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Math Help - Approximate the voltage

  1. #1
    Junior Member
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    Approximate the voltage

    Kirchoff's first law gives the relationship E(t) = L * (di/dt) + R*i where L is the inductance, R is the resistance and i is the current.

    <br />
\begin{tabular}{|c|c|c|c|c|c|}<br />
\hline<br />
$\emph{t}$ & 1.00 & 1.01 & 1.02 & 1.03 & 1.04\\<br />
\hline<br />
$\emph{i}$ & 3.10 & 3.12 & 3.14 & 3.18 & 3.24\\<br />
\hline<br />
\end{tabular}<br /> <br />

    Suppose t is measured in seconds, i is in amperes, the inductance L is a constant 0.98 henries and R is 0.142 ohms. Approximate the voltage E(t) when t = 1.00, 1.01, 1.02, 1.03, 1.04.

    How do I solve di/dt? That's the derivative of i with respect to t I think but there's no formula for i. So I'm not even sure what E(1.00) is supposed to be.
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  2. #2
    MHF Contributor

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    Since the derivative is the slope of the tangent line, you can approximate it by the slope of the secant lines- the lines between two consecutive points.

    For example, here, when t= 1.00, i= 3.10 and when t= 1.01, i= 3.12. The slope of that line is \frac{3.12- 3.10}{1.01- 1.00}= \frac{.02}{.01}= 2. You can approximate di/dt at t= 1.00 by 2.
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  3. #3
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    The official solution is.
    <br />
\begin{tabular}{|c|c|c|c|c|c|}<br />
\hline<br />
$\emph{t}$ & 1.00 & 1.01 & 1.02 & 1.03 & 1.04\\<br />
\hline<br />
$\emph{E(t)}$ & 2.400 & 2.403 & 3.386 & 5.352 & 7.320\\<br />
\hline<br />
\end{tabular}<br />

    Using 0.98*di/dt+0.142*i.
    di/dt = 2, i = 3.10 I get 2.400.
    di/dt = (2 + 2) / 2, i = 3.12 I get 2.403.
    di/dt = (2 + 4) / 2, i = 3.14 I get 3.386.
    di/dt = (4 + 6) / 2, i = 3.18 I get 5.352.

    Now the last one is di/dt = 7 but how do I get that 7? For the other ones, I took the slope from both sides and divided by 2.

    Is there a way get some expression in terms of t for di/dt. So I can use a three-point formula like \frac{1}{2h}[f(x_{0}+h) - f(x_{1}-h)]. Substitute f(x) for E(t).

    Thanks.
    Last edited by Makall; April 1st 2010 at 07:03 PM.
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