Rates of Change: Tangents, Normals and Exponential Growth

The Gradient of a Curve at a Point

The gradient of a tangent to a curve at a given point is found by evaluating the derivative of the function at that point. This is one of the most fundamental applications of differential calculus in the NSW HSC Mathematics Advanced course.

If y = f(x), then the derivative f'(x) (also written as dy/dx) gives a gradient function — a formula that outputs the instantaneous gradient at any value of x on the curve.

To find the gradient at a specific point, substitute the x-coordinate of that point into the derivative.

• Step 1: Differentiate f(x) to obtain f'(x).
• Step 2: Substitute the given x-value into f'(x).
• Step 3: The result is the gradient m of the tangent at that point.

Worked Example: Find the gradient of the tangent to y = x³ − 4x + 1 at the point where x = 2.

• Differentiate: dy/dx = 3x² − 4
• Substitute x = 2: dy/dx = 3(2)² − 4 = 3(4) − 4 = 12 − 4 = 8
• Verification: 3 × 4 = 12; 12 − 4 = 8. Confirmed.
• The gradient of the tangent at x = 2 is 8.

Note: the gradient of a curve at a point refers to the gradient of the tangent at that point — the two phrases are used interchangeably in HSC contexts.

Equation of the Tangent Line

Once the gradient m of the tangent is known, the equation of the tangent line can be found using the point-gradient formula:

y − y₁ = m(x − x₁)

where (x₁, y₁) is the point of tangency (the point on the curve where the tangent touches).

Worked Example: Find the equation of the tangent to y = x³ − 4x + 1 at the point where x = 2.

• Find the y-coordinate: Substitute x = 2 into the original curve:
  y = (2)³ − 4(2) + 1 = 8 − 8 + 1 = 1
  Verification: 8 − 8 = 0; 0 + 1 = 1. Confirmed. Point of tangency: (2, 1).
• Gradient: From the previous section, m = 8.
• Apply point-gradient formula:
  y − 1 = 8(x − 2)
  y − 1 = 8x − 16
  y = 8x − 15

The equation of the tangent is y = 8x − 15.

In some HSC questions the point of tangency is given directly as a coordinate pair, and you only need to verify it lies on the curve before proceeding. Always substitute back to check: at x = 2, y = 8(2) − 15 = 16 − 15 = 1. This matches the curve value, confirming the equation is correct.

Tangent lines are linear. Their equation is always in the form y = mx + b or equivalently ax + by + c = 0. The HSC typically accepts either form unless the question specifies otherwise.

Equation of the Normal Line

The normal to a curve at a point is the straight line that is perpendicular to the tangent at that same point. Because perpendicular lines have gradients that are negative reciprocals of each other, if the tangent has gradient m, the normal has gradient:

m_normal = −1/m   (provided m ≠ 0)

Special cases:

• If the tangent is horizontal (m = 0), the normal is vertical, with equation x = x₁.
• If the tangent is vertical (i.e. the curve has an undefined derivative at that point), the normal is horizontal.

The equation of the normal is then found using the same point-gradient formula, but with m_normal.

Worked Example: Using the same curve y = x³ − 4x + 1 at x = 2, find the equation of the normal.

• Tangent gradient: m = 8 (calculated above).
• Normal gradient: m_normal = −1/8
• Point of tangency: (2, 1)
• Apply point-gradient formula:
  y − 1 = −(1/8)(x − 2)
  Multiply through by 8:
  8y − 8 = −x + 2
  x + 8y − 10 = 0
• Verification: substitute (2, 1): 2 + 8(1) − 10 = 2 + 8 − 10 = 0. Confirmed.

The equation of the normal is x + 8y − 10 = 0, or equivalently y = −x/8 + 5/4.

The Derivative as an Instantaneous Rate of Change

Beyond geometry, the derivative carries a powerful physical interpretation: it measures the instantaneous rate of change of one quantity with respect to another at a specific moment or value.

In general, if Q = f(t), then dQ/dt is the instantaneous rate at which Q is changing per unit of t at time t. This is distinct from the average rate of change, which is calculated over an interval:

Average rate = (f(b) − f(a)) / (b − a)

The derivative gives the limiting value of this average as the interval shrinks to zero — hence instantaneous.

Common contexts in HSC Mathematics Advanced:

• Position and velocity: If x(t) is displacement in metres at time t seconds, then v(t) = dx/dt is instantaneous velocity in m/s.
• Volume filling rates: If V(t) is volume in litres, dV/dt is the rate at which volume is changing in L/s (or L/min depending on units of t).
• Population or biological growth: dP/dt gives the instantaneous growth rate of a population.
• Cost and revenue (marginal analysis): dC/dx is the marginal cost — the rate at which cost increases per additional unit produced.

When interpreting the derivative in context, always state the units and the direction of change (positive = increasing, negative = decreasing).

Worked Example — Volume Rate: A spherical balloon is being inflated so that its volume at time t seconds is V(t) = (4/3)π(2t + 1)³ cm³. Find the rate at which the volume is increasing when t = 3 s.

• V is given directly as a function of t, so differentiate with respect to t using the chain rule on the inner function 2t + 1.
• dV/dt = (4/3)π × 3(2t + 1)² × 2 = 8π(2t + 1)²
• At t = 3: 2t + 1 = 7, so dV/dt = 8π(7)² = 8π(49) = 392π cm³/s
• Verification: 8 × 49 = 392. Confirmed.
• The volume is increasing at 392π ≈ 1231.5 cm³/s when t = 3 s.

Exponential Growth and Decay: dQ/dt = kQ

Many quantities change at a rate that is proportional to their current size: a bacterial colony with more cells produces more new cells, and a larger mass of radioactive material loses more atoms each second. This is written as the differential equation

dQ/dt = kQ

where k is a constant called the growth (or decay) constant.

• If k > 0, the rate has the same sign as Q, so a positive quantity grows faster and faster: exponential growth.
• If k < 0, the rate opposes Q, so a positive quantity shrinks, and shrinks more slowly as it gets smaller: exponential decay.

The solution Q = Ae^kt. You are expected to verify that this function satisfies the equation, by substitution:

• Differentiate: dQ/dt = A × ke^kt = k(Ae^kt)
• Since Ae^kt = Q, this gives dQ/dt = kQ, as required.
• At t = 0: Q = Ae⁰ = A, so A is the initial amount, often written Q₀.

Graphs and the asymptote. Every curve Q = Ae^kt has the horizontal asymptote Q = 0 (the t-axis) and Q-intercept A. The curve never crosses the axis, because e^kt > 0 for all t.

Finding k from a half-life or doubling time. The half-life T is the time for a decaying quantity to halve; the doubling time is the time for a growing quantity to double. Neither depends on the starting amount.

• Half-life: Ae^kT = A/2 ⇒ e^kT = 1/2 ⇒ kT = ln(1/2) = −ln 2 ⇒ k = −(ln 2)/T
• Doubling time: e^kT = 2 ⇒ k = (ln 2)/T

Leave k in exact form (for example −(ln 2)/6) for as long as possible. Rounding k early is the most common cause of answers that are slightly out.

Worked Example — Medication Model: A patient is given 400 mg of a medication. The amount Q mg remaining in the body t hours later satisfies dQ/dt = kQ, and the half-life of the medication is 6 hours.

(a) Find the exact value of k.

• Q = 400e^kt, and Q = 200 when t = 6: e^6k = 1/2, so k = −(ln 2)/6 (approximately −0.1155).

(b) Find the amount remaining after 15 hours, correct to one decimal place.

• Q(15) = 400e^{−15(ln 2)/6} = 400 × 2^−2.5 = 400 ÷ 5.6569… ≈ 70.7 mg
• Verification: 2^2.5 = 4 × √2 ≈ 5.657, and 400 ÷ 5.657 ≈ 70.7. Confirmed.

(c) Find the rate at which the medication is being removed when 200 mg remains.

• dQ/dt = kQ = −(ln 2)/6 × 200 ≈ −23.1 mg/h
• The amount is decreasing at about 23.1 mg per hour at that moment.

(d) The medication is only effective while at least 50 mg remains. A doctor plans to give the second dose 20 hours after the first. Is this plan appropriate?

• Solve 400e^kt = 50: e^kt = 1/8, so t = ln(1/8)/k = (−3 ln 2) ÷ (−(ln 2)/6) = 18 hours.
• Verification: 18 hours is exactly three half-lives, and 400 → 200 → 100 → 50. Confirmed.
• Conclusion in context: the amount falls below the effective level 18 hours after the first dose, so a second dose at 20 hours would leave the patient without effective medication for about 2 hours. The plan is not appropriate; the second dose should be given no later than 18 hours after the first.

Contextual Rate of Change Problems: Setting Up and Interpreting

HSC extended response questions frequently embed rate of change problems in real-world scenarios. A systematic approach is essential to avoid errors in setup and interpretation.

Four-step approach for contextual rate problems:

• Identify quantities: Name both the dependent variable (what is changing) and the independent variable (with respect to what). Establish units clearly.
• Form or accept the function: The question usually provides Q = f(t) or a similar relationship.
• Differentiate: Find dQ/dt using the appropriate differentiation rules.
• Evaluate and interpret: Substitute the given value, state the numerical result with correct units, and describe what it means in plain language.

Worked Example — Displacement and Velocity: A particle moves along a straight line so that its displacement from the origin at time t seconds is x = 2t³ − 9t² + 12t metres, for t ≥ 0.

(a) Find the velocity at t = 1 s.

• v(t) = dx/dt = 6t² − 18t + 12
• At t = 1: v = 6(1) − 18(1) + 12 = 6 − 18 + 12 = 0 m/s
• Verification: 6 − 18 = −12; −12 + 12 = 0. Confirmed.
• Interpretation: The particle is momentarily at rest at t = 1 s.

(b) Find the velocity at t = 3 s and state whether the particle is moving toward or away from the origin.

• At t = 3: v = 6(9) − 18(3) + 12 = 54 − 54 + 12 = 12 m/s
• Verification: 54 − 54 = 0; 0 + 12 = 12. Confirmed.
• Position at t = 3: x = 2(27) − 9(9) + 12(3) = 54 − 81 + 36 = 9 m
• Verification: 54 − 81 = −27; −27 + 36 = 9. Confirmed.
• Interpretation: The particle is 9 m to the right of the origin, moving away from it at 12 m/s (positive velocity, positive displacement).

Sign conventions: In motion problems, positive velocity means movement in the positive direction; negative velocity means movement in the negative direction. A particle is at rest when v = 0.

The second derivative d²Q/dt². The second derivative is the rate of change of the rate of change. Its sign tells you whether the rate dQ/dt is itself increasing or decreasing:

A news report saying “unemployment is still rising, but more slowly” is a statement that dU/dt > 0 and d²U/dt² < 0. In motion, d²x/dt² is the acceleration.

Recovering Q from a given rate. If a question gives the rate dQ/dt rather than Q itself, integrate (find the primitive) to obtain Q plus a constant, then use an initial condition to find the constant.

Worked Example — Filling a Tank: Water flows into a tank at a rate of dV/dt = 20 − 2t litres per minute, for 0 ≤ t ≤ 10. Initially the tank holds 50 L.

• Integrate: V = 20t − t² + C
• Initial condition V(0) = 50: 0 − 0 + C = 50, so C = 50 and V = 20t − t² + 50.
• Volume after 10 minutes: V(10) = 200 − 100 + 50 = 150 L. Verification: 200 − 100 = 100; 100 + 50 = 150. Confirmed.
• Second derivative: d²V/dt² = −2 < 0, so the flow rate is decreasing throughout. Since dV/dt = 20 − 2t ≥ 0 on the domain, the tank is filling, but more and more slowly, and the inflow stops at t = 10.

Summary of Key Procedures and HSC Exam Strategy

Pulling together the techniques covered in this note, the table below organises the key procedures for quick reference under exam conditions.

Exam strategy tips:

• Always find the y-coordinate of the point of tangency if not given — a very common source of lost marks is using only the x-value.
• When writing the equation of a line, expand and simplify fully unless the question asks for a specific form.
• In rate of change questions, always write a sentence interpreting your answer in context — HSC marking guidelines award marks for contextual interpretation.
• In growth and decay questions, keep k exact (such as (ln 2)/6) until the final line, and check that the sign of k matches the situation: negative for decay, positive for growth.
• When a question says “show that Q = Ae^kt satisfies…”, differentiate and substitute; do not solve the equation from scratch.
• Include units in every answer that involves a physical quantity (metres, seconds, litres per second, etc.).