Unit No: DT5X 33 DG4H 33

Unit Title: Mathematics for Engineering 1:

Outcome 1

Additional Notes

Problem Solving involving Powers

and Logs(Evaluation using a scientific

calculator,Transposition and Indices)

Engineering and Construction

1 RD/PKN

25/8/2009 RD/PKN

Mathematics for Engineering

Applying Algebraic Methods in Problem Solving

Section 1 - Revision

It is assumed that you will have acquired some knowledge of indices and of logarithmand exponential expressions in previous mathematics courses. There follows a briefsummary of the work covered previously. More information and exercises are available.

Laws of Indices : Example

Multiplication am % an = a(m+n) x3 % x2 = x5

Division am

an = a(m−n) x2

x5 = x−3

Powers (am )n = amn (x2 )3 = x6

Roots amn = n am 3 x6 = x

63 = x2

Reciprocals ka−n = kan 5x−3 = 5

x3

Zero index a0 = 1 (2x3 )0 = 1

3 x 8 = 28 SHIFT x1y 3 = 2

3 8 or 813

( (−) 3 ) xy (−) 4 = 0.123 3 +/ − xy 4 +/ − = 0.0123

(−3)−4

3 xy ( (−) 2 ) = 0.1.

3 xy 2 +/ − = 0.1.

3−2

TYPE CCasio S-VPAM, All Graphicscalculators

TYPE BCasio VPAM, SharpDAL

TYPE ASome Casio fx, Texas,Texet, early Sharp

Note: Your x y key may be displayed as , y x or ^ or a b.x»

Try these ones

a. 4-3 = 0.0156 b. = 0.1975 c. = 0.3333( 23 )4 81− 1

4

25/8/2009 1 RD/PKN

2. The Constant ' 'e

Like , is a naturally occurring constant and has a value of 2.71828... . To�, echeck this, enter 1 and use the button, ie: (to 10 sf). ex e1 = 2.718281828

2.718281828459045235360287471352662497757247093699959574966967627724........

Note, like it is a transcendental number and although it looks as though it might�repeat ..1828.. it doesn’t. What follows looks completely random.

Functions containing are used in 2 closely related ways:ex

(i) as the base of Natural (or Naperian) logarithms

(ii) in the laws of Natural growth and decay----------------------------------------------------------------------------------------------------------------------

3. Definition of a Logarithm

If is a positive number such that: where is positive, then is called theN N = ax a xLOGARITHM of to BASE . It is written as N a logaN = x

N = ax e logaN = x

Remember: NUMBER = BASELOGARITHM

Example

If then 125 = 53 log5125 = 3----------------------------------------------------------------------------------------------------------------------

4. Base for Logarithms

Any positive number can be the base but the usual bases are:

(i) Base 10 written as log10x

Use the LOG button on the calculator.

(ii) Base e

Logarithms to base are known as Natural (or Naperian) logs. Written as e logexor usually . (John Napier a.k.a. Lord Merchiston, 1550 - 1617, Edinburgh)ln x

Use the ln button on the calculator.----------------------------------------------------------------------------------------------------------------------

transposes to transposes to y = log10x x = 10y y = 10x x = log10y transposes to transposes to y = ln x x = ey y = ex x = ln y

25/8/2009 2 RD/PKN

Examples Check using the calculator: (all answers given correct to 4 sig fig)

1. 2.e1.6 = 4.953 e−0.3 = 0.7408

3. 4.log1018 = 1.255 ln 4.3 = 1.459

Note: ln 1 = 0 logee = 1 log101 = 0 log1010 = 1

ln x = logex exp x h ex

----------------------------------------------------------------------------------------------------------------------

5. Laws of Logarithms

1. log(A % B) = log A + log B

2. log AB = log A − log B

3. log An = n log A

4. logay =logbylogba

----------------------------------------------------------------------------------------------------------------------

Examples Express as a single log:

Note that the base is not specified - they can’t be evaluated

(i). Use Law 1 (ii). Use Laws 1 & 2

log 5 + log 4

= log(5 % 4)

= log 20

log 32 − log 8 + log 24

= log 32 % 348

= log 96

Now using Law 3 then Law 1

(iii) (iv)

2 log 5 + 3 log x

= log 52 + log x3

= log 25x3

log 2x + k log x

= log 2x + log xk

= log((2x)(xk ))

= log(2xk+1 )

Example (iv) used Indices Rule 1

25/8/2009 3 RD/PKN

6. Graphs of Exponential and Logarithm Functions

(a) the natural growth curve (b) the exponential decay curve

y = ex y = e−x

We note the following points of importance in the 2 graphs.

(i) each graph is continuous(ii) and are each positive for all values of ex e−x x(iii) one graph is the mirror image of the other in the -axisy(iv) the gradient of is positive and increases rapidly as increasesex x(v) the gradient of is negative and decreases rapidly as increasese−x x

(c) the natural logarithm curve

y = logex

The graph, exists for only. y = logex x > 0The gradient is positive and decreases as x increases.

Note: the graph of is the mirror image of in the line y = logex y = ex y = x

25/8/2009 4 RD/PKN

10 8 6 4 2 0 2 4 6 8 10

2

1

12

3

45

6

78

9

1010

2−

ex

1010− x

10 8 6 4 2 0 2 4 6 8 10

2

1

12

3

45

6

78

9

1010

2−

ex−

1010− x

2 1 0 1 2 3 4 5 6 7 8 9 10

5

4

3

2

1

1

2

3

4

55

5−

ln x( )

102− x

10 8 6 4 2 0 2 4 6 8 10

2

1

12

3

45

6

78

9

1010

2−

ex

ln x( )

x

1010− x

An interesting fact which you will meet later in the calculus unit is: the function is theex

only mathematical function which does not change on differentiation, That is d

dx (ex) = ex

(gradient at any point is the same as original function value).

(d) y = 1 − e−x

In this graph, since decreases as increases, the value of y tends toe−x xapproach 1. As increases, becomes smaller and smaller, and getsx e−x 1 − e−x

closer and closer to 1.----------------------------------------------------------------------------------------------------------------------

Revision Exercise

1. Evaluate correct to 4 significant figures:

(i) (ii) (iii) (iv)3.5e2.8 65 e−1.5 6(1 − e−0.4 ) 3 ln 1.6

(v) 13 ln 0.7

2. Change to exponential form:

(i) (ii) (iii) (iv)log3p = q ln 2.6 = x log10y = b 4 = logx16

3. Change to logarithmic form:

(i) (ii) (iii) (iv)ex = 1.2 10x = 3 0.3 = e−x p = aq

4. Express as a single log:

(i) (ii) (iii)log 2 + log 3 + log 4 3 log 2 − 2 log 3 2 log x − log 3

(iv) (v) (vi)12 log 16 − 2

3 log 8 log(a + 2) + log a a log x + 3 log x

25/8/2009 5 RD/PKN

0 1 2 3 4 5 6 7 8 9 10

0.1

0.2

0.30.4

0.5

0.60.7

0.8

0.91

1.1

1.21.2

0

1 ex−−

1

100 x

Section 2 - Natural Laws of Growth and Decay

The Natural laws of growth and decay are of the form or where y = Aekx y = A(1 − ekx ) Aand are constants which can be either positive or negative . They relate quantities inkwhich the rate of increase of is proportional to itself for the growth law and the ratey yof decrease of is proportional to itself for the decay law.y y

These laws occur frequently in Engineering and Science. Examples include:

(i) Linear Expansion

A rod of length at temperature 0oC and having a positive coefficient of linear lexpansion of will become longer when heated. The natural growth law is:�

( is length of rod at 0oC, and is the temperature)l = l0e�t l0 t

(ii) Atmospheric Pressure

The pressure at height above ground level is given by where isp h p = p0e− hc p0

pressure at ground level and is a constant.c

(iii) Tension in Belts

The relationship between the tension in a belt round a pulley wheel and itsT1

angle of lap is of the form where is the coefficient of friction� T1 = T0em� mbetween belt and pulley and and are the tension on the tight and slackT1 T0

sides of the belt respectively.

Other Examples Include:

(iv) Change of Electrical Resistance with Temperatur e:R = R0e�t

(v) Discharge of a Capacitorq = Qe− t

CR

(vi) Newton's Law of Coolings = soe−kt

(vii) Radioactive DecayN = N0e−t

25/8/2009 6 RD/PKN

Examples of Growth and Decay Curves

Example 1The decay of voltage, v volts, across a capacitor at time t seconds is given by:

. The graph is drawn over the first 6 seconds . The natural decay curve is v = 250e− t3

.v = 250e− t3

Note: the rate of change can be calculated by finding the gradient at any point on thecurve

Example 2

The growth of current in the circuit is opposed by the emf induced in the coil ofinductance . The instantaneous current amperes after a time seconds is given by: L i t

.i = ER 1 − e

−RtL

This is the graph off the function . When is large, is very small andi = ER 1 − e

−RtL t e

−RtL

which represents the final steady current.i = ER

t

iER_

25/8/2009 7 RD/PKN

0 2 4 6

100

200

300250

33.834

250e

t−3

60 t

EVALUATION OF EXPRESSIONS

We have seen how exponential functions, and hence logarithmic functions form thebasis for rules for growth, decay, cooling, atmospheric pressure to name but a few. Weneed to be able to evaluate such expressions.

Examples

Evaluate to 3 decimal places:

1. when and ln( p1p2

) p1 = 7000 p2 = 8000

ln( p1p2

) = ln( 70008000 ) = −0.1335

2. i0(1 − e−20t ) when i0 = 2 and t = 0.06

i0(1 − e−20t ) = 2(1 − e(−20%0.06) ) = 1.398

Exercise 1

Evaluate to 3 decimal places:

1. ln( p1p2

) when p1 = 2500 and p2 = 4935

2. i0(1 − e−5t ) when i0 = 3.5 and t = 0.06

3. p0e− 0.15h when p0 = 100000 and h = 5

4. N0e− lt when N0 = 1500, l = 0.25 and t = 7.9

5. A ln( c1c2

) when c1 = 50, c2 = 38 and A = 15.6

6. KRT log10A when K = 5, R = 0.4, T = 56 and A = 315

----------------------------------------------------------------------------------------------------------------------

25/8/2009 8 RD/PKN

SECTION 3: INDICIAL, LOGARITHMIC AND EXPONENTIAL EQ UATIONS

When solving Indicial, Exponential and Logarithm equations it is usually necessary tore-arrange the equation by changing its form (i.e. Carry out trasposition). Certainrecognised methods can be used. Simple examples of these are as follows:

(a) Type k = ex

Working: (Remember NUMBER = BASELOG)k = ex

or (Change to form to "undo" fromln k = x logek = x log xthe index position)

----------------------------------------------------------------------------------------------------------------------

(b) Type You might see either formlogex = q ln x = q

Working: (Change to exponential form to "undo"logex = q ln x = q from log position)x

x = eq x = eq

----------------------------------------------------------------------------------------------------------------------

(c) Type loga p + logaq = logax

Working: loga p + logaq = logax

(Use first log law)loga(p % q) = logax

pq = x----------------------------------------------------------------------------------------------------------------------

(d) Type (ie, with base other than 10 or e)px = q

Working: (Taking log of both sides use base 10px ==== qor base e)

log10px = log10q

Log law 3x log10p = log10q

x =log10qlog10p

-------------------------------------------------------------------------------------------------------------------

25/8/2009 9 RD/PKN

Worked Examples

(a) Type k = ex

Solve for :x

(i) 5e− 0.3x = 14.1

(i)

5e− 0.3x = 14.1

e− 0.3x = 14.15

− 0.3x = ln 14.15

x = ln( 14.15 )

−0.3

x = −3.456

divide by 5

loge both sides

divide by − 0.3

4 sf

---------------------------------------------------------------------------------------------------------------------

(ii) 220(1 − e−0.5x )

(ii)

220(1 − e−0.5x ) = 48

1 − e−0.5x = 48220

1 − 48220 = e−0.5x

ln 1 − 48220 = −0.5x

ln(1 − 48220 )

−0.5 = x

0.4922 = x

divide by 220

add e−0.5x to both sides

loge both sides

divide by − 0.5

4 sf

----------------------------------------------------------------------------------------------------------------------

Try the following examples:

Exercise 2

Solve for (to 4 significant figures):x

(i) e2x = 5.6 (ii) 5e0.5x = 4.1 (iii) 1 − ex = 0.6 (iv) 100(1 − e−0.3x ) = 80----------------------------------------------------------------------------------------------------------------------

25/8/2009 10 RD/PKN

(b) Type logex = q log10x = q

Solve for :x

(i) log10(1 + x) = 2.3

(i)

log10(1 + x) = 2.3

1 + x = 102.3

x = 102.3 − 1

x = 198.5

10x both sides

− 1

(ii) 2 ln x + ln 5.3 = 4.6

(ii)

2 ln x + ln 5.3 = 4.6

ln x2 + ln 5.3 = 4.6

ln x2 = 4.6 − ln 5.3

x2 = e(4.6−ln 5.3)

x = e(4.6−ln 5.3)

= 4.333

ex both sides

only + ve soln. for real answer

(iii) 4 ln 6.8x = 3.2

(iii) is under the line!

4 ln 6.8x = 3.2

ln 6.8x = 0.8

6.8x = e0.8

6.8 = e0.8 % x

6.8e0.8 = x

3.055 = x

x

---------------------------------------------------------------------------------------------------------------------Exercise 3Solve for x

(i) log10x = 3 (ii) logx81 = 4 (iii) log10(3 + x) = 0.8 (iv) 3 ln x − ln 5 = 2.4

(v) 6 log106.2x = 8.4

-------------------------------------------------------------------------------------------------------------------

25/8/2009 11 RD/PKN

(c) Type log ap + log aq = log ax

Example

Solve for x:

(i) log(x − 3) + log x = log 4

Working: Use first log law

log((x − 3)x) = log 4

(x − 3)x = 4

x2 − 3x − 4 = 0

This is a quadratic equation which can be solved either by (i) factorisation or (ii)quadratic formula.

(i) Factorisation Method

x2 − 3x − 4 = 0

(x + 1)(x − 4) = 0

x + 1 = 0

x = −1

x − 4 = 0

x = 4

But log(-1) does not exist so x = 4

(ii) Quadratic Formula

For formula is ax2 + bx + c = 0 x = −b ! b2 − 4ac2a

For equation a = 1 b = -3 c = -4x2 − 3x − 4 = 0

x =−(−3) ! ((−3)2 − 4 % 1 % (−4))

2 % 1

= 3 ! 252

= 3 ! 52

x = 4 or x = −1

Only feasible answer is x = 4----------------------------------------------------------------------------------------------------------------------Exercise 4

Solve for (3 decimal places):x

(i) (ii)log(x − 5) + log x = log 6 2 log(3x) − log x = log 1.6

25/8/2009 12 RD/PKN

---------------------------------------------------------------------------------------------------------------------d. Type p x= q

Solve for :x

(i) 3x = 5.6

Using and 3rd log law

3x = 5.6

log103x = log105.6

x log103 = log105.6

x =log103

log105.6

x = 1.568

log10

(ii)(ii) 5(x−1) = 3.1x

Use natural logs this time (for a change). (You could use base 10 logs instead).

Log law 3

5(x−1) = 3.1x

ln(5(x−1)) = ln(3.1x)

(x − 1) ln(5) = x ln(3.1)

x ln(5) − ln(5) = x ln(3.1)

x(ln(5) − ln(3.1)) = ln(5)

x =ln(5)

(ln(5) − ln(3.1))

x = 3.368---------------------------------------------------------------------------------------------------------------------

Exercise 5

Solve for (to 3 decimal places):x

(i) (ii) (iii)3x = 12 2.4 = 3.6(−2x) 1.8(x+3) = 6.52

(iv) (v)3.5 = 1x1.3 1.3(2x−1) = 6.3(x+1)

---------------------------------------------------------------------------------------------------------------------

25/8/2009 13 RD/PKN

SECTION 4: CHANGING THE SUBJECT OF FORMULAE

Sometimes it is necessary to change the subject of a formula. If the formula containslogarithm, exponential or indicial expressions, certain procedures are used.

A. Formula Containing Several Logarithms

Example 1: Re-arrange the formula lna - lnb = c to make a the subject

Second log law, “undo” with

ln a − ln b = c

ln ab = c

ab = ec

a = bec

ln ex

Example 2: Re-arrange ln (y + 1) = ln (x + 2) + ln A to make y the subject

First log law

ln(y + 1) = ln(x + 2) + ln A

= ln(A(x + 2))

y + 1 = A(x + 2)

y = A(x + 2) − 1

Example 3: Re-arrange 2 ln y = 5x + ln3y for y

Log law 3 then 2 and some algebra

3 ln y = 5x + ln(3y)

ln y3 = 5x + ln(3y)

ln y3 − ln(3y) = 5x

lny3

3y = 5x

y3

3y = e(5x)

yyy/3y/ = e(5x)

yy3 = e(5x)

y2 = 3e(5x)

y = (3e(5x) )

----------------------------------------------------------------------------------------------------------------------

25/8/2009 14 RD/PKN

B. Formula Containing a Logarithm of More Than One V ariable

Example: Re-arrange the formulae L = k log 10d2

d1

a. for d2 and b. for d1

Working for a:

L = k log10d2

d1

Lk = log10

d2

d1

10Lk = d2

d1

d110Lk = d2

continuing for b:

d110Lk = d2

d1 = d2

10Lk

----------------------------------------------------------------------------------------------------------------------

C. Formulae Containing Variable in Index Form

Example: Re-arrange the formula l = l e µq to make µ the subject0

Working: is the same as

I = I0e(�q)

II0

= e(�q)

logeII0

= �q

logeI

I0

q = �

loge ln

----------------------------------------------------------------------------------------------------------------------

25/8/2009 15 RD/PKN

Example: change the subject of to i = ER 1 − e − t

CR t

This is a difficult but common problem

i = ER 1 − e − t

CR

iR = E 1 − e − tCR

iRE = 1 − e − t

CR

e − tCR + iR

E = 1

e − tCR = 1 − iR

E

− tCR = ln 1 − iR

E

− t = CR ln 1 − iRE

t = −CR ln 1 − iRE

----------------------------------------------------------------------------------------------------------------------

Example: Change the subject of PV n = C to n

Working:

(Note: base is not or 10). Log law 3

pvn = c

vn = cp

log10(vn ) = log10

cp

n log10(v) = log10

cp

n =log10

( cp )

log10(v)

e

---------------------------------------------------------------------------------------------------------------------

25/8/2009 16 RD/PKN

Exercise 6: Changing the Subject of Formulae

Re-arrange the following to give a formula in the letter indicated in brackets.

1(i) (ii)a = log10x − log10y [x] 12 log10a + 3 log10b = c [b]

(iii) (iv)T = ln x + ln y [y] 2 ln x − ln y = z [x]

(v) (vi) p = 13 (ln x − ln y) [x] ln(y + 1) = sin x + ln A [y]

(vii) (viii) ln(y − 3) = ln(x + 2) + ln a [y] 2 ln y = 3x + ln(4y) [y]

---------------------------------------------------------------------------------------------------------------------

2(i) (ii)Q = k log10( p2

p1) [p2] p = A loge

( v2v1

) [v1 ]

(iii) (iv)m = lnT1

T2[T1] a = 1

2 ln( xy ) [y]

---------------------------------------------------------------------------------------------------------------------

3(i) (ii)v = we�t [t]T1

T2= e�s [s]

(iii) (iv)�1 = �oekt [t] i = Ie−kt [t]

(v) (vi)i = ER (1 − e−kt ) [t] P = Poe− h

c [c]

_____________________________________________________________________

25/8/2009 17 RD/PKN

SECTION 5: PROBLEMS INVOLVING EXPONENTIAL AND LOGA RITHMFUNCTIONS

1. The pressure p at height h above ground level is given by , where Po isP = Poe− hc

the pressure at ground level and c is a constant. Whe Po is pascals101.3 % 103

and the pressure at a height of 1570 metres is pascals, determine the98.71 % 103

value of c.

Working:

Use your solution to Exercise 6, 3 (vi) and substitute in the numbers

c = 60617.4----------------------------------------------------------------------------------------------------------------------

2. The temperature of an electrical conductor at time t seconds is given by: q2oC

, where is the initial temperature and seconds is a constant.q2 = q1(1 − e− tT ) q1 T

Determine:

a. when = 50oC, = 30 seconds and = 80sq1 q2 t T

b. the time for to fall to half the value of if remains at 80st q2 q1 T

Working:

a. b.

q2 = q1(1 − e− tT )

q2

(1 − e− tT ) = q1

501 − e − 30

80

= 159.892..

q1 = 159.9oC

q2 = 12 q1

12 q1 = q1(1 − e− t

T )

12 = 1 − e− t

T

e− tT = 1 − 1

2

= 12

− tT = ln 1

2

t = −T ln 12

= −80 ln 12

= 55.45

It takes 55.45 seconds for q2 to fall to ½ value of q1 ----------------------------------------------------------------------------------------------------------------------

25/8/2009 18 RD/PKN

Example 3: Calculate T 1 from the Formula when q = 1� loge

T2

T1

� = 0.15, T2 =125, q = 13.3

q = 1� loge

T2

T1

q� = logeT2

T1

e(q�) = T2

T1

T1e(q�) = T2

T1 = T2

e(q�)

= 125e(13.3%0.15)

= 17.0017..

= 17.00----------------------------------------------------------------------------------------------------------------------

Problems can often be solved graphically as well as theoretically.

Example 4

The formula gives the relationship between the instantaneous current, i = 2(1 − e−10t) iamperes, and the time, seconds, in an inductive circuit. Plot a graph of against ,t i ttaking values of from 0 to 0.3 seconds at intervals of 0.05 seconds.t

Hence find:

a. the initial rate of growth of the current when , and;t = 0

b. the time taken for the current to increase from 1 to 1.6 amperes

Verify these results using theoretical methods.

Solution by Graphical Method

A table of values of from 0 to 0.3 seconds is drawn up and the corresponding valuestfor calculated.i

1.90 1.84 1.72 1.56 1.26 0.78 0 2(1 − e−10t)0.30 0.25 0.20 0.15 0.10 0.05 0 t

A graph of the results is shown below.

25/8/2009 19 RD/PKN

a. When t = 0, the initial rate of growth will be given by the gradient of the tangent at0. The tangent at 0 is the line OM and its gradient may be found by using asuitable right angled triangle MNO and finding the ratio .MN

ON

Initial rate of growth of: = i = MNON

1 amperes0.05 seconds

= 20 amperes per second

b. Point P on the curve corresponds to a current of 1.0 amperes and the time atwhich this occurs is read from the t scale and is 0.07 seconds.

Similarly point Q corresponds to current of 1.6 amperes at a time of 0.16seconds.

Hence time between P and Q is 0.16 - 0.07 = 0.09 seconds, ie the time for thecurrent to increase from 1 to 1.6 amperes is 0.09 seconds.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -continued on next page

25/8/2009 20 RD/PKN

N

M

O

Solution by Theoretical Methods

a. The rate of growth of current can be found by using calculus. [Students who havealready completed HN Unit Calculus 1 will understand the solution]. This methodof solution is not a required part of the present unit.

For interest only:

a. Differentiating to find the rate of change of current with time:

i = 2(1 − e(−10t) ) = 2 − 2e(−10t)

didt = 20e(−10t)

When t = 0

didt = 20e(−10t) = 20

Hence when t = 0 seconds, the rate of change of current is 20 amp/second.

b. When When i = 1.6 A i = 1.0 A

i = 2(1 − e(−10t) )

i2 = 1 − e(−10t)

e(−10t) = 1 − i2

− 10t = ln 1 − i2

t = ln(1 − i2 )

−10

t = ln(1 − 1.62 )

−10

t = 0.160943...

= 160.9 % 10−3 seconds

t = ln(1 − 1.02 )

−10

= 0.0693147....

t = 69.31 % 10−3 seconds

Time taken for current to increase from 1.0 A to 1.6 A is 0.092 seconds (92 ms).----------------------------------------------------------------------------------------------------------------------

25/8/2009 21 RD/PKN

Exercise 7: Practical Problems

1. In an experiment involving Newton's law of cooling the temperature s(oC) is givenby: . Find the volume of constant when = 56.6oC, = 16.5oC� = �oe(−k t) k �o �and = 83.0 seconds.t

2. The length metres of a metal bar at temperature toC is given by: ,l ToC l = l0e�T

where and are constants. Determine:l0 �

a. the value of when = 1.993m, = 1.894m and = 250oC and;� l l0 T

b. the value of when = 2.416, = 310oC and l0 l T � = 168.2 % 10−6

3. Given , find the value of when = 4300, = 11900 and W = nBx x W B n = 1.296 % 10−3

4. The equation gives the relationship between the instantaneousi = 2.4e(−6t)

current, mA, and the time, seconds. Plot a graph of against for values of i t i t tfrom 0 to 0.6 seconds at 0.1 second intervals. Use the curve obtained to find:

a. The rate at which the current is decreasing when = 0.2 seconds and;t

b. The time when the current is 0.54 mA.

Verify your answer to (b) by a calculation method.

5. expresses the decay of radioactive material used in a nuclearN = N0e−x

moisture/density meter for soil compaction control. If , calculate theN = 12 N0

value of .x

6. The atmospheric pressure at a height (in km) above sea level obeys the rule:P h

where is the pressure at sea level and equals 100000 pascals.P = P0e−0.15h P0

a. Find the pressure at a height of 2 km.

b. At what height is the pressure equal to half the value at sea level?

c. At what height is the pressure equal to one-tenth of the value at sea level?

25/8/2009 22 RD/PKN

7. The current in a circuit is described by the law:i

where is equilibrium current and is the time in seconds.i = i0(1 − e−20t ) i0 t

Find the value of when = 2 amperes if = 2.5 amperes.t i i0

8. The work done in the isothermal expansion of a gas from pressure to isp1 p2

given by the formula:

w = w0 lnp1p2

If the initial pressure is 7000 pascals find the final pressure if .p1 p2 w = 3w0

9. If calculate the value of when = 1600, = 0.56, f =p�t2

43 loge

Dd + 1 D p t

= 1.4 and = 5060.d f

10. The relation between pressure p and volume of a gas is . Determine thev pvn = cvalue of given that = 200 when = 0.8 and , the constant is 147.3.n p v c

11. Quantities and are related by the equation , where and p q p = 7.413 1 − ekqt k

are constants. Find when , = 712.8 and = -98.3.t t k = 37 % 10−3 q p

25/8/2009 23 RD/PKN

Answers

Revision exercise

1(i) 57.56 (ii) 0.2678 (iii) 1.978 (iv) 1.410

(v)-0.1189

2(i) (ii) (iii) (iv)p = 3q 2.6 = ex y = 10b x4 = 16

3(i) (ii) (iii) (iv)x = ln 1.2 x = log103 x = − ln 0.3 logap = q

4(i) (ii) (ii) (iv)log 24 log( 89 ) log x2

3 log 1 = 0

(v) ) (vi)log(a(a + 2)) log(x(a+3) )

Exercise 1

1. -0.6801 2. 0.9071 3. 47240 (4 sf)

4. 208.1 5. 4.281 6. 279.8

Exercise 2

(i) 0.8614 (ii) -0.3969 (iii) -0.9163 (iv) 5.365

Exercise 3

(i) 1000 (ii) 3 (iii) 3.310 (iv) 3.806

(v) 0.2468

Exercise 4

(i) 6 (ii) 0.1778

Exercise 5

(i) 2.262 (ii) -0.3417 (iii) 0.1897 (iv) 0.3815 (v)-1.598

25/8/2009 24 RD/PKN

Exercise 6

1(i) (ii) x = 103y b = 310(2c)

a

(iii) y = (iv) y = e t

x x = yez

(v) (vi)x = y e3p y = Ae(sin x) − 1

(viii) (viii) y = a(x + 2) + 3 y = 4e3x

2(i) (ii)p2 = 10Qk p1 v1 = v2

epA

(iii) (iv)T1 = T2em y = xe(−2a)

3(i) (ii)t = 1� loge

( VW ) s = 1

� lnT1

T2

(iii) (iv)k = 1t ln( �1

�0) t = ( −1

k) ln( i

I )

(v) (vi)t = ( 1k

) ln(1 − ( RiE )) c = −h

lnp

po

Exercise 7

1 2.14.85 % 10−3 � = 203.8 % 10−6, l0 = 2.293

3. 1.600 4. , 0.2486−4.337 mAs−1

5. 0.6931 6. 74080 pascals, 4.621 km, 15.35 km

7. 8. 348.580.47 % 10−3

9. 6.843 10. 1.371

11. 9.924

25/8/2009 25 RD/PKN