Additional Mathematics Synthesis Guide 4: Calculus, Algebra and Kinematics

Function → rate of change → derivative → stationary behaviour → accumulation → integral → motion → interpretation.

1. Differentiation measures local change

For a straight line, gradient is constant. For a curve, gradient changes from point to point. Differentiation gives a new function whose value tells us the gradient of the original curve at each point.

If y = xⁿ, then dy/dx = nxⁿ⁻¹ for the polynomial powers used here. The notation matters: dy/dx is not a decorative label. It names the rate at which y changes with x.

2. Algebra is the carrying structure of calculus

Before differentiating, simplify where it makes the structure clearer. Expand only when useful. Preserve exact coefficients. Watch negative signs. A student can understand differentiation perfectly and still lose the question through fragile algebra.

For example, if y = 3x⁴ − 5x² + 2x, then

dy/dx = 12x³ − 10x + 2.

The calculus rule is short. Reliable execution still depends on the algebra around it.

3. Tangents connect differentiation back to coordinate geometry

If the curve is y = f(x), then the tangent gradient at x = a is f′(a). Once the gradient and point are known, the tangent becomes an ordinary straight-line equation.

This is an important cross-topic bridge: calculus supplies the local gradient; coordinate geometry constructs the line.

4. Worked Example 1 — equation of a tangent

Find the equation of the tangent to y = x³ − 2x at x = 2.

Differentiate:

dy/dx = 3x² − 2.

At x = 2, gradient = 12 − 2 = 10. The point is y = 8 − 4 = 4, so (2, 4).

Using point-gradient form:

y − 4 = 10(x − 2), therefore y = 10x − 16.

5. Stationary points occur where the derivative is zero

A stationary point occurs when the tangent is horizontal, so dy/dx = 0. This condition identifies candidates. We still need to determine what kind of stationary point each candidate is.

Possible methods include inspecting the sign of the first derivative on either side or using the second derivative when appropriate.

6. Worked Example 2 — stationary points and classification

Find and classify the stationary points of y = x³ − 3x² − 9x + 5.

Differentiate:

dy/dx = 3x² − 6x − 9 = 3(x − 3)(x + 1).

Set dy/dx = 0, giving x = −1 or x = 3.

Coordinates:

• At x = −1, y = −1 − 3 + 9 + 5 = 10.
• At x = 3, y = 27 − 27 − 27 + 5 = −22.

Second derivative: d²y/dx² = 6x − 6.

• At x = −1, second derivative = −12 < 0, so (−1, 10) is a local maximum.
• At x = 3, second derivative = 12 > 0, so (3, −22) is a local minimum.

7. Optimisation is modelling plus calculus

An optimisation question usually has three stages. First build the quantity to be maximised or minimised as a function of one variable. Then differentiate and find stationary candidates. Finally interpret and verify the candidate in the original constraints.

Geometry or context → one-variable objective → derivative zero → candidate → check → interpretation.

8. Worked Example 3 — maximise an area

A rectangle is symmetric about the y-axis with its upper corners on the curve y = 12 − x² for x > 0 and its base on the x-axis. Find the maximum area.

The width is 2x and the height is 12 − x². Therefore

A = 2x(12 − x²) = 24x − 2x³.

Differentiate:

dA/dx = 24 − 6x².

Set to zero: 24 − 6x² = 0, so x² = 4 and x = 2 because x > 0.

Maximum area = 2(2)(12 − 4) = 32 square units. The negative second derivative for x > 0 confirms the maximum.

9. Integration reverses differentiation, but not completely

If differentiation removes constants, reversing it cannot recover which constant was originally present. Therefore indefinite integration includes an arbitrary constant C.

For example:

∫(6x² − 4x + 3) dx = 2x³ − 2x² + 3x + C.

Differentiating the answer returns the integrand, which is an immediate check.

10. Worked Example 4 — recover a function from its derivative

Given f′(x) = 6x − 4 and f(1) = 3, find f(x).

Integrate:

f(x) = 3x² − 4x + C.

Use f(1) = 3:

3 − 4 + C = 3, so C = 4.

Therefore f(x) = 3x² − 4x + 4.

11. Definite integration and signed area

A definite integral measures signed accumulation. Where the graph lies above the x-axis, contribution is positive; below the x-axis, contribution is negative. Therefore a definite integral is not automatically the same as total geometric area.

For a function that remains non-negative on the interval, the area under the curve from a to b is ∫_a^b f(x) dx.

12. Worked Example 5 — area under a curve

Find the area under y = 4x − x² from x = 0 to x = 2.

The function is non-negative on this interval, so

Area = ∫₀²(4x − x²)dx

= [2x² − x³/3]₀²

= 8 − 8/3 = 16/3 square units.

13. Kinematics gives calculus a physical interpretation

If displacement is s(t), then

• v = ds/dt is velocity.
• a = dv/dt = d²s/dt² is acceleration.

Integration runs the other way. If acceleration is known, integrate to recover velocity; integrate velocity to recover displacement. Initial conditions determine the missing constants.

14. Velocity is signed; speed is not

A negative velocity does not mean the particle is moving slowly. It means the motion is in the negative chosen direction. Speed is |v|. This distinction is essential when calculating total distance travelled.

A particle can have zero velocity at an instant and still have non-zero acceleration. That means it is momentarily at rest while its velocity is changing.

15. Worked Example 6 — displacement, velocity, acceleration and total distance

A particle has displacement

s = t³ − 6t² + 9t + 2

for 0 ≤ t ≤ 4.

Velocity:

v = 3t² − 12t + 9 = 3(t − 1)(t − 3).

Acceleration:

a = 6t − 12.

The particle is at rest at t = 1 and t = 3. To find total distance, evaluate position at the endpoints and turning times:

• s(0) = 2
• s(1) = 6
• s(3) = 2
• s(4) = 6

Distance = |6 − 2| + |2 − 6| + |6 − 2| = 12 units.

By contrast, displacement from t = 0 to 4 is s(4) − s(0) = 4 units. Distance and displacement answer different questions.

16. Worked Example 7 — recover motion from acceleration

A particle has acceleration a = 6t − 4. At t = 0, its velocity is 5 and displacement is 2. Find v and s.

Integrate acceleration:

v = 3t² − 4t + C.

Since v(0) = 5, C = 5. Thus v = 3t² − 4t + 5.

Integrate again:

s = t³ − 2t² + 5t + K.

Since s(0) = 2, K = 2. Therefore s = t³ − 2t² + 5t + 2.

17. Common failure patterns

• Differentiating correctly but substituting into the wrong expression.
• Finding x-coordinates of stationary points but not their y-coordinates.
• Assuming dy/dx = 0 automatically means a maximum.
• Forgetting +C during indefinite integration.
• Using an initial condition before integrating instead of after the constant appears.
• Confusing velocity with speed.
• Calculating displacement when the question asks for total distance.
• Ignoring sign changes in velocity.
• Reporting an optimisation candidate without checking the domain or whether it gives the required maximum/minimum.
• Premature approximation. Exact algebra is often safer until interpretation requires a decimal.

18. A reliable calculus routine

1. Identify the quantity and variable.
2. Simplify the algebra only as far as useful.
3. Differentiate or integrate carefully.
4. Write the condition being imposed: tangent point, stationary point, initial value, boundary or time.
5. Solve the resulting algebra.
6. Recover any missing coordinate or constant.
7. Classify or interpret the result.
8. Check by differentiating an integral, substituting a condition or inspecting the sign/graph.
9. Return to the units and meaning of the original problem.

19. Practice set

1. Differentiate y = 5x⁴ − 3x² + 7.
2. Find the gradient of y = 2x³ + x at x = 2.
3. Find the tangent to y = x² + 3x − 1 at x = 1.
4. Find the stationary x-values of y = x³ − 6x² + 9x.
5. Integrate 8x³ − 6x + 2.
6. Given f′(x) = 4x + 3 and f(0) = −2, find f(x).
7. Evaluate ∫₀³2x dx.
8. A particle has s = 2t³ − 3t² + 4. Find v and a.
9. For the motion in Question 8, find the times when the particle is at rest.
10. If a = 4t and v(0) = 3, find v.
11. Explain why total distance may exceed the magnitude of displacement.
12. Explain why +C is necessary after indefinite integration.

Answers

1. 20x³ − 6x.
2. dy/dx = 6x² + 1, so gradient 25.
3. At x = 1, point (1, 3), gradient 5; tangent y = 5x − 2.
4. dy/dx = 3(x − 1)(x − 3), so x = 1 and x = 3.
5. 2x⁴ − 3x² + 2x + C.
6. f(x) = 2x² + 3x − 2.
7. 9.
8. v = 6t² − 6t; a = 12t − 6.
9. t = 0 and t = 1.
10. v = 2t² + 3.
11. Because distance adds the lengths of all travelled segments, while displacement compares only final and initial positions with direction.
12. Because differentiation removes constants, so the antiderivative represents a family of functions until a condition fixes the constant.

20. What mastery looks like

Mastery means the student can explain derivative as rate and tangent gradient, construct and classify stationary points, build an optimisation function from a context, integrate with the correct constant, distinguish signed accumulation from geometric area, and move fluently among displacement, velocity and acceleration.

The decisive transfer test is whether the student can survive a change of wrapper. A calculus question may arrive as coordinate geometry, motion, optimisation or a graph. If the learner still identifies the quantity that changes, the relevant rate or accumulation, the necessary algebra and the correct interpretation, calculus has become part of a connected mathematical system.

---

Continue through the BTT Additional Mathematics system

• Additional Mathematics Directory.
• BTT Singapore Mathematics Hub.
• Guide 1: Quadratics, Coordinate Geometry and Tangency.
• Guide 2: Exponentials, Logarithms, Linear Law and Modelling.
• Guide 3: Trigonometric Functions, Identities, Equations and Geometry.