1. ## 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.

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

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.

2. 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 $\displaystyle \frac{3.12- 3.10}{1.01- 1.00}= \frac{.02}{.01}= 2$. You can approximate di/dt at t= 1.00 by 2.

3. The official solution is.
$\displaystyle \begin{tabular}{|c|c|c|c|c|c|} \hline$\emph{t}$& 1.00 & 1.01 & 1.02 & 1.03 & 1.04\\ \hline$\emph{E(t)}$& 2.400 & 2.403 & 3.386 & 5.352 & 7.320\\ \hline \end{tabular}$

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 $\displaystyle \frac{1}{2h}[f(x_{0}+h) - f(x_{1}-h)]$. Substitute f(x) for E(t).

Thanks.