Tangent and normal to a curve
A cyclist leaving a curved road at a bend goes along the tangent; the line perpendicular to it points toward the centre of the bend, like the normal.
At (x₁, y₁) on y = f(x) the tangent has slope m = f′(x₁); its equation is y − y₁ = m(x − x₁). The normal is perpendicular to the tangent, with slope −1/m, so y − y₁ = −(1/m)(x − x₁).
Rule 1: Tangent — Slope m = dy/dx at the point; y − y₁ = m(x − x₁).
Rule 2: Normal — Slope −1/m (if m ≠ 0); horizontal tangent ⇔ vertical normal.
How is the formula obtained? : y = x² − 3x + 2 at (2, 0): dy/dx = 2x − 3 = 1. Tangent: y = x − 2. Normal slope −1: y = −x + 2.
- Easy : y = x³ at (1, 1): slope 3; tangent y − 1 = 3(x − 1), i.e. y = 3x − 2.
- Medium : Normal of y = x³ at (1, 1): slope −1/3, so y − 1 = −(1/3)(x − 1).
- Exam-type : Tangent to y = x² − 3x + 2 is parallel to y = 5x at 2x − 3 = 5, i.e. x = 4.
Parallel to the x-axis means dy/dx = 0; parallel to the y-axis means dy/dx is infinite.
Using the normal slope as m instead of −1/m.
Find the point first (put x in y), then the slope, then the line. Memory clue : "Point, slope, line."
If the point is not given on the curve, first check that it lies on it.
CET-style example : Slope of the normal to y = x³ at (1, 1) is (A) 3 (B) −3 (C) 1/3 (D) −1/3
Answer : (D). Time-saving method : Slope of tangent 3, so normal is −1/3.
Spot the difference
| Line | Slope at (x₁, y₁) |
|---|---|
| Tangent | f′(x₁) |
| Normal | −1/f′(x₁) |
Practice questions
- 1Slope of tangent to y = x² at x = 3?
- 2Slope of tangent to y = x³ at x = 1?
- 3Slope of normal to y = x³ at x = 1?
- 4Tangent to y = x² − 3x + 2 at (2, 0).
- 5Normal to y = x² − 3x + 2 at (2, 0).
- 6Where is the tangent to y = x² − 3x + 2 parallel to y = 5x?
- 7Tangent to y = x² − 4x + 3 horizontal at which point?
- 8Tangent to y = x³ at x = 2 meets the x-axis where?
Show answers (try them yourself first)
- 16.
- 23.
- 3−1/3.
- 4y = x − 2.
- 5y = −x + 2.
- 6x = 4, point (4, 6).
- 72x − 4 = 0 ⇒ x = 2; point (2, −1).
- 8y = 8 + 12(x − 2); y = 0 ⇒ x = 4/3.
Increasing and decreasing functions
Water level in a tank that is filling is an increasing function of time; when the tap is closed and the tank drains it is decreasing.
f is increasing on an interval if f′(x) > 0 there and decreasing if f′(x) < 0. Points where f′(x) = 0 separate such intervals.
Rule 1: Test — Find f′(x), solve f′(x) = 0, then check the sign of f′ in each interval.
How is the formula obtained? : f = x³ − 6x² + 9x: f′ = 3x² − 12x + 9 = 3(x − 1)(x − 3). Positive for x < 1 and x > 3; negative for 1 < x < 3.
- Easy : f = x² is decreasing for x < 0 and increasing for x > 0.
- Medium : f = 2x − 5 has f′ = 2 > 0, so increasing everywhere.
- Exam-type : f = x³ − 6x² + 9x increases on (−∞, 1) and (3, ∞); decreases on (1, 3).
Sign of f′ decides; the value of f does not.
Choosing the interval where f is positive instead of where f′ is positive.
Draw a sign line for f′ using its roots. Memory clue : "Derivative positive, function climbs."
A point where f′ = 0 may not change sign (f = x³ at 0 is increasing both sides).
CET-style example : f(x) = x³ − 6x² + 9x is decreasing in (A) (1, 3) (B) (0, 1) (C) (3, 5) (D) (−∞, 1)
Answer : (A). Time-saving method : Factor f′ and test signs.
Spot the difference
| f′(x) | f is |
|---|---|
| > 0 | increasing |
| < 0 | decreasing |
| = 0 | stationary point |
Practice questions
- 1Is f = 3x + 1 increasing or decreasing?
- 2f = −2x: increasing or decreasing?
- 3Where is f = x² increasing?
- 4Find f′ for f = x³ − 6x² + 9x.
- 5Where is f = x³ − 6x² + 9x decreasing?
- 6Where is f = x³ − 3x increasing?
- 7Show f = x³ + x is increasing for all x.
- 8Find k so that f = x² − kx is increasing for x > 2.
Show answers (try them yourself first)
- 1Increasing; f′ = 3.
- 2Decreasing; f′ = −2.
- 3x > 0.
- 43x² − 12x + 9 = 3(x − 1)(x − 3).
- 51 < x < 3.
- 6f′ = 3(x² − 1) > 0 ⇒ x < −1 or x > 1.
- 7f′ = 3x² + 1 > 0 always.
- 8f′ = 2x − k ≥ 0 for x > 2 ⇒ k ≤ 4.
Maxima and minima
A farmer with 100 m of fencing along a wall wants the largest possible enclosed area. The best size is at a peak of the area function.
A stationary point has f′(x) = 0. It is a local maximum if f″ < 0 and a local minimum if f″ > 0 (second-derivative test); if f″ = 0 use the sign change of f′.
Rule 1: First derivative test — f′ changes + to − ⇒ maximum; − to + ⇒ minimum.
Rule 2: Second derivative test — f′(c) = 0 and f″(c) < 0 ⇒ maximum; f″(c) > 0 ⇒ minimum.
How is the formula obtained? : f = x³ − 6x² + 9x: f′ = 0 at x = 1, 3. f″ = 6x − 12: f″(1) = −6 < 0 (maximum, f = 4), f″(3) = 6 > 0 (minimum, f = 0).
- Easy : f = x² − 4x: f′ = 2x − 4 = 0 ⇒ x = 2; f″ = 2 > 0 ⇒ minimum value −4.
- Medium : Two numbers add up to 20; product x(20 − x) is largest at x = 10, product 100.
- Exam-type : Fence 100 m on three sides of a field along a wall: 2x + y = 100, A = x(100 − 2x); A′ = 100 − 4x = 0 ⇒ x = 25, y = 50, A = 1250 m².
Write the quantity to optimise as a function of one variable first.
Stopping at f′ = 0 without checking whether it is a maximum or a minimum.
Use the constraint to eliminate one variable. Memory clue : "Second derivative negative: hill (max)."
Check the end points too if the interval is closed.
CET-style example : Maximum area with 100 m of fence on three sides along a wall (A) 1250 (B) 625 (C) 2500 (D) 1000
Answer : (A). x = 25, y = 50. Time-saving method : Differentiate A = x(100 − 2x).
Spot the difference
| f″(c) | Conclusion |
|---|---|
| < 0 | local maximum |
| > 0 | local minimum |
| = 0 | test fails; use sign of f′ |
Practice questions
- 1Stationary points of f = x² − 4x?
- 2Minimum value of f = x² − 4x?
- 3f = 5 − x²: maximum value?
- 4f = x³ − 6x² + 9x: find local maximum value.
- 5Same f: local minimum value.
- 6Two numbers with sum 20 and largest product?
- 7Fence 100 m on three sides: maximum area?
- 8Rectangle with perimeter 40 cm: largest area?
Show answers (try them yourself first)
- 1x = 2.
- 2−4 at x = 2.
- 35 at x = 0.
- 44 at x = 1.
- 50 at x = 3.
- 610, 10; product 100.
- 71250 m² (25 m by 50 m).
- 8A = x(20 − x); x = 10; area 100 cm².
Approximations and mean value theorems
If the speedometer reads 50 km/h you can estimate the distance covered in the next few seconds without a new measurement, using the current rate.
For small Δx: f(x + Δx) ≈ f(x) + f′(x)Δx. Rolle's theorem: if f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b) then f′(c) = 0 for some c. Lagrange's theorem: f′(c) = [f(b) − f(a)]/(b − a) for some c in (a, b).
Rule 1: Approximation — f(x + Δx) ≈ f(x) + f′(x)Δx, with x chosen so that f(x) is easy.
Rule 2: Mean value — The conditions (continuity, differentiability) must be checked before the theorem is used.
How is the formula obtained? : ∛8.1: f(x) = x^{1/3}, x = 8, Δx = 0.1, f′(8) = 1/(3·4) = 1/12. So ∛8.1 ≈ 2 + 0.1/12 ≈ 2.0083 (calculator: 2.00830).
- Easy : √25.2 ≈ 5 + 0.2/(2·5) = 5.02.
- Medium : Rolle on f = x² − 4x + 3, [1, 3]: f(1) = f(3) = 0; f′ = 2x − 4 = 0 ⇒ c = 2.
- Exam-type : Lagrange on f = x², [1, 3]: slope (9 − 1)/2 = 4 = 2c ⇒ c = 2.
Approximation works best when Δx is small.
Applying Rolle's theorem when f(a) ≠ f(b).
Choose the nearest perfect square or cube. Memory clue : "Nearby easy value plus slope times the gap."
Rolle's theorem fails for |x| on [−1, 1] because it is not differentiable at 0.
CET-style example : Approximate value of ∛8.1 (A) 2.0083 (B) 2.1 (C) 2.01 (D) 2.083
Answer : (A). Time-saving method : f′(8) = 1/12.
Spot the difference
| Theorem | Conclusion |
|---|---|
| Rolle | f′(c) = 0 |
| Lagrange | f′(c) = [f(b) − f(a)]/(b − a) |
Practice questions
- 1Approximate √25.2.
- 2Approximate √16.1.
- 3Approximate ∛8.1.
- 4Approximate (1.02)².
- 5Find c in Rolle for f = x² − 4x + 3 on [1, 3].
- 6Find c in Lagrange for f = x² on [1, 3].
- 7Find c in Lagrange for f = x³ on [0, 2].
- 8Why does Rolle fail for f = |x| on [−1, 1]?
Show answers (try them yourself first)
- 15.02.
- 24 + 0.1/8 = 4.0125.
- 32.0083.
- 41 + 2(0.02) = 1.04 (exact 1.0404).
- 5c = 2.
- 6c = 2.
- 73c² = (8 − 0)/2 = 4 ⇒ c = 2/√3 ≈ 1.155.
- 8f is not differentiable at x = 0.