Problem 14 - Calculus

Question

An electric charge, \(Q,\) in a circuit is given as a function of time, \(t,\) by $$ Q=\left\\{\begin{array}{ll} C & \text { for } t \leq 0 \\ C e^{-t / \hbar C} & \text { for } t>0 \end{array}\right. $$ where \(C\) and \(R\) are positive constants. The electric current, \(I\), is the rate of change of charge, so 1 $$ I=\frac{d Q}{d t} $$ (a) Is the charge, \(Q,\) a continuous function of time? (b) Do you think the current, \(I\), is defined for all times, t? [Hint: To graph this function, take, for example, \(C=1 \text { and } R=1 .]\)

Step-by-Step Solution

Answer

(a) Yes, \(Q\) is continuous. (b) Yes, \(I\) is defined for all \(t\).

Step 1: Determine Continuity of Charge at t=0

The function is given as piecewise:For \(t \leq 0\), \(Q = C\).For \(t > 0\), \(Q = C e^{-t / \hbar C}\).To check continuity at \(t=0\), evaluate the limit as \(t\) approaches 0 from both sides. For \(t \leq 0\), \(Q = C\). For \(t > 0\), \(\lim_{t \to 0^+} Q = \lim_{t \to 0^+} Ce^{-t / \hbar C} = C \cdot 1 = C\). The two limits equal \(C\), so \(Q\) is continuous at \(t=0\).

Step 2: Calculate the Current for t ≤ 0

For \(t \leq 0\), \(Q = C\) implies that the derivative with respect to time \(t\) is:\[ I = \frac{dQ}{dt} = \frac{d}{dt}(C) = 0 \]Thus, the current \(I\) is zero for \(t \leq 0\).

Step 3: Calculate the Current for t > 0

For \(t > 0\), \(Q = Ce^{-t / \hbar C}\), differentiate using the chain rule:\[ I = \frac{dQ}{dt} = \frac{d}{dt}(Ce^{-t / \hbar C}) = -\frac{1}{\hbar} e^{-t / \hbar C} \]Thus, the current \(I\) is \(-\frac{1}{\hbar}e^{-t/\hbar C}\) for \(t > 0\).

Step 4: Consider Current Definition Across t = 0

Current \(I\) is defined clearly for both intervals: \(I=0\) for \(t \leq 0\) and \(I = -\frac{1}{\hbar}e^{-t/\hbar C}\) for \(t > 0\). At \(t=0\), \(I\) transitions from a constant zero to an exponential decay. It is defined for all \(t\), including \(t=0\).

Key Concepts

Electric Current

Electric current is the rate at which electric charge flows. When you have a circuit, imagine it as a road, and electric charges are the cars traveling on it.
In mathematical terms, the current, denoted by \( I \), is the derivative of the electric charge \( Q \) with respect to time \( t \). This means it is how fast or slowly the charge is changing over time:

• For \( t \leq 0 \), the charge \( Q \) is constant at \( C \). Therefore, the current \( I \) equals zero because there's no change in the charge.
• For \( t > 0 \), the charge \( Q \) starts to decrease following an exponential decay pattern, which results in a non-zero current. Specifically, it is \( I = -\frac{1}{\hbar} e^{-t / \hbar C} \).

This describes how electricity starts flowing in the circuit after \( t = 0 \), initially at a rate that slows down over time due to the nature of the exponential decay pattern.

Piecewise Functions

Piecewise functions are special functions that have different expressions based on different conditions along the domain. They are essentially assembled from "pieces" that apply to certain intervals of the input.
For this exercise, the charge \( Q \) is a piecewise function:

• When \( t \leq 0 \), \( Q = C \). This means for any time zero or below, the charge is just a constant value \( C \).
• When \( t > 0 \), \( Q = C e^{-t / \hbar C} \). This suggests that immediately after time zero, the charge begins to follow an exponential decay.

Understanding piecewise functions is crucial because it teaches us to analyze different behaviors of the same function at different parts of the domain.
Graphically, it shows us how a function might change its properties at certain key points, like going from constant to changing, as in our electric charge example.

Continuity

Continuity is about whether a function has no breaks, jumps, or holes at a particular point or over an interval. In calculus, a function is continuous at a point if the limit of the function approaching from the left equals the function value and the limit from the right.
For the charge \( Q \), continuity at \( t=0 \) is significant. Let's see why:

• For \( t \leq 0 \), the function is \( Q = C \), and for \( t > 0 \), it is \( Q = C e^{-t / \hbar C} \).
• As \( t \) approaches zero from both sides, the limits are both \( C \).
• This equal limit from either side shows no jumps or breaks in the function's value at \( t = 0 \).

Therefore, \( Q \) is continuous at \( t = 0 \), ensuring that the transition in the behavior of the charge from one piece to another does not have sudden interruptions, making it smooth and predictable.