Ryuuken
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Homework Statement
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 />
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.
Homework Equations
Three-point formula: \frac{1}{2h}[f(x_{0}+h) - f(x_{1}-h)]
The Attempt at a Solution
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 and di/dt = (y_{1} - y_{0})/(x_{1} - x_{0})
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. The only way I can think of is to add all the previous slopes together and divide by 2, (2 + 2 + 4 + 6) / 2 = 7 but I'm not sure if that makes sense or why that would work.
Also, 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) to get \frac{1}{2h}[E(t_{0}+h) - E(t_{1}-h)].
Thanks.