Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - physics related calc

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    39

    physics related calc

    Well I am once again at a stand still. Its basically a simple circuit problem, but not so simple to me. Heres where im at:

    e^(t/RC) q = CV e^(t/RC) + K

    All letters in capital are constant. So what i need to do from here is show that if q(0) = 0, then K = -CV.

    It is probably easy but im new to the natural base honestly so if someone could explain the process a little bit, id really appreciate it.
    This is the 1st of 2 parts
    Thanks guys
    Chris
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chrsr345 View Post
    Well I am once again at a stand still. Its basically a simple circuit problem, but not so simple to me. Heres where im at:

    e^(t/RC) q = CV e^(t/RC) + K

    All letters in capital are constant. So what i need to do from here is show that if q(0) = 0, then K = -CV.

    It is probably easy but im new to the natural base honestly so if someone could explain the process a little bit, id really appreciate it.
    This is the 1st of 2 parts
    Thanks guys
    Chris
    begin by dividing through by e^{t/RC}:

    \Rightarrow q(t) = CV + Ke^{-t/RC}

    now plug in t = 0 and q(0) = 0 and solve for K
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2007
    Posts
    39
    Boy i feel silly, thanks Jhevon.
    Welp from there i had to derive:

    CV(1 - e^(-t/RC))

    What id did was plug in -K and executed the product rule. From there i used the quotient rule on the riased power of (-t/RC) to get t/RC.

    My final derivative was dq/dt = K(e^(-t/RC) t/RC)

    Did i execute this correctly?
    Last edited by chrsr345; October 16th 2007 at 11:26 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chrsr345 View Post
    Boy i feel silly, thanks Jhevon.
    Welp from there i had to derive:

    CV(-1 + e^(1-t/RC)
    check this to make sure you typed the equation correctly. you are missing parentheses and i don't know what else.

    What id did was plug in -K and executed the product rule. From there i used the quotient rule on the riased power of (-t/RC) to get t/RC.

    My final derivative was dq/dt = K(e^(1-t/RC) t/RC)

    Did i execute this correctly?
    why would you do that. we went through all the trouble of finding out that K = -CV. shouldn't we plug that in for K and see what we get?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2007
    Posts
    39
    Ahh sorry, i corrected my last post. both equations were written wrong.

    But i plugged K back in so i would have 1 constant to work with for the product rule rather than 2 constants. But maybe should plug the -CV for K at the end for further simplification?

    dq/dt = K(e^(-t/RC) t/RC) --------> dq/dt = -V(e^(-t/RC) t/R)

    what do you think?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chrsr345 View Post
    Ahh sorry, i corrected my last post. both equations were written wrong.

    But i plugged K back in so i would have 1 constant to work with for the product rule rather than 2 constants. But maybe should plug the -CV for K at the end for further simplification?

    dq/dt = K(e^(-t/RC) t/RC) --------> dq/dt = -V(e^(-t/RC) t/R)

    what do you think?
    i think it's wrong.

    q(t) = CV - CVe^{-\frac t{RC}}

    \Rightarrow \frac {dq}{dt} = \frac {CV}{RC}e^{- \frac t{RC}} = \frac {V}{R}e^{- \frac t{RC}}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2007
    Posts
    39
    Hmm, yes your right...i think i actually took a tougher way which apparently is the wrong way. I see what you did.
    So for t > 0, dq/dt is getting larger and more negative?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chrsr345 View Post
    Hmm, yes your right...i think i actually took a tougher way which apparently is the wrong way. I see what you did.
    So for t > 0, dq/dt is getting larger and more negative?
    i don't know. it depends on what signs R and C have. (and to some extent the sign of V matters as well, not as t goes to infinity though)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Sep 2007
    Posts
    39
    double post, woops
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Sep 2007
    Posts
    39
    Hmm, or maybe its possible to figure out from the q(t)

    q(t) = CV - CVe^{-\frac t{RC}}


    Where C is capacitance, V is voltage and R is resistance. All these values should be positive i believe. The dq/dt is actually the formala for i(t) which is the current at time (t). Does this make any more sense to you? Im certainly no electrical engineer..
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chrsr345 View Post
    Hmm, or maybe its possible to figure out from the q(t)

    q(t) = CV - CVe^{-\frac t{RC}}


    Where C is capacitance, V is voltage and R is resistance. All these values should be positive i believe. The dq/dt is actually the formala for i(t) which is the current at time (t). Does this make any more sense to you? Im certainly no electrical engineer..
    if all the variables are positive, then dq/dt gets small as t gets large, since e raised to a negative number goes to zero as the negative number gets more negative
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Sep 2007
    Posts
    39
    Thank you again Dan. I didnt have enough time last night to thoroughly try out one of your answers. I tried it now, and actually i dont see the process..
    What did you do to get
    \Rightarrow \frac {dq}{dt} = \frac {CV}{RC}e^{- \frac t{RC}} = \frac {V}{R}e^{- \frac t{RC}}

    Im deriving the 1st CV as 1, then the rest as a prodect rule but not getting the above answer...
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by chrsr345 View Post
    Thank you again Dan.
    Ummm... No problem. But Jhevon's helping you on this one.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  14. #14
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by topsquark View Post
    Ummm... No problem. But Jhevon's helping you on this one.

    -Dan
    ...well, i guess i should be flattered if someone is mistaking me for you... it must mean i'm doing something right, after all
    Follow Math Help Forum on Facebook and Google+

  15. #15
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chrsr345 View Post
    Thank you again Dan. I didnt have enough time last night to thoroughly try out one of your answers. I tried it now, and actually i dont see the process..
    What did you do to get
    \Rightarrow \frac {dq}{dt} = \frac {CV}{RC}e^{- \frac t{RC}} = \frac {V}{R}e^{- \frac t{RC}}

    Im deriving the 1st CV as 1, then the rest as a prodect rule but not getting the above answer...
    t is the variable, so CV is a constant, it's derivative is zero
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. physics related rates problem.
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 21st 2010, 03:57 PM
  2. Physics related question
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: September 23rd 2010, 01:42 PM
  3. Physics calc/surface area
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: August 31st 2009, 02:13 AM
  4. Replies: 0
    Last Post: May 8th 2008, 03:20 PM
  5. Homework Prob, physics related
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2006, 10:07 AM

Search Tags


/mathhelpforum @mathhelpforum