simple calc

Discussion in 'Homework Help' started by matty204359, Jan 5, 2014.

  1. matty204359

    Thread Starter Member

    Apr 6, 2011
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    I feel retarded because i'm not getting this equation despite getting A's in all my math classes

    i(t) = \frac{dq(t)}{dt}

    some how becomes

    q(t) = \int^{t}_{t_{0}} i(t)dt + q(t_{0})

    I'm guessing we apply the integral operator \int dt to both sides. so where is the q(t_{0}) coming from?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Presumably the charge present in the "system" prior to and including t(0-).
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    It sounds like your math classes focused solely on the mechanics and not on the fundamental understanding underlying the math. Don't feel too bad, it is not your fault and it is an increasingly common phenomonen. But it is something you need to overcome.

    What is an integral? It is basically a summation over some variable.

    So let's start with

    <br />
i(t)=\frac{dq(t)}{dt}<br />

    This means that the current is the time-rate-of-change of charge, or roughly speaking the instantaneous current is equal to how much charge is flowing per unit time at a particular instant in time.

    Think of the charge as being the total amount of water in a bucket and the current as being the rate at which water is flowing into the bucket. If the current is negative, it means that water is flowing out of the bucket.

    If we reorganize this, we have

    <br />
dq( \tau )=i( \tau ) \cdot d \tau <br />

    This means that the change in charge, dq, from time t to t+dτ is equal to the current at time t times the amount of time that that current is flowing, namely dτ. This is an approximation that becomes exact as dτ approaches zero.

    To get the total amount of change in charge from τ=To to t, we sum up all of the dq(τ) elements, which is the process of integration.

    <br />
\Delta q = \int_{q_0}^{q_1} dq( \tau ) = \int_{T_0}^t i( \tau ) d \tau<br />

    Notice the emphasis I've been placing on the fact that the integral only tells us about the change in the charge. If I want to know how much water is in the bucket at t=10s and I know what the change in the amount of water has been from t=5s to t=10s, then I have to add this change to the amount of water that was in the bucket at t=5s. The same holds here. The integral of the current from To to t only tells us how much the charge has changed from To to t and we need to add that change to how much charge we had at To.

    <br />
q(t) = q(T_0) + \Delta q<br />
\ <br />
q(t) = q(T_0) + \int_{T_0}^t i( \tau ) d \tau<br />

    Another less rigorous way to think of it is the following:

    <br />
\int_{q(T_0)}^{q(t)} dq( \tau ) = q(t) - q(T_0)<br />

    Solving for q(t)

    <br />
q(t) = \int_{q(T_0)}^{q(t)} dq( \tau ) + q(T_0)<br />
     
    atferrari, #12 and matty204359 like this.
  4. matty204359

    Thread Starter Member

    Apr 6, 2011
    105
    3
    okay makes sence they are just replacing the + c with the initial conditions of the system and moving it to the RHS...

    Thanks WBahn, very well written, I had to read it over a couple times but your explanation makes intuitive sense to me.
     
    Last edited: Jan 5, 2014
  5. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    It is called Constant of Integration: http://en.wikipedia.org/wiki/Constant_of_integration

    Basically. Consider some function f(x). You take the derivative of f(x), the new function is A(x). How do you find the original f(x)? To find original f(x) you take the integral of A(x). But! What if your f(x) had a constant in it? Say f(x)=x+2. Derivative of x is 1, derivative of 2 is zero, so A(x)=1. When you integrate A(x) you will get the x from the 1, but you can not bring back the 2, it is lost forever.

    So to account for any constants lost when a derivative was taken we add Constant of Integration when we do integration.
     
  6. matty204359

    Thread Starter Member

    Apr 6, 2011
    105
    3
    well its funny before I even went to college I had this circuit analysis book. but its heavy in to calculus and matrix algebra. so I put it aside. three semesters later I aced pre calculus, derivatives, and integrals. about to go in to differential equations. some things like mesh/node analysis and solving Wheatstone bridges I haven't done in almost a year. I figured I'd pick that book up and brush up on my circuit analysis. of course page 8 and I'm stumped on simple current-charge relationship....
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Yeah, but there is much you won't get stumped on now. So we can help get you past the stumbling points that remain and you will be on your way. In many cases it will just be a need to look at the math from a physics or an application perspective. After you've done this a few times, you will get much better at doing it all on your own and before long you will be doing it automatically without even being aware of it.
     
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