📚 Calculus Applications in More Functions | 更多函数中的微积分应用
This article explores how calculus extends beyond simple polynomials into implicit, inverse, parametric, polar, and other advanced function types. We focus on the concepts and techniques most commonly tested in the IB Mathematics curriculum, using clear examples to strengthen your understanding.
本文探讨微积分如何从简单多项式延伸至隐函数、反函数、参数方程、极坐标等更复杂的函数类型。我们聚焦 IB 数学课程中常见的概念与技巧，并通过清晰示例加深你的理解。
1. Implicit Differentiation | 隐函数求导
When a curve is defined implicitly by an equation such as x² + y² = 25 or x³ + y³ = 6xy, we cannot always solve explicitly for y. Implicit differentiation treats y as a function of x and applies the chain rule to every y-term.
当曲线由形如 x² + y² = 25 或 x³ + y³ = 6xy 的方程隐式定义时，我们往往无法显式解出 y。隐函数求导将 y 看作 x 的函数，并对每个含 y 的项应用链式法则。
For example, differentiate x² + y² = 25 with respect to x:
2x + 2y · dy/dx = 0
Hence dy/dx = −x/y. This gives the slope of the tangent at any point (x, y) on the circle. To find the tangent line at (3, 4), the slope is −3/4, so the equation is y − 4 = −3/4(x − 3).
例如，对 x² + y² = 25 关于 x 求导：
2x + 2y · dy/dx = 0
因此 dy/dx = −x/y。这给出了圆上任意一点 (x, y) 处切线的斜率。要求点 (3, 4) 处的切线，斜率为 −3/4，故切线方程为 y − 4 = −3/4(x − 3)。
Implicit differentiation is also essential for related rates problems. For instance, if two variables change with time and are linked by an implicit relation, differentiating both sides with respect to t connects their rates of change.
隐函数求导在相关变率问题中同样关键。例如，若两个变量随时间变化且由隐式关系联系，两边对 t 求导即可建立它们变化率之间的关系。
2. Derivatives of Inverse Functions | 反函数的导数
If y = f⁻¹(x), then applying f to both sides gives f(y) = x. Differentiating implicitly:
f′(y) · dy/dx = 1
so dy/dx = 1 / f′(y). This result is particularly useful for inverse trigonometric and logarithmic functions.
若 y = f⁻¹(x)，两边同时施加 f 得 f(y) = x。隐式求导得：
f′(y) · dy/dx = 1
所以 dy/dx = 1 / f′(y)。这一结果对反三角函数和对数函数尤其有用。
Take y = arcsin x. Then sin y = x, so cos y · dy/dx = 1. Since cos y = √(1 − sin² y) = √(1 − x²), we obtain:
d/dx (arcsin x) = 1 / √(1 − x²)
Similarly, d/dx (arctan x) = 1 / (1 + x²). These derivatives frequently appear in integrals and differential equations.
以 y = arcsin x 为例，则 sin y = x，因此 cos y · dy/dx = 1。由于 cos y = √(1 − sin² y) = √(1 − x²)，可得：
d/dx (arcsin x) = 1 / √(1 − x²)
类似地，d/dx (arctan x) = 1 / (1 + x²)。这些导数经常出现在积分和微分方程中。
When dealing with functions like y = logₐ x, you can rewrite as x = aʸ and differentiate: 1 = aʸ ln a · dy/dx, so dy/dx = 1 / (x ln a).
处理 y = logₐ x 这类函数时，可改写为 x = aʸ 再求导：1 = aʸ ln a · dy/dx，因此 dy/dx = 1 / (x ln a)。
3. Parametric Equations and Differentiation | 参数方程与求导
A parametric curve is defined by x = x(t) and y = y(t). To find the slope of the tangent, use the chain rule:
dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0
参数曲线由 x = x(t) 与 y = y(t) 定义。要求切线斜率，使用链式法则：
dy/dx = (dy/dt) / (dx/dt)，前提是 dx/dt ≠ 0
For example, consider the curve x = 2 cos t, y = 2 sin t for 0 ≤ t ≤ 2π. Then dx/dt = −2 sin t and dy/dt = 2 cos t, so dy/dx = −cot t. At t = π/4, the slope is −1.
例如，考虑曲线 x = 2 cos t，y = 2 sin t，其中 0 ≤ t ≤ 2π。则 dx/dt = −2 sin t，dy/dt = 2 cos t，所以 dy/dx = −cot t。在 t = π/4 处，斜率为 −1。
The second derivative can also be found without eliminating the parameter:
d²y/dx² = d/dx (dy/dx) = (d/dt (dy/dx)) / (dx/dt)
二阶导数也可以在保留参数的情况下求得：
d²y/dx² = d/dx (dy/dx) = (d/dt (dy/dx)) / (dx/dt)
This is useful for analysing concavity and determining the nature of stationary points.
这对于分析凹凸性和判断驻点性质非常有用。
4. Polar Curves: Tangents and Area | 极坐标曲线：切线与面积
In polar coordinates, a curve is given by r = f(θ). To find the slope of the tangent, convert to Cartesian: x = r cos θ, y = r sin θ. Then:
dy/dx = (r′ sin θ + r cos θ) / (r′ cos θ − r sin θ)
在极坐标中，曲线由 r = f(θ) 给定。为求切线斜率，可转换为直角坐标：x = r cos θ，y = r sin θ。于是：
dy/dx = (r′ sin θ + r cos θ) / (r′ cos θ − r sin θ)
The area enclosed by a polar curve between θ = α and θ = β is:
A = ½ ∫αβ r² dθ
极坐标曲线在 θ = α 与 θ = β 之间围成的面积为：
A = ½ ∫αβ r² dθ
Take the cardioid r = 1 + cos θ. The total area is ½ ∫₀2π (1 + cos θ)² dθ = 3π/2. This integral uses the identity cos² θ = (1 + cos 2θ)/2.
以心形线 r = 1 + cos θ 为例，其总面积为 ½ ∫₀2π (1 + cos θ)² dθ = 3π/2。该积分使用了恒等式 cos² θ = (1 + cos 2θ)/2。
Be careful with symmetries: often you can compute half or quarter of the area and multiply. This saves time in exams.
注意利用对称性：通常可以先求一半或四分之一面积再乘以倍数。这能在考试中节省时间。
5. Integration by Substitution for Compound Functions | 复合函数的换元积分
Many integrals involving more complex functions require substitution. The chain rule in reverse allows us to rewrite an integral in terms of a new variable u.
许多涉及更复杂函数的积分需要换元法。链式法则的逆用使我们得以用新变量 u 重新改写积分。
For example, evaluate ∫ x e^(x²) dx. Let u = x², so du = 2x dx. Then:
∫ x e^(x²) dx = ½ ∫ eᵘ du = ½ eᵘ + C = ½ e^(x²) + C
例如，计算 ∫ x e^(x²) dx。令 u = x²，则 du = 2x dx。于是：
∫ x e^(x²) dx = ½ ∫ eᵘ du = ½ eᵘ + C = ½ e^(x²) + C
More generally, if the integrand contains f′(x) g(f(x)), the substitution u = f(x) often simplifies it. For definite integrals, remember to change the limits of integration when u is introduced.
更一般地，若被积函数包含 f′(x) g(f(x))，令 u = f(x) 通常能简化积分。对于定积分，引入 u 时必须相应改变积分上下限。
Another common substitution is trigometric: ∫ sin x cos x dx can be done with u = sin x, giving ½ sin² x + C.
另一个常见换元是三角换元：∫ sin x cos x dx 可令 u = sin x，得到 ½ sin² x + C。
6. Integration by Parts for Products | 分部积分处理乘积
For products of functions, integration by parts is derived from the product rule:
∫ u dv = uv − ∫ v du
对于函数乘积，分部积分由乘积法则导出：
∫ u dv = uv − ∫ v du
A classic IB example is ∫ x ln x dx. Choose u = ln x and dv = x dx. Then du = 1/x dx and v = x²/2. Hence:
∫ x ln x dx = (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C
一个经典 IB 例题是 ∫ x ln x dx。选择 u = ln x，dv = x dx。则 du = 1/x dx，v = x²/2。因此：
∫ x ln x dx = (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C
Sometimes integration by parts must be applied twice, as in ∫ eˣ sin x dx. This leads to an equation involving the original integral, which can be solved algebraically.
有时需要连续使用两次分部积分，例如 ∫ eˣ sin x dx。这会得到一个含原积分式子的方程，可通过代数方法解出。
Also, for definite integrals remember to evaluate uv at both limits and subtract the remaining integral.
此外，定积分中需要在上下限处计算 uv，再减去剩余积分。
7. Partial Fractions for Rational Functions | 有理函数的部分分式积分
Integrating rational functions often requires rewriting them as partial fractions. For example:
1 / [(x − 1)(x + 2)] = A/(x − 1) + B/(x + 2)
对有理函数积分通常需要先改写为部分分式。例如：
1 / [(x − 1)(x + 2)] = A/(x − 1) + B/(x + 2)
Solving gives A = 1/3 and B = −1/3. Therefore:
∫ dx / [(x − 1)(x + 2)] = (1/3) ln|x − 1| − (1/3) ln|x + 2| + C
解得 A = 1/3，B = −1/3。因此：
∫ dx / [(x − 1)(x + 2)] = (1/3) ln|x − 1| − (1/3) ln|x + 2| + C
For repeated linear factors, such as 1/[x(x − 1)²], include terms for both (x − 1) and (x − 1)²:
1/[x(x − 1)²] = A/x + B/(x − 1) + C/(x − 1)²
对重复线性因子，如 1/[x(x − 1)²]，需要同时包含 (x − 1) 与 (x − 1)² 的项：
1/[x(x − 1)²] = A/x + B/(x − 1) + C/(x − 1)²
Notice that the integral of 1/(x − 1)² is −1/(x − 1), not a logarithm.
注意 1/(x − 1)² 的积分是 −1/(x − 1)，而不是对数。
8. Trigonometric Integrals and Identities | 三角积分与恒等式
Trigonometric functions appear frequently in IB calculus. Key strategies include using identities to reduce powers:
sin² x = (1 − cos 2x)/2, cos² x = (1 + cos 2x)/2
三角函数在 IB 微积分中频繁出现。关键策略是利用恒等式降幂：
sin² x = (1 − cos 2x)/2, cos² x = (1 + cos 2x)/2
For example, ∫ sin² x dx = ∫ (1 − cos 2x)/2 dx = x/2 − sin 2x/4 + C.
例如，∫ sin² x dx = ∫ (1 − cos 2x)/2 dx = x/2 − sin 2x/4 + C。
Products like sin 3x cos x can be converted using product-to-sum formulae:
sin A cos B = ½ [sin(A + B) + sin(A − B)]
像 sin 3x cos x 这类乘积可借助积化和差公式转换：
sin A cos B = ½ [sin(A + B) + sin(A − B)]
Thus ∫ sin 3x cos x dx = ½ ∫ (sin 4x + sin 2x) dx = −cos 4x/8 − cos 2x/4 + C.
因此 ∫ sin 3x cos x dx = ½ ∫ (sin 4x + sin 2x) dx = −cos 4x/8 − cos 2x/4 + C。
For ∫ sec x dx, multiply numerator and denominator by (sec x + tan x): the result is ln|sec x + tan x| + C.
对于 ∫ sec x dx，可将分子分母同乘 (sec x + tan x)，结果为 ln|sec x + tan x| + C。
9. Area, Volume and Arc Length | 面积、体积与弧长
For a function y = f(x) ≥ 0, the area under the curve from a to b is:
A = ∫ab f(x) dx
对于函数 y = f(x) ≥ 0，从 a 到 b 的曲线下面积为：
A = ∫ab f(x) dx
The volume of revolution about the x-axis is:
V = π ∫ab [f(x)]² dx
绕 x 轴旋转所得旋转体体积为：
V = π ∫ab [f(x)]² dx
As an example, find the volume generated by rotating y = √x from x = 1 to x = 4 about the x-axis. Then V = π ∫₁⁴ x dx = π [x²/2]₁⁴ = 15π/2 cubic units.
例如，求曲线 y = √x 在 x = 1 到 x = 4 间绕 x 轴旋转所得体积。V = π ∫₁⁴ x dx = π [x²/2]₁⁴ = 15π/2 立方单位。
Arc length for a smooth curve y = f(x) is given by:
L = ∫ab √(1 + (dy/dx)²) dx
光滑曲线 y = f(x) 的弧长为：
L = ∫ab √(1 + (dy/dx)²) dx
For parametric curves, the arc length formula becomes L = ∫ √((dx/dt)² + (dy/dt)²) dt. These formulas appear in both Paper 2 and Paper 3 (HL).
对于参数曲线，弧长公式变为 L = ∫ √((dx/dt)² + (dy/dt)²) dt。这些公式在试卷 2 和 HL 的试卷 3 中都会出现。
10. Optimisation with More Complex Functions | 更复杂函数的优化问题
Optimisation involves finding maximum or minimum values of a function in a given context. With more complex functions, the same derivative rules apply, but simplification and domain awareness become more important.
优化问题涉及在给定情境中寻找函数的最大值或最小值。对于更复杂的函数，求导规则依然相同，但化简和定义域的判断变得更加重要。
For example, find the minimum of f(x) = x ln x for x > 0. We compute:
f′(x) = ln x + 1
Setting f′(x) = 0 gives ln x = −1, hence x = e⁻¹. The second derivative is f″(x) = 1/x > 0, so the stationary point is a minimum.
例如，求 f(x) = x ln x 在 x > 0 上的最小值。计算得：
f′(x) = ln x + 1
令 f′(x) = 0，得 ln x = −1，因此 x = e⁻¹。二阶导数为 f″(x) = 1/x > 0，所以该驻点为极小值。
When the derivative involves fractions or exponentials, always check endpoints and any points where the derivative does not exist. For example, f(x) = x²e−x has critical points at x = 0 and x = 2; the second derivative test distinguishes them.
当导数涉及分式或指数时，务必检查区间端点和导数不存在的点。例如，f(x) = x²e−x 的临界点为 x = 0 和 x = 2；可用二阶导数判别法区分它们是极大还是极小。
In modelling problems, translated word conditions into equations first, then optimise using calculus. This is a key assessment objective in IB.
在建模问题中，首先将文字条件转化为方程，然后用微积分进行优化。这是 IB 的重要考核目标。
11. Differential Equations with More Functions | 更复杂函数相关的微分方程
Many growth and decay models lead to differential equations. The simplest is dy/dx = k y, whose solution is y = C e^(kx). But more complex models may involve logistic equations:
dy/dt = k y (1 − y/L)
许多增长与衰减模型会导出微分方程。最简单的是 dy/dx = k y，其解为 y = C e^(kx)。但更复杂的模型可能包含逻辑斯谛方程：
dy/dt = k y (1 − y/L)
This equation can be solved by separating variables and using partial fractions:
∫ dy/[y(1 − y/L)] = ∫ k dt
该方程可通过分离变量并利用部分分式求解：
∫ dy/[y(1 − y/L)] = ∫ k dt
The solution gives a sigmoid curve, often used to model population growth with a carrying capacity L.
其解为 S 形曲线，常用于描述具有环境容量 L 的种群增长。
Another important type is first-order linear differential equations of the form dy/dx + P(x)y = Q(x). These require an integrating factor R(x) = e^(∫ P(x) dx).
另一重要类型是一阶线性微分方程 dy/dx + P(x)y = Q(x)。这需要使用积分因子 R(x) = e^(∫ P(x) dx)。
12. Summary | 总结
Calculus in more functions extends the familiar rules of differentiation and integration into new contexts. Master implicit differentiation, inverse functions, parametric and polar forms, and the various integration techniques. Then apply these to area, volume, arc length, optimisation, and differential equations.
更多函数中的微积分将熟悉的求导和积分法则拓展到新的情境。掌握隐函数求导、反函数、参数方程与极坐标、以及各种积分技巧，再将其应用到面积、体积、弧长、优化和微分方程中。
Regular practice with past IB questions will help you recognise which technique to use. Keep a formula sheet nearby, but understand why each formula works. This deep understanding is the key to success in IB mathematics.
通过定期练习 IB 历年真题，你会逐步学会判断该用哪种技巧。随身携带公式表，但更要理解每个公式背后的原理。这种深度理解正是 IB 数学取得成功的关键。
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