KT Classroom Applications of Differentiation

MATHEMATICS Learner’s Study and Revision Guide for Grade 12

APPLICATIONS OF DIFFERENTIATION

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle

Examination Questions by the Department of Basic Education

Applications of differentiation

Contents Unit 11 Definition of a gradient of a function

3

Definition of an increasing/decreasing function

4

Stationary points

6

The second derivative

7

Exercise 11

10

Answers

11

Examination questions with solution hints and answers

12

More questions from examination papers – applications to sketching cubic functions

18

Answers

22

More questions from past examination papers – maxima and minima of physical problems

24

Answers

28

How to use this revision and study guide 1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic in class or from a textbook.

2. “Warm-up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.

3. The notes and exercises are followed by questions from past examination papers.

4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.

5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.

6. What follows next are more questions taken from past examination papers.

7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.

8. Finally, don’t be a loner. Work through this guide in a team with your classmates.

2

Applications of differentiation

REVISION UNIT 11: APPLICATIONS OF DIFFERENTIATION Definition of the gradient of a function The gradient at any point on the curve of y = f ( x ) is defined by

dy = f ' (x ) dx which by definition is the gradient of the tangent at any point on the curve. Understanding this will help in establishing the criteria for: • •

increasing and decreasing functions to be studied in this Unit and maximum and minimum turning points also to be studied in this Unit.

NOTE: Important to note that the gradient of a curve is defined at a point on the curve.

Illustration

y

T 70

Tangent at point P on the curve 60

50

P

40

30

20

10

x 1

2

3

4

5

A graph of a function y = f ( x ) is shown in the diagram above. Also shown is the tangent PT at a point P on the curve.

Thus the gradient of the tangent PT = the gradient of the curve at the point P. If the equation of the line is y = mx + c then

dy = f ' (x ) = m . dx

Another way of illustrating that the gradient of a curve at a point is given by the gradient of the tangent at that point is to enlarge the area around the point by zooming in on that point. Repeatedly zooming on the point P will make the curve at that point look like a straight line. The gradient of this straight line will be the gradient of the curve at that point.

3

Applications of differentiation

Definition of an increasing/decreasing function Some obvious facts about all graphs is that: •

the x -coordinates increase in a horizontal direction from the left to the right which in the diagrams below we will indicate by an arrowed line → .

•

the y -coordinates increase in an upward direction ↑ and decrease in a downward direction ↓ .

Illustrations A straight line with negative gradient

A straight line with positive gradient

y

y

↓ →

y=f(x) x

y=f(x)

↑ →

x

Look at the direction of the arrows.

Look at the direction of the arrows.

As x increases so does y also increase and we say

As x increases so does y decreases and we say

In this case as can be seen from the diagram we have

In this case as can be seen from the diagram we have

that y = f ( x ) is an increasing function of x .

that y = f ( x ) is a decreasing function of x .

m>0 or

m<0

dy = f ' (x ) > 0 dx

or

dy = f ' (x ) < 0 dx

Conclusions

dy > 0. dx dy < 0.

Decreasing function: a characteristic of a decreasing function is that dx Increasing function: a characteristic of an increasing function is that

4

Applications of differentiation

Example 1: Apply the above characteristics to the parabola shown below.

y 9 8 7

y=f(x)

6

↓ → -2

5 4 3 2 1

-1

1

2

3

4

↑ → 5

-1 -2 -3 -4

A

-5 -6 -7 -8 -9

Point A is the lowest point on the curve and is called a minimum turning point.

The directions of the arrows in the diagram tell the story: •

on the left side of A or for x < 2 the parabola is a decreasing function

•

on the right side of A or for x > 2 the parabola is an increasing function.

The same story is told by examining the gradient of the parabola on either side of the vertical line of symmetry through A shown in the diagram. Use a ruler to indicate the tangent line at any point along the parabola: •

to the left of A the tangents have negative gradients

dy = f ' (x ) < 0 confirming that the dx

parabola is a decreasing function for x < 2 •

to the right of A the tangents have positive gradients parabola is an increasing function for x > 2 .

5

dy = f ' (x ) > 0 confirming that the dx

x

Applications of differentiation

Stationary points (or turning points) What happens at the lowest point A? At this point the function is neither increasing nor decreasing and the point is said to be a stationary point (or a minimum turning point as you can see from the diagram).

At this point

dy dy = f ' (x ) = 0 . is neither greater than or less than zero and so dx dx

Example 2: We now apply the characteristics to the cubic y = f ( x ) = − x 3 + 27 x − 2 .

y

B

50

y=f(x)

→ ↓

40

30

20

10

↑ →

→ ↓ x

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

-10

-20

-30

-40

-50

A

Need we say more? Again the arrows in the diagram tell the story.

• •

the function is decreasing for x < −3 and x > 3 the function is increasing for − 3 < x < 3 .

We can test these conclusions by differentiation:

y = − x 3 + 27 x − 2 ∴

dy = −3 x 2 + 27 → factorise and test to see for what values of x it is positive or negative. dx

(

)

− 3 x 2 + 27 = −3 x 2 − 9 = −3( x − 3)( x + 3) Construct a sign table and draw conclusions: -3

x −3 x+3 − 3(x − 3)(x + 3) Conclusions (Answer)

x < −3

−3< x < 3

x>3

-

+ +

+ + -

Decreasing Increasing function function

Decreasing function

confirming the answers obtained by observing the arrows in the diagram above.

6

Applications of differentiation

The second derivative Example 3:

Take y = x 4 + 3 x 2 + 2 x Then

dy = f ′( x) = 4 x 3 + 6 x + 2 which is the first derivative. dx

Clearly,

Thus

Write

dy is also a function called the gradient function which can be differentiated. dx

(

)

d  dy  d 4 x 3 + 6 x + 2 = 12 x 2 + 6  = dx  dx  dx

d2y d  dy  as which is the second derivative and is pronounced “dee squared y dee x   dx 2 dx  dx 

squared”.

Interpretation of the second derivative We can extend the interpretation of the sign positive or negative of the first derivative to the second derivative as follows: •

d2y dy d  dy  as a function is an increasing function if > 0 or f ′′( x ) > 0   > 0 or dx dx 2 dx  dx 

• •

d2y dy d  dy  as a function is a decreasing function if < 0 or f ′′( x ) < 0   < 0 or dx dx 2 dx  dx 

Application of the second derivative Let us explain this diagrammatically.

Example 3: Minimum turning point Look at what is happening as x increases from left

y 9 8

dy to right. What you will notice is that changes dx

7 6 5

Positive gradient

4

from negative through zero to positive values.

Negative gradient

3 2 1

x -13

dy Thus is an increasing function and so dx

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

-1 -2 -3 -4 -5 -6

d  dy  d y = f ′′(x ) > 0  = dx  dx  dx 2 2

-7 -8 -9

7

↑

Stationary point Gradient=0

7

8

9

10

11

12

13

Applications of differentiation

Example 4: Maximum turning point Look at what is happening as x increases from left to right. What you will notice is that

y 9 8

dy changes dx

7

↑

5 4

from positive through zero to negative values.

3

Positive 2 gradient 1 -13

Thus

Stationary point Gradient=0

6

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

dy is an decreasing function and so dx

-1

Negative gradient 1

2

3

4

5

6

7

x 8

9

10

11

12

13

-1 -2 -3 -4 -5 -6

d  dy  d y = f ′′(x ) < 0  = dx  dx  dx 2

-7

2

-8 -9

Summary Below we repeat the criteria for distinguishing between the various scenarios:

dy >0 dx dy Decreasing function: A function is said to be decreasing where its gradient is negative, that is if <0 dx

Increasing function: A function is said to be increasing where its gradient is positive, that is if

Maximum value or maximum turning point

Minimum value or minimum turning point

dy =0 dx

dy =0 dx

d2 y <0 dx 2

d2 y >0 dx 2

d2 y = 0 . Such dx 2 dy a point is called a point of inflection. What distinguishes a point of inflection is that the sign of dx

Point of inflection: The above doesn’t tell us what happens at a point on a curve where

remains the same on either side of the point even though the curve undergoes a change in its curvature.

At a point of inflection,

But

When

d2y = 0 always. dx 2

dy dy = 0 or ≠ 0. dx dx

dy = 0 we have what is called a horizontal point of inflection and the tangent will cross the dx

curve (very unlike what you have always understood about tangents only touching curves!).

8

Applications of differentiation

EXERCISE 11 1. Given the function y = x 3 â&#x2C6;&#x2019; 6 x 2 + 9 x

dy from first principles dx

1.1

find

1.2

calculate the average gradient between x = 1 and x = 3

1.3

for what values of x is the function a decreasing function?

1.3

for what values of x is the function an increasing function

1.4

calculate the turning points of the curve to this function and distinguish between them;

1.5

determine the coordinates of the point of inflection

1.6

determine the coordinates of the x and y intercepts

1.7

sketch the curve.

2. A rectangular block has a base x cms square. Its total surface area is 150 cm 2 . Prove that the volume of the block is

(

)

1 75 x â&#x2C6;&#x2019; x 3 cm 3 .

2

Calculate (a) the dimensions of the block when its volume is maximum; (b) this maximum volume. Show that your answer is a maximum rather than a minimum.

9

Applications of differentiation

ANSWERS EXERCISE 11 11.1 As required by the question, this must be done using first principles:

dy = 3 x 2 − 12 x + 9 dx

11.2 Average gradient between (1;4) and (3;0) is -2 11.3 Function is decreasing for 1 < x < 3 11.4 Function is increasing for x < 1 and x > 3 11.5 Maximum turning point is (1;4) and minimum turning point is (3;0) 11.6 Point of inflection is at 11.7 The graph cuts the x -axis at (0;0), touches the x -axis at (3;0); the y -intercept is at (0;0) 11.8 Sketch: y 9 8 7 6 5 4 3 2 1

x -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

-1 -2 -3 -4 -5 -6 -7 -8 -9

QQuestion 11.2 In questions of this type always sketch a diagram if one is not given to you.Denote total surface area by S whose formula is

S = 2 x 2 + 4hx

Denote the volume by V whose formula is

V = x2h You need to replace h using the formula for S .

1

(

)

1

(

V =x ⋅ 75 − x = 75 x − x Why is this the formula?

2 2x From this formula, solve for h and use this to Only now can you differentiate V w.r.t. x replace h in the formula for V in the box opposite.

Answers: (a) Maximum when the dimensions are 5cms by 5cms by 5cms. (b) Maximum volume = 125cc Well, to show that the answer is indeed a maximum you must show that the second derivative is negative for the value of x which makes the first derivate zero. Go ahead and show it.

2

10

2

3

)

Applications of differentiation

PAPER 1 QUESTION 11

DoE/ADDITIONAL EXEMPLAR 2008

PAPER 1 QUESTION 9

DoE/NOVEMBER 2008

11

Applications of differentiation

PAPER 1 QUESTION 11

DoE/ADDITIONAL EXEMPLAR 2008

Number Hints and answers 11.1 Solve f ( x ) = 0 to obtain x -intercepts.

Work out the solutions in the boxes below

This requires you to factorise f ( x ) .

f ( x ) is a cubic so use the factor theorem

to factorise.

Find a value a for which f (a ) = 0 .

That makes x − a a factor of f ( x ) .

Divide f ( x ) by x − a to find the other factor which will be quadratic.

Factorise the quadratic.

Now you have three linear factors which allow you to write the x -intercepts.

The y -intercept is given by f (0 ) .

Answers: x -intercepts are (1;0) and (-3;0) y -intercept is (0;3) 11.2

At the turning points, f ' ( x ) =0 Solve this equation to obtain the x coordinates and thereafter the y coordinates of the turning points.

Answers:

5 3 Solution of f ' ' ( x ) =0 gives the x -coordinate of the point of inflection.

Solve this equation for x . Answer: x =-1/3 (1;0) and ( − ;9,48)

11.3 11.4

From the above, you have gathered all the information you need to sketch the graph of f ( x )

.

Sketch the graph on the grid above.

12

Applications of differentiation

PAPER 1 QUESTION 9 Number 9.1

DoE/NOVEMBER 2008

Hints and answers g (x ) is factorised already making it easy to tell the coordinates of A and B.

Work out the solutions in the boxes below

Use them and the distance formula to determine the length of AB. 9.2

Answer: AB = 4,5 units T is a turning point.

Perform the necessary differentiation and use the condition for a turning point to find the x coordinate of T. T is a turning point.

9.3

Answer: x = 1 The gradient at a point on a curve is defined by the gradient of the tangent at that point.

Find the gradient function by differentiation and use it to find the gradient at P. Now write down the equation of the tangent and simplify as far as possible leaving y on the left hand side.

Answer: y = −24 x − 61

9.4

This question requires you to do very little working out. All you need to work out is the y coordinate of T. Then the answer to the question is simply obtained by inspection of the sketch shown in the question. However, the answer requires to be explained at great length which is what follows below.

Suppose we have the following two simultaneous equations:

y = g (x )

and

y=k

We can solve for x and y algebraically using a method of substitution. Using this method we will have

g (x ) = k And then try to apply the remainder factor theorem to determine the roots. We say ‘try’ because you may not be able to get the factors.

13

Applications of differentiation

Number

Hints and answers

Work out the solutions in the boxes below

Or, we can solve for x and y using a graphically method by drawing the graphs of

y = g (x )

and

y=k

on the same set of axes.

The graphical method will certainly work.

The graph of y = k is a straight line parallel to the x -axis.

The graph of y = g ( x ) is a cubic which is sketched in the question.

Where the line and the cubic meet will give a solution to the simultaneous equations, or the roots of

g (x ) = k There are three possible scenarios for the line and cubic graphs to meet. Either: 1. They meet at one and only one point in which case there is one real root; or 2. They meet at one point and touch each other at another point in which case there are two distinct roots; or 3. They meet at three points in which case there are three distinct roots.

The latter scenario will give the answer we seek. You can move a ruler horizontally up and down the sketch given in the question and you will see that for the line to meet the cubic at three distinct points, the value of k must lie between the y coordinates of A and T. Now work out the y - coordinate of T using the answer you determined in 9.2.

Answer: You write the answer down given by a range of values for k .

9.5

What is the condition for a point of inflection?

Use this condition to solve for x .

Answer: x = â&#x2C6;&#x2019;

1 2

14

Applications of differentiation

PAPER 1 QUESTION 12

DoE/ADDITIONAL EXEMPLAR 2008

PAPER 1 QUESTION 10

DoE/NOVEMBER 2008

15

Applications of differentiation

PAPER 1 QUESTION 12

DoE/ADDITIONAL EXEMPLAR 2008

Number

Hints and answers

Work out the solutions in the boxes below

12.1

Write down the formula for the volume V of a cylinder, and replace the height by 2 x which is what is given in the diagram.

Use Pythagoras Theorem to obtain the radius r in terms of x .

Substitute for r in the expression for V .

You now will have V = 150πx − 2πx 2 .

12.2

You must find the value of x which makes the volume V a maximum.

That is find

dV and equate to zero. dx

dV = 0 gives the value of x dx that makes V a maximum.

Solution of

Answer: Height = 10cm

16

Applications of differentiation

PAPER 1 QUESTION 10

DoE/NOVEMBER 2008

Number Hints and answers 10.1 Write down the formula for the volume V of a cylinder.

Make h the subject of the formula.

Replace V by its given value.

10.2

Note that this is a drinking glass so that one end is open and thus does not contribute to the surface area S .

The surface area is the curved side plus the circular area of the closed end.

Use the above to write the formula for the total surface area S .

Replace h by the expression in 10.1.

Simplify if possible.

10.3

You must find the value of r which makes the volume S a minimum.

That is find

dS and equate to zero. dr

dS = 0 gives the value of r dr that makes S a minimum.

Solution of

Answer:

r = 3,99 cm

17

Work out the solutions in the boxes below

Applications of differentiation

MORE QUESTIONS FROM PAST EXAMINATION PAPERS

Sketching cubic functions and use of factor theorem Exemplar 2008

18

Applications of differentiation

Preparatory Examination 2008

Feb â&#x20AC;&#x201C; March 2009

November 2009 (1)

19

Applications of differentiation

November 2009 (Unused paper)

20

Applications of differentiation

Feb â&#x20AC;&#x201C; March 2010

21

Applications of differentiation

ANSWERS Exemplar 2008   11 10.1 A (− 1;36 ) and B  ;−14,81  3   17 10.2  ;14,81   3 10.3 Average rate of change = -10,89 10.4 y = −16 x + 32 10.5 x = 2 10.6 k > 36 or k < −14,81  4 286  10.7  ;  or (1,33;10,59 )  3 27 

Preparatory Examination 2008 10.1 y -intercept: (0; 3) x -intercepts: (1; 0) and (-3; 0) 10.2 Turning points at (1;0 ) and  5 256   − ;  3 27  10.3 Sketch:

Feb/March 2009 3 12.1 a = and b = 6 2 12.2 Average gradient =

12.3 y = −12 x − 22 1 12.4 x = 2 12.5 p > 3,5 or p < −10 November 2009(1) 11.1 x -intercepts: (2; 0) and (-3; 0) 11.2 Turning points at (2; 0) and (-1,33; 18,52)  4 500  or  − ;−   3 27  11.3 Sketch:

11.4 10.4 y = 11x − 17 10.5 Turning points at (2;0 ) and   2  − ;9,48    3 1 10.6 x = − 3

9 2

x=

1 3

 4 581  (2; -3) and  − ;  or  3 27  (− 1,33;−21,52)

11.5

22

Applications of differentiation

Feb/March 2010 11.1 A(2; 0)

November 2009 (Unused papers) 10.1 x = 1 and/or x = 2 10.2 Local maximum at x = 1 Local minimum at x = 2 10.3 x = 1,5 10.4 Sketch:

11.2 Use given information to prove this.

11.3 B(1; -4) 11.4 Prove this by showing that m BC = 0 11.5 x = 0

23

11.6

k < â&#x2C6;&#x2019;4 or k > 0

11.7

â&#x2C6;&#x2019;1 < x < 1

Applications of differentiation

MORE QUESTIONS FROM PAST EXAMINATION PAPERS Applications of differentiation: maxima and minima Exemplar 2008

Preparatory Examination 2008

24

Applications of differentiation

Feb â&#x20AC;&#x201C; March 2009

25

Applications of differentiation

November 2009 (Unused paper)

26

Applications of differentiation

November 2009 (1)

Feb â&#x20AC;&#x201C; March 2010

27

Applications of differentiation

ANSWERS Exemplar 2008 11.1 Area = − x 2 + 3 x + 10 1 11.2 x = 3 Preparatory Examination 2008 x2 x 11.1 A(x ) = − + +3 2 2 1 11.2 x = 2 11.3 A = 3,125 sq units

November 2009(1) 12.1 s(0) =100m 12.2 s ' (t ) = −80 metres per minute 12.3 height of the car above sea level is decreasing at the rate of 80 metres per minute and is travelling downwards.

12.4 t = 4,33 minutes

12.1

Length of sides of square =

Feb/March 2009 13.1 AB 2 = a 4 + a 2 − 6a + 9 a =1 13.2

12.2

Area = 1 −

12.3

x = 1,76m

November 2009 (Unused paper) 52π 2 11.1 A( x ) = 8 x − x 9 11.2 x = 0,22m 11.3 Area of circles = 0,88m 2

Feb/March 2010

28

4− x 4

1 1  2 1 x+ + x 2  16 4π 