Instantaneous Rates of Change and Tangents

Key Terms

Instantaneous rate of change: : f′(a) is the instantaneous rate of change of f at x = a. It equals the gradient of the tangent to the curve at that point.
Tangent line at (a, f(a)): : the line that just touches the curve at x = a with the same gradient. Equation: y − f(a) = f′(a)(x − a).
Normal line at (a, f(a)): : the line perpendicular to the tangent at the same point. Its gradient is −1⁄f′(a) (provided f′(a) ≠ 0).
Steps for tangent/normal: : (1) differentiate to find f′(x); (2) substitute x = a to find the gradient; (3) find y-coordinate: f(a); (4) use point-slope form y − y₁ = m(x − x₁).

📚 Key Formulas
Tangent at x = a: y − f(a) = f′(a)(x − a)
Normal gradient: m_normal = −1 ⁄ f′(a)
Normal at x = a: y − f(a) = −¹⁄_f′(a) · (x − a)

Hot Tip The tangent and normal are always perpendicular to each other at the point of contact. If the tangent gradient is m, the normal gradient is −1/m. If m = 0 (horizontal tangent), the normal is vertical (undefined slope).

Worked Example 1 — Equation of a Tangent

Question: Find the equation of the tangent to y = x² − 3x + 2 at x = 2.

Step 1 — Differentiate: f′(x) = 2x − 3

Step 2 — Gradient at x = 2: f′(2) = 2(2) − 3 = 1

Step 3 — y-coordinate: f(2) = 4 − 6 + 2 = 0, so the point is (2, 0).

Step 4 — Point-slope form: y − 0 = 1(x − 2)

Answer: y = x − 2

Worked Example 2 — Equation of a Normal

Question: Find the equation of the normal to y = x³ at x = 1.

Step 1 — Differentiate: f′(x) = 3x²

Step 2 — Tangent gradient at x = 1: f′(1) = 3(1)² = 3

Step 3 — Normal gradient: m_normal = −1⁄3

Step 4 — y-coordinate: f(1) = 1³ = 1, so the point is (1, 1).

Step 5 — Point-slope form: y − 1 = −¹⁄₃(x − 1)

Answer: y = −x⁄3 + 4⁄3

Introduction: Average vs Instantaneous Rate of Change

Imagine a car driving along a motorway. The average speed over a 2-hour journey is total distance divided by 2. But your speedometer shows your instantaneous speed — the speed at one precise moment. The derivative gives us exactly this: the instantaneous rate of change at a single point.

For f(x), the average rate of change from x = a to x = b is:

(f(b) − f(a)) ⁄ (b − a) — this is the gradient of a secant (chord).

As b → a, this approaches f′(a) — the instantaneous rate of change, and the secant becomes the tangent.

What the Tangent Line Is

The tangent line at x = a is the straight line that:

passes through the point (a, f(a)) on the curve, and
has gradient equal to f′(a).

It represents the “best linear approximation” to the curve near x = a.

Step-by-Step Method: Tangent Line

Find f′(x) by differentiating.
Substitute x = a to find the gradient m = f′(a).
Find the y-coordinate: y₁ = f(a).
Write the tangent equation: y − y₁ = m(x − a).
Simplify to y = mx + c form if required.

Step-by-Step Method: Normal Line

Follow steps 1–3 above.
Normal gradient = −1⁄f′(a) (negative reciprocal).
Write: y − f(a) = (−1⁄f′(a))(x − a).

Worked Example — Tangent to a Cubic

Find the equation of the tangent to y = x³ − 2x at x = −1.

Step 1: f′(x) = 3x² − 2

Step 2: f′(−1) = 3(1) − 2 = 1

Step 3: f(−1) = (−1)³ − 2(−1) = −1 + 2 = 1. Point: (−1, 1).

Step 4: y − 1 = 1(x − (−1)) → y = x + 2

Worked Example — Tangent Parallel to a Given Line

Find the point(s) on y = x³ where the tangent is parallel to y = 3x − 1.

Step 1: Parallel lines have equal gradients. Target gradient = 3.

Step 2: f′(x) = 3x². Set 3x² = 3 → x² = 1 → x = ±1.

Step 3: f(1) = 1, f(−1) = −1. Points: (1, 1) and (−1, −1).

Worked Example — Tangent through a Point not on the Curve

Find the equation of the tangent to y = x² that passes through (0, −4).

Step 1: Let the point of tangency be (a, a²). f′(x) = 2x, so tangent gradient = 2a.

Step 2: Tangent equation: y − a² = 2a(x − a) → y = 2ax − a².

Step 3: Substitute (0, −4): −4 = 2a(0) − a² → a² = 4 → a = ±2.

Step 4: a = 2: y = 4x − 4; a = −2: y = −4x − 4. Two tangent lines.

💡 Key Reminder: The gradient of the tangent is f′(a), not f(a). Always differentiate first, then substitute.

Summary

The derivative f′(a) is simultaneously: (1) the instantaneous rate of change of f at x = a, and (2) the gradient of the tangent to the curve y = f(x) at x = a. Use this to find tangent equations (same gradient, same point) and normal equations (negative reciprocal gradient, same point).

Mastery Practice

See Answers ➔

Fluency
Given y = x² − 4x, find f′(x) and evaluate the instantaneous rate of change at each value.
1. (a) x = 1
2. (b) x = 3
3. (c) x = 0
Fluency
Find the equation of the tangent to each curve at the given point.
1. (a) y = x² at x = 3
2. (b) y = 2x³ at x = 1
3. (c) y = x² − 2x + 1 at x = 2
4. (d) y = 4⁄x at x = 2
Fluency
Find the equation of the normal to each curve at the given point.
1. (a) y = x² + 1 at x = 2
2. (b) y = x³ − x at x = 1
Understanding
Find the point(s) on y = x² − 6x where the tangent is parallel to y = 2x − 1.
Understanding
A tangent to y = x³ − 3x passes through the origin. Find the point(s) of tangency.

Hint: If the tangent at x = a passes through the origin, its equation y = f′(a)·x must also satisfy f(a) = f′(a)·a.
Understanding
Find the x-value(s) where the curves y = x² and y = −x² + 4x have equal gradients (parallel tangents).
Understanding
A ball is launched upward with height (in metres) given by h = −5t² + 20t, where t is time in seconds.
1. (a) Find the velocity function v(t) = h′(t).
2. (b) Find the velocity at t = 2 s. What does this tell you about the ball at this moment?
3. (c) At what time is the instantaneous velocity zero?
Problem Solving
The tangent to y = ax² at x = 2 has equation y = 8x − 8. Find the value of a.
Problem Solving
Find the equations of the tangent and normal to y = x² at x = 1. Then find the coordinates of the point where the tangent and normal intersect each other (other than at the point of tangency).
Problem Solving
A curve y = x³ + bx passes through the point (2, k) and has a tangent gradient of 15 at that point. Find the values of b and k.

Instantaneous Rates of Change and Tangents

Key Terms

Worked Example 1 — Equation of a Tangent

Worked Example 2 — Equation of a Normal

Introduction: Average vs Instantaneous Rate of Change

What the Tangent Line Is

Step-by-Step Method: Tangent Line

Step-by-Step Method: Normal Line

Summary

Mastery Practice

Given y = x² − 4x, find f′(x) and evaluate the instantaneous rate of change at each value.

Find the equation of the tangent to each curve at the given point.

Find the equation of the normal to each curve at the given point.

Find the point(s) on y = x² − 6x where the tangent is parallel to y = 2x − 1.

A tangent to y = x³ − 3x passes through the origin. Find the point(s) of tangency.

Find the x-value(s) where the curves y = x² and y = −x² + 4x have equal gradients (parallel tangents).

A ball is launched upward with height (in metres) given by h = −5t² + 20t, where t is time in seconds.

The tangent to y = ax² at x = 2 has equation y = 8x − 8. Find the value of a.

Find the equations of the tangent and normal to y = x² at x = 1. Then find the coordinates of the point where the tangent and normal intersect each other (other than at the point of tangency).

A curve y = x³ + bx passes through the point (2, k) and has a tangent gradient of 15 at that point. Find the values of b and k.