Question

http://books.google.com/books?id=CUbdMDUHRlUC&pg=PA101&dq=skydiver+125+terminal+velocity+15+seconds&hl=en&sa=X&ei=puNDT7a5HsWatwf2mq2BDA&ved=0CEwQ6AEwAw#v=onepage&q=skydiver%20125%20terminal%20velocity%2015%20seconds&f=false

number 32 .... hmmm

Answer 1

show that q(t) is a solution to the differential equation.

do I try to undo the diffy Q? which I havent been able to do yet.

or try yo up the qt into it?

Answer 2

I can't see!

Answer 3

its in braille :)

Answer 4

just feel around

Answer 5

Oh. Ok. Lemme skim around with my computer screen.

Answer 6

Aw. all I feel is my computer screen getting smudgy fingerprints everywhere.

Answer 7

...take this pebble from my hand, grasshopper

Answer 8

hmmm... i can't see that small exponent

Answer 9

1/Ck_2?

Answer 10

\[(k_1+k^2t)q'+\frac{1}{C}q=E(t)\]

E(t) = a constant; q(0) = \(q_0\), another constant

\[q(t)=E_0C+(q_0-E_0 C)\left(\frac{k_1}{k_1+k_2t}\right)^{(1/Ck_2)}\]

Answer 11

amistre why do you have to post ling I know nothing about
what is C?

Answer 12

is C=C(t)?

Answer 13

lingo*

Answer 14

C is capacitance; its constant as well

Answer 15

Resistance is variable

Answer 16

ok cool let me try

Answer 17

i think ive been looking at it too hard; trying to undo the diffyQ is a nightmare so I think it wasnt me to up the q(t) into it as proof

Answer 18

wants .. wasnt

at least i got the same keys this time :)

Answer 19

\[q(t)=\frac{\int\limits_{}^{}e^{\int\limits_{}^{}\frac{1}{k_1+k_2t} dt} E(t) dt +C}{e^{\int\limits_{}^{}\frac{1}{k_1+k_2t} dt}}\]

Answer 20

\[q(t)=E_0C+(q_0-E_0 C)\left(\frac{k_1}{k_1+k_2t}\right)^{(1/Ck_2)}\]
\[q'(t)=(q_0-E_0 C)\left(\frac{k_1}{C(k_1+k_2t)}\right)^{\frac{1-Ck_2}{Ck_2}}\]

Answer 21

yeah i missed my C

Answer 22

\[q(t)=\frac{\int\limits\limits_{}^{}e^{\int\limits\limits_{}^{}\frac{1}{C(k_1+k_2t)} dt} E(t) dt +C}{e^{\int\limits\limits_{}^{}\frac{1}{C(k_1+k_2t)} dt}}\]

Answer 23

oops and that other C is a different C

Answer 24

E(t) is constant, so it can be pulled out

Answer 25

\[q(t)=\frac{\int\limits\limits\limits_{}^{}e^{\int\limits\limits\limits_{}^{}\frac{1}{C(k_1+k_2t)} dt} E(t) dt +C_1}{e^{\int\limits\limits\limits_{}^{}\frac{1}{C(k_1+k_2t)} dt}}\]

Answer 26

that other "C" cant be put there until you int up the top anyhoos

Answer 27

unless of course your just putting it there as a primer lol

Answer 28

..when im done, this will go here ...

Answer 29

\[q(t)=\frac{E(t) \int\limits_{}^{} e^{\int\limits_{}^{} \frac{1}{C(k_1+k_2t)} dt} dt+C_1}{e^{\int\limits_{}^{}\frac{1}{C(k_1+k_2t)} dt}}\]

Answer 30

\[(k_1+k_2t)(q_0-E_0 C)\left(\frac{k_1}{C(k_1+k_2t)}\right)^{\frac{1-Ck_2}{Ck_2}}+E_0+C^{-1}(q_0-E_0 C)\left(\frac{k_1}{k_1+k_2t}\right)^{(1/Ck_2)}=E_0 \]maybe?

Answer 31

\[q(t)=\frac{E_0 \int\limits_{}^{}e^{\ln(k_1+k_2t)^\frac{1}{k^2}} dt+C_1}{e^{\ln(k_1+k_2t)^\frac{1}{k^2}}}\]

Answer 32

did i make a mistake?

Answer 33

\[1/Ck_2\]

but looks right so far

Answer 34

\[q(t)=\frac{E_0 \int\limits_{}^{}(k_1+k_2t)^\frac{1}{k^2} dt+C_1}{(k_1+k_2t)^\frac{1}{k^2}}\]

Answer 35

darn i'm missing my C again

Answer 36

\[q(t)=\frac{E_0 \int\limits\limits_{}^{}(k_1+k_2t)^\frac{1}{Ck^2} dt+C_1}{(k_1+k_2t)^\frac{1}{Ck^2}}\]

Answer 37

are we to assume CK^2 is never -1 ?

Answer 38

\[q(t)=E_0C+(q_0-E_0 C)\left(\frac{k_1}{k_1+k_2t}\right)^{(1/Ck_2)}\]
\[q'(t)=(q_0-E_0 C)\left(\frac{k_1}{k_1+k_2t}\right)^{(1-Ck_2)/Ck_2}C^{-1}k^{-2}\frac{-k_1}{(k_1+k_2t)^2}\]

ugh thats a mess

Answer 39

problem doesnt say we can make that assumption

Answer 40

well wouldn't we have two possible cases then
for when 1/Ck^2=-1 and then for whenever it equals anything else?

Answer 41

it looks like answer assumes it is not -1

Answer 42

Ck_2 is just a constant within the formula

Answer 43

there is no Ck^2 :)

Answer 44

not an arbitrary constant tho; one that is a result within the equation

Answer 45

Capacitance

Answer 46

whatever

Answer 47

that is what i meant i just got distracted

Answer 48

but still if that constant=-1 then we have another case to worry about unless i made a mistake

Answer 49

id have to look at it fresh later on, at the moment all i see is headache :/

Answer 50

\[\int\limits_{}^{}u^{-1} du=\ln|u|+C ; \int\limits_{}^{}u^n du =\frac{u^{n+1}}{n+1}+C , n \neq -1\]

Answer 51

i have a headache too
thanks a lot
i do want to help with your problem though
so i will continue to look at it
and i think i will do it on paper first
because typing it on screen i lose a lot of numbers and letters

Answer 52

yeah, typing aint condusive to mathing

Answer 53

http://books.google.com/books?id=BnArjLNjXuYC&pg=PA91&lpg=PA91&dq=suppose+an+rc+circuit+has+a+variable+resistor+E(t)+q(t)&source=bl&ots=_dS_jauZTi&sig=GD45MrxPWHcUomPEEotPHFYMlNg&hl=en&sa=X&ei=v1NET9avOJDVgAea9p2hBA&ved=0CEgQ6AEwBw#v=onepage&q=suppose%20an%20rc%20circuit%20has%20a%20variable%20resistor%20E(t)%20q(t)&f=false

this has the same problem, but includes pages with material about it as well

Answer 54

i think i'm getting somwhere

Answer 55

nope i'm having trouble ...

Answer 56

that constant I put in there
they have no new constants except the q_0...

Answer 57

i can get this:
\[q=E_0C+ \frac{q_0}{(k_1+k_2t)^\frac{1}{Ck_2}}\]

Answer 58

i might got something ...

\[R(t)=(k_1+k_2t)\ R'(t) = k_2\]
\[R(t)q'+\frac{1}{C}q=E(t)\]
\[q'+\frac{1}{CR(t)}q=\frac{E(t)}{R(t)}\]
\[exp(\int \frac{1}{CR(t)}dt)q=\int \frac{E(t)}{R(t)}exp(\int \frac{1}{CR(t)}dt)dt\]
\[exp(\frac{1}{Ck_2}ln(R(t))\ )q=\int ...\]
\[R(t)^{1/Ck_2}\ q=\int \frac{E(t)}{R(t)}R(t)^{1/Ck_2} dt\]
\[R(t)^{1/Ck_2}\ q=E(t)\int R(t)^{(1-Ck_2)/Ck_2} dt\]
\[R(t)^{1/Ck_2}\ q=\frac{E(t)}{k_2}\left(\frac{R(t)^{(1-Ck_2+Ck_2)/Ck_2} }{(1-Ck_2+Ck_2)/Ck_2}+C_1 \right)\]
\[R(t)^{1/Ck_2}\ q=\frac{E(t)}{k_2}\left(\frac{R(t)^{1/Ck_2} }{1/Ck_2}+C_1 \right)\]
\[R(t)^{1/Ck_2}\ q=\frac{E(t)Ck_2R(t)^{1/Ck_2}}{k_2}+C_1\frac{E(t)}{k^2} \]
\[R(t)^{1/Ck_2}\ q=E(t)CR(t)^{1/Ck_2}+C_1\frac{E(t)}{k^2} \]
\[ q=E(t)C\ \cancel{R(t)^{1/Ck_2}R(t)^{-1/Ck_2}}+C_1\frac{E(t)}{k^2}R(t)^{-1/Ck_2}\ \]
\[ q=E(t)C+C_1\frac{E(t)}{k^2}R(t)^{-1/Ck_2}\ \]
is what i got so far

Answer 59

ok i will type what i did

Answer 60

\[q(0)=E_0C+C_1\frac{E_0}{k_2}(k_1+k_2(0))^{-1/Ck_2}=q_0\]
\[C_1\frac{E_0}{k_2}(k_1)^{-1/Ck_2}=q_0-E_0C\]
\[C_1(k_1)^{-1/Ck_2}=\frac{(q_0-E_0C)k_2}{E_0}\]
\[C_1=\frac{(q_0-E_0C)k_2}{E_0}(k_1)^{1/Ck_2}\]

Answer 61

\[q=E(t)C+C_1\frac{E(t)}{k_2}R(t)^{-1/Ck_2}\]
\[q=E_0C+\frac{(q_0-E_0C)k_2}{E_0}(k_1)^{1/Ck_2}\frac{E_0}{k_2}R(t)^{-1/Ck_2}\]
\[q=E_0C+(q_0-E_0C)k_2(k_1)^{1/Ck_2}\frac{1}{k_2}R(t)^{-1/Ck_2}\]
\[q=E_0C+(q_0-E_0C)(k_1)^{1/Ck_2}R(t)^{-1/Ck_2}\]
almost got it :)

Answer 62

\[q'+\frac{1}{C(k_1+k_2t)}q=\frac{E_0}{k_1+k_2t}\]
\[v=e^{\int \frac{1}{C(k_1+k_2t)}dt} \]
\[v=e^{\frac{1}{C} \cdot \frac{1}{k_2} \ln(k_1+k_2t) }\]
\[v=(k_1+k_2t)^{\frac{1}{Ck_2}}]\]
So we have
\[(vq)'=v \cdot \frac{E_0}{k_1+k_2t}\]
Integrate both sides
\[vq=\int\limits_{}^{}E_0 \cdot (k_1+k_2t)^\frac{1}{Ck_2}(k_1+k_2t)^{-1} dt \]
\[vq=E_0 \int\limits_{}^{}(k_1+k_2t)^{\frac{1}{Ck_2}-1} dt\]
\[vq=\frac{E_0}{k_2} \cdot \frac{(k_1+k_2t)^{\frac{1}{Ck_2}}}{\frac{1}{Ck_2}}+C_1\]

Answer 63

\[q=E_0C+(q_0-E_0C)\frac{(k_1)^{1/Ck_2}}{R(t)^{1/Ck_2}}\]
\[q=E_0C+(q_0-E_0C)\left(\frac{(k_1)}{R(t)}\right)^{1/Ck_2}\]
\[q=E_0C+(q_0-E_0C)\left(\frac{k_1}{k_1+k_2t}\right)^{1/Ck_2}\]

Answer 64

\[vq=E_0C(k_1+k_2t)^\frac{1}{Ck_2}+C_1\]
\[q=E_0C+\frac{C_1}{(k_1+k_2t)^\frac{1}{Ck_2}}\]

Answer 65

oh and i see we are given q(0)=q_0

Answer 66

yep :)

Answer 67

i completely ignored that part at first

Answer 68

well it looks like you got it...
great job :(
i wanted to win

Answer 69

ok i have to go back to my studies
peace

Answer 70

thnx for the support ;)

Answer 71

the finished product ... yay!!