What is calculus in HSC Mathematics Advanced?

Calculus in HSC Mathematics Advanced covers differentiation (finding the rate of change of a function) and integration (finding the area under a curve, and reversing differentiation). Together they form roughly half of the HSC Mathematics Advanced course and a similar share of the exam paper.

What is the difference between differentiation and integration?

Differentiation finds the instantaneous rate of change (the slope of the tangent) of a function. Integration is the reverse process; given a rate of change, integration recovers the original function. The two are connected by the Fundamental Theorem of Calculus, which states that integration is the inverse of differentiation up to a constant.

What are the most important differentiation rules to memorise?

The product rule, quotient rule, chain rule, and the standard derivatives of polynomials, trigonometric functions ($\\sin$, $\\cos$, $\\tan$), exponentials ($e^x$, $a^x$), and logarithms ($\\ln x$). Strong students can recall these in under 5 seconds; weaker students lose time looking up rules during the exam.

What integration techniques are in HSC Mathematics Advanced?

HSC Mathematics Advanced covers basic integration (reversing polynomials, exponentials, trig functions), substitution (also called change of variable, only the simple cases), and the use of integration to find areas under curves and volumes of revolution. Integration by parts is not in Advanced; it is in Extension 1.

What is a typical calculus exam question?

Most extended-response calculus questions in the HSC paper involve multi-step problems combining differentiation with an application. A common pattern is to differentiate a function, find its stationary points, classify them as maxima or minima, and use that to optimise a quantity (volume of a box, profit, distance). Integration questions often ask for areas under curves or volumes of revolution.

How much of the HSC exam is calculus?

Calculus content makes up roughly 40-50% of HSC Mathematics Advanced exam marks. Differentiation typically takes up around 25%; integration around 20%; applications using both are spread across the longer extended-response questions in Section II.

NSWMaths Advanced

HSC Mathematics Advanced calculus: differentiation and integration (2026 guide)

A complete 2026 HSC Mathematics Advanced calculus guide. Differentiation rules, integration techniques, applications (rates of change, optimisation, areas, volumes), step-by-step worked examples, common traps, embedded practice questions, and how the topic is examined.

Generated by Claude Opus 4.822 min readNESA Mathematics Advanced Stage 6 Syllabus (2017), Year 12 Calculus: MA-C2 Differential Calculus, MA-C3 Applications of Differentiation, MA-C4 Integral CalculusUpdated 2026-05-25

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What calculus actually does
Differentiation: the rate of change
Applications of differentiation
Integration: the reverse process
Applications of integration
Common HSC calculus traps
How calculus is examined
Practice strategy
Check your knowledge

What calculus actually does

Calculus is the maths of change. It answers two questions:

What is the rate of change of a function? (Differentiation)
What is the total accumulation of a quantity over an interval? (Integration)

In HSC Mathematics Advanced, calculus is roughly half the course and roughly half the exam. Students who master it confidently typically score 85+ on the exam paper; students who do not typically struggle to break 70.

The good news: calculus rewards practice more than any other HSC maths topic. The differentiation and integration rules are finite and memorisable. The applications repeat every year in slightly different clothing.

Differentiation: the rate of change

The derivative of a function $f(x)$ , written $f'(x)$ or $\frac{df}{dx}$ , gives the slope of the tangent line at every point on the graph of $f$ .

The fundamental rules

Memorise these. Strong students recall them in under five seconds.

Power rule: If $f(x) = x^n$ , then $f'(x) = n x^{n-1}$ .
Constant rule: If $f(x) = c$ (a constant), then $f'(x) = 0$ .
Sum and difference: $(f + g)' = f' + g'$ and $(f - g)' = f' - g'$ .
Product rule: $(fg)' = f'g + fg'$ .
Quotient rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ .
Chain rule: If $y = f(g(x))$ , then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$ .

Standard derivatives

Memorise these by feel, not by lookup:

$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}(\tan x) = \sec^2 x$
$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(\ln x) = \frac{1}{x}$
$\frac{d}{dx}(a^x) = a^x \ln a$
$\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$

Differentiate

y = \ln(\sin(2x))

using the chain rule

Step 1. Identify the nested layers. The outermost function is $\ln$ , the middle layer is $\sin$ , and the innermost layer is $2x$ . The chain rule will give one factor per layer.

Step 2. Differentiate the outer layer, leaving the inside untouched. The derivative of $\ln u$ is $\frac{1}{u}$ , evaluated at $u = \sin(2x)$ .

\frac{d}{dx}\ln(\sin(2x)) = \frac{1}{\sin(2x)} \cdot \frac{d}{dx}\sin(2x)

Step 3. Differentiate the middle layer. The derivative of $\sin v$ is $\cos v$ , evaluated at $v = 2x$ , times the derivative of $2x$ .

\frac{d}{dx}\sin(2x) = \cos(2x) \cdot 2

Step 4. Multiply the factors together.

\frac{dy}{dx} = \frac{1}{\sin(2x)} \cdot \cos(2x) \cdot 2 = \frac{2\cos(2x)}{\sin(2x)}

Step 5. Simplify using $\frac{\cos}{\sin} = \cot$ .

\frac{dy}{dx} = 2\cot(2x)

Answer. $\dfrac{dy}{dx} = 2\cot(2x)$ .

Students lose marks here by forgetting one layer of the chain (the inner $2x$ derivative). Train your chain rule until you naturally see every layer.

Applications of differentiation

Tangents and normals

The tangent line to $y = f(x)$ at $x = a$ has slope $f'(a)$ and passes through $(a, f(a))$ . Equation: $y - f(a) = f'(a)(x - a)$ .

The normal line is perpendicular to the tangent. Its slope is $-\frac{1}{f'(a)}$ .

Stationary points and turning points

Stationary points occur where $f'(x) = 0$ . Classify them with the second derivative:

$f''(x) > 0$ : local minimum
$f''(x) < 0$ : local maximum
$f''(x) = 0$ : inconclusive (check sign of $f'$ on either side)

Optimisation: the highest-yield application

Optimisation questions appear every year. The pattern:

Set up the quantity to maximise or minimise as a function of one variable.
Differentiate, set the derivative to zero, solve.
Verify it is a max or min via the second derivative.
Check endpoints if the variable is constrained.

An open-top box with a square base has volume

250\text{ cm}^3

. Find the dimensions that minimise surface area

Step 1. Introduce variables and write the constraint: Let the base side be $x$ cm and the height be $h$ cm. The fixed volume gives $x^2 h = 250$ .
Step 2. Express the height in terms of $x$: Solving the constraint, $h = \dfrac{250}{x^2}$ .
Step 3. Write the surface area as a function of $x$ only: An open box has one square base and four rectangular sides, so $S = x^2 + 4xh$ . Substituting the constraint,

S(x) = x^2 + 4x \cdot \frac{250}{x^2} = x^2 + \frac{1000}{x}

Step 4. Differentiate and set to zero to find stationary points.

\frac{dS}{dx} = 2x - \frac{1000}{x^2} = 0

So $2x^3 = 1000$ , giving $x^3 = 500$ and $x = \sqrt[3]{500} \approx 7.94$ cm.

Step 5. Confirm it is a minimum using the second derivative.

\frac{d^2 S}{dx^2} = 2 + \frac{2000}{x^3} > 0 \text{ for all } x > 0

Positive everywhere, so this stationary point is a local minimum.

Step 6. Find the height and state the answer in context. $h = \dfrac{250}{(7.94)^2} \approx 3.97$ cm.

Answer. The minimum-surface-area box has base $7.94 \times 7.94$ cm and height $3.97$ cm.

Optimisation under constraint - rectangle inscribed in a semicircle of radius 5

A rectangle is inscribed in the semicircle $x^2 + y^2 = 25$ , $y \geq 0$ , with its base on the x-axis. Find the maximum area.

Step 1. Set up the rectangle in terms of one variable: Let the top-right corner of the rectangle be $(x, y)$ on the semicircle, so the rectangle has width $2x$ and height $y$ , with $0 \leq x \leq 5$ .
Step 2. Use the constraint to remove $y$: From the circle equation, $y = \sqrt{25 - x^2}$ .
Step 3. Write the area as a function of $x$

A(x) = 2x \sqrt{25 - x^2}

Step 4. Differentiate using the product rule. Let $u = 2x$ and $v = \sqrt{25 - x^2} = (25 - x^2)^{1/2}$ . Then $u' = 2$ and, by the chain rule,

v' = \frac{1}{2}(25 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{25 - x^2}}

\frac{dA}{dx} = 2\sqrt{25 - x^2} + 2x \cdot \frac{-x}{\sqrt{25 - x^2}} = \frac{2(25 - x^2) - 2x^2}{\sqrt{25 - x^2}} = \frac{50 - 4x^2}{\sqrt{25 - x^2}}

Step 5. Set the derivative to zero. The numerator must vanish: $50 - 4x^2 = 0$ , so $x^2 = 12.5$ and $x = \sqrt{12.5} = \dfrac{5}{\sqrt{2}}$ .

Step 6. Compute the maximum area. Then $y = \sqrt{25 - 12.5} = \sqrt{12.5} = \dfrac{5}{\sqrt{2}}$ , so

A = 2x \cdot y = 2 \cdot \frac{5}{\sqrt{2}} \cdot \frac{5}{\sqrt{2}} = \frac{50}{2} = 25

Step 7. Verify it is a maximum. At $x = 0$ and $x = 5$ the area is $0$ , so the interior stationary point must be a maximum.

Answer. The maximum area is $25$ square units, achieved when $x = \dfrac{5}{\sqrt{2}}$ and $y = \dfrac{5}{\sqrt{2}}$ .

Rates of change and related rates

Related rates problems link two changing quantities. The pattern: write an equation connecting them, differentiate implicitly with respect to time.

Air is pumped into a spherical balloon at

100\text{ cm}^3/\text{s}

. How fast is the radius increasing when

r = 10

cm?

Step 1. Write the geometric relationship. The volume of a sphere is

V = \frac{4}{3}\pi r^3

Step 2. Differentiate both sides with respect to time. Because $r$ depends on $t$ , the chain rule attaches a $\dfrac{dr}{dt}$ to the right.

\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

Step 3. Substitute the known values. Given $\dfrac{dV}{dt} = 100$ and $r = 10$ :

100 = 4\pi (10)^2 \frac{dr}{dt} = 400\pi \frac{dr}{dt}

Step 4. Solve for $\dfrac{dr}{dt}$ .

\frac{dr}{dt} = \frac{100}{400\pi} = \frac{1}{4\pi} \approx 0.0796 \text{ cm/s}

Answer. The radius is increasing at $\dfrac{1}{4\pi} \approx 0.0796$ cm/s.

Related rates - shadow length from a walking pedestrian

A 1.8 m tall person walks away from a 5 m streetlight at 1.5 m/s. How fast is the tip of their shadow moving along the ground?

Step 1. Draw the geometry and label variables. Let $x$ be the person's distance from the base of the lamp post, and $y$ be the distance from the lamp post to the tip of the shadow. Both depend on time.

Step 2. Use similar triangles to relate $x$ and $y$ . The large triangle (lamp top, base, shadow tip) has height $5$ and base $y$ . The small triangle (head, feet, shadow tip) has height $1.8$ and base $y - x$ . Similar triangles give

\frac{5}{y} = \frac{1.8}{y - x}

Step 3. Solve for $y$ in terms of $x$ . Cross-multiplying, $5(y - x) = 1.8 y$ , so $5y - 5x = 1.8y$ and $3.2 y = 5x$ , giving $y = \dfrac{5}{3.2} x = \dfrac{25}{16} x$ .

Step 4. Differentiate with respect to time.

\frac{dy}{dt} = \frac{25}{16} \frac{dx}{dt}

Step 5. Substitute the walking speed. With $\dfrac{dx}{dt} = 1.5$ ,

\frac{dy}{dt} = \frac{25}{16} \cdot 1.5 = \frac{37.5}{16} \approx 2.34 \text{ m/s}

Answer. The tip of the shadow moves at $\approx 2.34$ m/s. Notice it does not depend on $x$ , so the speed is constant.

Integration: the reverse process

Integration recovers a function from its derivative. The indefinite integral $\int f(x) \, dx$ is a family of primitive functions differing by a constant.

The fundamental theorem

\frac{d}{dx} \int_a^x f(t) \, dt = f(x) \quad \text{and} \quad \int_a^b f'(x) \, dx = f(b) - f(a)

This connects differentiation and integration. Memorise it; quote it; use it.

Standard integrals

$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \ne -1$
$\int \frac{1}{x} \, dx = \ln|x| + C$
$\int e^x \, dx = e^x + C$
$\int \sin x \, dx = -\cos x + C$
$\int \cos x \, dx = \sin x + C$
$\int \sec^2 x \, dx = \tan x + C$

Integration by substitution

For HSC Advanced, only simple substitutions appear. The pattern: if you see $f(g(x)) \cdot g'(x)$ in the integrand, let $u = g(x)$ .

Find

\int 2x \sin(x^2) \, dx

by substitution

Step 1. Spot the inner function. The integrand contains $\sin(x^2)$ , and the factor $2x$ outside is exactly $\dfrac{d}{dx}(x^2)$ . That is the signature of a substitution.

Step 2. Let $u$ be the inner function and find $du$ . Set $u = x^2$ , so $du = 2x \, dx$ .

Step 3. Rewrite the integral entirely in $u$ .

\int 2x \sin(x^2) \, dx = \int \sin u \, du

Step 4. Integrate. $\int \sin u \, du = -\cos u + C$ .

Step 5. Substitute back to $x$ .

-\cos(x^2) + C

Answer. $\int 2x \sin(x^2) \, dx = -\cos(x^2) + C$ .

Substitution with a missing constant factor - find

\int x \sqrt{x^2 + 1}\, dx

Step 1. Identify a candidate inner function. Set $u = x^2 + 1$ , so $du = 2x\, dx$ and $x\, dx = \dfrac{1}{2}\, du$ .

Step 2. Rewrite the integral. $\sqrt{x^2 + 1} = \sqrt{u} = u^{1/2}$ , so

\int x\sqrt{x^2 + 1}\, dx = \int u^{1/2} \cdot \frac{1}{2}\, du = \frac{1}{2}\int u^{1/2}\, du

Step 3. Integrate.

\frac{1}{2} \cdot \frac{u^{3/2}}{3/2} = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2}

Step 4. Substitute back and add $C$ .

\frac{1}{3}(x^2 + 1)^{3/2} + C

Answer. $\int x\sqrt{x^2 + 1}\, dx = \dfrac{1}{3}(x^2 + 1)^{3/2} + C$ .

Definite integral via substitution - evaluate

\int_0^1 \dfrac{x}{x^2 + 4}\, dx

Step 1. Pick the substitution: Set $u = x^2 + 4$ , so $du = 2x\, dx$ and $x\, dx = \dfrac{1}{2}\, du$ .
Step 2. Update the limits: When $x = 0$ , $u = 4$ . When $x = 1$ , $u = 5$ .
Step 3. Rewrite the definite integral in $u$

\int_0^1 \frac{x}{x^2 + 4}\, dx = \int_4^5 \frac{1}{u} \cdot \frac{1}{2}\, du = \frac{1}{2}\int_4^5 \frac{1}{u}\, du

Step 4. Integrate. $\frac{1}{2}[\ln|u|]_4^5 = \frac{1}{2}(\ln 5 - \ln 4) = \frac{1}{2}\ln\frac{5}{4}$ .

Answer. $\int_0^1 \dfrac{x}{x^2 + 4}\, dx = \dfrac{1}{2}\ln\dfrac{5}{4}$ .

Applications of integration

Areas under curves

The area between $y = f(x)$ and the x-axis from $x = a$ to $x = b$ is:

A = \int_a^b f(x) \, dx \quad \text{(when } f(x) \geq 0\text{)}

If the curve dips below the x-axis, split the integral at the zero crossings and take the absolute value of each segment.

Areas between curves

A = \int_a^b [f(x) - g(x)] \, dx

where $f(x) \geq g(x)$ on $[a, b]$ .

Find the area enclosed between

y = x^2

and

y = 2x

Step 1. Find the intersection points: Set $x^2 = 2x$ , so $x^2 - 2x = 0$ and $x(x - 2) = 0$ . The curves meet at $x = 0$ and $x = 2$ .
Step 2. Decide which curve is higher on the interval: At $x = 1$ , $y = x^2 = 1$ and $y = 2x = 2$ . So the line $y = 2x$ is above the parabola on $[0, 2]$ .
Step 3. Set up the area integral

A = \int_0^2 (2x - x^2)\, dx

Step 4. Evaluate.

A = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = \left( 4 - \frac{8}{3} \right) - 0 = \frac{12 - 8}{3} = \frac{4}{3}

Answer. The enclosed area is $\dfrac{4}{3}$ square units.

Volumes of revolution

Rotating $y = f(x)$ about the x-axis from $x = a$ to $x = b$ produces a solid of volume:

V = \pi \int_a^b [f(x)]^2 \, dx

For rotation about the y-axis, swap the variables.

Common HSC calculus traps

Forgetting the +C in indefinite integrals: Costs 1 mark per occurrence. Always include it.
Sign errors in differentiating $\cos x$: $\frac{d}{dx}(\cos x) = -\sin x$ , not $+\sin x$ . Easy to slip on under exam pressure.
Chain rule layers: Differentiating $y = \ln(\cos(3x))$ requires three nested applications: the outer $\ln$ , the middle $\cos$ , the inner $3x$ . Each is a separate step.
Mixing up product and chain rules: $\frac{d}{dx}[x \cdot e^x]$ is a product (use product rule). $\frac{d}{dx}[e^{x^2}]$ is a chain (use chain rule). They look similar, but the structure differs.
Setting up optimisation in two variables: You must express the optimisation quantity in one variable before differentiating. Beginners differentiate the two-variable equation and get nonsense.
Not checking max vs min: Stationary points can be maxima, minima, or horizontal points of inflection. Always verify.
Forgetting to interpret the answer: A calculus question that ends with "the box has dimensions $\approx 7.94 \times 7.94 \times 3.97$ cm" scores higher than one that ends with " $x = 500^{1/3}$ ". State the answer in the units the question asked.

How calculus is examined

In an HSC Mathematics Advanced paper:

Multiple-choice (Section I): 1-2 calculus questions, usually a straightforward differentiation or integration.
Section II early questions (worth 2-3 marks each): single-step differentiation or integration applications.
Section II middle questions (worth 5-7 marks): multi-step problems combining differentiation with optimisation, or integration with area.
Section II late questions (worth 8-12 marks): full extended problems testing the connection between differentiation and integration (a function's rate of change vs total accumulation).

Practice strategy

For HSC Mathematics Advanced calculus:

Week 1-4 (Term 4). Drill the rules. Memorise standard derivatives and integrals. Do 5-10 short practice questions per day.
Week 5-8. Apply rules to extended problems. One full past paper per week, with focus on calculus questions.
Week 9-12. Full past papers under timed 3-hour conditions. Mark yourself against the NESA marking guide.

The students who score 90+ on HSC Mathematics Advanced calculus are those who have done 8-10 past papers plus targeted practice on specific weaknesses. Past papers reveal the predictable structure of the questions.

Check your knowledge

Differentiate $y = (3x^2 + 1)^4$ .
Differentiate $y = x^2 \ln x$ .
Find the stationary points of $f(x) = x^3 - 6x^2 + 9x + 2$ and classify each as a maximum, minimum or horizontal point of inflection.
Evaluate $\displaystyle \int_0^{\pi/2} \cos x \, dx$ .
Use the substitution $u = 1 + x^3$ to evaluate $\displaystyle \int 3x^2(1 + x^3)^5 \, dx$ .
Find the area enclosed between the curves $y = x^2$ and $y = 4 - x^2$ .
Find the volume of the solid formed when $y = x + 1$ from $x = 0$ to $x = 2$ is rotated about the x-axis.
The radius of a circular oil spill is increasing at $0.5$ m/s. How fast is the area increasing when the radius is $20$ m?

Solutions to check-your-knowledge questions

The eight questions above cover chain rule, product rule, stationary points, definite integrals, substitution, area between curves, volumes of revolution, and related rates. Work through each in turn.

Step 1: Q1. Chain rule on $y = (3x^2 + 1)^4$

Identify the outer function (the power of 4) and the inner function $u = 3x^2 + 1$ . The chain rule says $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ .

\frac{du}{dx} = 6x, \qquad \frac{dy}{du} = 4u^3

\frac{dy}{dx} = 4(3x^2 + 1)^3 \cdot 6x = 24x(3x^2 + 1)^3

Step 2: Q2. Product rule on $y = x^2 \ln x$

Set $u = x^2$ and $v = \ln x$ . Their derivatives are $u' = 2x$ and $v' = \dfrac{1}{x}$ . The product rule gives $\dfrac{dy}{dx} = u'v + uv'$ .

\frac{dy}{dx} = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x = x(2\ln x + 1)

Step 3: Q3. Stationary points of $f(x) = x^3 - 6x^2 + 9x + 2$

Differentiate to find where the gradient is zero, then use the second derivative to classify each point.

f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)

So the stationary points are at $x = 1$ and $x = 3$ . Now $f''(x) = 6x - 12$ .

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

Step 4: Q4. Evaluate $\displaystyle \int_0^{\pi/2} \cos x \, dx$

The antiderivative of $\cos x$ is $\sin x$ . Apply the fundamental theorem by substituting the upper and lower limits.

\int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1

Step 5: Q5. Substitution on $\displaystyle \int 3x^2(1 + x^3)^5 \, dx$

Let $u = 1 + x^3$ . Differentiating gives $du = 3x^2 \, dx$ , which matches the factor in the integrand exactly.

\int 3x^2 (1 + x^3)^5 \, dx = \int u^5 \, du = \frac{u^6}{6} + C = \frac{(1 + x^3)^6}{6} + C

Step 6: Q6. Area between $y = x^2$ and $y = 4 - x^2$

First find where the curves meet by setting $x^2 = 4 - x^2$ , giving $x = \pm\sqrt{2}$ . On this interval the downward parabola $y = 4 - x^2$ lies above $y = x^2$ , so integrate top minus bottom.

A = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2)\, dx = \left[ 4x - \frac{2x^3}{3} \right]_{-\sqrt{2}}^{\sqrt{2}}

By symmetry: $A = 2 \left( 4\sqrt{2} - \dfrac{4\sqrt{2}}{3} \right) = 2 \cdot \dfrac{8\sqrt{2}}{3} = \dfrac{16\sqrt{2}}{3}$ square units.

Step 7: Q7. Volume of revolution for $y = x + 1$ from $x = 0$ to $x = 2$

Rotating about the x-axis gives $V = \pi \displaystyle \int_0^2 (x + 1)^2 \, dx$ . Use the substitution $u = x + 1$ (so $du = dx$ ) to shift the limits to 1 and 3.

V = \pi \int_1^3 u^2 \, du = \pi \left[ \frac{u^3}{3} \right]_1^3 = \pi \cdot \frac{27 - 1}{3} = \frac{26\pi}{3} \text{ cubic units}

Step 8: Q8. Related rate for a circular oil spill

The area of the spill is $A = \pi r^2$ . Differentiating both sides with respect to time gives $\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$ . Substitute $r = 20$ and $\dfrac{dr}{dt} = 0.5$ .

\frac{dA}{dt} = 2\pi \times 20 \times 0.5 = 20\pi \approx 62.8\ \text{m}^2/\text{s}

Final answer: the area of the spill is increasing at approximately $20\pi \approx 62.8\ \text{m}^2/\text{s}$ .