Applications of derivatives for NDA

This is where differentiation earns its keep. The NDA syllabus names increasing and decreasing functions and maxima and minima explicitly, and questions on the slope of a tangent or a rate of change use exactly the same skill. Each type follows a short fixed routine. Learn the routine, and a question that looks like a word problem becomes two lines of calculus.

Typical question types are:

• the slope or equation of a tangent or normal at a point;
• the point where the tangent is parallel or perpendicular to a given line;
• the interval where a function is increasing or decreasing;
• local maxima and minima, and the values there;
• the greatest and least values on a closed interval;
• word problems: largest area, largest volume, least sum;
• related rates, such as how fast an area grows as a radius grows.

Rate of change

dy/dx is the rate at which y changes per unit change in x. When two quantities both change with time, link them by the chain rule: dy/dt = (dy/dx) × (dx/dt). For a circle, A = πr², so dA/dt = 2πr × dr/dt. For a sphere, V = (4/3)πr³, so dV/dt = 4πr² × dr/dt.

Tangents and normals

At the point (x₁, y₁) on y = f(x), the tangent has slope m = f′(x₁).

• Tangent: y − y₁ = m(x − x₁).
• Normal (perpendicular to the tangent): y − y₁ = (−1/m)(x − x₁).
• The tangent is horizontal where f′(x) = 0, and vertical where dx/dy = 0.
• A tangent parallel to a line of slope k touches where f′(x) = k.

Increasing and decreasing functions

• f′(x) > 0 on an interval: f is increasing there.
• f′(x) < 0 on an interval: f is decreasing there.

A function can still be increasing when f′ is zero at isolated points. x³ has f′(0) = 0 but increases everywhere. To find the intervals, solve f′(x) = 0, mark the roots on a number line, and test the sign of f′ in each piece.

Maxima and minima

Critical points are where f′(x) = 0 (or where f′ does not exist). They are the candidates for local extremes.

First derivative test. Watch the sign of f′ as x passes through the critical point c:

Second derivative test. At a critical point c:

• f″(c) < 0: local maximum;
• f″(c) > 0: local minimum;
• f″(c) = 0: the test fails, so go back to the first derivative test.

Absolute (global) extremes on a closed interval [a, b]. Evaluate f at every critical point inside the interval and at both end points. The largest value is the absolute maximum and the smallest the absolute minimum. For f(x) = x³ − 3x on [0, 2]: the critical point x = 1 gives −2, and the end points give f(0) = 0 and f(2) = 2. The maximum is 2 and the minimum −2.

Standard optimisation results

These come up so often that it pays to know the answer before you start.

The last one is worth deriving once. V = x(a − 2x)², so V′ = (a − 2x)² − 4x(a − 2x) = (a − 2x)(a − 6x). The root x = a/2 gives zero volume, which leaves x = a/6.

Worked NDA-style MCQs

Q1. The equation of the tangent to y = x² − 2x + 3 at (2, 3) is:
(a) y = 2x − 1 (b) y = 2x + 1 (c) x + 2y = 8 (d) y = x + 1

dy/dx = 2x − 2, which is 2 at x = 2. Tangent: y − 3 = 2(x − 2), so y = 2x − 1. Option (c) is the normal, y − 3 = −½(x − 2). Answer: (a).

Q2. The point on y = x² where the tangent is parallel to y = 4x + 1 is:
(a) (4, 16) (b) (2, 4) (c) (1, 1) (d) (−2, 4)

We need 2x = 4, so x = 2 and y = 4. Answer: (b).

Q3. f(x) = 2x³ − 9x² + 12x + 5 is decreasing on:
(a) (−∞, 1) (b) (1, 2) (c) (2, ∞) (d) (0, 3)

f′(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2), which is negative between the roots. Answer: (b).

Q4. The local maximum value of f(x) = x³ − 6x² + 9x + 1 is:
(a) 1 (b) 3 (c) 5 (d) 9

f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), and f″(x) = 6x − 12. At x = 1, f″ = −6 < 0, so there is a maximum: f(1) = 1 − 6 + 9 + 1 = 5. (At x = 3 there is a minimum, f(3) = 1.) Answer: (c).

Q5. An open box is made from an 18 cm square sheet by cutting equal squares from the corners and folding up the sides. The greatest possible volume is:
(a) 216 cm³ (b) 432 cm³ (c) 486 cm³ (d) 324 cm³

The cut is 18/6 = 3 cm, leaving a 12 cm base. V = 3 × 12 × 12 = 432 cm³. Answer: (b).

Q6. The greatest value of f(x) = x e⁻ˣ for x > 0 is:
(a) 1 (b) e (c) 1/e (d) 2/e²

f′(x) = e⁻ˣ − x e⁻ˣ = e⁻ˣ(1 − x), which is zero at x = 1. f′ changes from + to − there, so f(1) = 1/e is the maximum. Answer: (c).

Common mistakes

• Using the tangent's slope for the normal. The normal's slope is −1/m.
• Stopping at f′ = 0. A critical point is only a candidate; test it.
• Ignoring the end points in closed-interval questions.
• Reporting x instead of f(x). Read whether the question asks for the point or the value.
• Dropping units or the chain factor in related-rates questions.

Practice set

1. The slope of the tangent to y = x³ at x = 2 is: (a) 8 (b) 6 (c) 12 (d) 4
2. The least value of x² − 6x + 10 is: (a) 10 (b) 1 (c) 3 (d) −1
3. The critical points of 2x³ − 6x are: (a) x = 0 (b) x = ±1 (c) x = ±√3 (d) x = 1 only
4. x² − 4x is decreasing for: (a) x > 2 (b) x < 2 (c) x < 0 (d) all x
5. The radius of a spherical balloon grows at 0.5 cm/s. When the radius is 10 cm, its volume grows at: (a) 200π cm³/s (b) 400π cm³/s (c) 100π cm³/s (d) 20π cm³/s
6. The least value of x + 1/x for x > 0 is: (a) 0 (b) 1 (c) 2 (d) 4
7. The slope of the normal to y = x² at (1, 1) is: (a) 2 (b) −2 (c) 1/2 (d) −1/2
8. f(x) = x³ + 3x + 5 is: (a) increasing for all x (b) decreasing for all x (c) increasing only for x > 0 (d) neither

Answers:

1. (c). 3x² at x = 2.
2. (b). f′ = 2x − 6 = 0 at x = 3; f(3) = 9 − 18 + 10 = 1.
3. (b). 6x² − 6 = 0.
4. (b). f′ = 2x − 4 < 0.
5. (a). 4π × 10² × 0.5 = 200π.
6. (c). At x = 1, by AM ≥ GM or by f′ = 1 − 1/x² = 0.
7. (d). The tangent slope is 2.
8. (a). f′ = 3x² + 3 > 0 for every x.

What to do next

• Write the optimisation table from memory and derive the open-box result once more.
• Solve 25 old NDA questions on this chapter, and write the routine (tangent, monotonic, extreme, rate) at the top of each before solving.
• If any derivative slowed you down, go back to differentiation. For least values of quadratics without calculus, compare the vertex method in quadratic equations.