Suppose the following table gives the voltage across the capacitor for the specified times. Assume, as suggested by the data, that the voltage increases by smaller amounts for the same amount of time later in its charging than earlier in its charging.

(seconds)	3.96	3.98	4.00	4.02	4.04	4.06
(Volts)	17.68	17.95	18.21	18.45	18.67	18.87
Find the best underestimate possible from this data for the instantaneous rate at which the voltage across capacitor is increasing 4 seconds after it started charging (in Volts per second).

To find the best underestimate for the instantaneous rate of change (i.e., the derivative) of the voltage at \( t = 4 \) seconds, we need to compute the difference quotients (average rates of change) around \( t = 4 \) seconds and choose the smallest of these values. The best underestimate will be the smallest of these values because the voltage increase is slowing down over time.

Let's calculate the difference quotients around \( t = 4 \) seconds:

### Data Points:
- \( t_1 = 3.96 \), \( V_1 = 17.68 \)
- \( t_2 = 3.98 \), \( V_2 = 17.95 \)
- \( t_3 = 4.00 \), \( V_3 = 18.21 \)
- \( t_4 = 4.02 \), \( V_4 = 18.45 \)
- \( t_5 = 4.04 \), \( V_5 = 18.67 \)
- \( t_6 = 4.06 \), \( V_6 = 18.87 \)

### Calculate the Difference Quotients:

The difference quotient between any two points \( (t_i, V_i) \) and \( (t_j, V_j) \) is:

\[
\frac{V_j - V_i}{t_j - t_i}
\]

We will compute the difference quotients around \( t = 4 \) seconds.

#### 1. Between \( t = 3.98 \) and \( t = 4.00 \):

\[
\frac{V_3 - V_2}{t_3 - t_2} = \frac{18.21 - 17.95}{4.00 - 3.98} = \frac{0.26}{0.02} = 13 \, \text{Volts per second}
\]

#### 2. Between \( t = 4.00 \) and \( t = 4.02 \):

\[
\frac{V_4 - V_3}{t_4 - t_3} = \frac{18.45 - 18.21}{4.02 - 4.00} = \frac{0.24}{0.02} = 12 \, \text{Volts per second}
\]

#### 3. Between \( t = 4.02 \) and \( t = 4.04 \):

\[
\frac{V_5 - V_4}{t_5 - t_4} = \frac{18.67 - 18.45}{4.04 - 4.02} = \frac{0.22}{0.02} = 11 \, \text{Volts per second}
\]

#### 4. Between \( t = 4.04 \) and \( t = 4.06 \):

\[
\frac{V_6 - V_5}{t_6 - t_5} = \frac{18.87 - 18.67}{4.06 - 4.04} = \frac{0.20}{0.02} = 10 \, \text{Volts per second}
\]

### Best Underestimate:

Since the voltage increases at a decreasing rate, the smallest difference quotient will provide the best underestimate of the instantaneous rate at \( t = 4 \) seconds.

The smallest difference quotient calculated is:

\[
10 \, \text{Volts per second}
\]

### Conclusion:

The best underestimate possible for the instantaneous rate at which the voltage across the capacitor is increasing 4 seconds after it started charging is **10 Volts per second**.

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Would you like more details, or do you have any questions?

Here are 5 related questions:
1. How would the calculation change if the voltage was decreasing over time?
2. What is the significance of an "underestimate" in this context?
3. How can we improve the estimate of the instantaneous rate of change using this data?
4. What is the role of the difference quotient in calculus?
5. How could we use a graph to visually find the best underestimate?

**Tip:** To find the instantaneous rate of change more accurately, consider using smaller time intervals or different interpolation methods.

Learn how to find the best underestimate for the instantaneous rate of change of voltage across a capacitor 4 seconds after charging starts. This problem involves calculating difference quotients around a specific time point to estimate the rate of increase. Suitable for college-level calculus students.

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