📚 Further Applications of Calculus | 微积分的进一步应用
In IB Mathematics, the study of calculus moves beyond basic rules of differentiation and integration to powerful real-world applications. These further applications include kinematics, related rates, optimisation, areas, volumes of revolution, and average values. Understanding how derivatives and integrals model physical, geometric, and economic situations deepens mathematical insight and prepares students for exam-style problem solving.
在 IB 数学中，微积分的学习从基本的微分与积分规则延伸到强大的现实应用。这些进一步的应用包括运动学、相关变化率、最优化、面积、旋转体体积以及函数平均值。理解导数与积分如何对物理、几何和经济情境进行建模，能加深数学洞察力，并为应对考试风格的题目打下基础。
1. Kinematics: Displacement, Velocity, and Acceleration | 运动学：位移、速度与加速度
Kinematics describes the motion of a particle along a straight line. If displacement s(t) is a function of time t, then instantaneous velocity is the first derivative v(t) = s'(t). Acceleration is the derivative of velocity, a(t) = v'(t), or equivalently the second derivative of displacement, a(t) = s”(t). These relationships allow us to determine how quickly position changes and how motion accelerates or decelerates.
运动学描述质点沿直线的运动。若位移 s(t) 是时间 t 的函数，则瞬时速度是一阶导数 v(t) = s'(t)。加速度是速度的导数 a(t) = v'(t)，也等于位移的二阶导数 a(t) = s”(t)。这些关系使我们能够确定位置变化的快慢以及运动如何加速或减速。
v(t) = ds/dt, a(t) = dv/dt = d²s/dt²
Conversely, given acceleration a(t) and initial conditions, we can recover velocity and displacement by integration. For instance, if a(t) = 2t + 1, then integrating yields v(t) = t² + t + C, and a second integration gives s(t) = (1/3)t³ + (1/2)t² + Ct + D. The constants are determined by the initial velocity and position.
反之，给定加速度 a(t) 和初始条件，我们可以通过积分恢复速度和位移。例如，若 a(t) = 2t + 1，积分得 v(t) = t² + t + C，再次积分得 s(t) = (1/3)t³ + (1/2)t² + Ct + D。常数由初始速度和初始位置确定。
| Quantity | Symbol | Relationship |
|---|---|---|
| Displacement | s(t) | – |
| Velocity | v(t) = s'(t) | ds/dt |
| Acceleration | a(t) = v'(t) = s”(t) | d²s/dt² |
A classic problem: given s(t) = t³ − 6t² + 9t, find when the particle is at rest. Compute v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3). Setting v(t) = 0 gives t = 1 s and t = 3 s as the times when velocity is zero. The acceleration at those instants can then be analysed to determine whether the particle is momentarily changing direction.
经典问题：已知 s(t) = t³ − 6t² + 9t，求质点何时静止。计算 v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3)。令 v(t) = 0 得 t = 1 s 和 t = 3 s 为速度为零的时刻。随后可分析这些时刻的加速度，以判断质点是否瞬间改变运动方向。
2. Related Rates: Connecting Changing Quantities | 相关变化率：联结变化的量
Related rates problems involve finding the rate at which one quantity changes by relating it to another quantity whose rate of change is known. The technique uses implicit differentiation with respect to time. For example, if a spherical balloon is inflated and its radius r increases at a constant rate dr/dt, we can find the rate of change of its volume V = (4/3)πr³ by differentiating: dV/dt = 4πr² (dr/dt).
相关变化率问题需要找到一个量变化的速率，方法是把它与另一个变化率已知的量联系起来。该技巧使用对时间 t 的隐函数求导。例如，如果给一个球形气球充气，其半径 r 以恒定速率 dr/dt 增大，我们可以通过对体积 V = (4/3)πr³ 求导得出体积变化率：dV/dt = 4πr² (dr/dt)。
dV/dt = 4πr² · (dr/dt)
Another common scenario is a ladder sliding down a wall. Let the foot of the ladder be x metres from the wall and the top y metres above the ground, with ladder length L constant. Then x² + y² = L². Differentiating with respect to time gives 2x(dx/dt) + 2y(dy/dt) = 0, allowing us to find how fast the top slides down when the bottom is pulled away at a known speed.
另一个常见情景是梯子沿墙壁滑下。设梯子底端距墙 x 米，顶端离地 y 米，梯长 L 恒定。则 x² + y² = L²。对时间求导得 2x(dx/dt) + 2y(dy/dt) = 0，从而可以在已知底端被匀速拉离时求出顶端下滑的速率。
In related rates problems, always identify the variables, write a geometric or physical equation, differentiate implicitly with respect to t, substitute known values, and solve for the unknown rate. Pay careful attention to units and signs (e.g., whether a length is increasing or decreasing).
在处理相关变化率问题时，应始终做到：确定变量，写出几何或物理方程，对 t 进行隐函数求导，代入已知值，求解未知变化率。要特别注意单位和正负号（例如长度是增加还是减少）。
3. Optimisation I: Maximising Area and Volume | 优化一：最大化面积与体积
Optimisation uses derivatives to find maximum or minimum values of a quantity under given constraints. A classic application is maximising the area of a rectangle inscribed in a curve. Suppose a rectangle has its base on the x-axis and its top corners on the parabola y = 12 − x². Express area A as a function of x, A(x) = 2x(12 − x²) = 24x − 2x³, differentiate, set A'(x) = 0, and verify the maximum with the second derivative test.
最优化利用导数寻找给定约束下一个量的最大值或最小值。一个经典应用是求内接于曲线的矩形面积最大。假定矩形底边在 x 轴上，上方两角点在抛物线 y = 12 − x² 上。将面积 A 表示为 x 的函数，A(x) = 2x(12 − x²) = 24x − 2x³，求导，令 A'(x) = 0，并用二阶导数检验确认最大值。
A'(x) = 24 − 6x² = 0 → x = 2 (since x > 0)
Many optimisation problems involve volumes. For example, an open box is made by cutting squares of side x from each corner of a 20 cm by 30 cm cardboard and folding up the sides. The volume V = x(30 − 2x)(20 − 2x). Finding the derivative, setting it to zero, and checking endpoints yields the x that maximises volume.
许多优化问题涉及体积。例如，将一张 20 cm × 30 cm 的纸板四个角各剪去边长为 x 的正方形，然后折起边缘做成无盖盒子。体积 V = x(30 − 2x)(20 − 2x)。求导，令导数为零，并结合端点检查，可得出使体积最大的 x。
Always identify the feasible domain of the variable. For the box problem, x must satisfy 0 < x < 10 so that dimensions remain positive. The global extremum may occur at a critical point or at a domain endpoint, so testing is crucial.
始终要确定变量的合理定义域。在盒子问题中，x 必须满足 0 < x < 10，才能保证各边长为正。全局极值可能出现在临界点或定义域端点，因此检验至关重要。
4. Optimisation II: Minimising Cost and Distance | 优化二：最小化成本与距离
Minimisation problems appear frequently in business and geometry. A company may want to minimise the cost of manufacturing a cylindrical can of fixed volume. Let the can have radius r and height h. With volume V = πr²h fixed, the cost function might be proportional to surface area S = 2πr² + 2πrh. Eliminate h using h = V/(πr²), then S(r) = 2πr² + 2V/r. Differentiate, set S'(r) = 0, and solve for the optimal radius.
最小化问题经常出现在商业和几何中。一个公司可能要在固定体积下使圆柱形罐头的制造成本最小。设底面半径为 r、高为 h。给定体积 V = πr²h，成本可能与表面积 S = 2πr² + 2πrh 成正比。用 h = V/(πr²) 消去 h，得 S(r) = 2πr² + 2V/r。求导，令 S'(r) = 0，解出最优半径。
S'(r) = 4πr − 2V/r² = 0 → r³ = V/(2π)
Another typical problem is finding the point on a curve that minimises distance to a given point. For instance, find the point on y = x² + 1 closest to (0,3). The distance squared D = (x − 0)² + (x² + 1 − 3)² = x² + (x² − 2)². Minimising D (or D² to avoid square roots) by differentiation gives critical points; checking second derivative confirms the minimum.
另一类典型问题是求曲线上离给定点距离最小的点。例如，求 y = x² + 1 上离 (0,3) 最近的点。距离的平方 D = (x − 0)² + (x² + 1 − 3)² = x² + (x² − 2)²。通过求导最小化 D（或 D² 以避免平方根），得到临界点；用二阶导数检验确认最小值。
In all optimisation tasks, clearly state the quantity to be optimised, express it in one variable using constraints, differentiate, find stationary points, classify them, and interpret the result in context. A well-labelled diagram is invaluable.
在所有最优化任务中，要清晰陈述待优化的量，利用约束将其表示为单变量函数，求导，找到驻点，进行分类，并在实际背景中解释结果。清晰标注的图形极有帮助。
5. Area Under a Curve and Between Curves | 曲线下与曲线间的面积
The definite integral ∫ab f(x) dx represents the signed area between the graph of y = f(x) and the x-axis from x = a to x = b. If f(x) ≥ 0, the area is positive; if f(x) dips below the x-axis, the integral gives net area. To compute the total geometric area, we partition the interval where the function changes sign and take absolute values.
定积分 ∫ab f(x) dx 表示从 x=a 到 x=b 曲线 y=f(x) 与 x 轴之间的有符号面积。若 f(x) ≥ 0，面积为正；若 f(x) 低于 x 轴，积分给出净面积。要计算总的几何面积，我们需要在函数变号的区间分段并取绝对值。
Area between two curves y = f(x) (top) and y = g(x) (bottom) on [a, b] is given by A = ∫ab [f(x) − g(x)] dx. For example, find the area enclosed by y = x² and y = 2x. The curves intersect at x = 0 and x = 2. On [0,2], the line y = 2x is above. So A = ∫02 (2x − x²) dx = [x² − (1/3)x³]02 = 4 − 8/3 = 4/3.
两条曲线 y=f(x)（上方）和 y=g(x)（下方）在 [a,b] 上的面积为 A = ∫ab [f(x) − g(x)] dx。例如求 y=x² 与 y=2x 所围面积。曲线交于 x=0, x=2。在 [0,2] 上，直线 y=2x 在上方。故 A = ∫02 (2x − x²) dx = [x² − (1/3)x³]02 = 4 − 8/3 = 4/3。
Sometimes the roles of x and y are reversed, integrating with respect to y. If curves are better expressed as x = f(y), the area between them is ∫cd [right − left] dy. Identifying the correct variable of integration can simplify the calculation significantly.
有时需要将 x 和 y 的角色互换，对 y 积分。如果曲线用 x = f(y) 表达更方便，那么面积就是 ∫cd [右曲线 − 左曲线] dy。识别正确的积分变量可以极大简化计算。
6. Volume of Revolution: Disk Method | 旋转体体积：圆盘法
When a region under a curve is rotated about the x-axis, the resulting solid’s volume can be computed by the disk method. If y = f(x) is continuous on [a, b] and revolved around the x-axis, the volume is V = π ∫ab [f(x)]² dx. This formula arises from summing thin disks of radius |f(x)| and thickness dx.
当曲线下的区域绕 x 轴旋转时，所得立体的体积可以用圆盘法计算。若 y = f(x) 在 [a, b] 上连续并绕 x 轴旋转，体积为 V = π ∫ab [f(x)]² dx。这个公式来源于许多半径为 |f(x)|、厚度为 dx 的薄圆盘的累加。
V = π ∫ab r² dx, where r = f(x)
For example, rotate the region bounded by y = √x, x = 0, x = 4, and the x-axis about the x-axis. The volume is V = π ∫04 (√x)² dx = π ∫04 x dx = π [½ x²]04 = 8π.
例如，将曲线 y=√x，x=0，x=4 与 x 轴围成的区域绕 x 轴旋转。体积为 V = π ∫04 (√x)² dx = π ∫04 x dx = π [½ x²]04 = 8π。
When rotating about the y-axis, the disk method uses x = g(y) and integrates with respect to y: V = π ∫cd [g(y)]² dy. The key is to identify the radius function that is perpendicular to the axis of rotation.
绕 y 轴旋转时，圆盘法使用 x=g(y) 并对 y 积分：V = π ∫cd [g(y)]² dy。关键是确定垂直于旋转轴的半径函数。
7. Volume of Revolution: Washer Method | 旋转体体积：圆环法
When the region between two curves is rotated, the resulting solid has a hole, and the washer method is used. If the region bounded by y = f(x) (outer) and y = g(x) (inner) from a to b is revolved about the x-axis, the volume is V = π ∫ab ([f(x)]² − [g(x)]²) dx. This subtracts the inner disk from the outer disk.
当两条曲线间的区域旋转时，得到的立体有一个洞，需使用圆环法。若由 y=f(x)（外）和 y=g(x)（内）所围区域从 a 到 b 绕 x 轴旋转，体积为 V = π ∫ab ([f(x)]² − [g(x)]²) dx。这相当于从外圆盘减去内圆盘。
V = π ∫ab (R² − r²) dx
Example: the region between y = x² and y = x from x = 0 to x = 1 is rotated about the x-axis. Here R = x, r = x², so V = π ∫01 (x² − x⁴) dx = π [⅓ x³ − ⅕ x⁵]01 = π(1/3 − 1/5) = 2π/15.
例子：y=x² 与 y=x 之间的区域从 x=0 到 x=1 绕 x 轴旋转。此时 R=x, r=x²，故 V = π ∫01 (x² − x⁴) dx = π [⅓ x³ − ⅕ x⁵]01 = π(1/3 − 1/5) = 2π/15。
If a region is revolved about the y-axis and it is not possible to express both curves as functions of y, the shell method can be an alternative, but IB typically emphasises disk and washer methods. Pay attention to whether the rotation axis coincides with a boundary to avoid incorrect radii.
如果绕 y 轴旋转，且不能将两条曲线都表示为 y 的函数，那么壳层法是一个替代，但 IB 一般强调圆盘和圆环法。要注意旋转轴是否与边界重合，以免半径出错。
8. Average Value of a Function | 函数平均值
The average value of a continuous function f on the interval [a, b] is defined as favg = (1/(b − a)) ∫ab f(x) dx. This is analogous to finding the arithmetic mean of finitely many numbers, extended to a continuum of function values. Graphically, favg is the height of a rectangle with base [a, b] whose area equals the definite integral.
连续函数 f 在区间 [a, b] 上的平均值定义为 favg = (1/(b − a)) ∫ab f(x) dx。这类似于求有限多个数字的算术平均，但被推广到了函数值的连续体。从图像上看，favg 是一个以 [a, b] 为底、面积等于定积分的矩形的高。
favg = (1/(b − a)) ∫ab f(x) dx
For instance, the average value of f(x) = x² on [0, 2] is (1/(2−0)) ∫02
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