chapter 3 differentiation

1

Calculus (Differentiation)

Analytical Methods 1

Level 4

© University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

Calculus Differentiation

The following presentation is an introduction to the Algebraic Methods – part one for level 4 Mathematics. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1st year undergraduate programme.

The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

Contents Calculus Terminology Slope of a straight line Slope of a tangent to a curve. Measuring the Slope of a curve. Generalised Analysis The meaning of dy/dx Another rule. Function of a function Differentiating a function of a function Successive Differentiation

Differentiation of Sin x, Cos x, ex, ln x Differentiation of ex Differentiation of ln x (natural log of x) Differentiation of a Product Differentiation of a Quotient Velocity and Acceleration Maxima and Minima Locating turning points on a graph Distinguishing between Maxima and Minima Summary Exercises Credits

In addition to the resource below, there are supporting documents which should be used in combination with this resource.

Please see:

KA Stroud & DJ Booth, Engineering Mathematics, 8th Edition, Palgrave 2008.

http://www.mathcentre.ac.uk/

Derive 6

3

CalculusCalculus deals with small changes in a function, increasing or decreasing.

Its name derives from “calcula” which is a small counting stone. First introduced into the UK by Newton in the 17th century and at the same time in Germany by Liebnitz.

There are two main areas of calculus – Differential Calculus or Differentiation and Integral Calculus or Integration.

Before we look at the process it is important that we understand some of the terminology.


4

TerminologyFunction

The area of a square (A) depends upon the length of its side (l). We say that A is a function of l.

A falling body travels a distance (s) which is dependent upon the length of time (t). s is a function of t.

In the equation y = 3x2 + 2x + 5 the value of (y) depends upon the value of (x). y is a function of x.

Dependent and Independent Variables.

In the above examples the one variable is dependent upon the value of the other. This variable is the dependent variable. The other variable is not dependent upon the dependent variable and is the independent variable.

5

TerminologyIn the examples:

1st l is the independent variable and A the dependent.

2nd t is the independent variable and s the dependent.

3rd x is the independent variable and y the dependent.

The Functional Notation.

The symbol (x) is used to denote “function of x”.

Thus if y = 3x2 + 2x + 5

We can write (x) = 3x2 + 2x + 5

If we wish to replace x by t we can say:

(t) = 3t2 + 2t + 5

If we want the solution when x = 2

(2) = 3x22 + 2x2 + 5 = 21

6

TerminologyCalculus notation for increasing functions.

If we have a function of x and we have a small increase in x then in calculus this is denoted by Δx (some texts use δx – both are Greek symbols for d Δ – uppercase, δ – lowercase).

Thus for a small increase in x the value would be x + Δx and a small decrease in x would give x – Δx.

If y is a function of x and x increases to x + Δx, y would increase by a corresponding small amount to y + Δy.


7

Slope of a straight lineThe graph shows a straight line graph.

0 x

y

x

y

Δx

ΔyP

Q

R

P and Q are points on the straight line.

The co-ordinates of

P are (x, y) and those of

Q are (x + Δx, y + Δy).

In the triangle PQR, PR is Δx and RQ is Δy.

The slope is QR/PR = Δy/Δx.

For a straight line this value is constant over it’s entire length. It also does not depend upon the distance between P and Q i.e. Δx and Δy.

8

Slope of a tangent to a curve.The slope of a curve varies depending upon the point at which it is measured. The slope at a point is the slope of a tangent at that point.

A tangent is a straight line that touches the curve only at that one point.

P

Q

Note that the slope of the tangent at P is smaller than the slope of the tangent at Q.

For a curve the slope is related to the value of x.


9

Measuring the Slope of a curve.

P

Q

Let us say we want to know the slope at point P on the curve.

P = (x, y)

We pick a point Q further up the curve.

The co-ordinates of Q are (x + Δx, y + Δy). The slope of the line is therefore Δy/Δx.

From the graph we can see that this is not an accurate value of the slope.

If we bring Q closer to P (reduce the value of Δx) then the slope becomes closer to the tangent.

Δx

Δy


10

Measuring the Slope of a curve.

Therefore as Δx 0 the slope of the line between P and Q the slope of the tangent.

As Δx approaches zero the value of Δy/Δx approaches what is called the limiting value.

We write Lim (Δy/Δx) = slope of the tangent of the curve.

Δx 0

Liebnitz used dy/dx as a short hand for the limit expression.


11

ExampleConsider the curve y = x2

Question – What is the slope of the curve at x = 2?

Point P = (2, 22) = (2, 4)

Δx = 1 Q = (3, 9)

Slope = (Δy/ Δx) = (9 – 4)/(3 – 2) = 5

Δx = 0.5 Q = (2.5, 6.25)

Slope = (Δy/ Δx) = (6.25 – 4)/(2.5 – 2) = 4.5

Δx = 0.1 Q = (2.1, 4.41)

Slope = (Δy/ Δx) = (4.41 – 4)/(2.1 – 2) = 4.1

Δx = 0.01 Q = (2.01, 4.0401)

Slope = (Δy/ Δx) = (4.0401 – 4)/(2.01 – 2) = 4.01

The limiting value is 4 and so dy/dx = 4Calculus Differentiation

12

Generalised AnalysisConsider the curve y = x2

Point P = (x, y) y = x2

Point Q = (x + Δx, y + Δy)

But y + Δy = (x + Δx)2 = x2 + 2xΔx + (Δx)2

Δy = x2 + 2xΔx + (Δx)2 – y = x2 + 2xΔx + (Δx)2 – x2

Δy = 2xΔx + (Δx)2

Slope = Δy/Δx = (2xΔx + (Δx)2)/Δx = 2x + Δx

As Δx 0 Slope 2x

Therefore dy/dx = 2x

When x = 2 slope = 4Calculus Differentiation

13

The meaning of dy/dxIt is called the first differential coefficient of y with respect to x.

It is the slope of the tangent to a given curve.

It is the rate of change of y with respect to x

It is possible to determine the value of dy/dx for a polynomial using a simple rule.

RULE

If y = axn Where a and n are constants

Then dy/dx = naxn-1


14

Worked examplesSpecial Cases

y = 6x = 6x1

dy/dx = (1x6)x0 = 6

y = 7 = 7x0

dy/dx = (0x7)x-1 = 0

Differentiate the following with respect to x

1. Y = 5x3

dy/dx = (3x5)x2 = 15x2

2. Y = 4/x2 = 4x-2

dy/dx = (-2x4)x-3 = -8/x3Calculus Differentiation

15

Worked examples3. Y = 4√x = 4x½

dy/dx = (½x4)x-½ = 2/√x

Examples

Differentiate with respect to the appropriate variable:

• Y = 8x5

• s = -15/t4

• l = 6√m


16

Another rule.If terms are linked by positive or negative signs the

differential is simply the combination of the individual differentiated terms.

e.g. if y = 5x4 + 2x2 – 3x

then dy/dx = 20x3 + 4x - 3

if y = 4x2 + 2x3

dy/dx = 8x + 6x2

if y = 3x4 – 7x3 + 2x – 5

dy/dx = 12x3 – 21x2 + 2


17

Examples1. Find dy/dx for each of the following cases

3

2

23

2

23

)3)(12()(

435)(

)3()(

31

4)(

x

xxyd

x

xxyc

xyb

xxya

2. Find the gradient of the curve y = 7x4 at the point where x = 1

3. If y = 3x2 – 6x + 4, find the co-ordinates of the points at which (i) dy/dx = 0, (ii) dy/dx = 6


18

Function of a functionMeaning:

Suppose y = 3(4x – 2)4

Let z = 4x – 2

Now y = 3z4

We can see that y is a function of z

But also we can see that z is a function of x

We can therefore say that:

y is a function of a function of x


19

Differentiating a function of a functionUse the example given: y = 3(4x – 2)4

Once again let z = 4x – 2 and y = 3z4

From the first equation: dz/dx = 4

From the second equation: dy/dz = 12z3 = 12(4x – 2)3

Now

--chain rule

Using this dy/dx = 12(4x – 2)3 times 4 = 48(4x – 2)3

What we can say is where we have brackets, the value of dy/dx is the differentiation of the expression taking the brackets as an entity times the differentiation of the inside of the brackets.

dx

dz

dz

dy

dx

dy


20

Worked examplesDifferentiate with respect to x

y = 4(5x2 - 1)2

dy/dx = 8(5x2 – 1) times 10x

dy/dx = 80x(5x2 – 1)

21

1221

1221

xxy

643

)13(1

)()23(7)(

x

bxxa

xtorespectwithfollowingtheateDifferenti

2times)12(41

21

xdxdy

122

1

xdxdy

21

Successive DifferentiationWhen we have a function e.g. y = (x) and we

differentiate it, then the first differential coefficient is written as:(x) or dy/dx

If it is differentiated again then the second differentail coefficient is written as: (x)or d2y/dx2

Note d2y/dx2 =d/dx(dy/dx) (differentiation of dy/dx with

respect to x.)

This is pronounced as dee two y by dee x squared.

If we differentiate a third time then we have the third differential coefficient: (x)or d3y/dx3

This process is known as successive differentiation.Calculus Differentiation

22

Worked Examples1. Successively differentiate y = x3 + 2x2 – 1

dy/dx = 3x2 + 4x

d2y/dx2 = 6x + 4

d3y/dx3 = 6

d4y/dx4 = 0

2. If y = 3x2 + 2x3 – 6 find (x) and (2)

(x) = 6x + 6x2

(x) = 6 + 12x

(2) = 6 + 12(2) = 30


23

Examples1. Successively differentiate

y = 5x4 – 2x2 + 8

2. If y = 3x3 + 2x – 4

Show that

30392

2

2

3

3

ydxdyx

dxyd

dxyd


24-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Differentiation of Sin x, Cos x, ex, ln xThe graph below shows a sine curve (y = sin x):

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 π/2 π 3π/2 2π

At 0, slope is +ive max

0 – π/2, slope reduces to 0

At π/2, slope is zero

π/2 – π, slope becomes –ive

At π, slope max –ive

π - 3π/2, slope reduces to 0

At 3π/2, slope is zero

3π/2 - 2π, slope increases +ively

The slope of y = sin x is cos x

dy/dx = cos x

25

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Sin x continuedThe graphs show sin x and sin 2x:

Note:

The slopes are greater for the sin 2x curve. This means that the value of dy/dx will be larger.

What we find is that for sin x the rate of change dy/dx will be cos x but for sin 2x the value is 2cos 2x

Generalised result y = sin ax dy/dx = a cos ax

Sin x

Sin 2x


26-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Differentiation of Cos xThe graph below shows a cosine curve (y = cos x):

0 π/2 π 3π/2 2π

At 0, slope is 0

0 – π/2, slope increases -ively

At π/2, slope is max -ive

π/2 – π, slope reduces to 0

At π, slope is 0

π - 3π/2, slope increases +ively

At 3π/2, slope is max +ive

3π/2 - 2π, slope reduces to 0

The slope of y = cos x is -sin x

dy/dx = -sin x

27

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Cos x continuedThe graphs show cos x and cos 2x:

Note:

The slopes are greater for the cos 2x curve. This means that the value of dy/dx will be larger.

What we find is that for cos x the rate of change dy/dx will be -sin x but for cos 2x the value is -2sin 2x

Generalised result y = cos ax dy/dx = -a sin ax

Cos xCos 2x


28

Differentiation of ex

The graph below shows an exponential curve (y = ex):

0

1

2

3

4

5

6

7

8

-2 -1 0 1 2

x y

-2 0.14

-1 0.37

0 1.00

+1 2.72

+2 7.37

1

2.72

If we measure the slope at a point e.g. x = 1 – then the value we end up with is 2.72/1 = 2.72 (the value of y at that point). This is true for all points on the curve.

If y = ex then the slope dy/dx = ex

29

0

5

10

15

20

25

-2 -1 0 1

ex continuedThe graphs below show curves (y = ex and e2x):

If we measure the slope at a point e.g. x = 1 – then the value we end up with 2.72 for the ex curve and 14.78 for the e2x curve. For the second curve this is twice the value of the function at this point.

Generalised result y = eax dy/dx = aeax

When x = 1

ex = 2.72

e2x = 7.39

e2x

ex

30

Differentiation of ln x (natural log of x)We can work out what the equation is by using the information we have just discovered.

y = ln x what is the value of dy/dx?

We can write y = ln x = loge x as x = ey

If we now differentiate this we get:

dx/dy = ey

But we want dy/dx so this the reciprocal of dx/dy

So dy/dx = 1/ey = 1/x

Generalised result y = ln x dy/dx = 1/x


31

Summary (standard derivatives)

y = f(x) dy/dx

axn naxn-1

sin ax acos ax

cos ax -asin ax

eax aeax

ln x 1/x

Note for sin and cos the angle must be in radians for the answer to be correct.


32

Examples1. Differentiate sin 5x with respect to x

2. Find dy/dx when y = cos(3x – 1)

3. If y = cos(π/2 – 3x), find dy/dx

4. If y = e3-4x, find dy/dx

5. If y = ln(8x + 5), find dy/dx

6. If y = sin8 x, find dy/dx


33

Differentiation of a ProductWe know that 3 x 4 = 12

12 is said to be the product of 3 and 4

Likewise if y = x5 sin x then y is the product of two functions x5 and sin x.

In order to differentiate products we must use the following rule.

If y = u v where u and v are functions of x, then

dx

duv

dx

dvu

dx

dy

-- product rule


34

Worked examplesDifferentiate y = x5 sin x

Always quote the rule ands say what u is and what v is.

dy/dx = u dv/dx + v du/dxu = x5 du/dx = 5x4

v = sin x dv/dx = cos x

Therefore dy/dx = x5 cos x + sin x 5x4

dy/dx = x5 cos x + 5x4 sin x

dy/dx = x4 (x cos x + 5 sin x)


35

Worked examplesDifferentiate y = ex(x3 + 1)

dy/dx = u dv/dx + v du/dx u = ex du/dx = ex

v = x3 + 1 dv/dx = 3x2

Therefore dy/dx = ex 3x2 + (x3 + 1)ex

dy/dx = ex(x3 + 3x2 + 1)

Exercises

1. y = x4ex

2. y = 5x3 cos x

3. y = x5 ln x

4. y = (x7 – 3) sin xCalculus Differentiation

36

Once again we know that 20 4 = 5

5 is said to be the quotient of 20 and 4

Likewise if then y is the quotient of two functions x2 + 1 and 2x

+ 1.

In order to differentiate quotients we must use the following rule.

If y = u/v where u and v are functions of x, then

Differentiation of a Quotient

2vdxdvu

dxduv

dx

dy

1212

xx

y

-- quotient rule


37

Worked examples

Differentiate

Always quote the rule ands say what is u and what is v. u = x2 + 1 du/dx

= 2x

v = 2x + 1 dv/dx = 2Therefore

1212

xx

y

2vdx

dvudxduv

dxdy

2

2

)12(2)1(2)12(

x

xxxdx

dy

2

2

2

22

)12(222

)12(2224

xxx

xxxx

dxdy


38

Worked examples

Differentiate

u = e3x du/dx = 3e3x

v = x3 + 1 dv/dx = 3x2

Therefore

Exercises

13

3

xe

yx

2vdx

dvudxduv

dxdy

23

233

23

2333

)1()1(3

)1(33)1(

xxxe

xxeex

dxdy xxx

xe

xy

xx

y

xx

y

xx

y

sin23.4

1.3

2323

.2

sin.1

3

4

2

2

39

Velocity and AccelerationIf a body (e.g. a car travelling along a straight road) moves a distance x metres in time t seconds at a constant velocity v metres per second, then

tx

vei ..takentimetravelledDistance

Velocity

or x = vt and a graph of x against t would give the following straight line.

x

t

A

B C

The slope of the graph is:

which is the velocity of the body

takentimeDistancea

..BCAC

ei

40

Velocity and AccelerationIf a body is not moving with constant velocity, the graph of distance travelled (x) against time (t) would be a curve, as shown:

x

t

A

B C

The slope of the curve (e.g. at P) would be given by the differential coefficient, dx/dt, and this slope represents:

i.e. an instant velocity.(BC)taken time

(AB) travelleddistancea

P


41

Velocity and AccelerationConsequentially, if we know x (the distance travelled) in terms of t (the time taken) in the form x = (t) (i.e. x is a function of t)

then dx/dt = V, the velocity at time t

and d2x/dt2 = a, the acceloration at any time t.

a is also given by dV/dt.

Summary

If x = (t)

Then velocity = V = dx/dt

And acceleration = a = dV/dt = d2x/dt2


42

ExampleThe distance in metres moved by a body in time t is given by x = t3 – 9t2 + 24t + 1

Find (i) V and a after t seconds and after 3 seconds

(ii) when the body comes to rest.

(iii) V when a = 12 ms-2

Solution (i)

x = t3 – 9t2 + 24t + 1

Velocity = V = dx/dt = 3t2 - 18t + 24

Acceleration = a = dV/dt = 6t – 18

When t = 3 V = 3t2 - 18t + 24 = 3(32) – 18(3) + 24 = -3ms-1

a = 6t – 18 = 6(3) – 18 = 0

43

ExampleSolution (ii)

The body comes to rest when the velocity V = 0.

3t2 - 18t + 24 = 03 t2 – 6t + 8 = 0

(t – 2)(t – 4) = 0 so at rest when t = 2seconds or 4 seconds

Solution (iii)

Time when the acceleration = 12ms-2

a = 6t – 18 = 12 6t = 30 t = 5 seconds

Velocity = 3t2 - 18t + 24 = 3(52) – 18(5) + 24 = 75 – 90 + 24

Velocity = 9ms-1


44

ExercisesA body moves a distance x metres along a straight line in time t seconds where x = 4t3 - 15t2 – 18t + 100

Find (a) the velocity and acceleration after t seconds

(b) the velocity and acceleration at the start (when t = 0)

(c) the time at which the body comes to rest

(d) the velocity when the acceleration is zero

(e) how far has the body travelled when it comes to rest.


45

Maxima and MinimaWe may be presented with a function of x, (x), and it may be necessary to determine when the function reaches a maximum or minimum value. We can do this using differentiation.

Consider the following graphs

x

yThis line is said to have a positive slope as when x increases in value y increases in value, i.e., dy/dx > 0.

x

yThis line is said to have a negative slope as when x increases in value y decreases in value, i.e., dy/dx < 0.

46

Locating turning points on a graphConsider a hypothetical curve y = (x) shown in the plot below.

x

y

At the points A, B, C and D the curve turns and these are called the turning points. If the tangents are drawn at the turning points it can be seen that they are horizontal i.e. zero slope or dy/dx = 0

The x values can be found by equating the first differential to zero and solving the equation.

A

B

C

D

(x)

47

Locating turning points on a graphThe corresponding y values can be found by using the x values in the original function.

e.g. Locate the turning points on the curve y = 3x2 – 6x + 2

Slope = dy/dx = 6x – 6

Equate to zero and solve 6x – 6 = 0 6x = 6 x = 1

Determine value of y y = 3x2 – 6x + 2 = 3(1)2 – 6(1) + 2 = -1

Therefore the single turning point occurs at (1, -1)


48

Distinguishing between Maxima and Minima

Consider the slopes just before and just after a turning point.

Maxima

x

y

+ive slope

-ive slope

zero slope

As we move from points below the maxima to points above the slope goes from positive through zero to negative. If we were to plot the slope we would have:

x

dy/dx The slope of this line is negative. So if the slope of the slope curve is negative we have a maxima

If d/dx(dy/dx) = d2y/dx2 < 0 (maxima)

49

Distinguishing between Maxima and Minima

Minima

x

y

+ive slope

-ive slope

zero slope

As we move from points below the minima to points above the slope goes from negative through zero to positive. If we were to plot the slope we would have:

x

dy/dx The slope of this line is positive. So if the slope of the slope curve is positive we have a minima

If d/dx(dy/dx) = d2y/dx2 > 0 (minima)


50

Summary

To locate and distinguish between turning points (local maxima and local minima) the following method should be used.

To find the location of the points

(i) Differentiate the function i.e. find dy/dx

(ii) Solve the equation dy/dx = 0

(iii) The solution will give us the x values

(iv) Substitute values of x into the original function equation to determine corresponding values of y.


51

Summary

To find the nature of each turning point

(i) Find the second derivative d2y/dx2 of the function

(ii) Substitute in the x values found in (iii) above.

(iii) If d2y/dx2 < 0, we have a maxima

if d2y/dx2 > 0, we have a minima

If you are asked to determine the maximum or minimum value of the function, it is the y value at the corresponding value of x.


52

Worked Example 1

Find the turning points on the curve y = 2x3 + 3x2 – 36x + 1 and distinguish between them. Sketch the curve.

y = 2x3 + 3x2 – 36x + 1

To find turning points dy/dx = 6x2 + 6x – 36

Equate to zero 6x2 + 6x – 36 = 0

Simplify x2 + x – 6 = 0 1 x -6 = -6

+6 -6 +3 -3

-1 +1 -2 +2

+5 -5 +1 -1

Result +3 -2

x2 + 3x – 2x -6

x(x + 3) – 2(x + 3)

(x – 2)(x + 3)Calculus Differentiation

53

Worked Example 1(x – 2)(x + 3) = 0

x = 2 or x = -3

When x = 2

y = 2x3 + 3x2 – 36x + 1 = 2(2)3 + 3(2)2 – 36(2) + 1

y = 16 + 12 – 36 + 1 = -43

One turning point is (2, -43)

When x = -3

y = 2x3 + 3x2 – 36x + 1 = 2(-3)3 + 3(-3)2 – 36(-3) + 1

y = -54 + 27 + 108 +1 = 82

One turning point is (-3, 82)


54

Worked Example 1Distinguish between maxima and minima

dy/dx = 6x2 + 6x – 36

d2y/dx2 = 12x + 6

When x = 2 d2y/dx2 = 12x + 6 = 12(2) + 6 = +30 minima

When x = -3 d2y/dx2 = 12x + 6 = 12(-3) + 6 = -30 maxima

The point (2, -43) is a minima

The point (-3, 82) is a maxima


55

Worked Example 1Sketch not necessarily an accurate graph

What points do we know on the curve?

1 It has a maxima at (-3, 82)

2 It has a minima at (2, -43)

3 It crosses the y axis at 1 (x = 0)

-60

-40

-20

0

20

40

60

80

100

-4 -3 -2 -1 0 1 2 3

56

Worked Example 2Find the maximum and minimum values of the curve

Y = 2x3 – 15x2 + 36x + 2

dy/dx = 6x2 -30x + 36

Equate to zero 6x2 -30x + 36 = 0

x2 -5x + 6 = 0 (x – 2)(x – 3)=0 x = 2 or x = 3

x = 2 y = 2x3 – 15x2 + 36x + 2 = 2(2)3 – 15(2)2 + 36(2) + 2 (2, 30)

x = 3 y = 2x3 – 15x2 + 36x + 2 = 2(3)3 – 15(3)2 + 36(3) + 2 (3, 29)

d2y/dx2 = 12x – 30

x = 2 d2y/dx2 = 12(2) – 30 = -6 (2, 30) maxima

x = 3 d2y/dx2 = 12(3) – 30 = +6 (3, 29) minima


57

Exercises1. Find the maxima or minima coordinates for the

following curve: y = 3x2 – 6x + 2

2. Find the maxima and minima coordinates for the following curve: y = x3 + 3x2 – 9x

3. Find the maxima or minima values for the following function: y = x3 – 9x2 + 15x - 7

4. The bending moment M at a distance x from one end of a beam is given by the equation:

Show that when W = 40 and L = 10, the maximum value of M = 500

)(2

xLWx

M


58

Practical Exercises1. A sheet of metal has the dimensions 100cm by

75 cm. Squares are to be cut out of each corner xcm by xcm and the sides folded up to make a tray. What size square must be removed on each corner to maximise the tray volume?

2. A 330ml can is to be produced from thin sheet metal. What must its dimensions equal to minimise the area of metal used?

3. A site has an area of 7200m2. It is bounded on one side by a shallow river. Find the minimum length of fencing needed to enclose this rectangular area.


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© 2009 University of Wales Newport

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chapter 3 differentiation

Education

factors x

x yx

value of b

c algebraic methods

b1 y algebraic methods

original quadratic

quadratic means

a1 y