MATH 1.1: Arithmetic and algebra 
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PPLATO / FLAP (Flexible Learning Approach To Physics) 


This module is about basic algebra and arithmetic. It assumes that you already have some familiarity with these topics and concentrates on giving you an opportunity to practise the skills you have already acquired, and to sharpen–up your use of the very precise terminology of mathematics.
As you work through this module remember that it has been designed as a tool to be used by you for your own benefit. You should not waste time on topics with which you are already fully familiar, but nor should you shy away from topics you have not completely mastered. Use the questions in the module to probe your own knowledge and ability, and use the text to rectify any deficiencies that you detect. By reinforcing your understanding of basic mathematics you will be preparing yourself for some of the more demanding mathematical challenges that you will inevitably meet as your studies progress.
The overall thrust of this module is towards problem solving – abstract principles and general definitions are certainly included, but their role is to support your development as a confident user of mathematics.
Section 1 contains the various Opening items that are a common feature of all FLAP modules. These are designed to give you a clear view of the module’s content and the prior knowledge that it assumes, so that you can assess for yourself the extent to which you need to study the module and the degree to which you are prepared for such a study. Section 2Sections 2 and Section 33 contain the basic teaching material of the module, dealing with mathematical expressions and with relations between expressions, particularly equality and inequality. Section 2 starts with a review of the basic operations of arithmeticoperations (addition, subtraction, multiplication and division) together with their symbolic representation and order of priority in written expressions. This topic is taken further in Subsection 2.2 which concentrates on brackets and their use in simplifying and expanding expressions. The manipulation of algebraic and arithmetic fractions (a common source of errors) is discussed in Subsection 2.3, and the topic of power_mathematicalpowers, roots and reciprocals is covered in Subsection 2.4. The writing and rearrangement of simple equations is dealt with in Subsection 3.1Subsections 3.1 and 3.2, while Subsection 3.2Subsection 3.3 looks at proportionality. (Both direct and inversely_proportionalinverse proportionality are considered, together with the equations that reflect such relationships.) Subsection 3.4 deals with the subject of inequalities and reviews the rules for their manipulation. As is usual in FLAP, the module ends with a section of Section 4Closing items that includes a Subsection 4.1Summary, a list of the Subsection 4.2Achievements that you should be able to do on completing the module, and an Subsection 4.3Exit test designed to let you assess your achievements and alert you to any remaining difficulties.
Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions given in. If not, proceed directly to the Subsection 1.3Ready to study? Subsection.
Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 4.1Module summary and the Subsection 4.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 4.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.
Question F1
Given z = [2xy − (x + y)^{2}]^{2}
(a) Evaluate z when x = 1 and y = 2.
(b) Expand the expression for z by removing all the brackets.
Answer F1
(a) z = [4 − (1 + 2)^{2}]^{2} = (4 − 9)^{2} = (−5)^{2} = 25
(b) z = 4x^{2}y^{2} − 4xy_{ }(x + y)^{2} + (x + y)^{4}
(b) z = 4x^{2}y^{2} − 4xy_{ }(x^{2} + 2xy + y^{2}) + (x^{2} + 2xy + y^{2})(x^{2} + 2xy + y^{2})
(b) z = 4x^{2}y^{2} − 4x^{3}y − 8x^{2}y^{2} − 4xy^{3} + x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4} = x^{4} + 2x^{2}y^{2} + y^{4}
Question F2
Write each of these expressions as a single fraction expressed in as simple a form as possible:
(a) $2mu  \dfrac 2u d$ (b) $\dfrac{2l^2}{3h} + \dfrac h2$ (c) $\dfrac{3l(h+f)^2}{2a^2} \div \dfrac{3a(h^2f^2)}{2l^2}$
Answer F2
(a) $2mu\dfrac2u d = \dfrac{2mu^22d}{u} = \dfrac{2(mu^2d)}{u}$
(b) $\dfrac{2l^2}{3h} + \dfrac h2 = \dfrac{4l^2+3h^2}{6h}$
(c) $\dfrac{3l(h+f)^2}{2a^2} \div \dfrac{3a(h^2f^2)}{2l^2} = \dfrac{3l(h+f)^2}{2a^2}\times\dfrac{(2l^2)}{3a(h^2f^2)} = \dfrac{6l^3(h+f)^2}{6a^3(h+f)(hf)} = \dfrac{l^3(h+f)}{a^3(hf)} = (l/a)^3\dfrac{(h+f)}{(hf)}$
Question F3
A rocket of mass m_{1} fired vertically at speed υ from the surface of a planet of mass m_{2} and radius r can escape from the planet provided its kinetic energy m_{1}υ^{2}/2 is greater than Gm_{1}m_{2}/r. Derive an inequality, with υ as its subject, that must be satisfied if the rocket is to escape.
Answer F3
The value of υ that will just permit escape is given by m_{1}υ^{2}/2 = Gm_{1}m_{2}/r. Cancelling a common factor of m_{1} on both sides and multiplying by 2 gives υ^{2} = 2Gm_{2}/r.
Taking the positive square root gives $\upsilon = \sqrt{2Gm_2/r}$.
The inequality $\upsilon \gt \sqrt{2Gm_2/r}$ gives the condition for escape.
Study comment Relatively little background knowledge is required to study this module, which deals with very basic topics in arithmetic and algebra. Nonetheless, it is assumed that you have previously performed a wide range of calculations involving the addition, subtraction, multiplication and division of positive and negative quantities, and that you have already encountered terms such as fraction, power, square, and square root. (In any case, these terms are defined in the Glossary.) In addition it is assumed that you are accustomed to using a calculator and that you have one available while working on this module.
Formulae from many different branches of physics have deliberately been used to illustrate the mathematical points made in this module. As a result you may find that you are asked to deal with some unfamiliar physical quantities. If this should happen, don’t worry. You will find that the physical properties of the quantities concerned are of no real importance in this module and that any unfamiliar quantity can be treated as a simple mathematical ‘unknown’. Moreover, if you are so inclined, you can consult the Glossary for further information about any unfamiliar quantity. i
The four basic operations of arithmeticoperations of arithmetic are addition (+), subtraction (−), multiplication (×) and division (÷ or /_{ }). Other operations include squaring (e.g. x^{2} = x × x) and taking square roots (e.g. $\sqrt{x^2} = x$). A mathematical expression is a combination of numbers and letters (which generally represent numbers) linked together by various operations. 5 × 4 − 3 and 3a − b/c are both examples of expressions i; the former is purely arithmetic (i.e. involving numbers only), the latter algebraic (i.e. involving numbers and letters).
When working with expressions it is crucially important that they should be unambiguous. Thus, when dealing with the expression 5 × 4 − 3, for example, you need to know whether it means ‘multiply 5 by 4 then subtract 3’ or ‘subtract 3 from 4 then multiply by 5’ – the outcome is different in the two cases. The potential ambiguity is removed by insisting on a standard order of priority for carrying out the basic operations.
Standard order of priority for basic operations:
multiplication and division, then addition and subtraction.
Question T1
Using the standard order of priority evaluate the following expressions:
(a) 3 + 6/2 − 4 (b) 4.2 × 3.0 − 1.4 (c) ab + c/2, where a = 2, b = 1/2 and c = −4
Answer T1
(a) 3 + 6/2 − 4 = 3 + 3 − 4 = 2
(b) 4.2 × 3.0 − 1.4 = 12.6 − 1.4 = 11.2
(c) ab + c/2 = 2 × 1/2 + (−4)/2 = 1 − 2 = −1
A note on terminology For future reference, here are the definitions of some words that are often used in connection with the basic operations.
Each of these words may be applied to algebraic expressions in the same way that it is applied to arithmetic expressions.
It is often useful to regard a given expression as consisting of several distinct terms each of which may be a number, a letter or even another expression. Strictly speaking, an expression should be treated as a sum of terms, so 5 × 4 − 3 should be viewed as a positive term 5 × 4 added to a negative term −3. However, in practice the word term is often used more loosely and an expression such as 3ab is spoken of as a product of terms.
The terms making up an expression can usually be ordered in a variety of different ways. For instance, 5 × 4 − 3 could be equally well written as −3 + 5 × 4, though it may not be written 3 − 5 × 4, and 3ab can be rewritten as 3ba or even a3b or ab3, though these last two would be regarded as ugly and unusual. The conventional way of ordering terms within an expression is best picked up by seeing examples such as those contained in this module. The basic rules that govern reordering are explored in the following question.
✦ Is the ordering of two terms, a and b, immaterial if they are:
(a) multiplied (b) added (c) subtracted (d) divided?
✧ (a) Yes, ab = ba for every choice of a and b
(b) Yes, a + b = b + a for every choice of a and b
(c) No, a − b ≠ b − a for every choice of a and b i
(d) No, a/b ≠ b/a for every choice of a and b
How would you write an expression corresponding to the instruction ‘subtract 3 from 4 then multiply the result by 5’? You certainly can’t write it as 4 − 3 × 5 because (according to the standard order of priority) that would mean ‘multiply 3 by 5 and then subtract the result from 4’. One way to get round this problem is to use brackets to separate parts of an expression and to give the evaluation of one part priority over another. The standard order of priority still applies within the brackets but the bracketed part of the expression is treated as a single entity and its evaluation is given precedence over all other operations. Thus, the instruction ‘subtract 3 from 4 then multiply the result by 5’ can be written (4 − 3) × 5, or more conventionally 5_{ }(4−3).
Quite often a bracketed expression includes a term that also involves brackets. In such a case one pair of brackets appears inside another pair as in 5_{ }(3 + 2_{ }(2 − 4)). The two pairs of brackets are said to be nested and are dealt with by evaluating the contents of the innermost brackets first and then moving outwards. Thus
5_{ }(3 + 2_{ }(2 − 4)) = 5_{ }(3 + 2_{ }(−2)) = 5_{ }(3 − 4) = 5_{ }(−1) = −5
Sometimes brackets of different shapes i are used to make it clearer which brackets are to be paired – the previous example might have been written 5_{ }[3 + 2_{ }(2 − 4)] – but this is not always done, even for complicated expressions.
Question T2
Evaluate the following expressions:
(a) (6 + 3/(9 − 7))/2
(b) 2_{ }[(3a − 4)b + 2a_{ }(3b + 1)], where a = 1 and b = −1
(c) x{[x − 2_{ }(x + 3)/(x − 1)] + [1 + (x + 3)/2]/x}, where x = 2
Answer T2
(a) (6 + 3/(9 − 7))/2 = (6 + 3/2)/2 = (6 + 1.5)/2 = 7.5/2 = 3.75
(b) 2 [(3a − 4)_{ }b + 2a_{ }(3b + 1)] = 2_{ }[(3 − 4)(−1) + 2_{ }(−3 + 1)] = 2_{ }[(−1)(−1) + 2_{ }(−2)] = 2_{ }(1 − 4) = −6.
(c) 2{[2 − 2_{ }(2 + 3)/(2 − 1)] + [1 + (2 + 3)/2]/2} = 2_{ }[(2 − 10) + (1 + 2.5)/2] = 2_{ }(−8 + 1.75) = −12.5
When expressions involve powers, such as squares and square roots, as well as brackets i the overall order of priority is as follows.
Overall order of priority for operations:
First deal with anything in brackets
then deal with powers (including roots and reciprocals)
then multiply and divide
then add and subtract.
Question T3
Evaluate the following expressions:
(a) 2 + 4/3^{2} (b) 2 + (4 − 3a)^{2}, where a = 1 (c) (2 + (2 + x)^{2})^{2}, where x = 2
Answer T3
(a) 2 + 4/3^{2} = 2 + 4/9 = 2 + 0.44 = 2.44
(b) 2 + (4 − 3a)^{2} = 2 + (4 − 3)^{2} = 2 + 1 = 3
(c) (2 + (2 + x)^{2})^{2} = (2 + (2 + 2)^{2})^{2} = (2 + 42)^{2} = (2 + 16)^{2} = 182 = 324
Many calculators are designed to carry out processes in the correct order if you key in a calculation exactly as it is written. Explore your own calculator to find out whether it does this – use the examples given above, or invent some for yourself and do the calculations ‘by hand’ as well as on your calculator.
You may think that if your calculator does it all for you, there is no need to take much notice of the order in which operations should be carried out. But when you are manipulating algebraic expressions, and rearranging equations, it is important to know in exactly what order you can do, or undo, processes. (Manipulating algebraic expressions features throughout this module.) Also, it is always possible to make a mistake when using a calculator, so it is often useful to carry out a rough calculation to make sure that the answer on your calculator is sensible.
In the previous subsection, brackets were used to separate one part of an expression from the rest. Sometimes it is useful either to rewrite an expression by removing brackets (in other words, to expand the expression), or to use brackets to group parts of an expression together (in other words to simplify the expression).
(3 + 4) × 5 means ‘add 3 to 4, then multiply by 5’. In this case, 3 and 4 have been added, and their sum multiplied by 5, so the calculation could have been expanded as 3 × 5 + 4 × 5. In cases like this, each and every term inside the brackets has to be multiplied by the factor outside. If the factor enclosed by brackets is a difference, rather than a sum, the expression can be expanded in exactly the same way. For example: 4 × (7 − 5) = 4 × 2 = 8, and this could be rewritten as 4 × 7 − 4 × 5 = 28 − 20 = 8.
If an expression is a product of two or more factors enclosed by brackets, it can still be expanded, but each term inside one pair of brackets must be multiplied by each term inside every other pair of brackets. For example: (2 − 3)(4 − 5) = 2 × 4 − 2 × 5 − 3 × 4 + 3 × 5. i
Note that when writing out an expansion it is advisable to treat it systematically, by working along from left to right as above, otherwise you risk forgetting or duplicating some of the terms.
Numbers were used in the examples above because they make it is easier to show what is happening, but in practice there is not often much point in expanding purely numerical expressions since it rarely makes the calculation any easier. Expanding algebraic expressions can, however, often help to simplify an expression or equation.
✦ Expand the following expressions:
(a) −3_{ }(4x − 2) (b) 2_{ }(3x + 2y) − 4y (c) 3_{ }(x + 2_{ }(2x + 5))
✧ (a) −12x + 6 (each term in the bracket is multiplied by −3)
(b) 2_{ }(3x + 2y) − 4y = 6x + 4y − 4y = 6x
(c) 3_{ }(x + 2_{ }(2x + 5)) = 3_{ }(x + 4x + 10) = 3_{ }(5x + 10) = 15x + 30
Brackets can be removed from a division calculation in the same way as for multiplication. Each term inside the brackets must be divided by the divisor outside. For example:
(6 + 9)/3 = 6/3 + 9/3 = 2 + 3 = 5
Question T4
Expand the following expressions so that in each case the resulting expression does not involve brackets:
(a) 3x_{ }(4 + 3y) + 2x (b) (x + 2y)^{2} (c) (x + y)(x − y) (d) (3x + 6y)/2 (e) (p + q)/q (f) (2xy)^{2}/2
Answer T4
(a) 3x_{ }(4 + 3y) + 2x = 12x + 9xy + 2x = 14x + 9xy
(b) (x + 2y)^{2} = (x + 2y)(x + 2y) = x^{2} + 2xy + 2yx + 4y^{2} = x^{2} + 4xy + 4y^{2}
(c) (x + y)(x − y) = x^{2} − xy + yx − y^{2} = x^{2} − y^{2}
(d) (3x + 6y)/2 = 3x/2 + 6y/2 = 3x/2 + 3y
(e) (p + q)/q = p/q + q/q = p/q + 1
(f) (2xy)^{2}/2 = 4x^{2}y^{2}/2 = 2x^{2}y^{2}
Terms with a common factor (that is, a number or expression that is a factor of each) can be collected together using brackets. This is the reverse of the process shown above, and can often be used to simplify an expression, or at least to rewrite it in a more compact form. For example, all the terms in the expression 8x + 6y + 10, have 2 as a common factor, so it could be written as 2_{ }(4x + 3y + 5). As another example: the expression 6x + 4x^{2} + 8xy has 2x as a common factor, so could be rewritten as 2x_{ }(3 + 2x + 4y). The terms 2x and 4y also share a common factor of 2, so nested brackets could be used to collect these terms together, 2x_{ }(3 + 2_{ }(x + 2y)). i
Question T5
By finding the common factors, simplify the following expressions:
(a) 3x^{2}y + 6xy (b) 8x^{3} + 4x^{2}y + 2xy^{2} + 2x
Answer T5
(a) 3x^{2}y + 6xy = 3xy_{ }(x + 2)
(b) 8x^{2} + 4x^{2}y + 2xy^{2} + 2x = 2x_{ }(4x^{2} + 2xy + y^{2} + 1)
The following expansions arise frequently and are worth remembering
(a + b)^{2} = a^{2} + 2ab + b^{2}(1)
(a−b)^{2} = a^{2} − 2ab + b^{2}(2)
(a + b)(a − b) = a^{2} − b^{2}(3)
The word fraction probably conjures up in your mind numbers like 1/2, 3/4 and 7/3. All these fractions are ratios of two whole numbers which can be rewritten as a single number by dividing the number on top (the numerator) by the one on the bottom (the denominator) to find their quotient. But in physics you also have to deal with fractions that are ratios of two algebraic expressions, such as Q/C, mυ^{2}/r, and q_{1}q_{2}/(4πε_{0}r^{2}). Algebraic fractions can be manipulated in exactly the same ways as arithmetic ones, and the terms numerator and denominator are still used for the expressions on top and underneath respectively.
Any number may be expressed as a fraction of any other number; 2, for instance, is 1/2 of 4, and 6 is 3/2 of 4. This observation provides the basis of the system of percentages in which one number is expressed as so many hundreths of another number. For example
$2 = \dfrac{50}{100} \times 4$, so we may say 2 is 50% (read as ‘50 per cent’) of 4
More generally, if we want to express a as a percentage of b that percentage is given by (a/b) × 100.
If the numerator and denominator of a fraction have a common factor, then that factor both multiplies and divides the rest of the fraction – and those two operations cancel each other out. cancellingCancelling therefore allows common factors to be removed from both the numerator and denominator of a fraction without having any net effect. For example, in the case of 6/10, the numerator and denominator each have 2 as a factor and 6/10 can therefore be written as 3/5. The same principle applies to algebraic fractions such as πr^{2}/(2πr), the ratio of the area of a circle to its circumference. The numerical constant π i and the radius r are both common factors of the numerator and the denominator, so they may be cancelled from both, leaving r/2.
Question T6
Simplify these fractions by cancelling the common factors in their numerators and denominators:
(a) $\dfrac{7}{42}$ (b) $\dfrac{6x}{2x^2}$ (c) $\dfrac{8(x + 3)}{2x}$
Answer T6
(a) 7/42 = 7/(6 × 7) = 1/6
(a) 6x/(2x^{2}) = (3 × 2x)/(2x × x) = 3/x
(a) 8_{ }(x + 3)/(2x) = 4_{ }(x + 3)/x
Question T7
By first using brackets to group terms in the numerator and/or denominator, simplify the following fractions by cancelling common factors:
(a) (x + 2)(x + 3)/(3x^{2}y + 6xy) (b) (3xy + 3y^{2})/(4x^{2}y + 4x^{3})
Answer T7
(a) The denominator is (3x^{2}y + 6xy) = 3xy_{ }(x + 2), so the numerator and denominator have a common factor (x + 2) and the fraction becomes (x + 3)/(3xy).
(b) The numerator is (3xy + 3y^{2}) = 3y_{ }(x + y), and the denominator is (4x^{2}y + 4x^{2}) = 4x^{2}_{ }(y + x), so cancelling a common factor of (x + y) in the fraction gives 3y/4x^{2}.
Notice that in Question T7 you were able to cancel factors that were not just single numbers or symbols, and that using brackets to write the various expressions as products made these factors much easier to spot.
All the standard mathematical operations can be applied to fractions. Multiplication and division are especially easy – provided the problem is written out clearly in the first place. In general algebraic terms;
$\dfrac ab \times \dfrac cd = \dfrac{ac}{bd}$(4)
and$\dfrac ab \div \dfrac cd = \dfrac{ad}{bc}$(5)
So, dividing by (c/d) is the same as multiplying by (d/c).
✦ Simplify the following fractions:
(a) (3/5) ÷ (5/2) (b) (mυ^{2}/2)/(r/2)
✧ (a) $\dfrac35 \div \dfrac52 = \dfrac{3\times 2}{5 \times 5} = \dfrac 6 {25} = 0.24$
(b) $\dfrac {m\upsilon^2}2 \div \dfrac r2 = \dfrac{m\upsilon^2\times 2}{2 \times r} = \dfrac {m\upsilon^2}r$
Adding and subtracting fractions is also straightforward provided they all have the same denominator – a so–called common denominator. Expressions such as 4/5 + 7/5, or 8/3 − 5/3 can be rewritten as (4 + 7)/5 or (8−5)/3. But if fractions that are to be added or subtracted do not share a common denominator, they can – and must be – rewritten in such a way as to give them one. For example, to add the fractions 4/5 and 8/3, you can write
$\dfrac45 + \dfrac83 = \dfrac{(4 \times 3)}{(5 \times 3)} + \dfrac{(8 \times 5)}{(3 \times 5)}$
Notice that the value of neither fraction has been changed, since all that has been done is to introduce a common factor into the numerator and denominator of each, also notice that the denominator of each fraction was used to rewrite the other. Now the addition can be written as
$\dfrac43 + \dfrac83 = \dfrac{(4 \times 3) + (8 \times 5)}{(5 \times 3)}$
This technique can, of course, also be used when fractions are to be subtracted, and algebraic fractions are treated in exactly the same way. Thus, in general algebraic terms:
$\dfrac ab + \dfrac cd = \dfrac{ad + cb}{bd}$(6)
and$\dfrac ab  \dfrac cd = \dfrac{ad  cb}{bd}$(7)
✦ Rewrite the fractions 3x/5 and 2/y so that they have a common denominator and then write (3x/5) − (2/y) as a single fraction.
✧ $\dfrac{3x}{5} = \dfrac{3xy}{5y} \quad\text{and}\quad\dfrac 2y = \dfrac{10}{5y}$
Thus$\dfrac{3x}{5}  2 = \dfrac{3xy}{5y}  \dfrac{10}{5y} = \dfrac{3xy10}{5y}$
The rules for manipulating fractions are fairly straightforward but under pressure some students panic when faced with fractions and make serious mistakes. Here is a short catalogue of common fallacies that you should try to avoid.
Forgetting the common denominator It is not uncommon to see totally bogus additions such as:
$\dfrac ab + \dfrac cd = \dfrac{a + c}{b + d}$  this looks plausible, but it’s WRONG.
Incomplete cancelling It is also quite common to see half–hearted cancelling along the following lines:
$\dfrac{1 + ax}x = 1 + a$  this too is WRONG. i
Wrongly ordered divisions If asked to divide (2ax/b) by (2a/b) most students will find the correct answer (x) without any difficulty. But different ways of presenting the same calculation can lead to mistakes. The following are all equal (if not equally attractive).
$(2ax/b)/(2a/b) = \dfrac{2ax/b}{2a/b} = \dfrac{\dfrac{2ax}{b}}{\dfrac{2a}{b}} = \dfrac{2ax}{b(2a/b} = \dfrac{2axb}{2ab} = x$ i
Question T8
Simplify the following, if possible:
(a) $\dfrac23 + \dfrac35$
(b) $\dfrac x3 + \dfrac x2$
(c) $(1/u) + (1/\upsilon)$
(d) $\dfrac{1+a}{3+a}$
(e) $\dfrac{2(1+ x)}{(1+x)^2}$
(f) $\dfrac{5/7}{2/3}$
(g) $\dfrac{3x}{x/3}$
(h) $\dfrac 3a + \dfrac{7}{2a}$
(i) $(−7x^2) \div (3x)$ i
Answer T8
(a) $ \dfrac23 + \dfrac35 = \dfrac{10}{15} + \dfrac{9}{15} = \dfrac{19}{15}$
(b) $\dfrac x3 + \dfrac x2 = \dfrac{2x}{6} + \dfrac{3x}{6} = \dfrac{5x}{6}$
(c) $ \dfrac1u + \dfrac1v = \dfrac{v}{uv} + \dfrac{u}{uv} = \dfrac{v+u}{uv}$
(d) $\dfrac{1+a}{3+a}$ cannot be simplified
(e) $\dfrac{2(1+x)}{(1+x)^2} = \dfrac{2}{(1+x)}$
(f) $\dfrac{5/7}{2/3} = \dfrac57 \times \dfrac32 = \dfrac{15}{14}$
(g) $\dfrac{3x}{x/3} = 3x \times \dfrac3x = \dfrac{9x}{x} = 9$
(h) $\dfrac3a + \dfrac{7}{2a}= \dfrac{6}{2a} + \dfrac{7}{2a}= \dfrac{13}{2a}$
(i) $\left(7x^2 \right) \div (3x) = \dfrac{7x^2}{3x} = \dfrac{7x}{3}$
A power (mathematical)power (often called an index) indicates repeated multiplication as in:
$\begin{align} &\text{squares} && a^2 = a \times a && \text{(read as ‘}a\text{ squared’)}\\ &\text{cubes} && a^3 = a \times a \times a && \text{(read as ‘}a \text{ cubed’)}\\ \text{or, generally}&& && &&\\& n\text{th powers} && a^n = \underbrace{a \times a \times a \ldots \times a}_{\color{purple}{\large n~{\rm factors}}} && \text{(read as ‘} a \text{ to the n’)}\\ \end{align}$
$\begin{align} &\text{squares} && a^2 = a \times a && \text{(read as ‘}a \text{ squared’)}\\ &\text{cubes} && a^3 = a \times a \times a && \text{(read as ‘}a \text{ cubed’)}\\ \text{or, generally}&& && &&\\& n\text{th powers} && a^n = \underbrace{a \times a \times a \ldots \times a}_{\color{purple}{\large n\text{ factors}}} && \text{(read as ‘} a \text{ to the n’)}\\ \end{align}$
(Note the use of the three dots (...), known as an ellipsis, to indicate that the sequence continues in a similar fashion.)
Powers (or indices) enable many expressions to be written neatly and compactly. For instance, Keplers laws of planetary motionKepler’s third law, which relates the time T that a planet requires to orbit the Sun to the mean distance R between that planet and the Sun, can be written
$T^2 = kR^3$(8)
where k is a constant. Relationships of this kind, that relate one quantity to a power of another are often called power laws.
The use of powers is simple and straightforward as long as the power is a positive whole number (1, 2, 3, etc.), but you will encounter many situations in which the power is neither positive nor a whole number. Fortunately, the general rules for manipulating powers are simply extensions of the rules that apply to positive whole number powers, so let’s start with those.
It follows directly from the definition of a^{ n} that if m and n are positive whole numbers
$a^m \times a^n = \underbrace{a \times a \times \ldots \times a}_{\color{purple}{\large m~{\rm factors}}} \times \underbrace{a \times a \times \ldots \times a}_{\color{purple}{\large n~{\rm factors}}}$
Now, the right–hand side of this equation consists of m + n factors of a, so it may be written a^{ m+n}. Thus,
a^{m} × a^{n} = a^{m+n}(9)
Similarly, raising a^{ m} to the power n gives:
$(a^m)^n = \underbrace{\overbrace{(a \times a \times \ldots \times a)}^{\color{purple}{\large m~{\rm factors~of~}a}} \times (a \times a \times \ldots \times a) \times \ldots \times (a \times a \times \ldots \times a)}_{\color{purple}{\large n~{\rm factors~of~}a^m}}$
The right–hand side now contains m×n factors of a. So,
(a^{ m})^{ n} = a^{ m× n} = a^{ m n}(10)
Question T9
Using the boxed equations given above, simplify and then evaluate the following expressions:
(a) 2^{2} × 2^{3} (b) 2^{3} × 2^{4} × 2^{5} (c) (−2)^{4} × (−2)^{2} (d) (2^{3})^{2} (e) ((−10.2)^{2})^{3} (f) ((−1)^{17})^{23}
Answer T9
(a) 2^{2} × 2^{3} = 2^{5} = 32
(b) 2^{3} × 2^{4} × 2^{5} = 2^{12} = 4 096
(c) (−2)^{4} × (−2)^{2} = (−2)^{4} = 64
(d) (2^{3})2 = 2^{6} = 64
(e) [(−0.2)^{2}]^{3} = (−0.2)^{6} = 64 × 10^{−6} = 6.4 × 10^{−5}
(f) [(−1)^{17}]^{23} = (−1)^{17 × 23} = (−1)^{391} = −1
We can make sense of negative powers by extending the pattern of behaviour that we find for a sequence of terms such as a^{2}, a^{3}, a^{4} ... and so on. For the sake of having a concrete example we will assume that a = 5, but the general results that we deduce will be independent of this particular choice and would be equally true for any value of a other than zero. Given our chosen value of a, the first few terms in the sequence a^{2}, a^{3}, a^{4}... are
5 × 5 5 × 5 × 5 5 × 5 × 5 × 5 ...
To go from one term to the next in the sequence a^{2}, a^{3}, a^{4}... you simply raise the power by 1, that’s to say you multiply the previous term by a, or 5 in this case.
✦ Now suppose that the sequence is written in reverse order: ... a^{4}, a^{3}, a^{2}. What must you now do to go from one term to the next, working from left to right?
✧ You have to reduce the power by 1, i.e. divide the previous number by a, or 5 in this case.
It is useful to continue this sequence of decreasing powers for a few steps:
... a^{3}, a^{2}, a^{1}, a^{0}, a^{−1}, a^{−2} ...
and to evaluate some of the terms using our chosen value, a = 5.
Dividing 5^{2} by 5 gives 5, so 5^{1} = 5
Dividing 5^{1} by 5 gives 1, so 5^{0} = 1
Dividing 5^{0} by 5 gives 1/5, so 5^{−1} = 1/5 = 0.2
Dividing 5^{−1} by 5 gives $\dfrac15$, so 5^{−2} = $\dfrac 1{5^2}$ = 0.04
You might like to repeat this process using some other value of a (especially a negative value such as a = −2) but as long as a is not zero the pattern should always be the same and you should always find the following:
a^{1} = a, a^{0} = 1, a^{−1} = 1/a, a^{−2} = 1/a^{2} i
Given any non–zero quantity a, the quantity 1/a is called the reciprocal of a. So our deductions have shown that the reciprocal of a quantity a can be written a^{−1}.
✦ Write down the reciprocal of a^{2} in two different ways.
✧ The reciprocal of a^{2} may be written as 1/a^{2} or (a^{2})^{−1}.
If the power notation is to make sense we must require these two alternative ways of writing the reciprocal to be equal. But we already know that 1/a^{2} may be written as a^{−2}, so (a^{2})^{−1} = 1/a^{2} = a^{−2}.
This is an example of a more general result that works for any value of n.
(a^{n})^{−1} = 1/a^{n} = a^{−n}(11)
Question T10
Evaluate the following:
(a) 2^{−4} (b) 10^{−3} (c) $\left(\dfrac12\right)^{3}$ (d) (0.2)^{−2}
Answer T10
(a) 2^{−4} = 1/2^{4} = 1/16 = 0.062 5
(b) 10^{−3} = 1/10^{3} = 0.001
(c) $\left(\dfrac12\right)^{3} = 2^3 = 8$
(d) (0.2)^{−2} = 1/(0.2)^{2} = 1/0.04 = 25
The use of negative powers to indicate reciprocals is consistent with the two general rules that were introduced earlier (Equations 9 and 10).
am × an = am + n(Eqn 9)
(a^{ m})^{ n} = a^{ m× n} = a^{ m n}(Eqn 10)
For example, we know that if m and n are positive whole numbers
a^{ m} × a^{ n} = a^{ m+n}
but even if m or n (or both) are negative, the rule still works
$2^3 \times 2^{5} = (2 \times 2 \times 2)\times \dfrac{1}{2 \times 2 \times 2 \times 2 \times 2} = \dfrac{1}{2 \times 2} = \dfrac{1}{2^2} = 2^{2}$
which is exactly the result we would expect from the general rule (Equation 9).
Similarly, for the second rule, (a^{ m})^{ n} = a^{ m n}
$(2^2)^{4} = \dfrac{1}{(2^2)^4} = \dfrac{1}{2^8} = 2^{8}\text{, which is also the result we would expect from the general rule (Equation 10)}.$
Question T11
Express each of the following in the form a^{ n}:
(a) $5^2 \times 5^{5}$ (b) $(5^2)^{3}$ (c) $(2^4)/(2^6)$ (d) $(3^{1})^{1}$ (e) $\left(\dfrac12\right)^{3}$ (f) $\left(\dfrac{2^5}{2^6}\right)^{1} \times 2 \times 2^3$
Answer T11
(a) 5^{2} × 5^{−5} = 5^{2−5} = 5^{−3}
(b) (5^{2})^{−3} = 5^{−6}
(c) 2^{4}/2^{6} = 2^{4} × 2^{−6}} = 2^{4−6} = 2^{−2}
(d) (3^{−1})^{−1} = 3^{1} (i.e. 3)
(e) $\left(\dfrac12\right)^{3}$ is already in the required form, with $a = \dfrac12$. However, it may be rewritten as $(2^{1})^{3} = 2^3$, or as $8^1$, if desired.
(f) $\left(\dfrac{2^5}{2^6}\right)^{1} × 2 × 2^3 = \left(\dfrac{2^6}{2^5}\right) \times 2 \times 2^3 = \dfrac{2^{10}}{2^5} = 2^{105} = 2^5$
Negative powers, or powers appearing in the denominator of an expression, are often referred to as inverse powers, so Newton’s law of gravitation, which asserts that the strength of the gravitational force attracting a mass m_{1} towards another mass m_{2} a distance r away is
$F = \dfrac{Gm_1 m_2}{r^2}$
is often described as an inverse square law.
Although Keplers laws of planetary motionKepler’s third law (T^{2} = kR^{3}) is a valuable result, it is often even more valuable to know the way in which T itself (rather than T^{2}) is related to k and R. Such a relation is easily obtained from Kepler’s law by taking square roots:
$T = \sqrt{kR^3}$
The root symbol ($\sqrt{\phantom{0}}$) can be used to indicate a variety of roots such as
$\begin{align} && \text{square roots:} && \sqrt{a} \sqrt{a} = a\\ && \text{cube roots:} && \sqrt[{\large\raise{2pt}3}]{a} \;\sqrt[{\large\raise{2pt}3}]{a}\;\sqrt[{\large\raise{2pt}3}]{a} = a\\ && \text{or, generally, }n\text{th roots:} && \underbrace{\sqrt[{\large\raise{2pt}n}]{a}\;\sqrt[{\large\raise{2pt}n}]{a} \ldots \sqrt[{\large\raise{2pt}n}]{a}}_{\color{purple}{\large n\,{\rm factors}}} = a \end{align}$
In fact, it is somewhat unusual to see cube (or higher) roots written in this form since most physicists prefer to make use of another notation based on fractional powers. According to this alternative notation
$\sqrt[{\large\raise{2pt}n}]{a} = a^{1/n}$(12)
Once again this new notation is consistent with the rules given earlier. For example, we know that 8^{1/3} = 2 since 2 × 2 ×2 = 8, but the second of our rules for powers ((a^{ m})^{ n} = a^{ m n}) would lead us to expect that
(8^{1/3})^{3} = 8^{(1/3)×3} = 8^{1} = 8 i
which is indeed the case since 8^{1/3} = 2 and 2^{3} = 8.
All this may seem fairly straightforward, but, in fact, a certain degree of care is needed when dealing with roots and fractional powers. In particular, the following points should be noted. i
It is true that 4 = 2 × 2, but it is equally true that 4 = (−2)×(−2). So, both 2 and −2 are square roots of 4. Thus we may write 4^{1/2} = 2 or 4^{1/2} = −2. These two statements may be combined in the form
4^{1/2} = ±2
where the symbol ± is to be read as ‘+ or −’. Some authors introduce a convention whereby a^{1/2} or a or both are always used to represent the positive square root of a – FLAP does not use such a convention, though negative square roots will be ignored when there is a physical reason for doing so. (For instance, we know physically that quantities such as mass and length cannot be negative.)
The need to remember that a^{1/n} may be positive or negative is not restricted to the case of square roots where n = 2 – it applies to all cases where n is an even whole number.
Some negative numbers certainly have roots for instance:
−27 = (−3) × (−3) × (−3) So (−27)^{1/3} = −3
But it is not possible to find any number a, positive or negative, such that
a^{2} = −4 or a^{2} = −0.1
Indeed, it is generally said that negative quantities do not have square roots. Nor, in the same spirit, is it possible to find the nth root of any negative quantity unless n is an odd whole number. i
Question T12
Evaluate the following:
(a) 9^{1/2} (b) 16^{1/2} (c) 16^{1/4} (d) 27^{1/3} (e) $\sqrt{64\os}$ (f) $\sqrt[{\large\raise{2pt}4}]{49\os}$
Answer T12
(a) 9^{1/2} = 3 or −3 (often written 9^{1/2} = ±3)
(b) 16^{1/2} = 4 or −4 (often written 16^{1/2} = ±4)
(c) 16^{1/4} = 2 or −2 (often written 16^{1/4} = ±2)
(d) 27^{1/3} = 3
(e) $\sqrt{64\os}$ = 8 or –8 (often written $\sqrt{64\os}$ = ±8)
(f) $\sqrt[{\large\raise{2pt}4}]{49\os} = \sqrt{7\os}$ = 2.65 or –2.65 (often written $\sqrt[{\large\raise{2pt}4}]{49\os}$ = ±2.65)
Question T13
Write down the following roots, or explain why they cannot be found:
(a) (−8)^{1/3} (b) $\left(\dfrac{1}{16}\right)^{1/2}$ (c) (−2)^{1/2}
Answer T13
(a) (−8)^{1/3} = −2
(b) $\left(\dfrac{1}{16}\right)^{1/2}$ cannot be found since it is the square root of a negative number.
(c) (−2)^{1/2} cannot be found. It too is the square root of a negative number.
We have seen that, subject to certain cautionary notes, the rules for manipulating negative and fractional powers are similar to those for powers that are positive whole numbers. In fact the general rules for manipulating powers are:
a^{ p} × a^{ q} = a^{ p + q}(13) i
and(a^{ p})^{ q} = a^{ pq}(14)
✦ Use the above rules to derive a similar rule expressing the quotient a^{ p}/a^{ q} as a power of a.
✧ a^{ p}divided by a^{ q} is identical to a^{ p} multiplied by a^{−q}.
So,a^{ p}/a^{ q} = a^{ p} × a^{−q} = a^{ p−q}
where we have used the rule for multiplication with q replaced by −q.
✦ How would you interpret a^{ p/q} in words?
✧ If we use the general rules we may equally well write
a^{ p/q} = (a^{ p})^{1/q} or a^{ p/q} = (a^{1/q})^{ p}
so(a^{ p})^{1/q} is the qth root of the pth power a
and (a^{1/q})^{ p} is the pth power of the qth root of a.
Question T14
Evaluate the following:
(a) 7^{4/9} (b) (2.7)^{3}/(2.7)^{2} (c) $\sqrt{2\os} \times \sqrt[{\large\raise{2pt}3}]{2\os}$ (d) $\dfrac{\sqrt{3\os}}{\sqrt[{\large\raise{2pt}5}]{3\os}}$
Answer T14
(a) 7^{4/9} = 2.37
(b) (2.7)^{3}/(2.7)^{2} = (2.7)^{3−2} = (2.7)^{1} = 2.7
(c) $\sqrt{2\os} \times \sqrt[{\large\raise{2pt}3}]{2\os} = 2^{1/2} \times \left(2^{1/3}\right) = 2^{1/2+1/3} = 2^{5/6} = 1.78$
(d) $\dfrac{\sqrt{3\os}}{\sqrt[{\large\raise{2pt}5}]{3\os}} = 3^{1/2} \times (3^{1/5})^{1} = 3^{1/21/5} = 3^{3/10} = 1.39$
Quite apart from numerical evaluations, the rules for powers allow algebraic relations and physical laws to be written in convenient forms. For example, if you are dealing with a formula that involves a term $r\sqrt{r}$ you might well find it more convenient (and elegant) to rewrite the term as r^{3/2} or even r^{1.5}. Such rewriting might also make the term easier to evaluate since many calculators have an x^{ y} button but none are likely to have an $r\sqrt{r}$ button.
Question T15
Rewrite the following in the form r^{ p}:
(a) $1/\sqrt{r\os}$ (b) $r^3\sqrt{r\os}$ (c) $r^5\sqrt{r\os}$
Answer T15
(a) $1/\sqrt{r\os} = 1/r^{1/2} = r^{1/2}$
(b) $r^3\sqrt{r\os} = r^3 \times r^{1/2} = r^{7/2}$
(c) $r^5/\sqrt{r\os} = r^5/r^{1/2} = r^5 \times r^{1/2} = r^{9/2}$
Question T16
Simplify the following expressions:
(a) 4x^{2}y^{2}z^{4} (b) $\dfrac{h^2n^2}{8m}\left(\sqrt[{\uproot2\large3}]{L^3}\right)^{2}$ (c) $\left( \dfrac{8\pi^2Z\sqrt{l^5}}{3\sqrt[{\uproot2\large3}]{Z\os}}\right)^{1/2}$
Answer T16
(a) 4x^{2}y^{2}z^{4} = (2xyz^{2})^{2}
(b) $\dfrac{h^2n^2}{8m} \left(\sqrt[{\large3}\uproot2]{L^3}\right)^{2} = \dfrac{h^2n^2L^{2}}{8m} = \dfrac{1}{8m} \left(\dfrac{hn}{L}\right)^2$
(c) $\left(\dfrac{8\pi^2Z\sqrt{l^5}}{3\sqrt[{\large3}\uproot2]{Z\os}}\right)^{1/2} = \left(\dfrac{8\pi^2Z^{2/3}l^{5/2}}{3}\right)^{1/2} = \dfrac{2\sqrt{2\os}\pi Z^{2/6}l^{5/4}}{\sqrt{3\os}} = \left(\dfrac83\right)^{1/2}\pi Z^{1/3}l^{5/4}$
In Section 2 we dealt with simplifying and rewriting expressions, but an expression on its own is not usually very useful. You are much more likely to deal with an equation – a statement that two expressions are equal, such as
$s = ut + \dfrac12 at^2$(15)
It is worth stressing that the = sign means ‘has the same value as’, so, while the expressions on either side of the equation may look very different, they actually represent exactly the same number. This holds true even for algebraic equations – the letters in such an equation simply stand for the numbers that make the two sides equal.
Equations generally involve constants (i.e. quantities with a fixed and pre–determined value) and variables (i.e. quantities that may take on any one of a range of values). Often, an equation will relate the value of one variable to the value (or values) of one or more other variables. Thus, in Equation 15, if the values of u, a and t are specified then the value of s may be determined. In this way, Equation 15 provides both a general relationship between the variables s, u, a and t, and a recipe for calculating the value of s when u, a and t have particular known values. In the case of Equation 15, s is said to be the subject of the equation since it is the single variable that is expressed in terms of the others.
In Section 2, you met various ways of using brackets and common factors to simplify expressions. These same techniques can be used to simplify equations, as the following example illustrates.
When a drop of water falls vertically through air three forces act on it: a downward gravitational force of strength F_{g}, an upward buoyancy force of strength F_{b} (due to air displacement), and another upward force (due to air resistance – or more properly viscosity) of strength F_{v}. The strengths of these forces are determined by the following equations:
$F_g = \dfrac43 \pi r^3\rho g\quad\quad F_b = \dfrac43\pi r^3 \sigma g\quad\quad F_s = 6\pi \eta r \upsilon$ i
where r is the radius of the water drop, ρ is its density, g is a constant known as (the magnitude of) the acceleration due to gravity, σ is the density of air, η is a constant known as the viscosity of air and υ is the speed of the drop.
✦ If the directions of the forces are taken into account, the strength of the total force acting on the falling drop is F = F_{g} − F_{b} − F_{v}. Express F in terms of r, g, ρ, σ, η and υ, and then simplify it as much as possible.
✧ $F = \dfrac43 \pi r^3\rho g  \dfrac43 \pi r^3\sigma g  6\pi \nu r \upsilon$.
If a common factor of $\dfrac43 \pi r^3 g$ is extracted from the first two terms on the right
$F = \dfrac43 \pi r^3 g (\rho  \sigma)  6\pi \nu r \upsilon$.
As a last step it is possible to extract a common factor of 2πr from each term on the right–hand side i
$F = 2\pi r \left[\dfrac23 \pi r^2 g (\rho  \sigma)  3\nu r \upsilon\right]$(16)
Question T17
Show (justifying each step) that
$\phi = \dfrac{b^2Vr\theta}{b^2a^2}  \dfrac{b^2Va^2\theta}{r(b^2a^2)}  \dfrac{a^2Ur\theta}{b^2a^2} + \dfrac{a^2Ub^2\theta}{r(b^2a^2)}$
may be simplified to give
$\phi = \left[b^2V\left(r\dfrac{a^2}{r}\right)  a^2U\left(r\dfrac{b^2}{r}\right)\right] \dfrac{\theta}{b^2a^2}$
Answer T17
If a common factor of $\dfrac{\theta}{b^a^2}$ is extracted from each term on the right–hand side
$\phi = \left(b^2Vr\dfrac{b^2Va^2}{r}a^2Ur\dfrac{a^2Ub^2}{r}\right)\dfrac{\theta}{b^2a^2}$
If a common factor of b^{2}V is extracted from the first two terms and a common factor of −a^{2}U from the second two terms
$\phi = \left[b^2V\left(r\dfrac{a^2}{r}\right)a^2U\left(r\dfrac{b^2}{r}\right)\right] \dfrac{\theta}{b^2a^2}$
You will often need to rearrange equations, perhaps to simplify them, or to ensure that they express relationships in a more illuminating way, or simply because you want to calculate the value of a particular variable that is not already the subject of the equation. Whatever the reason, the rearrangement should provide a different but equivalent relationship between the relevant variables.
A given equation can usually be rearranged in a variety of ways. The important principle to remember while performing the rearrangement is that both sides of the equation represent the same number. So, if you add or subtract a term to one side of an equation you must add or subtract an identical term to the other side of the equation. The same principle applies to multiplication and division and to other operations such as squaring or taking square roots. (In the latter case you must remember that a positive number has both a positive and a negative square root.) When rearranging equations you should bear in mind two important points:
When one side of an equation consists of the sum of several terms, each of those terms must be treated in the same way. (It’s no good just multiplying the first term on one side of an equation by some factor if you forget to do the same to all the other terms on that side.)
When dividing both sides of an equation by some divisor you must ensure that divisor is not equal to zero. (You wouldn’t expect to get a meaningful result if you divided a number by zero, so you shouldn’t try to do it to equations either.)
✦ The following ‘proof’ shows that 1 = 2. It is, of course, quite wrong. Spot the error. i
$\begin{align}&\text{Let }a\text{ and }b\text{ represent two numbers and suppose:}& & a = b\\&\text{Multiply both sides by }a\text{:} & & a^2 = ab\\&\text{Subtract }b^2\text{ from both sides:} & & a^2  b^2 = ab  b^2 \\&\text{Extract a common factor of }(a  b)\text{ from both sides:} & & (a  b)(a + b) = b(a  b) \\&\text{Divide both sides by }(ab)\text{:} & & (a + b) = b\\&\text{But, since }a = b\text{, this implies that:} & & 2b = b \\&\text{Finally, divide both sides by }b\text{:} & & 2 = 1\end{align}$
✧ From a mathematical point of view this is a ‘silly’ proof from the outset, but there is not actually anything wrong with it until both sides are divided by (a − b). Since a = b, this step amounts to dividing both sides by zero, which is not a legitimate procedure. The result of the division is quite bogus, as you can see by replacing a and b by some simple number such as 1 throughout the ‘proof’.
In general, if both sides of an equation are divided by an algebraic expression such as (a − b) then the result of that division (and anything deduced from that result) will only be valid if a ≠ b.
The following questions will give you some practice in rearranging equations. The first two make further use of the example of the falling water drop that was introduced in Subsection 3.1.
Question T18
Initially, as a drop of water falls through air its speed υ will increase and F_{v} will become larger. Consequently the value of F (= F_{g} − F_{b} − F_{v}) will decrease until υ becomes so large that F = 0. The speed at which this happens is called the terminal speed and is denoted υ_{t}. Find as simple an expression as possible for υ_{t} in terms of r, g, ρ, σ and η.
Answer T18
$F = 2\pi r \left[\dfrac23 r^2g(\rho\sigma)  3\eta \upsilon \right]$
So, when $F = 0$ and $\upsilon = \upsilon_t$ we have
$0 = \dfrac23 r^2g (\rho\sigma)  3\eta \upsilon_{\rm t}$
If $3\eta \upsilon_{\rm t}$ is added to both sides
$3\eta \upsilon_{\rm t}= \dfrac23 r^2g(\rho\sigma)  3\eta \upsilon_{\rm t} + 3\eta \upsilon_{\rm t}$
The last two terms on the right–hand side cancel, so if both sides are divided by $3\eta 2r^2$
$\upsilon_{\rm t} = \dfrac{2r^2}{9\eta}g(\rho\sigma)$
Question T19
The relative importance of F_{g}, F_{b} and F_{v} under various circumstances can be determined by studying the ratios F_{b}/F_{g}, F_{v}/F_{g} and F_{b}/F_{v}.
(a) Work out fully simplified equations relating each of these ratios to r, g, ρ, σ, η and υ.
(b) If F_{v} is ten times greater than F_{b} when the drop reaches its terminal speed (υ = υ_{t}) find an equation that expresses the radius of the drop in terms of σ, g, η and υ_{t}.
(c) If F_{b} is negligibly small when υ = υ_{t} then F_{g} = F_{v}. Under these circumstances, find an equation that expresses r in terms of ρ, g, η and υ_{t}.
Answer T19
(a) $\dfrac{F_{\rm b}}{F_{\rm g}} = \dfrac{(4/3)\pi r^3 \sigma g}{(4/3)\pi r^3 \rho g} = \dfrac{\sigma}{\rho}$
$\dfrac{F_{\rm v}}{F_{\rm g}} = \dfrac{6\pi \eta r \upsilon}{(4/3)\pi r^3 \rho g} = \dfrac{3\eta\upsilon}{2r^2\rho g/3} = \dfrac{9\eta\upsilon}{2r^2\rho g}$
$\dfrac{F_{\rm b}}{F_{\rm v}} = \dfrac{(4/3)\pi r^3 \sigma g}{6\pi \eta r \upsilon} = \dfrac{2r^2\sigma g}{9\eta\upsilon}$
(b) If $\dfrac{F_{\rm v}}{F_{\rm b}} = 10$ when $\upsilon = \upsilon_{\rm t}$
$10 = \dfrac{F_{\rm v}}{F_{\rm b}} = 1/(F_{\rm b}/F_{\rm v}) = \dfrac{9\eta\upsilon_{\rm t}}{2r^2\sigma g}$
If we multiply both sides by $\dfrac{r^2}{10}, r^2 = \dfrac{9\eta\upsilon_{\rm t}}{20\sigma g}$
Since r^{2} is positive we can take square roots of both sides, and since we know that r itself (a radius) is also positive we can neglect negative roots. Thus, $r = \sqrt{\dfrac{9\eta\upsilon_{\rm t}}{20\sigma g}}$
(c) If $F_{\rm g} = F_{\rm v} \left(\text{ i.e. }\dfrac{F_{\rm v}}{F_{\rm g}} = 1\right)\text{ when }\upsilon = \upsilon_{\rm t}\quad 1 = \dfrac{9\eta\upsilon_{\rm t}}{2r^2\rho g}$
If both sides are multiplied by r^{2}, $r^2 = \dfrac{9\eta\upsilon_{\rm t}}{2\rho g}$
If positive square roots are taken as before $r = \sqrt{\dfrac{9\eta\upsilon_{\rm t}}{2\rho g}}$
Question T20
Rearrange the following equations to make x the subject:
(a) y = 2ax + b^{2} (b) (y − b) + (x − a) = 5xh^{2} + 3 (c) $\dfrac{1}{xa}+\dfrac{1}{yb} = 3t^2$, (d) $t+a = \sqrt{\dfrac{3b\os}{xy}}$
Answer T20
(a) $\text{If } y = 2ax + b^2 \text{ then }x = \dfrac{y  b^2}{2a} \text{ provided }a \ne 0$
(b) If (y − b) + (x − a) = 5xh^{2} + 3
then(x − a) − 5xh^{2} = 3 − (y − b)
sox − 5xh^{2} = 3 − (y − b) + a
i.e.x_{ }(1 − 5h^{2}) = 3 − y + b + a
If both sides are divided by (1 − 5h^{2})
$x = \dfrac{3  y + a + b}{(15h^2)} \text{ provided }(1  5h^2) \ne 0$
(The condition (1 − 5h^{2}) ≠ 0 is required to make sure we have not unwittingly divided by zero.)
(c) $\text{If }\dfrac{1}{xa} + \dfrac{1}{yb} = 3t^2 \text{ then }\dfrac{1}{xa} = 3t^2  \dfrac{1}{yb}$
so$\dfrac{1}{xa} = \dfrac{3t^2(yb)1}{yb}$
If both sides are multiplied by ${(x  a)(y  b)}{3t^2(yb)1}$
$\dfrac{yb}{3t^2(yb)1} = xa \text{ provided } 3t^2(yb)  1 \ne 0$
thus$x = \dfrac{yb}{3t^2(yb)1} + a \text{ provided } 3t^2(yb)  1 \ne 0$
(d) If $t + a = \sqrt{\dfrac{3b}{(xy)}} \text{ then }(t + a)^2 = \dfrac{3b}{(xy)}$
If both sides are multiplied by $\dfrac{x  y}{(t + a)^2}$
$xy = \dfrac{3b}{(t+a)^2}\text{ provided that }(t + a) \ne 0$
thus$x = \dfrac{3b}{(t+a)^2}\text{ provided that }(t + a) \ne 0$
Once you have completed the rearrangement of an equation it is often a good idea to check that the result you have obtained is consistent with the original equation. This can often be done quite easily by substituting some simple numbers for the various algebraic quantities that enter the equation. For instance, suppose you start out with the equation
υ^{2} = u^{2} + 2as and you rearrange it to find
$s = \dfrac{\upsilon^2 u^2}{2a}$
As a simple check of the correctness of this rearrangement pick some simple values for u, a and s in the original equation and substitute them into the equation to find υ^{2}.
Note that these values don’t have to be realistic nor do they involve units of measurement – we’re just checking for mathematical consistency. Letting u = 2 (i.e. u^{2} = 4), a = 3 and s = 4, for example, gives υ^{2} = 2^{2} + 2 × 3 × 4 = 28. Now, if these values are substituted into the rearranged equation we get s = (28 − 4)/(2 × 3) = 4, which was exactly the value of s we chose. By showing that the same set of values for u^{2}, υ^{2}, a and s satisfy both of the equations we have not proved that the rearrangement is correct but at least we haven’t found any inconsistency, so there’s a good chance that the rearrangement is right.
✦ Use the method outlined above to show that υ^{2} = u^{2} + 2as may not be rearranged to give
$\dfrac{\upsilon^2}{2s} = u^2 + a$
✧ Using the values chosen earlier
$\dfrac{\upsilon^2}{2s} = \dfrac{28}{8} = 3.5\text{, but }u^2 + a = 4 + 3 = 7$
Since the same set of values cannot satisfy the original equation and its supposed rearrangement there is clearly something amiss.
Question T21
Rearrange the equation F = Gm_{1}m_{2}/r^{2} to make r the subject, and check your result using small whole numbers before looking at the answer.
Answer T21
If $F=\dfrac{Gm_1m_2}{r^2}\text{ then }r^2=\dfrac{Gm_1m_2}{F}$
All of the quantities involved are positive, so we can take positive square roots of both sides to find
$r = \sqrt{\dfrac{Gm_1m_2}{F}}$
To check this out with some simple numbers, let m_{1} = 2, m_{2} = 3, r = 4 and G = 5. (This is nothing like the correct value of the constant G, but that does not matter; we are not carrying out a physical calculation, we are just checking algebraic manipulation.) With the assumed values for G, m_{1}, m_{2} and r we find
$F = \dfrac{5 \times 2 \times 3}{16} = \dfrac{30}{16}$
If these same values are used in the rearranged equation
$r = \sqrt{\dfrac{5\times 2\times 3\os}{30/16}} = \sqrt{16\os} = 4$ which is consistent
Many important relationships in physics involve just two variables. For example, an object of fixed mass m travelling at speed υ has momentum of magnitude p given by
p = mυ(17)
Similarly, light travelling at fixed speed c through a vacuum is characterized by a frequency f and a wavelength λ that are related by
f_{ }λ = c(18)
In these two examples m and c are constants, so only two variables are involved in each case.
In the case of Equation 17, the variables p and υ are said to be directly proportional (or just proportional). This indicates that if υ is multiplied by some factor then p must be multiplied by the same factor. Or, to put it another way, the ratio p/υ is constant.
In Equation 18 the relationship between f and λ is very different. Now, it is the product f_{ }λ that is constant with the consequence that if f is multiplied by a given factor then λ must be divided by that same factor. This situation is described by saying that f and λ are inversely proportional.
Direct proportionality and inverse proportionality are closely related, as can be seen by dividing both sides of Equation 18 by λ to obtain
$f = \dfrac{c}{\lambda}$
This shows that if f and λ are inversely proportional then f and (1/λ) are directly proportional.
The symbol ∝ is used to mean ‘is directly proportional to’ i so the proportionalities embodied in Equations 17 and 18 can be written as
$p \propto \upsilon \quad\text{and}\quad f \propto \dfrac{1}{\lambda}$
Conversely, a proportional relationship may always be represented by an equation, as follows:
Ify ∝ x then y = Kx where K is a constant(19)
A constant introduced in this way is generally referred to as a constant of proportionality. In Equation 17, m is the constant of proportionality relating p to υ, while in Equation 18 c is the constant of proportionality relating f to 1/λ.
Of course, physical relationships often involve more than two variables. For example, the pressure P, volume V and temperature T of a low–density sample of gas obey (at least approximately) an equation of the form
PV = NkT
where N is the number of particles in the sample and k is a constant. i Despite the relative complexity of this relationship the idea of proportionality is still very useful in this context, provided we keep the full relationship in mind. For instance, it is quite correct to say that V is proportional to T provided P and N are kept constant, and it is equally correct to say that P and V are inversely proportional provided N and T are kept constant.
Question T22
Write down equations that embody the following relationships. (You will have to introduce constants of proportionality for yourself; choose an appropriate letter and make a note to remind yourself that the newly introduced letter represents a constant.)
(a) V ∝ I
(b) R ∝ L provided r is constant, and R ∝ 1/r^{2} provided L is constant.
(c) (E + ϕ) ∝ f
(d) F_{grav} ∝ 1/r^{2} provided m_{1} and m_{2} are constant, F_{grav} ∝ m_{1} provided r and m_{2} are constant and F_{grav} ∝ m_{2} provided r and m_{1} are constant. i
Answer T22
(a) V = IR where R is a constant
(b) $R = \dfrac{kL}{r^2}$ where k is a constant
(c) E + ϕ = hf where h is a constant
(d) $F_{\rm grav} = \dfrac{G m_1 m_2}{r^2}$ where G is a constant
(Of course you may have used different symbols for the constants.)
As you have seen, equations are often used to specify the precise value of a quantity, but it is sometimes useful to be able to express the range of values that a variable might take. For example, you might have a variable power–supply unit incorporating a circuit breaker that trips if the current reaches a certain maximum value I_{max}. The condition that the current must be less than I_{max} can be written as I < I_{max}: the symbol < means ‘is less than’. The symbol ≤ means ‘is less than or equal to’.
✦ What do you think the symbols > and ≥ mean?
✧ > means ‘is greater than’ and ≥ means ‘is greater than or equal to’.
A statement that uses any of the symbols <, ≤, > or ≥ to compare two values is called an inequality. Note that whichever symbol is used, the greater of the two quantities is always at the wider end of the symbol. So, you might correctly write 3 < 6 or 6 > 3 but not 6 < 3 or 3 > 6.
When using an inequality it is important to remember that any negative number is considered to be less than any positive number. In fact, numbers are treated as if they are laid out along a line, sometimes called the number line, with increasingly large positive numbers further and further to the right of zero, and increasingly large negative numbers further and further to the left of zero. Any given point on the number line then represents a number that is greater than the numbers represented by points to the left of the given point.
lesser −4 −3 −2 −1 0 1 2 3 4 greater
Question T23
Which of the following inequalities are correct?
(a) 3 > −4
(b) −4 > −6
(c) (−4)^{2} > 8
(d) −10 ≤ −8
(e) $\dfrac12 \le 0.5$
(f) $\dfrac12 \lt \dfrac18$
(g) $\sqrt{2\os} \gt 1.40$
(h) −1 > 0
Answer T23
(a), (b), (c), (d) and (e) are true, (f) and (h) are false, (g) is ambiguous since $\sqrt{2\os} = ±1.41$ (This is why some authors prefer to use the root to denote positive square roots only.)
Inequality symbols can also be used to express both ends of a range of values. For example the statement 3 < x < 10 is read as ‘x is greater than 3 and less than 10’, in other words ‘x is in the range 3 to 10 – excluding the ‘endpoints’, 3 and 10.’
✦ Use inequality symbols to represent the statement ‘x is in the range 3 to 10 – including the endpoint values 3 and 10’.
✧ 3 ≤ x ≤ 10
(Note that in writing statements of this kind it is conventional to put the lesser value on the left – just like the number line.)
In practice, purely arithmetic inequalities such as 2 > 1 are rarely written down since they are ‘obvious’. It is much more likely that you will have to deal with algebraic inequalities such as x > b or 2x ≥ b − y. It is quite likely that you will be called on to rearrange such inequalities, either to simplify them or possibly to change their subject. The procedure is broadly similar to that for rearranging equations, but much greater care is needed when dealing with inequalities.
The basic rules for manipulating inequalities that involve the symbol > are given below. Similar rules concerning the other inequality symbols may be obtained by replacing the > by ≥, < or ≤ throughout and simultaneously replacing the < by ≤, > or ≥ throughout.
Rules for manipulating inequalities:
If x > y and a is any number, then x + a > y + a
If x > y and k is a positive number, then > ky
If x > y and k is a negative number, then kx < ky
Pay particular attention to Rule 3: if both sides of an inequality are multiplied by the same negative quantity then the inequality must be reversed. (So, if 2 < 3 then (−2)(2) > (−2)(3), i.e. −4 > −6, which is correct.)
✦ What problems might arise from the following actions?
(a) Divide both sides of an inequality by (1 − a) when a = 1.
(b) Multiply both sides of an inequality by (1 − a) when a = 1.
(c) Square both sides of an inequality.
(d) Take square roots of both sides of an inequality.
✧ (a) Dividing by (1 − a) when a = 1 is just another way of dividing by zero. It will produce meaningless results and should be avoided.
(b) Multiplying both sides of an inequality by zero is also dangerous. It’s true that 3 < 6, but is not true that (0) × (3) < (0) × (6), i.e. 0 < 0.
(c) Even if a > b it is not generally true that a^{2} > b^{2}. For instance −3 > −4 but it is not true that (−3)^{2} > (−4)^{2}.
(d) Square roots also need care. Negative quantities don’t have square roots and positive quantities have two square roots, so although 9 > 4, it is not true that −3 > −2, though it is true that 3 > 2.
Obviously, the most important thing to do when manipulating inequalities is to think carefully about each step. Here are some questions that require you to do just that.
Question T24
Make x (alone) the subject of each of the following inequalities:
(a) 2x + 4 > 6 (b) 2_{ }(a − x) ≤ 7 (c) 2_{ }(a − x) ≤ 3x + b
Answer T24
(a) If 2x + 4 > 6 then 2x > 6 − 4
i.e. 2x > 2 so x > 1
(b) If $2(a − x) ≤ 7$ then $a  x \le \dfrac 72$, i.e. $x \le 27  a$
If both sides are multiplied by −1 and the inequality is reversed $x \ge a  \dfrac72$
(c) If 2_{ }(a − x) ≤ 3x + b then 2a ≤ 3x + b + 2x
i.e. $ 2ab \le 5x$ so $\dfrac{2ab}{5} \le x$
which may be rewritten $x \ge \dfrac{2a  b}{5}$
Question T25
The electrical resistance R of a piece of wire of length l and cross–sectional area A, made from a material of resistivity ρ is given by the expression ρl/A. Suppose you need a sample of wire with resistance R such that R > R_{0}. Write down inequalities expressing,
(a) the range of acceptable lengths for a sample of given ρ and A, and
(b) the range of acceptable areas, for a sample of given l and ρ.
Answer T25
If $R = \dfrac{\rho l}{A}$, then $R \gt R_0$ implies that $\dfrac{\rho l}{A} \gt R_0$
(a) It follows that if ρ and A are given $l \gt \dfrac{R_0 A}{\rho}$
(b) It also follows that if l and ρ are given $A \lt \dfrac{\rho l}{R_0}$
The overall order of priority for operations is
Section 2Expressions containing a product of bracketed factors can be Subsection 2.2expanded by multiplying each term inside one pair of brackets by all the terms in the other pair of brackets, e.g.
(a + b)(c + d) = ac + ad + bc + bd
(a + b)^{2} = a^{2} + 2ab + b^{2}
(a + b)(a − b) = a^{2} − b^{2}
(a − b)^{2} = a^{2} − 2ab + b^{2}
Expressions can be Subsection 2.2Subsection 2.2simplified by using brackets to group together terms with a common factor, e.g.
ab + ac = a_{ }(b + c)
Subsection 2.3Fractions may be multiplied and divided according to the following rules:
$\dfrac ab \times \dfrac cd = \dfrac{ac}{bd}$(Eqn 5)
$\dfrac ab \div \dfrac cd = \dfrac{ad}{bc}$(Eqn 6)
Subsection 2.3Fractions may be added or subtracted by rewriting them so that they have a common denominator:
$\dfrac ab + \dfrac cd = \dfrac{ad+cb}{bd}$(Eqn 7)
$\dfrac ab  \dfrac cd = \dfrac{adcb}{bd}$(Eqn 8)
Subsection 2.4Powers may be used to express Subsection 2.4roots and Subsection 2.4reciprocals. In particular, if x is greater than zero:
x^{1} = x, x^{0} = 1, x^{−1} = 1/x and $x^{1/n} = \sqrt[{\large\raise{2pt}n}]{x}$.
The rules for manipulating powers are:
a^{ p} × a^{ q} = a^{ p+q}(Eqn 13)
(a^{ p})^{ q} = a^{ pq}(Eqn 14)
a^{ p}/a^{ q} = a^{ p−q}
a^{ p}b^{ p} = (ab)^{ p}
An Section 3equation can be Subsection 3.2rearranged by performing exactly the same operations on the whole of both sides of the equation.
Two variables x and y are directly proportional (x ∝ y) if their ratio can be expressed wholly in terms that do not include x and y. They are inversely proportional (x∝1/y) if their product can be expressed wholly in terms that do not include x and y.
If x ∝ y then x = Ky where K is a constant of proportionality.
Subsection 3.4Inequality symbols <, ≤, > and ≥ may be used to express the range of values of a variable that fulfil particular conditions.
Subsection 3.4Inequalities can be manipulated in ways similar to equations, but more care is needed and the direction of an inequality is reversed if both sides are multiplied by the same negative quantity.
Having completed this module, you should be able to:
Define the terms that are emboldened and flagged in the margins of the module.
Decide the correct order of priority for operations in an arithmetic or algebraic expression and hence evaluate such expressions (using a calculator where appropriate).
Use powers to express roots and reciprocals, and, where appropriate, combine powers in arithmetic and algebraic expressions.
Manipulate and simplify fractions, both arithmetic and algebraic, including cases where there is initially no common denominator.
Expand expressions that include brackets, and use brackets to simplify expressions.
Rearrange equations in order to simplify them or to change their subject.
Recognize examples of direct and inverse proportionality, and combine two or more proportional relationships.
Rewrite proportional relationships as equations by introducing appropriate constants of proportionality.
Interpret, use and rearrange inequalities.
Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.
Study comment Having completed this module, you should be able to answer the following questions each of which tests one or more of the Achievements.
Question E1 (A2)
Write down the operations you would need to carry out, in the correct order, to evaluate the following expressions, e.g. for (a − b)/3 you subtract b from a and then divide by 3:
(a) 5_{ }(x + 3y^{2}) (b) E/(1 + r/R)
Answer E1
(a) Square y, multiply by 3, add x, multiply by 5.
(b) Divide r by R, add 1, divide E by the result (or, after adding 1, divide by E and find the reciprocal).
(Reread Subsection 2.1 if you had difficulty with this question.)
Question E2 (A3 and A5)
Expand the following expressions and write them in as simple a form as possible:
(a) (x + 5)(x + 7) (b) x_{ }(x + 2)(x^{2} − 1) − x^{2}(2x − 1)
Answer E2
(a) (x + 5)(x + 7) = x^{2} + 7x + 5x + 35 = x^{2} + 12x + 35
(b) x_{ }(x + 2)(x^{2} −1) − x^{2}(2x − 1) = x_{ }(x^{3} − x + 2x^{2} − 2) − 2x^{3} + x^{2} = x^{4} − x^{2} + 2x^{3} − 2x − 2x^{3} + x^{2} = x^{4} − 2x = x_{ }(x^{3} − 2)
(Reread Subsection 2.2 if you had difficulty with this question.)
Question E3 (A4)
Rewrite the following expression as a single fraction expressed in as simple a form as possible:
2p/q + 7pr/3
Answer E3
$\dfrac{2p}{q} + \dfrac{7pr}{3} = \dfrac{6p}{3q} + \dfrac{7pqr}{3q} = \dfrac{p}{3q}(6+7qr)$
(Reread Subsections 2.2 and 2.3 if you had difficulty with this question.)
Question E4 (A6)
The equation $\upsilon = \sqrt{E/\rho\os}$ relates the speed, υ, of sound in a solid to the elastic modulus E and density ρ. Rearrange the equation so that ρ is the subject.
Answer E4
$\upsilon = \sqrt{E/\rho\os}$. If both sides are squared $\upsilon^2 = E/\rho$. If this is then rearranged, $\rho = E/\upsilon^2$.
(Reread Subsections 2.4 and 3.2 if you had difficulty with this question.)
Question E5 (A4 and A6)
The equation (1/u) + (1/υ) = 1/f describes a relationship between the distance u of an object from a lens of focal length f and the distance υ from the lens to the image. Rearrange the equation so that f is the subject.
Answer E5
$\dfrac1f = \dfrac1u + \dfrac{1}{\upsilon} = \dfrac{u+\upsilon}{u\upsilon}$
So,$f = \dfrac{u\upsilon}{u+\upsilon}$
(Reread Subsections 2.3 and 3.2 if you had difficulty with this question.)
Question E6 (A4, A5 and A6)
The kinetic energy E of an object of mass m and speed υ is given by E = mυ^{2}/2. If the object increases its speed from an initial value υ_{1} to a final value υ_{2}, its kinetic energy increases from E_{1} to E_{2}. Derive expressions in terms of m, υ_{1} and υ_{2} for
(a) the ratio of the final to the initial kinetic energy E_{2}/E_{1},
(b) the increase in kinetic energy E_{2} − E_{1}, and
(c) the fractional increase in kinetic energy (E_{2} − E_{1})/E_{1}.
In each case, write the expression in as simple a form as possible.
Answer E6
$E_1 = \dfrac12 m\upsilon_1^2$, $E_2 = \dfrac12 m\upsilon_2^2$
(a) $E_2/E_1 = (m\upsilon_2^2/2)/(m\upsilon_1^2/2)\text{. If a factor of }m/2\text{ is cancelled, }E_2/E_1 = \upsilon_2^2/\upsilon_1^2\text{, which can be further simplified to}$
$E_2/E_1 = (\upsilon_2/\upsilon_1)^2$
(b) $E_2 E_1 = (m\upsilon_2^2/2)  (m\upsilon_1^2/2)\text{. If a common factor of }m/2\text{ is taken out and brackets are added:}$
$E_2  E_1 = \dfrac m2(\upsilon_2^2  \upsilon_1^2)$
(This equation could also be written $E_2  E_1 = m(\upsilon_2\upsilon_1)(\upsilon_2+\upsilon_1)/2$, but this involves more terms.)
(c) $(E_2  E_1)/E_1 = (E_2/E_1)  (E_1/E_1) = (E_2/E_1)  1$.
If the answer to (a) is used, $(E_2  E_1)/E_1 = (\upsilon_2/\upsilon_1)^2  1$.
(Reread Subsections 2.2 and 3.2 if you had difficulty with this question.)
Question E7 (A3)
Simplify the following expressions:
(a) $\left.\left(l\sqrt{l\os}\right)\middle/l^2\right.$
(b) $z\left[\left.\left(\sqrt[{\large\raise{2pt}5}]{z^{1}}x^{0.5}y^{0.1}\right)\middle/\left(x^{1/2}y^{1/5}\right)^2\right.\right]^{1/3}$
Answer E7
(a) $(l\sqrt{l\os})/l^2 = l^{3/2}/l^2 = l^{3/2}l^{2} = l^{1/2}$
(b) $z\left[\left(\sqrt[{\large\raise{2pt}5}]{z^{1}}x^{0.5}y^{0.1}\right)/\left(x^{1/2}y^{1/5}\right)^2\right]^{1/3}$
= $z\left[\left(z^{1/5}x^{1/2}y^{1/10}\right)/\left(x^{1}y^{2/5}\right)\right]^{1/3}$
= $z[\left(z^{1/5}x^{1/2}y^{1/10}xy^{2/5}\right)^{1/3}$
= $z[\left(z^{1/5}x^{3/2}y^{3/10}\right)^{1/3}$
= $z^{14/15}x^{1/2}y^{1/10}$
(Reread Subsection 3.4 if you had difficulty with this question.)
Question E8 (A7)
The strength F_{el} of the electrostatic force between two objects with positive charges q_{1} and q_{2}, separated by a distance r, is given by F_{el} = (q_{1}q_{2})/(4πε_{0}r^{2}) where ε_{0} is a constant known as the permittivity of free space.
Express the relationships between the following variables as proportionalities:
(a) F_{el} and q_{1} (b) F_{el} and r
Answer E8
(a) F_{el} ∝ q_{1} provided q_{2} and r are fixed.
(b) F_{el} ∝ 1/r^{2} provided q_{1} and q_{2} are fixed.
(Reread Subsection 3.3 if you had difficulty with this question.)
Question E9 (A8)
A pendulum of length l, swinging under the influence of gravity, completes a full swing in a time T. Observations in many different places, each characterized by a local value for g (the magnitude of the acceleration due to gravity), show that $T \propto \sqrt{l\os}$ for fixed g and $T \propto 1/\sqrt{g\os}$ for fixed l. Write down an equation relating T, g and l.
Answer E9
$T = k\sqrt{l/g\os}$ where k is a constant.
(Reread Subsection 3.3 if you had difficulty with this question.)
Question E10 (A9)
(a) Express in words the statement 3 < x ≤ 20.
(b) If a^{2}x^{2} − a^{2}x^{2}h ≥ 2y, find the condition under which $x^2 \le \dfrac{2y}{a^2(1h )}$
Answer E10
(a) x is greater than 3, but less than or equal to 20.
(b) If a^{2}x^{2} − a^{2}x^{2}h ≥ 2y then x^{2}a^{2}(1 − h) ≥ 2y
So,$x^2 \ge \dfrac{2y}{a^2(1h)}$ > if a^{2}(1 − h) > 0
and$x^2 \le \dfrac{2y}{a^2(1h)}$ if a^{2}(1 − h) < 0
However, a^{2} must be positive, so the condition a^{2}(1 − h) < 0 will be satisfied if h > 1.
(Reread Subsection 3.4 if you had difficulty with this question.)
Study comment This is the final Exit test question. When you have completed the Exit test go back and try the Subsection 1.2Fast track questions if you have not already done so.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to the following Ready to study? Subsection.
Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Section 4Closing items.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.