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GAC016 Mathematics III: Calculus & Advanced Applications

GAC Mathematics · Topic 3

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GAC016 Mathematics III: Calculus

An average speed does not tell you every instantaneous speed, and a signed displacement does not always equal total distance. Calculus makes those distinctions…

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3.1

What this module is, and how it is marked

An average speed does not tell you every instantaneous speed, and a signed displacement does not always equal total distance. Calculus makes those distinctions precise.

GAC016 develops differentiation 微分, integration 积分 and their applications. Your centre's current brief determines assessment tasks and grading requirements; these original practice sheets do not establish a university credit decision or an official examination pattern.

State the independent variable, domain and units. The notation $dy/dx$ is a derivative with respect to x; a dot over y normally denotes a derivative with respect to time. Those notations describe the same operation only when their independent variables agree.

For applications, describe what a rate or accumulated value means in the stated model. A negative volume-change rate of 3 cubic centimetres per second means volume is falling at that rate, rather than that the volume itself is negative.

Vocabulary Train
English
differentiation/ˌdɪfəˌrenʃɪˈeɪʃn/
integration/ˌɪntɪˈɡreɪʃn/
3.1

Differentiation

Syllabus

Unit 1 of 3 in GAC016 Mathematics III: Calculus & Advanced Applications (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

Module purpose: On completion of this module, students should be able to demonstrate a basic understanding of the principles of calculus and how they can be applied to the quantitative analysis of practical and financial situations.

The module outcomes this unit works towards:

Learning Objective GAC016.1: Determine the derivative (if it exists) of most mathematical functions and use the derivative to analyse functional behaviour.

Source: Cambridge International syllabus

A derivative 导数 is a rate of change: how fast $y$ changes as $x$ changes. Geometrically it is the gradient of the tangent 切线斜率 at a point.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

That limit is the definition. Differentiation from first principles 从定义求导 uses the difference quotient for nonzero h and then takes its limit. Unequal one-sided limits, as for $|x|$ at zero, prevent a derivative even when the function is continuous.

  • The power rule 幂法则, on a domain where the power is differentiable: if $y = x^n$ then $\dfrac{dy}{dx} = nx^{\,n-1}$.
  • The product rule 乘积法则: $(uv)' = u'v + uv'$.
  • The quotient rule 商法则, where $v\ne0$: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.
  • The chain rule 链式法则: if $y = f(g(x))$ then $\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)$.
  • Stationary points 驻点 occur where $f'(x) = 0$. The second derivative 二阶导数 then classifies a stationary point when it is nonzero: negative gives a local maximum 极大值 and positive a local minimum 极小值. A zero second derivative is inconclusive; use derivative signs or other evidence. A stationary inflection can be neither an extremum.

Worked example. a tangent slope differs from a secant slope

The curve y equals x squared has a tangent at (1,1) and a dashed secant joining (1,1) to (2,4).

Known: $f(x)=x^2$. The secant between x equal to 1 and 2 has slope 3, but the derivative at x equal to 1 is 2.

$$m_{sec}=\frac{f(b)-f(a)}{b-a}=\frac{4-1}{2-1}=3$$
$$f'(x)=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}=\lim_{h\to0}(2x+h)=2x$$
$$m_{tan}=f'(1)=2(1)=2$$

The average slope over a finite interval need not equal the instantaneous slope at either endpoint.

Use original practice sheet 3.1 to test the method, domain and interpretation against its solutions.

Vocabulary Train
English
derivative/dɪˈrɪvətɪv/
gradient of the tangent/ˈɡreɪdɪənt ɒvðə ˈtændʒənt/
differentiation from first principles/ˌdɪfəˌrenʃɪˈeɪʃn frɒm fɜːst ˈprɪnsɪplz/
power rule/ˈpaʊə ruːl/
product rule/ˈprɒdʌkt ruːl/
quotient rule/ˈkwəʊʃənt ruːl/
chain rule/tʃeɪn ruːl/
Stationary points/ˈsteɪʃənəri pɔɪnts/
second derivative/ˈsekənd dɪˈrɪvətɪv/
maximum/ˈmæksɪməm/
minimum/ˈmɪnɪməm/
3.2

Integration

Syllabus

Unit 2 of 3 in GAC016 Mathematics III: Calculus & Advanced Applications (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

The module outcomes this unit works towards:

Learning Objective GAC016.2: Use techniques of integration to find indefinite and definite integrals.

Source: Cambridge International syllabus

Indefinite integration finds antiderivatives. Definite integration measures signed accumulation, which can differ from geometric area. For a continuous integrand, the fundamental theorem of calculus 微积分基本定理 connects a definite integral to the difference of antiderivative values at its limits.

  • An indefinite integral 不定积分 has no limits and needs the constant of integration 积分常数: $\displaystyle\int x^n\,dx = \frac{x^{\,n+1}}{n+1} + c$ for $n \neq -1$.
  • A definite integral 定积分 has limits and gives a number: $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$.
  • Integration by substitution 换元积分法 reverses the chain rule.
  • ⚠ Forgetting $+c$ on an indefinite integral is the single most frequent lost mark in the module, and it is lost on questions you have otherwise answered correctly.

Worked example. negative and positive areas can cancel

The line y equals x crosses the axis at zero, with shaded regions on both sides over negative 1 to 1.

Known: $f(x)=x$ from negative 1 to 1. Its definite integral is zero because the negative and positive contributions cancel.

$$I=\int_{-1}^{1}x\,dx=\left[\frac{x^2}{2}\right]_{-1}^{1}=\frac12-\frac12=0$$

Geometric area instead adds the magnitudes of the two triangular regions.

$$A=-\int_{-1}^{0}x\,dx+\int_0^1x\,dx=\frac12+\frac12=1$$

Use original practice sheet 3.2 to test the method, domain and interpretation against its solutions.

Vocabulary Train
English
fundamental theorem of calculus/ˌfʌndəˈmentl ˈθɪərəm ɒv ˈkælkjʊləs/
indefinite integral/ɪnˈdefɪnət ˈɪntɪɡrəl/
constant of integration/ˈkɒnstənt ɒv ˌɪntɪˈɡreɪʃn/
definite integral/ˈdefɪnət ˈɪntɪɡrəl/
Integration by substitution/ˌɪntɪˈɡreɪʃn baɪ ˌsʌbstɪˈtjuːʃn/
3.3

Advanced applications

Syllabus

Unit 3 of 3 in GAC016 Mathematics III: Calculus & Advanced Applications (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

The module outcomes this unit works towards:

Learning Objective GAC016.3: Apply differentiation and integration techniques in a variety of practical problems.

Source: Cambridge International syllabus

  • Optimisation 最优化 finds the largest or smallest value of a quantity: write the quantity as a function of one variable on a stated feasible domain, differentiate, find candidates and justify the required optimum. Check included boundaries and integer constraints where relevant.
  • Rates of change 变化率 chain together: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$.
  • Area between two curves 两曲线间面积 is $\displaystyle\int_a^b (f(x) - g(x))\,dx$ where $f$ is the upper curve on that whole interval. Split at crossings if their order changes.
  • Marginal cost and revenue 边际成本与边际收益 are derivatives of stated continuous cost and revenue models. They are instantaneous rates, not automatically exact discrete-unit changes.

Worked example. a fixed perimeter leaves one area variable

A rectangle has width x and length 10 minus x under a fixed perimeter of 20 units.

Known: a rectangle has perimeter 20, so length plus width is 10. Let width be x; length is $10-x$ and $0<x<10$.

$$A(x)=x(10-x)$$
$$A'(x)=10-2x$$

The stationary candidate is $x=5$. Since $A''(x)=-2<0$ and the area approaches zero at both domain ends, this candidate is the global maximum. It is a 5 by 5 square with area 25.

Use original practice sheet 3.3 to test the method, domain and interpretation against its solutions.

Vocabulary Train
English
Optimisation/ˌɒptɪmaɪˈzeɪʃn/
Rates of change/reɪts ɒv tʃeɪndʒ/
Area between two curves/ˈeərɪə bɪˈtwiːn tuː kɜːvz/
Marginal cost and revenue/ˈmɑːdʒɪnl kɒst ænd ˈrevənjuː/

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