AQA Level 2 Further Maths · Level 2 Certificate · Chapter 8: Calculus

Differentiation, tangents and normals, and stationary points

About 90 minutesNo calculator113 marks

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Date

Do Now

The first one is what today needs. The others bring back earlier work.

A1
Needed today
Write $\frac{3 x ^{5} - 8}{2 x ^{2}}$ as a sum of terms of the form $a x^{n}$ .
[2 marks]
Answer

A2
From chapter 5
Work out the equation of the line through $(4, - 3)$ that is perpendicular to $y = \frac{2}{5} x + 1$ . Give it in the form $y = m x + c$ .
[2 marks]
Answer
A3
From chapter 4
Solve $3 x^{2} - 13 x - 10 \leq 0$ .
[2 marks]
Answer
A4
From chapter 2
Write $x^{2} - 14 x + 58$ in the form $(x - a)^{2} + b$ .
[2 marks]
Answer
A5
GCSE Higher
Solve $2 x = \frac{54}{x ^{2}}$ .
[2 marks]
Answer

Examples, then your turn

Level 2

Your teacher works through each example with you. Then try the one beside it on your own.

Example 1

The normal to the curve $y = x^{2} - 3 x + 2$ at the point $P (3, 2)$ meets the curve again at $Q$ .

Work out the coordinates of $Q$ .

Not drawn accurately

Hint

\frac{d y}{d x} = 2 x - 3 = 3

P

, so the normal has gradient

- \frac{1}{3}

y - 2 = - \frac{1}{3} (x - 3)

, which is

y = - \frac{1}{3} x + 3

. Where it meets the curve,

x^{2} - 3 x + 2 = - \frac{1}{3} x + 3

, so

3 x^{2} - 8 x - 3 = 0

, which is

(3 x + 1) (x - 3) = 0

x = 3

P

, so

Q

has

x = - \frac{1}{3}

and

y = \frac{1}{9} + 1 + 2 = \frac{28}{9}

B1
The normal to the curve $y = 2 x^{2} + x - 1$ at the point $A$ where $x = - 1$ meets the curve again at $B$ .
Work out the coordinates of $B$ .
[5 marks]
Not drawn accurately
Answer $B (\dots\dots, \dots\dots)$

Example 2

The curve $y = x^{3} + a x^{2} + b x + 4$ has a stationary point at $(3, - 5)$ .

Work out the values of $a$ and $b$ . Work out the coordinates of the other stationary point and decide whether each point is a maximum or a minimum.

Hint At

(3, - 5)

27 + 9 a + 3 b + 4 = - 5

, so

3 a + b = - 12

\frac{d y}{d x} = 3 x^{2} + 2 a x + b = 0

x = 3

6 a + b = - 27

. Subtracting,

3 a = - 15

a = - 5

b = 3

. Then

\frac{d y}{d x} = 3 x^{2} - 10 x + 3 = (3 x - 1) (x - 3)

, so the other point has

x = \frac{1}{3}

and

y = \frac{121}{27}

\frac{d ^{2} y}{d x ^{2}} = 6 x - 10

8 > 0

x = 3

(a minimum) and

- 8 < 0

x = \frac{1}{3}

(a maximum).

B2
The curve $y = a x^{3} + b x^{2} - 6 x$ has a stationary point at $(1, - 5)$ .
Work out the values of $a$ and $b$ . Work out the coordinates of the other stationary point and decide whether each point is a maximum or a minimum.
[6 marks]
Answer $a = \dots\dots$ , $b = \dots\dots$ ; $(\dots\dots, \dots\dots)$

Example 3

A closed box is a cuboid. Its base is $x$ cm by $2 x$ cm and its height is $y$ cm. Its volume is $72$ cm³.

Show that its surface area, $A$ cm², is $A = 4 x^{2} + \frac{216}{x}$ . Work out the least surface area and show that it is a minimum.

Not drawn accurately

Hint

2 x^{2} y = 72

, so

y = \frac{36}{x ^{2}}

A = 2 (2 x^{2}) + 2 (x y) + 2 (2 x y)

, which is

4 x^{2} + 6 x y = 4 x^{2} + \frac{216}{x}

\frac{d A}{d x} = 8 x - \frac{216}{x ^{2}} = 0

when

x^{3} = 27

, so

x = 3

\frac{d ^{2} A}{d x ^{2}} = 8 + \frac{432}{x ^{3}} > 0

, a minimum.

A = 36 + 72 = 108

B3
An open box has no lid. Its base is $x$ cm by $3 x$ cm and its height is $h$ cm. Its volume is $144$ cm³.
Show that the area of card used, $A$ cm², is $A = 3 x^{2} + \frac{384}{x}$ . Work out the least area of card and show that it is a minimum.
[6 marks]
Not drawn accurately
Answer $\dots\dots$ cm²

Practice

Level 2

Each question asks for something different. Show your working.

C1
$P (3, 15)$ is a point on the curve $y = x^{3} - 4 x$ . $Q$ is the point on the curve with $x$ -coordinate $3 + h$ .
Show that the gradient of the chord $P Q$ is $23 + 9 h + h^{2}$ . Hence write down the gradient of the curve at $P$ .
[3 marks]
Answer $\dots\dots$

C2
$y = \frac{( x ^{2} - 3 ) ^{2}}{x}$ . Work out $\frac{d y}{d x}$ .
[3 marks]
Answer $\frac{d y}{d x} = \dots\dots$
C3
Work out the coordinates of the point on the curve $y = 3 x + \frac{16}{x ^{2}}$ where the gradient is $7$ .
[3 marks]
Answer $(\dots\dots, \dots\dots)$
C4
The tangent to the curve $y = x^{2} + \frac{64}{x}$ at the point $P (4, 32)$ meets the $y$ -axis at $A$ and the $x$ -axis at $B$ .
Work out the area of triangle $O A B$ .
[4 marks]
Not drawn accurately
Answer $\dots\dots$
C5
Work out the equation of the normal to the curve $y = x^{3} - 4 x^{2} + 7$ at the point where $x = 3$ . Give your answer in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.
[4 marks]
Answer $\dots\dots$
C6
Work out the values of $x$ for which $y = 4 x^{3} - 3 x^{2} - 18 x + 1$ is a decreasing function.
[3 marks]
Answer $\dots\dots$
C7
$f (x) = x^{3} - 6 x^{2} + 15 x - 4$ . Use differentiation to show that $f (x)$ is an increasing function for all values of $x$ .
[3 marks]
C8
$y = (2 x - 1)^{2} (x + 3)$ . Work out $\frac{d ^{2} y}{d x ^{2}}$ . Hence work out the rate of change of the gradient of the curve at the point where $x = - 2$ , and say whether the gradient is increasing or decreasing there.
[3 marks]
Answer $\dots\dots$
C9
$y = x^{3} + \frac{48}{x}$ for $x > 0$ . Work out the coordinates of the stationary point and decide whether it is a maximum or a minimum.
[4 marks]
Answer $(\dots\dots, \dots\dots)$

Exam-style questions

AQA exam style

Answer every question in the space given. Show your working: most of the marks are for method.

D1
Leo is finding the stationary points of $y = x^{4} - 4 x^{3}$ . He writes:
" $\frac{d y}{d x} = 4 x^{3} - 12 x^{2} = 0$ . Dividing by $4 x^{2}$ gives $x - 3 = 0$ , so the curve has only one stationary point, where $x = 3$ ."
Leo's conclusion is wrong.
(a)
Explain Leo's mistake.
[1 mark]
(b)
Work out the coordinates of both stationary points. Decide whether each is a maximum, a minimum or neither. You must show your working.
[5 marks]
Answer $(\dots\dots, \dots\dots)$ and $(\dots\dots, \dots\dots)$
Total for D1: 6 marks

D2
The curve $y = x^{2} - 6 x + 5$ passes through the point $A (4, - 3)$ .
Not drawn accurately
(a)
Work out the equation of the tangent to the curve at $A$ . Give your answer in the form $y = m x + c$ .
[3 marks]
Answer $y = \dots\dots$
(b)
$B$ is the point on the curve where the tangent is perpendicular to the tangent at $A$ . Work out the coordinates of $B$ .
[3 marks]
Answer $B (\dots\dots, \dots\dots)$
Total for D2: 6 marks
D3
A particle moves along a straight line. Its displacement from a fixed point $O$ after $t$ seconds is $s$ metres, where $s = t^{3} - 9 t^{2} + 24 t$ for $t \geq 0$ .
(a)
Work out the times at which the particle is at rest.
[3 marks]
Answer $t = \dots\dots$
(b)
Write down the values of $t$ for which the displacement is decreasing.
[1 mark]
Answer $\dots\dots$
(c)
The acceleration of the particle is $\frac{d ^{2} s}{d t ^{2}}$ m/s². Work out the acceleration when $t = 5$ .
[2 marks]
Answer $\dots\dots$ m/s²
Total for D3: 6 marks
D4
A closed cylinder has radius $r$ cm and height $h$ cm. Its total surface area is $54 π$ cm².
Not drawn accurately
(a)
Show that the volume of the cylinder, $V$ cm³, is $V = 27 π r - π r^{3}$ .
[2 marks]
(b)
Use calculus to work out the greatest volume of the cylinder. Show that it is a maximum. Give your answer in terms of $π$ .
[4 marks]
Answer $\dots\dots$ cm³
Total for D4: 6 marks
D5
A curve has equation $y = a x + \frac{b}{x ^{2}}$ , where $a$ and $b$ are constants. The curve passes through the point $(1, 10)$ , and its gradient at that point is $- 14$ .
(a)
Work out the values of $a$ and $b$ .
[3 marks]
Answer $a = \dots\dots$ , $b = \dots\dots$
(b)
Hence work out the coordinates of the stationary point on the curve and decide whether it is a maximum or a minimum.
[2 marks]
Answer $(\dots\dots, \dots\dots)$
Total for D5: 5 marks
D6
$y = x^{3} + k x^{2} + 3 x$ , where $k$ is a constant. $\frac{d y}{d x} > 0$ for all values of $x$ .
Work out the range of possible values of $k$ .
[4 marks]
Answer $\dots\dots$
D7
The line $y = 9 x - 16$ is a tangent to the curve $y = x^{3} + k x$ at the point $P$ , where $k$ is a constant.
Work out the value of $k$ and the coordinates of $P$ .
[5 marks]
Answer $k = \dots\dots$ , $P (\dots\dots, \dots\dots)$
D8
$y = \frac{x ^{2} + 2}{x}$ . Which of these is $\frac{d y}{d x}$ ? Circle your answer.
$1 - \frac{2}{x ^{2}} 1 + \frac{2}{x ^{2}} 1 - \frac{2}{x} 2 x$
[1 mark]
Answer
D9
$y = f (x)$ is a cubic function.
$f (x) < 0$ for $x < - 1$ and $f (x) \geq 0$ for $x \geq - 1$ .
The function is increasing for $x < 1$ , decreasing for $1 < x < 4$ and increasing for $x > 4$ .
Draw a possible sketch of $y = f (x)$ for values of $x$ from $- 2$ to $6$ .
[3 marks]

Extension

Stretch

No route is given. Spend five minutes on a problem before you look at a hint. The hints are on the last page of the worksheet, and you can use one at a time.

E1
Two tangents to the curve $y = 2 x^{2} + 8$ pass through the origin. Work out the equations of the two tangents.
[4 marks]
Hint 1 · What to try
Call the point of contact $(p, 2 p^{2} + 8)$ . The tangent there has gradient $4 p$ .
Hint 2 · The first line
A line through the origin with gradient $4 p$ is $y = 4 p x$ . It passes through the point of contact, so $2 p^{2} + 8 = 4 p \times p$ .
Hint 3 · The full method
$2 p^{2} = 8$ , so $p = 2$ or $p = - 2$ . The gradients are $8$ and $- 8$ : $y = 8 x$ and $y = - 8 x$ .
Answer $y = \dots\dots$ and $y = \dots\dots$

E2
The curve $y = x^{3} + p x^{2} + q x$ has stationary points where $x = - 2$ and where $x = 4$ .
Work out the values of $p$ and $q$ . Work out the distance between the two stationary points. Give your answer as a surd in its simplest form.
[5 marks]
Hint 1 · What to try
$\frac{d y}{d x} = 3 x^{2} + 2 p x + q$ is zero at $x = - 2$ and $x = 4$ , so it is $3 (x + 2) (x - 4)$ .
Hint 2 · The first line
$3 (x + 2) (x - 4) = 3 x^{2} - 6 x - 24$ , so $2 p = - 6$ and $q = - 24$ . The points are $(- 2, 28)$ and $(4, - 80)$ .
Hint 3 · The full method
The distance is $6^{2} + 10 8^{2} = 11700 = 900 \times 13 = 3013$ .
Answer $p = \dots\dots$ , $q = \dots\dots$ , distance $\dots\dots$
E3
Work out the shortest distance from the origin $O$ to the curve $y = x^{2} - 4$ . Give your answer in its simplest exact form.
[5 marks]
Hint 1 · What to try
A point on the curve is $P (x, x^{2} - 4)$ . Work with $D = O P^{2}$ : the shortest distance gives the least $D$ .
Hint 2 · The first line
$D = x^{2} + (x^{2} - 4)^{2} = x^{4} - 7 x^{2} + 16$ , and $\frac{d D}{d x} = 4 x^{3} - 14 x = 2 x (2 x^{2} - 7)$ .
Hint 3 · The full method
$x = 0$ gives $D = 16$ ; $x^{2} = \frac{7}{2}$ gives $D = \frac{15}{4}$ , which is smaller. The shortest distance is $\frac{15}{4} = \frac{15}{2}$ .
Answer $\dots\dots$

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can work out the gradient of a curve by differentiating, including negative powers of x and expressions I expand or divide first.
I can find the equation of a tangent or a normal to a curve, and the points where tangents meet conditions.
I can find where a function is increasing or decreasing, and use the second derivative.
I can find stationary points, decide their nature, and solve problems about a greatest or least value.

Answers and mark scheme

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