Sec 4 · Additional Mathematics

Applications of differentiation

Practise applications of differentiation with step-by-step help. Free account.

1. Tangents and normals

Line at $(x_1, y_1)$	Gradient	Equation
tangent	$m = \dfrac{dy}{dx}$ at $x_1$	$y - y_1 = m(x - x_1)$
normal	$-\dfrac{1}{m}$	$y - y_1 = -\dfrac{1}{m}(x - x_1)$

y = x² + 1 at P(1, 2): gradient 2

\displaystyle \dfrac{dy}{dx} = 2x

The curve y = x² + 1 with the tangent at P(1, 2) and the dashed normal at P, at right angles to the tangent.

e.g.

\displaystyle y - 2 = 2(x - 1)

\displaystyle y = 2x

normal: gradient $-\dfrac{1}{2}$ , so $y = -\dfrac{1}{2}x + \dfrac{5}{2}$

📖 Worked example

Find the equation of the normal to the curve $y = x^{3} - 2x^{2} + 1$ at the point where $x = 1$ .

Step 1 of 5

Differentiate to get the gradient function

\dfrac{dy}{dx} = 3x^{2} - 2(2)x

= 3x^{2} - 4x

Practise this →

2. Rates of change

Rule	Formula
connected rates	$\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$
sphere	$V = \dfrac{4}{3}\pi r^{3}$
	$\dfrac{dV}{dr} = 4\pi r^{2}$
circle	$A = \pi r^{2}$
	$\dfrac{dA}{dr} = 2\pi r$

Sphere: volume increasing at 10 cm³ per second, radius 3 cm

\displaystyle \dfrac{dr}{dt} = \dfrac{dV}{dt} \div \dfrac{dV}{dr}

e.g.

\displaystyle \dfrac{10}{4\pi(3)^{2}}

\displaystyle = \dfrac{5}{18\pi}

📖 Worked example

The volume of a sphere is increasing at a rate of $10$ cm³/s. Find the rate of increase of its radius when the radius is $3$ cm, leaving your answer in terms of $\pi$ .

Step 1 of 4

Write the formula for V in terms of r, and differentiate it

V = \dfrac{4}{3}\pi r^{3}

\dfrac{dV}{dr} = 4\pi r^{2}

Practise this →

3. Increasing and decreasing functions

Function	Condition
increasing	$\dfrac{dy}{dx} > 0$
decreasing	$\dfrac{dy}{dx} < 0$

y = −2x³ − 6x² + 18x + 1 is increasing

\displaystyle \dfrac{dy}{dx} = -6x^{2} - 12x + 18

e.g.

\displaystyle -6x^{2} - 12x + 18 > 0

\displaystyle x^{2} + 2x - 3 < 0

\displaystyle (x + 3)(x - 1) < 0

\displaystyle -3 < x < 1

📖 Worked example

Find the range of values of $x$ for which $y = -2x^{3} - 6x^{2} + 18x + 1$ is increasing.

Step 1 of 6

Differentiate

\dfrac{dy}{dx} = -2(3)x^{2} - 6(2)x + 18(1)

= -6x^{2} - 12x + 18

Practise this →

4. Stationary points

At a stationary point	Maximum	Minimum	Point of inflexion
$\dfrac{dy}{dx} = 0$	$\dfrac{d^{2}y}{dx^{2}} < 0$	$\dfrac{d^{2}y}{dx^{2}} > 0$	$\dfrac{d^{2}y}{dx^{2}} = 0$ (test fails)
sign of $\dfrac{dy}{dx}$	$+$ , $0$ , $-$	$-$ , $0$ , $+$	$+$ , $0$ , $+$ or $-$ , $0$ , $-$

y = x³ − 6x² + 9x + 3

\displaystyle \dfrac{dy}{dx} = 3x^{2} - 12x + 9

e.g.

\displaystyle 3x^{2} - 12x + 9 = 0

\displaystyle x^{2} - 4x + 3 = 0

\displaystyle (x - 1)(x - 3) = 0

\displaystyle x = 1 \text{ or } x = 3

$\dfrac{d^{2}y}{dx^{2}} = 6x - 12$
$x = 1$ : $-6 < 0$ , maximum $(1, 7)$
$x = 3$ : $6 > 0$ , minimum $(3, 3)$

📖 Worked example

Find the coordinates of the maximum point of the curve $y = x^{3} - 6x^{2} + 9x + 3$ .

Step 1 of 5

Differentiate

\dfrac{dy}{dx} = 3x^{2} - 6(2)x + 9(1)

= 3x^{2} - 12x + 9

Practise this →

5. Maximum and minimum problems

Step	Example: $A = 36x - 2x^{2}$
1. write the quantity in one variable	$A = x(36 - 2x)$
2. differentiate and set to $0$	$36 - 4x = 0$
	$x = 9$
3. check with the second derivative	$\dfrac{d^{2}A}{dx^{2}} = -4 < 0$ : maximum
4. find the value	$A = 162$

📖 Worked example

A farmer has $36$ m of fencing to make a rectangular pen against a long straight wall, so the fencing forms only three sides. The two sides perpendicular to the wall are each $x$ m long. Find the maximum area of the pen.

Step 1 of 5

Write the quantity in terms of x, then expand

\text{The side parallel to the wall} = 36 - 2x

A = x(36 - 2x)

A = 36x - 2x^{2}

Practise this →

Try a question: Applications of differentiation

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Practise applications of differentiation with step-by-step help

Endless questions from basic to exam level. Get one wrong and see where it went wrong. Free account.

Start free

Practice questions by skill

Sec 4 · Additional Mathematics

Applications of differentiation

Practise applications of differentiation with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Tangents and normals

Line at $(x_1, y_1)$	Gradient	Equation
tangent	$m = \dfrac{dy}{dx}$ at $x_1$	$y - y_1 = m(x - x_1)$
normal	$-\dfrac{1}{m}$	$y - y_1 = -\dfrac{1}{m}(x - x_1)$

y = x² + 1 at P(1, 2): gradient 2

\displaystyle \dfrac{dy}{dx} = 2x

e.g.

\displaystyle y - 2 = 2(x - 1)

\displaystyle y = 2x

normal: gradient $-\dfrac{1}{2}$ , so $y = -\dfrac{1}{2}x + \dfrac{5}{2}$

📖 Worked example

Find the equation of the normal to the curve $y = x^{3} - 2x^{2} + 1$ at the point where $x = 1$ .

Step 1 of 5

Differentiate to get the gradient function

\dfrac{dy}{dx} = 3x^{2} - 2(2)x

= 3x^{2} - 4x

Practise this →

2. Rates of change

Rule	Formula
connected rates	$\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$
sphere	$V = \dfrac{4}{3}\pi r^{3}$
	$\dfrac{dV}{dr} = 4\pi r^{2}$
circle	$A = \pi r^{2}$
	$\dfrac{dA}{dr} = 2\pi r$

Sphere: volume increasing at 10 cm³ per second, radius 3 cm

\displaystyle \dfrac{dr}{dt} = \dfrac{dV}{dt} \div \dfrac{dV}{dr}

e.g.

\displaystyle \dfrac{10}{4\pi(3)^{2}}

\displaystyle = \dfrac{5}{18\pi}

📖 Worked example

The volume of a sphere is increasing at a rate of $10$ cm³/s. Find the rate of increase of its radius when the radius is $3$ cm, leaving your answer in terms of $\pi$ .

Step 1 of 4

Write the formula for V in terms of r, and differentiate it

V = \dfrac{4}{3}\pi r^{3}

\dfrac{dV}{dr} = 4\pi r^{2}

Practise this →

3. Increasing and decreasing functions

Function	Condition
increasing	$\dfrac{dy}{dx} > 0$
decreasing	$\dfrac{dy}{dx} < 0$

y = −2x³ − 6x² + 18x + 1 is increasing

\displaystyle \dfrac{dy}{dx} = -6x^{2} - 12x + 18

e.g.

\displaystyle -6x^{2} - 12x + 18 > 0

\displaystyle x^{2} + 2x - 3 < 0

\displaystyle (x + 3)(x - 1) < 0

\displaystyle -3 < x < 1

📖 Worked example

Find the range of values of $x$ for which $y = -2x^{3} - 6x^{2} + 18x + 1$ is increasing.

Step 1 of 6

Differentiate

\dfrac{dy}{dx} = -2(3)x^{2} - 6(2)x + 18(1)

= -6x^{2} - 12x + 18

Practise this →

4. Stationary points

At a stationary point	Maximum	Minimum	Point of inflexion
$\dfrac{dy}{dx} = 0$	$\dfrac{d^{2}y}{dx^{2}} < 0$	$\dfrac{d^{2}y}{dx^{2}} > 0$	$\dfrac{d^{2}y}{dx^{2}} = 0$ (test fails)
sign of $\dfrac{dy}{dx}$	$+$ , $0$ , $-$	$-$ , $0$ , $+$	$+$ , $0$ , $+$ or $-$ , $0$ , $-$

y = x³ − 6x² + 9x + 3

\displaystyle \dfrac{dy}{dx} = 3x^{2} - 12x + 9

e.g.

\displaystyle 3x^{2} - 12x + 9 = 0

\displaystyle x^{2} - 4x + 3 = 0

\displaystyle (x - 1)(x - 3) = 0

\displaystyle x = 1 \text{ or } x = 3

$\dfrac{d^{2}y}{dx^{2}} = 6x - 12$
$x = 1$ : $-6 < 0$ , maximum $(1, 7)$
$x = 3$ : $6 > 0$ , minimum $(3, 3)$

📖 Worked example

Find the coordinates of the maximum point of the curve $y = x^{3} - 6x^{2} + 9x + 3$ .

Step 1 of 5

Differentiate

\dfrac{dy}{dx} = 3x^{2} - 6(2)x + 9(1)

= 3x^{2} - 12x + 9

Practise this →

5. Maximum and minimum problems

Step	Example: $A = 36x - 2x^{2}$
1. write the quantity in one variable	$A = x(36 - 2x)$
2. differentiate and set to $0$	$36 - 4x = 0$
	$x = 9$
3. check with the second derivative	$\dfrac{d^{2}A}{dx^{2}} = -4 < 0$ : maximum
4. find the value	$A = 162$

📖 Worked example

Step 1 of 5

Write the quantity in terms of x, then expand

\text{The side parallel to the wall} = 36 - 2x

A = x(36 - 2x)

A = 36x - 2x^{2}

Practise this →

Try a question: Applications of differentiation

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Practise applications of differentiation with step-by-step help

Endless questions from basic to exam level. Get one wrong and see where it went wrong. Free account.

Start free