1

Applications of the Calculus

The calculus is a mathematical process with many applications.

Of interest are those aspects of calculus that enable us to calculate the maximum and minimum value of a function.

For example, the maximum value of this function is to be calculated: xxy 12003 2

200

321200

2

ab

xWe know this occurs at

2

x

y

200

tangents

The slope of tangents to any curve can be calculated using differentiation.

3

At the functions maximum value, the tangent is horizontal and so its slope is 0.

This is true for all maximum and minimum values.

Using calculus then, involves searching for the places where the slope of the tangent is zero.

4

Rules for Differentiation

Notation: The derivative of a function is denoted by

dxdy y xf

5

nxy .1

3 . xyeg

cy .2

3 . yeg

1 nnxy

0 y

23 xdxdy

0 y

Rules for Differentiation

6

xcfy .3

25 .1 xyeg

56

.2 x

yeg

xfcy

xxdxdy

1025 1

56

156

0 xdxdy

Rules for Differentiation

7

xxfy g .4

xxyeg 2 .1

24 24 .2 xxyeg

33 416 xx

33 2244 xxy

xgxfy

12 xy

Rules for Differentiation

8

Special Cases:

axey .5

xy ln .6

axaey

xy

1

Rules for Differentiation

9

xxy 12003 2

120061120023 So 01 xxxy

represents the slope of the tangent to the curve y.y

1200 ,0 . yxeg

60012001006 ,100 yx

The introductory example has

10

When the tangent is horizontal (slope = 0) then .0y

120060 x

2006

1200 x

The maximum occurs at x = 200.

11

Using differentiation to find the maximum and minimum points is a more general method in that it “works” for all curves.

Using is a shortcut method that works

for quadratic equations (parabolas) only.ab

x2

The equation of the tangent is not difficult to determine. At a point on a curve, the slope is calculated using differentiation and the result is used in the equation 11 xxmyy

11, yx

12

Example: Find the equation of the tangent to the curve when x = 2.

When x = 2,

and the point on the curve is (2,9)

523 xxy

9522 23 y

xxy 23 2

822232 2 y

13

So using 11 xxmyy

289 xy

1689 xy

78 xy

14

Example: Find all the points on the curve 123 xxxxf

where the slope is 6.

123 2 xxxf

6 xf5230 2 xx

5 ,2 ,3 cba

32

53422 2 x 1or 35 x

1236 2 xx

15

Example: Find the derivative of the following function

23 xxy

962 xxxy

xxxy 96 23

9123 2 xxy

16

Example: Find the derivative of the following function

xx

xf2

xx

xxf

2

121 xxf

22 212 xxxf2

2or x

17

Example: Find the derivative of the following function

xxx

y32

21

32 xxxy

212 62

xxxy

21

23

62 xxy

2

12

1

21

623

2 xxy

xxxxy

3333 2

12

1

18

The Derivative as a Measure of Rate of Change

The derivative as a measure of the gradient of a tangent to a curve has been discussed.

A second interpretation of this process describes the rate of change of some variable.

19

Consider the following physics example:

A car’s position (s) is described by the equation , where s is in metres and t is in seconds.

2ts

After 2 seconds, (t = 2), .422 s

The car is 4 metres from some starting point.

At t = 10, s = 102 = 100 and so on.

20

Looking more closely at the case when t = 10, s = 100, the question is asked “How fast was the car travelling?”

One answer is 100m in 10 seconds equals 100/10 = 10 m/s. This describes the average velocity (velocity is a measure of the rate of change of position(s) ). But as with any car journey, the velocity is always changing.

Differentiation allows us to calculate the velocity at any time instant.

21

dtds

sv or Velocity

tdtds

ts 2 then If 2

smdtds

t 422 ,2 Then when

smdtds

t 20102 ,10when

22

t (secs) s (m)

1 1 2

3 9 6

5 25 10

7 49 14

10 100 20

v s m s

tdtds

svts 2 ,2

Average velocity = 100/10 = 10 m/s

The derivative describes the rate of change of position (instantaneous velocity) at any time.

In summary:

timetotaldistance total

velocityaverage

23

Consider the following situation:

The total cost function is given by

20.5 10,000 1000C q q

1000 ,0 Cq

210, 0.5 10 10,000 10 1000

$100,950

q C

2

1000, 0.5 1000 10,000 1000 1000

$9,501,000

q C

24

The average cost for producing q items is given by

( is the usual symbol for average)

That is,

Cc

q

100,95010, $10,095 per item

10q c

9,501,0001000, $9501 per item

1000q c

produced items no. totalcost total

cost average

25

How can the derivative be used and interpreted?

describes how quickly cost is changing

A comparison shows that cost is changing at a faster rate when q = 10 than when q = 1000.

2If 0.5 10,000 1000C q q then 1 10,000C q

C

10 1 10 10,000 9990C

1000 1 1000 10,000 9000C

26

These figures also give an approximate cost to produce the next item.

The 11th item will cost approximately $9990 to produce.

The 1001st item will cost approximately $9000 to produce.

When C describes the total cost, then represents the marginal cost.

.999010 C

.90001000 C

dqdC

C or

C is called the marginal cost function.

27

Marginal cost function:

Marginal cost

77005.46.003.0 23 qqqC

5.42.109.0 2 qqdqdC

5.15.4102.11009.0 ,10 2 dqdC

q

5.165.4202.12009.0 ,20 2 dqdC

q

5.7845.41002.110009.0 ,100 2 dqdC

q

Example:

28

Total cost function

Marginal cost function:

2 70000.001 0.3 40c q q

q

CC

c cqq

7000403.0001.0

7000403.0001.0

23

2

qqq

qq

qqC

406.0003.0 2 qqdqdC

Example:

29

406.0003.0 2 qqdqdC

3.3440106.010003.0 ,10 2 dqdC

q

2.2940206.020003.0 ,20 2 dqdC

q

30

There is an equivalent interpretation for the revenue function.

If R represents total revenue then is called the marginal revenue.

Marginal revenue describes two things:

(1) How quickly revenue is changing.

(2) Approximate revenue received by selling the next unit.

dqdR

R

31

Example:

Marginal revenue:

22.0601.0302 qqqqR

qdqdR

4.060

56104.060 ,10 dqdR

q

52204.060 ,20 dqdR

q

32

The following conclusions can be drawn from these results:

(1) Revenue is changing faster at q = 10 than at q = 20.

(2) The 11th item will generate approximately $56 and the 21st item will earn $52