D I F F E R E N T I A L C A L C U L U Sgifs.africa/wp-content/uploads/2020/06/Grade-12-Ma...MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 3 1 D I F F E R E N T I A L C A L C U L

MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 3

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D I F F E R E N T I A L C A L C U L U S

OPTIMISATION & RATE OF CHANGE

IMPORTANT NOTES & THEORY

Differential Calculus is primarily about the rate of change (i.e. the speed at which

something is changing).

The derivative of a function ( ( ) or

) gives the gradient (or rate of change) of

that function at any point.

(The derivative may also be referred to as the ‘gradient function’)

Make sure that you clearly understand that for any function :

( ) is a y - value on the graph of associated with a specific x - value

( ) is the gradient of associated with a specific x – value

OPTIMISATION (Maximum and minimum values)

We use calculus to find the x - value that will minimise or maximise a quantity like

area, volume, etc.

It is important to remember that the gradient of a function is zero at the minimum and

maximum values.

To determine where a function will have a maximum / minimum, solve for

( ) .

To calculate the actual minimum / maximum value itself, substitute the answer of

( ) into ( ).

Follow these steps:

If necessary calculate the missing dimensions on the given diagram

(E.g. height in terms of the radius)

Determine a formula in terms of one variable (e.g. x) for whatever must be

minimised or maximised, like

Area………... ( ) = …. or

Cost………... ( ) = …. or

Volume……. ( ) = ….

If your formula contains a value like , then keep it as and only in the final

answer convert it to a value/number

Now determine the first derivative of your formula (e.g. ( ))

Solve for x in ( )

Test the x - values obtained. It will maximise or minimise the area, volume,

cost and so forth. REJECT the x - value that is not valid (e.g. a negative area

etc.)

In Mathematics: Rate of change = gradient

(m)


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Remember that these questions could be asked about distances and shapes on the

Cartesian plane. Then height is given by the y - values and the horizontal distances

are given by the x - values

Make sure you KNOW the formulas for the surface areas and volumes of right

prisms. (Will not be given in an exam)

For other shapes (e.g. cones and pyramids), formulae will be given. However, you

will be expected to select the applicable formula.

The following formulae will assist you when dealing with optimisation problems

involving mensuration:

Shape (2D) Perimeter Area

Triangle

P = a + b + c

Area =

base Perp. Height

Area =

acsin

Square P = 4s Area = s2

Rectangle P = 2l + 2b Area = l b

Parallelogram P = 2l + 2b Area = base perp. Height

Trapezium P = sum of four sides Area =

(sum of // sides) h

Circle C = r = D Area =

Object (3D) Volume Surface area

Rectangular solid V = lbh TSA = 2lb + 2lh + 2bh

(for a closed box)

Cube V = s3 TSA = 6s

2

Sphere V =

Surface Area =

Cylinder V = h Area of curved surface =

TSA =

a

b

𝜃


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RATE OF CHANGE & CALCULUS OF MOTION

Memorise the following:

If an equation ( ) is given with distance (s) and time (t), then…

( ) represents distance (displacement) at time t

( ) represents speed (velocity) at time t

( ) represents acceleration at time t

Follow this easy example:

Given a height (or distance) equation: ( ) with distance (s) in meters

and time (t) in seconds. (Like a stone thrown up into the air)

( ) is the distance after 2 seconds

( ) is the distance after 5 seconds

( ) is the speed after 2 seconds

( ) is the speed after 5 seconds

( ) is the distance EVEN BEFORE THE MOTION HAS STARTED

Maximum height will be at ( ) (the turning point of s)

After how many seconds will there be a maximum height? Solve for:

( )

What is the maximum height? (Substitute t = 5 into s)

( ) ( ) ( )

( ) When (at what time) does it hit the ground again?

(When speed = 0 again, ( ) )

Use the bigger value because it will be the end of the ‘journey’ of the stone.

Speed with which it hits the ground? (Calculate ( ))

( ) ( )

The negative indicates that the stone hits the ground when in DOWNWARD

motion.

SPECIAL CASES

Suppose TWO moving bodies are involved, one going up and on coming down, and

then ask:

When (at what time) will they pass each other?

This will happen when the two heights are equal.

Therefore, equate the two height equations of the bodies and solve for t.

When (at what time) will their speeds be equal?

Now you equate the two speed equations (s’ = s’) and solve for t.


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EXERCISE 1:

1. A stone falls from a cliff 320 metres high. The distance covered by the stone while

falling is given by the function ( ) , where s is the distance covered in metres

and t is the time in seconds.

1.1 Determine the speed of the falling stone after 3 seconds.

1.2 How long will it take before the stone hits the bottom of the cliff?

1.3 What will the speed of the stone be when it hits the bottom of the cliff?

2. Sales of a new product grow rapidly and then level off with time. This situation is

represented by the equation ( ) where t represents time in

months and ( ) represents sales.

2.1 Determine the rate of change of sales during the third month.

2.2 Determine after how many months a maximum sale is obtained.


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3. The mass of a baby in the first 30 days of life is given by 00039 23 tt)t(M ;

300 t . t is the time in days and M is the mass of the baby in grams.

3.1 Write down the mass of the baby at birth.

3.2 A baby's mass usually decreases in the first few days after birth.

On which day will the baby's mass return to its birth mass?

3.3 On which day will this baby have a minimum mass?

On which day will the baby's mass be decreasing the fastest?

4. A Petrol tank at BP Depot has both the inlet and the outlet pipes which are used

to control the amount of petrol it contains. The depth of the tank is given by

( )

where is in metres and is in hours that are measured

from 9h00.

4.1 Determine the rate at which the depth is changing at 12h00, and then tell

whether there is and increase or decrease in depth. Answer correct to two

decimal digits.

4.2 At what time will the inflow of petrol be the same as the outflow?

5. A stone is thrown vertically upwards. Its height (in metres) above the ground at time t

(in seconds) is given by: 2( ) 2 12 32h t t t .

5.1 Determine the initial height of the stone above the ground.

5.2 Determine the time taken to reach the maximum height.

5.3 How fast was the stone travelling when it hit the ground?

5.4 Determine the acceleration of the stone.

6. PQRS is a rectangle with P on the curve ( ) and with the x-axis and the line

as boundaries.

6.1 Show that the area of rectangle PQRS can be expressed as: A .

6.2 Determine the largest possible area for rectangle PQRS.

Show all your calculations.


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7. A wooden block is made as shown in the diagram below.

The ends are right-angled triangles having sides .

The length of the block is y.

The total surface area of the block is 3 600 .

7.1 Show that:

7.2 Determine the value of x for which the block will have a maximum volume.

8. A rectangular box has a length of 5x units, breadth of )29( x units and its height

of x units.

8.1 Show that the volume (V) of the box is given by 32 1045V xx

8.2 Determine the value of x for which the box will have maximum volume.


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9. A 340 ml can of cool drink with height and radius is shown below.

9.1 Determine the height of the can in terms of the radius .

9.2 Show that the surface area of the can be written as

.

9.3 Determine the radius of the can in , if the surface area of the can has to be

as small as possible.

10. ABCD is a square with sides 20 mm each. PQRS is a rectangle that fits inside

the square such that .

10.1 Prove that the area of ( )

10.2 Determine the value of for which the area of is a maximum.


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Exercise 2:

QUESTION 1

After flying a short distance, an insect came to rest on a wall. Thereafter the insect started

crawling on the wall. The path that the insect crawled can be described by

)632)(6()( 2 tttth , where h is the height (in cm) above the floor and t is the time

(in minutes) since the insect started crawling.

1.1 At what height above the floor did the insect start to crawl? (1)

1.2 How many times did the insect reach the floor? (3)

1.3 Determine the maximum height that the insect reached above the floor. (4)

[8]

QUESTION 2

A cone with radius r cm and height AB is inscribed in a

sphere with centre O and a radius of 8 cm. OB = x.

2.1 Calculate the volume of the sphere. (1)

2.2 Show that 22 64 xr . (1)

2.3 Determine the ratio between the largest volume of this cone and the volume of the sphere. (7)

[9]

Volume of sphere =

Volume of cone =


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QUESTION 3

A right circular cone with radius p and height t is machined (cut out) from a solid sphere (with

centre C) with a radius of 30 cm, as shown in the sketch.

3.1 From the given information, express the following:

3.1.1 AC in terms of t. (1)

3.1.2 2p , in its simplest form, in terms of t . (3)

3.2 Show that the volume of the cone can be written as .3

120)( 32 tttV

(1)

3.3 Calculate the value of t for which the volume of the cone will be a maximum. (3)

3.4 What percentage of the sphere was used to obtain this cone having maximum volume? (4)

[12]

QUESTION 4

Given: xxxf 33 .

Calculate the value of q for which qxf will have a maximum value of .9

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[6]

QUESTION 5

A farmer wanted to determine the amount of water he used every

week during the drought. He measured the amount of water drawn

from a Jojo water tank used on the farm. He determined that the

volume of water, in litres, weeks after he started measuring was:


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5.2.1 After how many weeks was the volume at a maximum? (3)

5.2.2 After how many weeks will the tank be empty? (3)

5.2.3 Determine the rate at which the volume changes with respect to the time at

?

(2)

[15]

QUESTION 6

In the diagram below, TUVW is a rectangular picture. The picture is framed such that there is

a 3 cm space around the picture. The perimeter of the rectangle PQRS is 70 cm.

PQ = x units and QR = y units.

Calculate the maximum area of the picture TUVW. [8]

QUESTION 7

A hunter was standing at point A, along the fence of a rectangular game enclosure,

when he spotted a deer standing at point B, the corner of the rectangular enclosure.

The distance from A to B is 1200m. At exactly the same time as the hunter started to

move in an easterly direction towards B, the deer started to move in a southerly

direction towards D. The hunter moves at 4metres per second and the deer moves at

5metres per second. After t seconds, the hunter is at a point H and the deer is at point

D.

A H B

D

The hunter tries to shoot the deer but with his caliber rifle he must be at most 800m

from the deer.


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7.1 Show that the distance between the hunter and the deer (HD) attseconds after

they both started moving can be written as: 2HD(t) 41 9 600 1 440 000t t

(4)

7.2 How long after they started walking, were they the nearest to one another?

Show all calculations.

(3)

7.3 The calibre of the hunter’s rifle allows him to be at most 800m from his target.

Was the hunter within shooting range of the deer at the time when they were

nearest to each other? Show all calculations.

(3)

[10]

QUESTION 8

An open rectangular box is made of a very thin sheet of metal. The volume is 128 3cm and the

base of the box has a width of cmx and a length of cmx)4( .

8.1 Determine an expression for the height of the box in terms of x . (2)

8.2 Show that the total surface area of the box can be written as 22 3204 cm

xx

.

(3)

8.3 Calculate the height of the box for which the surface area is a minimum. (4)

[9]


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QUESTION 9

The diagram below shows a cylindrical can that fits into circle O with radius

32R units.

9.1 Express r in terms of x. (2)

9.2 Calculate the value of x for which the volume of the cylinder is a

maximum. (5)

9.3 Calculate the maximum volume of the can in terms of π. (2)

[9]

QUESTION 10

A car speeds along a 1 kilometre track in 25 seconds. Its distance (in metres) from the start

after t seconds is given by ttts 15)( 2

10.1 Write down an expression for the speed (the rate of change of distance with

respect to time) of the car after t seconds. (1)

10.2 Determine the speed of the car when it crosses the finish line. (1)

10.3 Write down an expression for the acceleration (the rate of change of speed with

respect to time) of the car after t seconds. (2)

10.4 Hence, or otherwise, calculate the acceleration of the car after 5 seconds. (1)

10.5 Calculate the speed of the car when it is 250 metres down the track. (4)

[9]

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D I F F E R E N T I A L C A L C U L U Sgifs.africa/wp-content/uploads/2020/06/Grade-12-Ma...MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 3 1 D I F F E R E N T I A L C A L C U L