\]

Answer 4

seen this method before? integrating factor?

Answer 5

yeah i have

Answer 6

k, got it from here?

Answer 7

think so, thank you

Answer 8

@Algebraic! i dont have this as much as i thought i did!

Answer 9

did you find V?

Answer 10

from the step where we multiplied through by u:

Answer 11

yeah, ive done that.

Answer 12

\[Ce ^{\frac{ kt }{ m }} \frac{ dV }{ dt } + Ce ^{\frac{ kt }{ m }} \frac{ k }{ m }V = Ce ^{\frac{ kt }{ m }} g\]

Answer 13

how far have you gotten?

Answer 14

ya i have that

Answer 15

LHS now looks like the result of an application of the product rule... ie V' *u +V*u'

Answer 16

rewrite as (V*u)' = u*g

Answer 17

integrate both sides

Answer 18

\[ Ce ^{\frac{ kt }{ m }}V =\int\limits_{ }^{ } Ce ^{\frac{ kt }{ m }} g\]

Answer 19

ok wait ..

Answer 20

why do we insert u*g ?

Answer 21

we found u so that the LHS would look like the result of a product rule : (V*u)'
once we multiplied the *whole* equation by u

Answer 22

so I didn't 'insert' ug , we multiplied everything by u...

Answer 23

uV' + k/m *u*V = gu

Answer 24

right ok

Answer 25

ok, so how did you know
(vu)' = gu ?

Answer 26

i think i'm slow . idk why i'm taking so long to get this

Answer 27

when we integrate both sides for the last part its by dt right?

Answer 28

ok.. so i've made it to

\[Ce^(kt/m) * v = mg/k * Ce^(kt/m) \]

Answer 29

@amistre64 @Hero could you possibly help me with where i should go with this next step?

Answer 30

right so..sorry I got called away abruptly.

Answer 31

thats ok, thanks for coming back

Answer 32

but you're integrating from t=0 to t=t, so you get\[\huge Ce ^{\frac{ kt }{ m }} V = \frac{ mg }{ k }(Ce ^{\frac{ kt}{ m }} -Ce ^{\frac{ k0}{ m }} )\]

Answer 33

divide through by C*e^(kt/m) and you've found V

Answer 34

integrate V(t) to get x(t)

Answer 35

ok .. trying that

Answer 36

if you ever come back, i'd love your help

Answer 37

I'm here, what's up?

Answer 38

AH, sorry, i missed you!!
Ok, so i've done everything you said.
Ce^kt(0)m is just a constant when you integrate it right?

Answer 39

well...

Answer 40

did you divide through by Ce^(kt/m) yet?

Answer 41

ya .. so it isn't when you go to integrate it because its still over the other equation

Answer 42

go up to the last post I made with an equation. that's after integrating.
divide both sides by Ce^(kt/m)
now you have V

Answer 43

get V and then we'll go on and get x(t)

Answer 44

wait .. we have to find v, then x(t) ? i thought by dividing through we found v, then we had to integrate that .

Answer 45

to get x(t)

Answer 46

you do, but you're skipping steps or something

Answer 47

you're evidently not getting the right V(t). tell me what you get for V(t)

Answer 48

hmm .. well if you just divide (Ce^kt/m) through i got ..
v = mg (Ce^(kt/m) - Ce^(kt(0)/m)) // kCe^(kt/m)

Answer 49

then we're supposed to integrate that, right?

Answer 50

simplify it

Answer 51

Ce^0 = C

C/(Ce^(kt/m) ) = e^(-kt/m)

Answer 52

(Ce^(kt/m) )/(Ce^(kt/m) ) =1

Answer 53

\[\huge V = \frac{ mg }{ k } (1-e ^{\frac{-kt }{m} })\]

Answer 54

why is it -k?

Answer 55

because it's more fun.

Answer 56

than k

Answer 57

ok, so now we integrate that?

Answer 58

yep

Answer 59

wait, can i ask why its 1 - e^(kt/m)

Answer 60

you could ask that, but it's not 1 - e^(kt/m)

so it would be fruitless...

Answer 61

hmm ok... i'm not sure if i integrated properly. but i did it on the interval 0 - t again because of the second initial condition ... right?
and got

mgt/k - (m^2g/k^2)(e^kt/m) - m^2g/k^2

Answer 62

naw...

Answer 63

what are you integrating?

Answer 64

\[\huge \frac{ mg }{ k } \int\limits_{0}^{t} (1-e ^{\frac{ -kt }{m }}) dt\]

Answer 65

yeah. thats what i integrated, but i left k postiivive.. can i do that?

Answer 66

no.

Answer 67

oh so it has to be ive? its not just for the fun of it?

Answer 68

ngative i meant aha.

Answer 69

C/(Ce^(kt/m) ) = e^(-kt/m)

Answer 70

so that happens for anything you divide into the numerator then?

Answer 71

\[\huge \frac{ 1 }{ x ^{a} } =x ^{-a}\]

Answer 72

idk why this is so tough for me, thanks for being so patient.
ok, so i'm integrating that now.

Answer 73

cool:)

Answer 74

so i would get t + m/ke^(-kt/m)

Answer 75

right?! then evaluate it?

Answer 76

SO AS AN ANSWER
we would get
x(t) = t - (m/k)(e^(-kt/m)) - m/k

Answer 77

RIGHT?!

Answer 78

please tell me i`m right. i dont know what i'll do if i'm not

Answer 79

ya pretty close
(mg/k) (t + (m/k)(e^(-kt/m)) - m/k)

if that sign in front of the exponential isn't positive, then at t=0 you're not at x=0

Answer 80

k?

Answer 81

oh ya thats what i actually had in my notes .. thank goodness.
thank you so much for everything!

Answer 82

Sorry, one more question.
In this, m = mass, gravity = g and k = acceleration right?

Answer 83

so if its asking what the terminal velocity is of the falling object, i just make g = 9.8 and k = 0 right?

Answer 84

I am not getting notifications from you... the server must be stuffed.

Answer 85

you'll have to PM me if you want attention..

Answer 86

yeah m = mass g= acceleration k is just a constant relating to aerodynamics, cross section and shape etc

Answer 87

so would i evaluate x, or find it's first der. then evaluate with those constantS?

Answer 88

you have it all now, x(t), V(t) you can find a(t) easily (derivative of V(t) )...
you can do whatever you want....!

Answer 89

you already have V(t) so don't bother taking a derivative of x(t) unless you're a masochist.

Answer 90

i try not to be

Answer 91

so plug them into v then

Answer 92

w. k being 0, it's undefined .. right? .. that doens't seem right

Answer 93

k=0?

Answer 94

lol

Answer 95

for acceleration? no?

Answer 96

or is k resistance?

Answer 97

if that's true the whole problem was pretty pointless...