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In this task, we are asked to derive the functions \( Q(t) \) (charge) and \( I(t) \) (current) based on the provided graph, and then to graphically represent \( I(t) \) for the given values of \( Q_1 \), \( t_1 \), \( t_2 \), \( t_3 \), and \( \tau \).

### Step-by-Step Solution

1. **Analyzing the Graph and Data:**
   - For \( t < 0 \), \( Q(t) = 0 \).
   - Between \( t = 0 \) and \( t = t_1 = 1 \) s, \( Q(t) \) appears to increase in a parabolic fashion.
   - From \( t_1 = 1 \) s to \( t_2 = 2 \) s, \( Q(t) \) remains constant at \( Q_1 = 1 \) As.
   - From \( t_2 = 2 \) s to \( t_3 = 3.5 \) s, \( Q(t) \) decreases linearly, reaching \( Q = -Q_1 \).
   - For \( t > t_3 = 3.5 \) s, \( Q(t) \) decays exponentially towards zero, with a time constant \( \tau = 0.5 \) s.

2. **Formulating \( Q(t) \):**
   Let's define \( Q(t) \) in pieces according to the time intervals:

   - **For \( 0 \leq t < t_1 \):**
     \[
     Q(t) = Q_1 \frac{t^2}{t_1^2} = 1 \frac{t^2}{1^2} = t^2 \quad \text{(parabolic increase)}
     \]

   - **For \( t_1 \leq t < t_2 \):**
     \[
     Q(t) = Q_1 = 1 \, \text{As} \quad \text{(constant)}
     \]

   - **For \( t_2 \leq t < t_3 \):**
     \[
     Q(t) = Q_1 \left(1 - 2\frac{t - t_2}{t_3 - t_2}\right) = 1 \left(1 - 2\frac{t - 2}{1.5}\right)
     \]
     Simplifying, this gives:
     \[
     Q(t) = 3 - \frac{4}{3}t
     \]

   - **For \( t \geq t_3 \):**
     \[
     Q(t) = -Q_1 e^{-\frac{t - t_3}{\tau}} = -1 e^{-\frac{t - 3.5}{0.5}} = -e^{-2(t - 3.5)}
     \]

3. **Determining \( I(t) \):**
   Since \( I(t) = \frac{dQ(t)}{dt} \), we differentiate \( Q(t) \) piece by piece.

   - **For \( 0 \leq t < t_1 \):**
     \[
     I(t) = \frac{d}{dt}(t^2) = 2t
     \]

   - **For \( t_1 \leq t < t_2 \):**
     \[
     I(t) = \frac{d}{dt}(Q_1) = 0
     \]

   - **For \( t_2 \leq t < t_3 \):**
     \[
     I(t) = \frac{d}{dt} \left(3 - \frac{4}{3}t\right) = -\frac{4}{3}
     \]

   - **For \( t \geq t_3 \):**
     \[
     I(t) = \frac{d}{dt} \left(-e^{-2(t - 3.5)}\right) = 2e^{-2(t - 3.5)}
     \]

4. **Summary of \( I(t) \):**
   \[
   I(t) = \begin{cases} 
      2t & \text{for } 0 \leq t < t_1 \\
      0 & \text{for } t_1 \leq t < t_2 \\
      -\frac{4}{3} & \text{for } t_2 \leq t < t_3 \\
      2e^{-2(t - 3.5)} & \text{for } t \geq t_3 
   \end{cases}
   \]

Let me know if you need further details or a graphical representation of \( I(t) \).

### Related Questions:
1. How would the solution change if \( Q_1 \) had a different value?
2. What is the physical interpretation of each section of \( I(t) \)?
3. How does changing \( \tau \) affect the exponential decay in \( Q(t) \) and \( I(t) \)?
4. Could we derive \( Q(t) \) and \( I(t) \) if \( t_1 \), \( t_2 \), and \( t_3 \) were different?
5. What is the total charge transported over time, and how does it integrate across these intervals?

**Tip:** For exponential decay problems, understanding the time constant \( \tau \) helps predict how quickly the function approaches zero.

Determine the functions Q(t) and I(t) based on the given graph, and graphically represent I(t) for the specified values.

Explore the functions Q(t) and I(t) based on a piecewise time interval graph. This solution includes parabolic, constant, linear, and exponential decay sections for a charge function, providing insights into current changes over time.

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