GAC016 Mathematics III: Calculus & Advanced Applications

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.

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.

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

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.

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

Worked example. the chain and product rules together

Differentiate $y=x^2(3x-1)^4$.

Let $u=x^2$ and $v=(3x-1)^4$. Then $u'=2x$ and, by the chain rule, $v'=4(3x-1)^3\cdot3=12(3x-1)^3$.

Taking out the common factor $2x(3x-1)^3$ makes the stationary points easy to read: $x=0$, $x=\tfrac13$ and $x=\tfrac19$.

Worked example. find and classify stationary points

For $f(x)=x^3-6x^2+9x+2$, $f'(x)=3x^2-12x+9=3(x-1)(x-3)$, so the stationary points are at $x=1$ and $x=3$.

$f''(x)=6x-12$. At $x=1$, $f''(1)=-6<0$: a local maximum at $(1,6)$. At $x=3$, $f''(3)=6>0$: a local minimum at $(3,2)$.

Always give both coordinates: substitute back into $f$, not into $f'$.

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

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

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

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

Worked example. integration by substitution

Find $\displaystyle\int 6x(x^2+1)^5\,dx$.

Let $u=x^2+1$, so $\dfrac{du}{dx}=2x$ and $6x\,dx=3\,du$.

Check by differentiating: $\dfrac{d}{dx}\dfrac{(x^2+1)^6}{2}=3(x^2+1)^5\cdot2x=6x(x^2+1)^5$. ✓

For a definite integral, either change the limits to $u$-values or substitute back before using the original $x$-limits; never mix them.

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

• 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

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

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.

Worked example. connected rates of change

A spherical balloon is inflated at $\dfrac{dV}{dt}=50$ cubic centimetres per second. Find how fast the radius grows when $r=5$ cm.

$V=\tfrac43\pi r^3$, so $\dfrac{dV}{dr}=4\pi r^2=100\pi$ at $r=5$.

The same inflation rate gives a slower radius change as the balloon grows, because each extra centimetre of radius needs more volume.

Worked example. area between two curves

Find the area enclosed by $y=x^2$ and $y=2x$.

They meet where $x^2=2x$, so $x=0$ and $x=2$. On $0<x<2$, the line is above the parabola (at $x=1$: $2>1$).

Common mistakes.

• Using the secant slope when the question asks for an instantaneous rate, or the reverse.
• Forgetting the inner derivative in the chain rule: $\dfrac{d}{dx}(3x-1)^4\ne4(3x-1)^3$.
• Writing $(uv)'=u'v'$. The product rule has two terms.
• Swapping the numerator order in the quotient rule: it is $u'v-uv'$, not $uv'-u'v$.
• Classifying a stationary point with $f''=0$ as a point of inflection without checking the sign of $f'$ on both sides.
• Dropping $+c$, or treating a definite integral of a sign-changing function as an area.
• Not checking the ends of the feasible domain in an optimisation question.
• Leaving the answer to a related-rates question without units or a sign that shows increase or decrease.

Practice: the whole module.

Answer each before reading the answers below.

1. Differentiate (a) $y=5x^4-\dfrac{3}{x}+2\sqrt{x}$; (b) $y=(2x+3)^5$; (c) $y=x\,e^{2x}$; (d) $y=\dfrac{x^2}{x+1}$.
2. Use first principles to show that the derivative of $f(x)=3x^2$ is $6x$.
3. Find the equation of the tangent to $y=x^3-2x$ at $x=2$.
4. Find and classify the stationary points of $y=2x^3-3x^2-12x+5$.
5. Find (a) $\displaystyle\int(4x^3-6x+1)\,dx$; (b) $\displaystyle\int_1^4\sqrt{x}\,dx$; (c) $\displaystyle\int 2x(x^2-3)^4\,dx$.
6. Evaluate $\displaystyle\int_0^3(x-2)\,dx$, then find the total area between $y=x-2$ and the $x$-axis for $0\le x\le3$. Explain why the two answers differ.
7. An open box is made from a $12$ cm by $12$ cm square sheet by cutting a square of side $x$ cm from each corner. Show that $V=x(12-2x)^2$, state the domain and find the maximum volume.
8. A firm's cost is $C(q)=0.02q^2+4q+500$ dollars for $q$ units. Find the marginal cost at $q=100$ and explain what it means.
9. The radius of a circular oil spill increases at $0.5$ m per minute. Find the rate of increase of its area when $r=20$ m.
10. Find the area enclosed by $y=4-x^2$ and $y=x+2$.

Answers.

1. (a) $20x^3+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}$. (b) $10(2x+3)^4$. (c) $e^{2x}(1+2x)$. (d) $\dfrac{2x(x+1)-x^2}{(x+1)^2}=\dfrac{x^2+2x}{(x+1)^2}$.
2. $\dfrac{3(x+h)^2-3x^2}{h}=\dfrac{6xh+3h^2}{h}=6x+3h$ for $h\ne0$; as $h\to0$ this tends to $6x$.
3. At $x=2$, $y=4$ and $\dfrac{dy}{dx}=3x^2-2=10$. Tangent: $y-4=10(x-2)$, so $y=10x-16$.
4. $y'=6x^2-6x-12=6(x-2)(x+1)$. $y''=12x-6$. At $x=-1$: $y=12$, $y''=-18<0$, local maximum $(-1,12)$. At $x=2$: $y=-15$, $y''=18>0$, local minimum $(2,-15)$.
5. (a) $x^4-3x^2+x+c$. (b) $\left[\tfrac23x^{3/2}\right]_1^4=\tfrac{16}{3}-\tfrac23=\tfrac{14}{3}$. (c) With $u=x^2-3$: $\dfrac{(x^2-3)^5}{5}+c$.
6. $\left[\tfrac{x^2}{2}-2x\right]_0^3=\tfrac92-6=-\tfrac32$. The line is below the axis on $0<x<2$ (area $2$) and above on $2<x<3$ (area $\tfrac12$), so the total area is $\tfrac52$. The integral counts the part below the axis as negative.
7. The base is $(12-2x)$ by $(12-2x)$ and the height is $x$; the domain is $0<x<6$. $V'=(12-2x)^2-4x(12-2x)=(12-2x)(12-6x)$, so $x=2$ (as $x=6$ is excluded). $V(2)=2\cdot8^2=128$ cm³; $V\to0$ at both ends, so this is the maximum.
8. $C'(q)=0.04q+4$, so $C'(100)=8$ dollars per unit: near $q=100$, one more unit costs about $8$ dollars more.
9. $\dfrac{dA}{dt}=2\pi r\cdot\dfrac{dr}{dt}=2\pi(20)(0.5)=20\pi\approx62.8$ m² per minute.
10. $4-x^2=x+2$ gives $x^2+x-2=0$, so $x=-2$ or $x=1$. The parabola is above on this interval: $\displaystyle\int_{-2}^{1}(2-x-x^2)\,dx=\left[2x-\tfrac{x^2}{2}-\tfrac{x^3}{3}\right]_{-2}^{1}=\tfrac76-\left(-\tfrac{10}{3}\right)=\tfrac92$.