📚 Investigation 1 - Fluid filling functions | 探究1：液体填充函数
In many physical contexts, a liquid is poured into a container at a constant volumetric flow rate. The central question of this investigation is: how does the height of the liquid surface change over time? The answer is not always linear – it depends crucially on the shape of the vessel. By combining geometric volume formulas with calculus, we can model the filling process, set up differential equations, and predict the height-time relationship. This exploration is typical for an IB Mathematics Internal Assessment, as it connects modelling, integration, and rates of change in a tangible setting.
在许多物理情境中，液体以恒定的体积流速注入容器。这个探究的核心问题是：液面高度如何随时间变化？答案不总是线性的——它关键取决于容器的形状。通过结合几何体积公式与微积分，我们可以对填充过程建模，建立微分方程，并预测高度-时间关系。这个探索是IB数学内部评估的典型题目，它将建模、积分和变化率联系在一个具体的场景中。
1. Understanding the Problem | 理解问题
We consider a rigid container of known shape. Liquid enters through an inlet at a constant volume flow rate, often denoted as Q or dV/dt. The container is initially empty or partially filled up to a certain height. We aim to find a function h(t) that gives the height of the liquid surface above the base at any time t. Because the rate at which the height increases depends on how much volume is required to raise the level by one unit, the cross-sectional area at the current level is the decisive factor.
我们考虑一个形状已知的刚性容器。液体以恒定的体积流速（常记作Q或dV/dt）通过入口注入。容器最初是空的，或已经填充到某个高度。我们的目标是找到一个函数h(t)，给出任意时刻t液面距离底部的高度。由于高度上升的快慢取决于液面上升一个单位需要多少体积，当前液面处的横截面积就成了决定性因素。
2. Defining Variables and Assumptions | 变量定义与假设
k. The container is axisymmetric about a vertical axis in most simple cases, allowing us to express the cross-sectional area A as a function of height: A(h). We also assume no evaporation, no overflow, and negligible surface tension effects. The liquid is treated as incompressible, and the flow is steady.
设V(t)为时刻t容器中的液体体积，h(t)为对应的液面高度。填充过程假设具有恒定的流入速率k（例如cm³ s⁻¹），从而dV/dt = k。在大多数简单情形中，容器关于竖直轴对称，这允许我们将横截面积A表示为高度的函数：A(h)。我们还假设没有蒸发、没有溢出，且表面张力效应可忽略。液体视为不可压缩，且流动是稳定的。
3. Volume-Height Relationship for Common Vessels | 常见容器的体积-高度关系
k of radius R, taking the origin at the lowest point and filling upward, the volume of a spherical cap of height h is V = (πh²/3)(3R - h). These formulas are summarised in the table below.
在进行动力学分析之前，我们需要静态关系V(h)。对于半径为R的竖直圆柱，体积为V = πR² h。对于一个总高度为H、底半径为R的倒立圆锥，由截面相似性可得r(h) = (R/H)h，因此V(h) = (1/3)πr²h = (πR²/(3H²)) h³。对于半径为R的球形烧瓶，以最低点为原点并向上填充，高度为h的球冠体积为V = (πh²/3)(3R - h)。这些公式总结于下表中。
| Shape | V(h) | dV/dh |
|---|---|---|
| Cylinder | πR²h | πR² |
| Inverted cone | (πR²/(3H²))h³ | (πR²/H²)h² |
| Sphere (cap) | πRh² - (π/3)h³ | 2πRh - πh² |
These expressions are the starting points for the rate-of-change equations that follow.
这些表达式是接下来变化率方程的出发点。
4. Rate of Change of Height: The Core Differential Equation | 高度变化率：核心微分方程
dV/dt = k (constant), we obtain the fundamental relationship governing the liquid height:
dh/dt = k / (dV/dh)
. This makes intuitive sense: where the vessel is narrow, the height rises quickly; where it is wide, the height rises slowly.
根据链式法则，dV/dt = (dV/dh) · (dh/dt)。因为dV/dt = k（常数），我们得到控制液面高度的基本关系式：dh/dt = k / (dV/dh)。由于dV/dh就是容器在高度h处的横截面积A(h)，我们也可以写成dh/dt = k / A(h)。这直观上有意义：容器狭窄处液面上升快，宽敞处上升慢。
5. Example 1: Vertical Cylinder | 示例1：竖直圆柱体
o h₀, the solution becomes h(t) = h₀ + (k/(πR²)) t. This is the simplest filling function.
对于圆柱体，A(h) = πR²为常数，因此dh/dt = k/(πR²)为常数。如果圆柱最初是空的（h₀ = 0），积分得h(t) = (k/(πR²)) t。高度随时间线性增长。如果圆柱已有高度为h₀的液体，解为h(t) = h₀ + (k/(πR²)) t。这是最简单的填充函数。
The constant rate of height increase is independent of the current water level, so a graph of h versus t is a straight line whose slope is k/(πR²). This serves as a baseline for comparison with more complex vessels.
高度增加的恒定速率与当前水位无关，故高度随时间变化的图形是一条斜率为k/(πR²)的直线。这为更复杂容器的对比提供了基准。
6. Example 2: Inverted Right Circular Cone | 示例2：倒立圆锥
nce dh/dt = k / [ (πR²/H²) h² ] = k H² / (πR² h²). The differential equation is separable:
h² dh = (k H² / (πR²)) dt
+ C. With initial condition h(0) = 0 (empty), C = 0, so h³ = (3k H² / (πR²)) t. Therefore,
h(t) = [ (3k H²) / (πR²) t ]1/3 = ( 3k t / (π tan²θ) )1/3
where θ is the half-angle of the cone. The height grows proportionally to the cube root of time. The rate of increase is large when h is small (near the tip) and diminishes as the cone widens.
对于倒立圆锥，由相似性可得A(h) = π(R/H)² h² = (πR²/H²) h²。因此dh/dt = k / [ (πR²/H²) h² ] = k H² / (πR² h²)。微分方程是可分离的：h² dh = (k H² / (πR²)) dt。两边积分得(1/3)h³ = (k H² / (πR²)) t + C。由初始条件h(0) = 0（空），C = 0，故h³ = (3k H² / (πR²)) t。因此h(t) = [ (3k H²) / (πR²) t ]1/3 = ( 3k t / (π tan²θ) )1/3，其中θ是锥体的半角。高度按时间的立方根增长。在h很小时（靠近尖端）上升速率大，随着锥体变宽而减小。
7. Example 3: Spherical Flask | 示例3：球形烧瓶
= πRh² - (π/3) h³, so dV/dh = 2πRh - πh² = πh(2R - h). The differential equation becomes:
dh/dt = k / [ πh(2R - h) ]
Separating variables and integrating with h(0)=0 gives the time as a function of height:
t = (π/k) ( R h² - h³/3 )
2R), the area shrinks again and the rate increases. The graph of h against t is S-shaped.
考虑半径为R的球形烧瓶，从底部填充。球冠体积V = πRh² - (π/3) h³，所以dV/dh = 2πRh - πh² = πh(2R - h)。微分方程为 dh/dt = k / [ πh(2R - h) ]。分离变量并以h(0)=0积分，得到时间关于高度的函数：t = (π/k) ( R h² - h³/3 )。我们无法轻易反解出初等形式的h(t)，但它清晰地反映了形状：在底部附近（h很小），A(h) ≈ 2πRh（小），因此高度快速上升；在赤道附近（h ≈ R），A(h)最大，高度增加变慢；接近顶部（h → 2R），面积再次缩小，速率增大。h对t的图形呈S形。
8. Generalising the Approach: Separation of Variables | 通用方法：变量分离
alytically, the problem reduces to solving a first-order separable ODE. From dh/dt = k / V'(h), we write dt = (1/k) V'(h) dh. Integrating provides t as a function of h:
t = (1/k) ∫ V'(h) dh + C = (1/k) V(h) + C
since ∫ V' mply the volume that has been poured in divided by the flow rate, a restatement of conservation of mass.
对于任何可用解析式表达V(h)的轴对称容器，问题可归结为求解一阶可分离常微分方程。由dh/dt = k / V'(h)，我们写为dt = (1/k) V'(h) dh。积分得t关于h的函数：t = (1/k) ∫ V'(h) dh + C = (1/k) V(h) + C，因为∫ V'(h) dh就是V(h)。常数C由初始高度确定。这揭示了一个深刻的物理洞见：达到高度h所需的时间就是注入的体积除以流速，这是质量守恒的重述。
9. Numerical and Graphical Insights | 数值与图形理解
When an explicit h(t) cannot be obtained, we can still plot h against t by computing t as a function of h and swapping axes, or by numerical integration. Using software like GeoGebra, Excel, or Python, we can generate comparative curves for different vessel shapes under the same flow rate. A typical plot shows the cylinder giving a straight line, the inverted cone giving a concave down curve, and the sphere giving an S-shaped curve with an inflection point at the maximum cross-sectional area.
当无法得到h(t)的显式表达式时，我们仍然可以通过计算t作为h的函数并交换坐标轴，或通过数值积分来绘制h-t图。使用GeoGebra、Excel或Python等软件，可以为相同流速下的不同容器形状生成对比曲线。典型的图形显示：圆柱为直线，倒圆锥为向下凹的曲线，球体为S形曲线，其拐点出现在横截面积最大处。
here occurs when d²h/dt² = 0. Since dh/dt = k/A(h), differentiating gives d²h/dt² = - (k A'(h) / A²(h)) dh/dt. Setting A' ), which for a sphere is at h = R. This analytical finding matches the shape of the graph.
球体的拐点出现在d²h/dt² = 0时。因dh/dt = k/A(h)，微分得d²h/dt² = - (k A'(h) / A²(h)) dh/dt。令A'(h)=0找到A(h)的极值点，对于球体即h = R处。这一解析发现与图形形状吻合。
10. Considering a Varying Cross-Section (General Vessel) | 考虑变截面的通用容器
π/k) [f⁻¹(h)]². Integration may still be possible numerically or symbolically. This approach can be applied to a flask with a cylindrical neck, a conical base, or even an irregular flask described by a piecewise function.
如果容器不是简单的旋转体，而是具有已知轮廓y = f(x)绕竖直轴旋转而成，则横截面积为A(h) = π [f⁻¹(h)]²，其中f⁻¹是给出半径作为高度函数的反函数。那么dt/dh = (1/k) A(h) = (π/k) [f⁻¹(h)]²。积分仍然可以通过数值或符号方法进行。该方法可应用于带圆柱瓶颈的烧瓶、带锥底的烧瓶，甚至由分段函数描述的不规则烧瓶。
For a compound vessel, the function V(h) may be defined piecewise. The overall h(t) is obtained by linking the solutions across the height ranges where the shape changes, ensuring continuity of h and t. This is a nice extension for an IB Exploration.
对于复合容器，函数V(h)可以分段定义。总的h(t)可通过连接形状改变处各高度区间的解来获得，并确保h和t连续。这是IB探究的一个很好拓展。
11. Real-World Modifications and Non-Constant Flow | 实际修正与非恒定流速
In laboratory practice, the inflow rate might not be perfectly constant, or the vessel might not be perfectly axisymmetric. More advanced modelling could incorporate a time-dependent inflow k(t), or account for the fact that as the liquid height increases, the driving pressure from a header tank may drop. However, for most IB investigations, the constant flow assumption is sufficient and yields rich mathematical behaviour. Another realistic factor is the meniscus effect, which slightly alters the effective height reading, though it is usually ignored.
在实验室实践中，流入速率可能并非完全恒定，或容器并非完美轴对称。更高级的建模可以纳入时变流入量k(t)，或考虑随着液面升高，高位水箱的驱动压力可能下降。然而对于大多数IB探究，恒定流速假设已足够，并能产生丰富的数学行为。另一个现实因素是弯月面效应，它会轻微改变有效高度读数，但通常被忽略。
Students can also consider the emptying process driven by a hole at the bottom (Torricelli’s law), where the outflow rate depends on √h. Combining filling and emptying creates a more complex ODE, suitable for HL explorations.
学生还可以考虑底部小孔驱动的排液过程（托里拆利定律），其中流出速率与√h有关。将填充与排空结合起来会得到一个更复杂的常微分方程，适合高水平（HL）探究。
12. Summary and IA Writing Tips | 总结与IA写作提示
nd t(h) or h(t), and interpret the results graphically. Always link the mathematical features (curvature, inflection, asymptotes) back to the shape of the vessel. A strong IA will include a personalised example, a clear statement of assumptions, a validation against experiment, and a reflection on limitations.
流体填充函数的探究处于几何、微积分和物理建模的交汇处。关键步骤是：定义几何形状，建立V(h)和A(h)，列出dh/dt = k / A(h)，积分求出t(h)或h(t)，并以图形解释结果。始终将数学特征（曲率、拐点、渐近线）与容器形状联系起来。一份优秀的内部评估（IA）会包含一个个性化示例、清晰的假设陈述、与实验的验证，以及对局限性的反思。
- Choose a vessel with an interesting but manageable V(h).
- Show all calculus steps explicitly.
- 使用技术绘制并讨论图形。
- Comment on the physical reasonableness of the solution.
- 选择形状有趣但可处理的容器。
- 显式展示所有微积分步骤。
- Use technology to plot and discuss graphs.
- 评论解的物理合理性。
This investigation elegantly demonstrates how calculus transforms a static shape into a dynamic prediction, making it a rewarding topic for any IB Mathematics student.
这个探究优雅地展示了微积分如何将静态形状转化为动态预测，对任何IB数学学生来说都是一个有价值的选题。
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