Question: S. 12 Nr. 3 (Bestandsgraph darunterzeichnen, wie in Beispiel A, ggfs. Tabelle als Hilfe) S. 13 Nr. 5 (mit Bestandsgraph/t-s-Diagramm) S. 15 Nr. 11 Übungen 3 Zufluss bekannt, Bestand gesucht Rekonstruieren Sie aus dem Graphen der Zuflussrate von Wasser in ein Becken den Graphen der Bestandsfunktion (Wassermenge im Becken in Abhängigkeit von der Zeit). Das Becken ist zu Beginn leer. f1(x): top-left, piecewise constant inflow rate with y=1 for 06leq;x6leq;2 and y=2 for 26leq;x6leq;4; corresponding stock graph starts at 0 and rises linearly with slopes 1 then 2. f2(x): center, piecewise constant inflow rate with y=2 for 06leq;x6leq;1, y=-1 for 16leq;x6leq;2, and y=1 for 26leq;x6leq;4; corresponding stock graph starts at 0, rises linearly, then falls, then rises again. f3(x): top-right, inflow rate is constant y=2 from 06leq;x6leq;1, then decreases linearly to y=1 from x=1 to x=3, then stays constant y=1 from 36leq;x6leq;4; corresponding stock graph starts at 0 and increases, first with steep linear rise, then with decreasing slope, then with constant slope 1.

1. **Problem statement:** We are given inflow rate functions $f_1(x)$, $f_2(x)$, and $f_3(x)$ describing the rate of water flowing into a tank over time $x$. The tank is initially empty, and we want to reconstruct the stock function (water volume in the tank) over time by integrating the inflow rate. 2. **Key formula:** The stock function $S(x)$ is the integral of the inflow rate $f(x)$ from the start time to $x$: $$S(x) = \int_0^x f(t) \, dt$$ Since the tank is empty at $x=0$, $S(0) = 0$. 3. **Important rules:** - For piecewise constant inflow rates, the stock function is piecewise linear with slope equal to the inflow rate on each interval. - For piecewise linear inflow rates, the stock function is piecewise quadratic or linear depending on the slope. 4. **Function $f_1(x)$:** - $f_1(x) = 1$ for $0 \leq x \leq 2$ - $f_1(x) = 2$ for $2 < x \leq 4$ Calculate $S_1(x)$: - For $0 \leq x \leq 2$: $$S_1(x) = \int_0^x 1 \, dt = x$$ - For $2 < x \leq 4$: $$S_1(x) = S_1(2) + \int_2^x 2 \, dt = 2 + 2(x-2) = 2x - 2$$ 5. **Function $f_2(x)$:** - $f_2(x) = 2$ for $0 \leq x \leq 1$ - $f_2(x) = -1$ for $1 < x \leq 2$ - $f_2(x) = 1$ for $2 < x \leq 4$ Calculate $S_2(x)$: - For $0 \leq x \leq 1$: $$S_2(x) = \int_0^x 2 \, dt = 2x$$ - For $1 < x \leq 2$: $$S_2(x) = S_2(1) + \int_1^x (-1) \, dt = 2 + (-1)(x-1) = 3 - x$$ - For $2 < x \leq 4$: $$S_2(x) = S_2(2) + \int_2^x 1 \, dt = (3 - 2) + (x-2) = 1 + x - 2 = x - 1$$ 6. **Function $f_3(x)$:** - $f_3(x) = 2$ for $0 \leq x \leq 1$ - $f_3(x)$ decreases linearly from 2 to 1 on $1 < x \leq 3$, so slope $m = \frac{1-2}{3-1} = -\frac{1}{2}$ - $f_3(x) = 1$ for $3 < x \leq 4$ Express $f_3(x)$ on $1 < x \leq 3$: $$f_3(x) = 2 - \frac{1}{2}(x-1) = \frac{5}{2} - \frac{x}{2}$$ Calculate $S_3(x)$: - For $0 \leq x \leq 1$: $$S_3(x) = \int_0^x 2 \, dt = 2x$$ - For $1 < x \leq 3$: $$S_3(x) = S_3(1) + \int_1^x \left(\frac{5}{2} - \frac{t}{2}\right) dt = 2 + \left[ \frac{5}{2}t - \frac{t^2}{4} \right]_1^x$$ Calculate the integral: $$= 2 + \left( \frac{5}{2}x - \frac{x^2}{4} - \left( \frac{5}{2} \cdot 1 - \frac{1}{4} \right) \right) = 2 + \frac{5}{2}x - \frac{x^2}{4} - \frac{9}{4} = \frac{-1}{4} + \frac{5}{2}x - \frac{x^2}{4}$$ - For $3 < x \leq 4$: $$S_3(x) = S_3(3) + \int_3^x 1 \, dt$$ Calculate $S_3(3)$: $$S_3(3) = \frac{-1}{4} + \frac{5}{2} \cdot 3 - \frac{9}{4} = \frac{-1}{4} + \frac{15}{2} - \frac{9}{4} = \frac{-1 - 9}{4} + \frac{15}{2} = -\frac{10}{4} + \frac{15}{2} = -2.5 + 7.5 = 5$$ Then: $$S_3(x) = 5 + (x - 3) = x + 2$$ 7. **Summary:** - $S_1(x) = \begin{cases} x & 0 \leq x \leq 2 \\ 2x - 2 & 2 < x \leq 4 \end{cases}$ - $S_2(x) = \begin{cases} 2x & 0 \leq x \leq 1 \\ 3 - x & 1 < x \leq 2 \\ x - 1 & 2 < x \leq 4 \end{cases}$ - $S_3(x) = \begin{cases} 2x & 0 \leq x \leq 1 \\ \frac{-1}{4} + \frac{5}{2}x - \frac{x^2}{4} & 1 < x \leq 3 \\ x + 2 & 3 < x \leq 4 \end{cases}$ These stock functions represent the water volume in the tank over time reconstructed from the inflow rates.