All skills Math

Learn Index · Calculus Grade 11–12

Tangent lines and instantaneous rates of change

The slope of the tangent at a point is the instantaneous rate of change.

Read

The tangent line

The tangent line to a curve at a point just touches the curve there and has the same direction. Its slope equals the derivative: the instantaneous rate of change at that point.

Finding the equation of a tangent line

  1. Find the point of tangency (x₁, f(x₁)).
  2. Find the slope m = f′(x₁).
  3. Use point-slope form: y − y₁ = m(x − x₁).

Example: f(x) = x² − 3x + 1 at x = 2

  1. f(2) = 4 − 6 + 1 = −1 → point (2, −1)
  2. f′(x) = 2x − 3 → m = f′(2) = 1
  3. y + 1 = 1(x − 2) → y = x − 3

Normal lines

The normal line is perpendicular to the tangent at the point of tangency. Its slope is the negative reciprocal −1/m.

From the example: normal slope = −1 → y + 1 = −(x − 2) → y = −x + 1

Horizontal tangents

A horizontal tangent has slope 0. Solve f′(x) = 0.

Example: f(x) = x³ − 3x

  • f′(x) = 3x² − 3 = 0 → x = ±1
  • Horizontal tangents at (1, −2) and (−1, 2)

These are often local maximums or minimums.

Tangents with a given slope

Find where y = x² has a tangent parallel to y = 6x − 1.

  • 2x = 6 → x = 3 → point (3, 9)

Instantaneous rate of change in context

Example: A ball’s height is h(t) = −4.9t² + 20t + 1.5 (m).

h′(t) = −9.8t + 20 → h′(1) = 10.2 m/s (rising)

h′(3) = −9.4 m/s (falling)

Linear approximation

Near the point of tangency, the tangent line is a good estimate for the curve. This is linear approximation.

Example: Estimate √4.1 using f(x) = √x at x = 4.

  • f(4) = 2; f′(x) = 1/(2√x) → f′(4) = 0.25
  • Tangent: y = 2 + 0.25(x − 4)
  • At 4.1: y ≈ 2 + 0.025 = 2.025 (actual 2.0248…)

Worked examples

Example 1: Tangent to y = x³ at x = 1. Point (1, 1), slope 3 → y = 3x − 2.

Example 2: Slope of y = eˣ at x = 0? 1.

Example 3: Where does y = x² − 4x have a horizontal tangent? 2x − 4 = 0 → x = 2 (point (2, −4)).

Common mistakes

  • Using f(x) instead of f′(x) for the slope.
  • Using the general derivative instead of evaluating it at the point.
  • Forgetting to find the y-coordinate.

Try it yourself

  1. Tangent to y = 2x² at x = 1.
  2. Normal to y = 2x² at x = 1.
  3. Where does y = x³ − 12x have horizontal tangents?
  4. Use linear approximation to estimate √9.2.

Answers: 1) y = 4x − 2 2) y = −¼x + 2.25 3) x = 2 and x = −2 4) ≈ 3.033

Quick recap

  • The tangent line slope at a point is f′(a), the instantaneous rate of change.
  • Use point-slope form with the point of tangency.
  • The normal line has slope −1/f′(a); tangents give linear approximations.