Mathematics Revision Questions for the University of Bristol School of Physics

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Mathematics Revision Questions for the University of Bristol School of Physics You will not be surprised to find you have to use a lot of maths in your stu of physics at university!

... University physics lecturers will assume a familiarity with some basic mathematical tools and techniques.

1 Mathematics Revision Questions for the University of Bristol School of Physics You will not be surprised to find you have to use a lot of maths in your stu of physics at university! You need to be completely familiar with all the maths you learnt at A-level and be able to apply it quickly and easily. This will involve algebra, calculus (differentiation and integration), vectors,.... University physics lecturers will assume a familiarity with some basic mathematical tools and techniques. All this requires a lot of practice so we have provided here a brief set of revision notes and revision questions which cover the most important techniques for the physics lectures at the beginning of the first year. In addition, you will be asked to complete some electronic exercises, testing these skills, throughout the first semester. We advise that you practice your Maths skills regularly throughout your first year. If you find that you can t do a question or if there is a particular area with which you have difficulties, do not worry. We suggest you first look through your A level notes but do not panic if you are completely stuck! Simply make a note of the topic or question and ask your physics tutor in tutorial sessions. Another possible source for more information may be the recommended textbook Physics for Scientists and Engineers by Tipler and Mosca. In recent years, first year students have found it helpful to be given these A-level maths revision notes and questions before the start of term to give you an opportunity to refresh your memory before lectures begin. Please note that there is no need to complete all the questions or hand them in. They are for your benefit, so that you can understand what will be expected of you, what you have forgotten and what you may need to learn. 1 Algebra Revision Material Quadratic equations of the form ax + bx + c = where a, b and c are real valued constants (in mathematical notation this written as a, b, c R) are solved by the following equation, x = b ± b 4ac. (1) a Polynomial long division allows for the solving of problems such as, x 3 + 3x + 3x + 1 x = x + x + 1,

2 and can be solved in multiple ways. Two common schemes for solving such expressions are, x + x + 1 or x + 1 ) x 3 + 3x + 3x + 1 x 3 x 1 1 x + 3x x x x + 1 x 1 For partial fractions, there are a number of rules which are useful to know: For every linear factor such as (ax+b) in the denominator, there will be a partial fraction of the form A/(ax + b). For every repeated factor such as (ax + b) in the denominator, there will be two partial fractions: A/(ax + b) and B/(ax + b). For higher powers there will be correspondingly more terms. For quadratic factors in the denominator e.g.(ax +bx+c), there will be a partial fraction of the form: (Ax + B)/(ax + bx + c). The exponents of variables are combined as follows, Logarithms follow the below rules, a m a n = a m+n a m a n = am n. a x = n x = log a n n = a log a n ; log a 1 = ; log a a = 1 ( ) m log a (mn) = log a m + log a n; log a = log n a m log a n; log a (m p ) = p log a m

3 Practice Problems 1.1. Solve the following quadratic equations by factorisation: (a) x + 8x + 1 = (b) x + x = (c) 4x + 8x 1 = (d) x x 6 = (e) 3x 11x = (f) x + 3 = x (g) 4x 15x + 9 = (h) 36x 48x + 16 = (i) 3(x + x) = 9 (j) x + 4x 5 = (k) x 7x 4 = (l) x 63 x = 1.. Complete the square for the following quadratic equations i.e. rewrite them in the form (x + a) + b =, and hence solve for x: (a) x 4x 5 = (b) x 6x + 9 = (c) x 1x + 35 = (d) x 3x 5 = (e) x 6x + 1 = (f) x + 8x + 17 = (g) 3x 3x = (h) 3x 6x 1 = 1.3. Solve the following quadratic equations using the quadratic formula given in Equation (1): (a) x = 8 3x (b) (x )(x 1) = 3 (c) (x + ) 3 = (d) x + 4 x = 7 (e) 3x + x = 6 ( (f) x 1 ) = 1 4 (g) x(x + 4) = 1 (h) 4x(x ) = Solve the following cubic equations: (a) x 3 x 15x =. (b) x 3 4x + x + 6 =, given that x = 1 is a solution. (c) x: x 3 19x + 3 =, given x = 3 is a solution. 3

4 (d) x: x 3 + x x 4 =, given x = is a solution. (e) x 3 9x + 15x + 5 =, given x = 5 is a solution Solve the following equations: (a) x 1/3 3x 1/3 =. (b) (x x) 9(x x) + 18 =. (c) 3 x x = (.67) 1/x. (d) 3 x 3 x+1 + =. (e) 5 x = 7 x Express the following in partial fractions: (a) (b) 4 x 4 x + 1 (x 1)(x + ) (c) 1 x + 5x + 6 (d) x (x + 1)(x 1) (e) 1 x 3 1 (f) x + x + 1 x + x (g) 5x3 + x + 5x x 4 1 7x + (h) (x + ) (x ) (i) 1(x + 1) (x + 3)(x + 1) (j) 4x + 1 (x 3)(x + x + 1) (k) x + 7 (x + ) (x 3) (l) x4 + 5x 3 + 9x + 6x + 5 (x + )(x + 1) 1.7. Simplify the following expressions (a) (x3 ) y 4 x 6 y a b (b) abc 4 (c) 5 4 3n+1 8 n (d) 36x y 4 (e) 3 7x 6 y 3 4

5 1.8. Evaluate the following expressions: (a) 1 3/ 1 1/ (b) (16 1/4 ) Solve the following expressions for x: (a) x = 4 (b) 4 x = (c) ( 1 ) x = 1.1. Questions on logarithms (a) If log a q = 5 + log a b, and c = a 4, prove that q = abc. (b) Simplify (i) a log a x ; (ii) a log a x. (c) Write down the values of log 16 and log 8. (d) Given that log b a = c and log c b = a, prove that log c a = ac. (e) If log x 1.4 =, find x. (f) Without using a calculator, simplify and evaluate the following expression: ) ) ) log ( log ( 6 7 log ( 5 8 (g) Solve the following equation: log a (x + 3) log a x = log a. 5

6 Trigonometric Equations Revision Material Basic definitions: Pythagorean identities: Addition formulae: Double angle formulae: tan θ = sin θ cos θ ; cosec θ = 1 sin θ ; sec θ = 1 cos θ ; cot θ = 1 tan θ. sin θ + cos θ = 1 sec θ = 1 + tan θ cosec θ = 1 + cot θ sin(a + B) = sin A cos B + cos A sin B sin(a B) = sin A cos B cos A sin B cos(a + B) = cos A cos B sin A sin B cos(a B) = cos A cos B + sin A sin B tan A + tan B tan(a + B) = 1 tan A tan B tan A tan B tan(a B) = 1 + tan A tan B sin A = sin A cos A cos A = cos A sin A cos A = cos A 1 cos A = 1 sin A tan A = tan A 1 tan A 6

7 Sum and difference formulae: ( ) ( ) A + B A B sin A + sin B = sin cos ( ) ( ) A + B A B sin A sin B = cos sin ( ) ( ) A + B A B cos A + cos B = cos cos ( ) ( ) A + B A B cos A cos B = sin sin Formulae using t = tan x : sin x = t 1 t t ; cos x = ; tan x = 1 + t 1 + t 1 t. Transformation: a cos x + b sin x = R cos(x α), where R = a + b and tan α = b/a. Practice Problems.1. Find all angles x in the range x 36 which satisfy the following equations: x (a) tan x =.456 (c) cos = tan (b) sin 3x =.5678 (d) sin 3x = cot Find all values of x ( x 36 ) which satisfy the following equations: (a) 6 sin x 5 cos x = (b) 4 tan x cot x = 5cosec x (c) tan(45 + x) + cot(45 + x) = 4 (d) cos x + 7 sin x + 3 = (e) cos 3x 3 cos x = cos x Using Trigonometric Formulae: 1 + cot A cot B (a) Use addition formulae to show that cot(a B) = cot B cot A. (b) Show that sin(a + B + C) = cos A cos B cos C(tan A + tan B + tan C tan A tan B tan C). 7

8 (c) Show that: i. sin 3A = 3 sin A 4 sin 3 A; ii. cos 3A = 4 cos 3 A 3 cos A. (d) Show that sin x(cos x + cos 4x + cos 6x) = sin 3x cos 4x. (e) Show that sin (A + B) sin (A B) = sin A sin B. (f) Find the values of x between and 18 that satisfy: cos x = cos x + cos 4x. (g) Without using a calculator, evaluate cos sin Problems with tan x : (a) If t = tan x, find the values of t that satisfy: (a + ) sin(x) + (a 1) cos x = a + 1 where a is a non-zero content. Hence find two acute angles satisfying the equation when a = 3. (b) If tan x = cosec x sin x, prove that tan x = ± 5. (c) If sec θ tan θ = x, prove that tan θ = (1 x)/(1 + x). (d) Find (1 + sin θ)(3 sin θ + 4 cos θ + 5) in terms of tan θ..5. Transformations: (a) Express cos x + sin x in the form R cos(x α) given the values of R and α. Hence find the maximum and minimum values of cos x + sin x. (b) Find the values of x between and 36 which satisfy the equations: (i) 3 cos x + sin x = 1 (ii) 8 cos x + 9 sin x = 7.5 (iii) cos x + 7 sin x = 5 (iv) 7 cos x 6 sin x = (v) 5 sin x 6 cos x = 4 (vi) 1 cos x 5 sin x + 3 = (c) Express y = W (sin α+µ cos α) in the form R cos(α β), giving R and tan β. Show that the maximum value of y is W 1 + µ and that it occurs when tan α = 1/µ 8

9 .6. Solve the following: (a) Find a positive value of x satisfying the equation tan 1 (x) + tan 1 (3x) = 1 4 π. (b) Solve the equation cos 1 (x 3) + sin 1 x = 1 π. 3 Differentiation Revision Material For a general function y = f(x), the derivative of y with respect to (sometimes abbreviated to w.r.t) x is denoted by: and is defined by: or y or f (x) { } f(x + δx) f(x) = lim δx δx () While this expression may generally be used to find any derivative from first principles, it is more usual to rely on standard results, such as given in the following table: 9

10 Function Derivative Comment y = x n y = e x y = ln x y = sin x y = cos x y = tan x y = sin 1 x y = tan 1 x y = f(x).g(x) y = f(x) g(x) y = f[g(x)] = nxn 1 = ex = 1 x = cos x = sin x = sec x = 1 1 x = x = g(x)f (x) + f(x)g (x) = g(x)f (x) f(x)g (x) g (x) = df dg dg n may be positive, negative or fractional. ln x log e x The Product rule The Quotient rule The Chain rule (A function of a function) 1

11 Practice Questions 3.1. Differentiate w.r.t. x: (a) x 3 (b) x + 4x + 5 (c) x 4 x 3.. Differentiate w.r.t. x: (d) (e) 1 x 1 x 4 (a) (b) 1 (1 + x) 1 (1 x) (c) 1 (3x + ) 3.3. Differentiate w.r.t. x: (a) sin 3x (b) x + sin x cos 5x (c) 5 (d) sec x (e) tan x 3.4. Differentiate w.r.t. x: (a) (1 x)( 3x + x ) (b) 5x3 3x x (c) (x + 1) x 3.5. Solve the following problems: (a) If y + x = cos x, find. (b) If y = x 4x, find the value of x for which =. (c) Differentiate 3x x w.r.t. x and evaluate the derivative when x = 3. (d) If y(x + 1) = 4, find. 11

12 3.6. Differentiate w.r.t. x: (a) x 5 + 5x 4 + 1x (b) 3x 5 cos x + (c) 3x 1 x (d) x 1 + x Differentiate w.r.t. x: (a) (x + )(x + 4) (b) (1 + x )(1 3x ) (c) (x ) (d) (x 1) /x (e) (1 x) 3 (f) (3x + 4) 3 (g) (3 + x)(4 x) 3.8. Differentiate w.r.t. x: (a) sin x cos x (b) x sin x + cos x (c) 1 (x ) sin x + x cos x (d) (x + 1) tan x (e) sec x tan x (f) cosec x cot x (g) x sec x + cos x 3.9. Differentiate w.r.t. x: (a) x 1 x (b) + x 1 x 3x (c) (x 1)(x + ) (d) (3 x ) 1 (e) x 4 x Differentiate w.r.t. x: (a) sin x x (b) 3 sin x 1 + cos x (c) sin x sin x 1 (d) + 7 cos x cos x sin x (e) 1 + tan x

13 3.11. Differentiate w.r.t. x: (a) (x 3) 4 (b) x 3 (c) cos 5x (d) sin(x ) (e) tan(x + 5) (f) sec(x 3 ) (g) sin x cos x (h) cosec x ( (i) x x (j) 1 + sin x 1 sin x (k) sin 3 x sin 3x (l) 4x 1 + x ) Differentiate w.r.t. x: (a) sin 1 x (b) 3 tan 1 x Differentiate w.r.t. x: (a) x(ln x 1) (b) ln(1/x) (c) ln(sec x) (d) x ln x (c) x 1 x + sin 1 x (d) tan 1 (sin x) (e) ln ( ) sin x + cos x sin x cos x (f) e x ln 3x (g) e x ln(sec x) (h) x e x tan 1 x Solve the following problems: (a) If x = aθ and y = aθ, find / in terms of θ. (b) If x = b(θ sin θ) and y = b(1 cos θ), show that ( ) ( ) θ = cosec 1. (c) If x = 3t + t 3 and y = 3 t 5/ express / in terms of t and prove that, when d y/ =, x has one of the values, ±6 3. (d) Find d y and d3 y when (i) y = 3 x5 + x and (ii) y = sin 4x. (e) Find the maximum and minimum values of the function (1 + x )e x. 13

14 4 Integration Revision Material A brief list of standard integrals: x n = xn+1 n C, n 1, (ax + b) n cos(ax + b) = 1 sin(ax + b) + C, a = (ax + b)n+1 (n + 1)a + C, n 1 sin(ax + b) = 1 cos(ax + b) + C a sec (ax + b) = 1 tan(ax + b) + C, a cosec (ax + b) = 1 cot(ax + b) + C a a + x = 1 a tan 1 x a + C, f (x) f(x) = ln f(x) + C, a x = sin 1 x a + C e ax = eax a + C Change of variable (integration by substitution): Special cases: Useful substitutions: φ(x) = φ(ax + b) = 1 a xφ(x ) = 1 φ(x) du du φ(u) du, u = ax + b φ(u) du, u = x If the integrand contains (a x ), x = a sin u or x = a cos u may be useful. If the integrand contains (a + x ), x = a tan u may be useful. If the integrand contains sin m x cos n x, where m and n are positive integers and at least one of them is odd: If n is odd, sin x = u should be used; If m is odd, cos x = u should be used. 14

15 Integration by parts: b a u dv b = [uv]b a a v du Practice Problems 4.1. Evaluate the following integrals: (a) (b) (c) (d) (e) (4x x + 9) ( x x 5 ) ( sin x + 5cosec x) (1 x ) x x x 4.. Evaluate the following integrals: (a) (b) (c) (d) (e) π/ 1 sin(1 x) cos 5x 1 9x (x 3) 3/ [ ] 1 (x 1) 1 3 ( x) 3 (f) (g) (h) (i) (f) (g) (h) (i) 1/ 1/ ( 4x + 3x 4 + 5x + 3 x x x + 1 ) 3 1 x x 4 + x + 1 x + x x x x 4x 3x 4x

16 4.3. Integrate the following functions w.r.t. x using appropriate substitutions: (a) x 1 x (b) x(1 + x ) 3/ (c) x x + 1 (d) [ x x ] 4.4. Integrate the following using the suggested substitutions: (a) (b) x a3 + x 3, using a3 + x 3 = u x x 8, using x4 = tan u (c) tan x sec x, using tan x = u (d) (e) (f) x 1 x 6, using x3 = sin u x x 4 + 9, using x = 3 tan u x + 1 x + x 9, using x + x 9 = u 4.5. Integrate by substitution or otherwise: (a) cos 3 x (b) sin 5 x (c) sin x cos x (d) cos 4 x sin 3 x (e) sin 3 x cos 5 x (f) sin 3 x (g) sin3 x cos x (h) cos x sin x 4.6. Evaluate the following integrals by parts, or otherwise: (a) cos 3x cos x π (e) x cos x (b) sin 3x cos x (f) x 1 + x (c) (d) π/ sin 3x sin x cos 4x sin x (g) (h) 1 x 1 + x sin 1 x 16

17 (i) (j) (k) π x sin 1 x 1 x x sin x x sin x (l) (m) (n) (o) π (π x) sin 3x x ln x x ln(x + 4) x 3 ln 5x 4.7. Evaluate the following integrals: (a) 1 e 3x (e) x e x (b) (c) (d) 1 (e x e x ) (x 1)e x e x+ (f) (g) (h) 1 x 3 e x [ e x + e x ] e x e x x e 1 + e x 1 x e + e x Vectors Revision material In the context of this course, a vector is a line segment from a fixed point, called the origin and written O, to another point in two- and three-dimensional space. We will write vectors as bold italic, e.g. v, r. In handwritten script, this is often denoted with a straight underline, v or a curly underline v. Some books and lecturers will use other notation for a vector, such as an overarrow, v. Vectors are used to denote quantities which inherently involve position in space or direction: position, velocity, acceleration, force, momentum,... are all vectorial quantities. Quantities without a sense of direction are called scalars; examples include energy, temperature, and speed. Vectorial quantities can have physical dimensions and units; for instance the position vector r is a length (measured in metres), whereas velocity v has the same units as speed (m s 1 ), and acceleration a is measured in m s, etc. 17

18 Multiplying a vector by a number (i.e. a scalar) gives a vector parallel to the original vector, but with its length changed. For instance, the vector r lies in the same direction as the line from O to r, but is twice as far from the origin. Vectors can also be added together, and by the parallelogram rule, the order of addition not not matter, a + b = b + a. In general, vectors can be added together and multiplied by scalars, which looks a bit like the algebra of scalars. However, vector multiplication is different (as we will see) and in particular you can t divide by a vector! Although there is a simple geometric meaning to vector addition and scalar multiplication, there is no obvious geometric construction for vector multiplication or division. Expressing a vector as a sum of other vectors is sometimes called resolving the vector. Often, we write a vector v as a product of its magnitude (or modulus) v and its direction v. The magnitude v is a scalar that carries the sense of the length of the vector; it has the same units as the vector itself. The direction v is defined as v = v v, i.e. the vector divided by its scalar magnitude. v is therefore a dimensionless vector of length 1 (said to be a unit vector), pointing in the same direction as v. Different vectors, measuring different physical quantities, may point in the same direction (e.g. force and acceleration in the vectorial version of Newton s law F = ma in this case, they have the same direction, i.e. F = â. Putting these ideas together, we can define the direction vector i in the direction of the x-coordinate, j in the y-coordinate, and k in the z-coordinate. These vectors are defined to be unit vectors, so i = î, j = ĵ, k = k, i = j = k = 1. Therefore we can express a coordinate vector r with components x, y and z in their respective directions as r = xi + yj + zk. The velocity vector v, with components v x, v y and v z, can similarly be written v = v x i + v y j + v z k. The x, y, z axes are sometimes called Cartesian axes (after the French scientist and philosopher René Descartes), and the vectors i, j, k pointing along these directions are sometimes called Cartesian basis vectors. 18

19 If we consider vectors in the xy-plane, we can see how some of this fits together. Let s say r = xi + yj. By Pythagoras theorem, we can see that the modulus of the direction vector r = x + y, since the x, y axes are mutually perpendicular. To keep the notation simple, we will write r = r. If the vector r makes an angle φ with respect to the x-axis (i.e. the i direction), so that if φ = 9 then it is parallel to the y-axis (the j-direction), then we can use straightforward trigonometry to show that r = cos φ i + sin φ j. Since r = r r, we conclude that x = r cos φ, y = r sin φ. r is sometimes called the radial coordinate, since it is the radius of the circle centred on O as the angle φ varies. It is conventional to define the angle as here as the direction angle (also called azimuthal angle, and often denoted by θ) of the two-dimensional vector. It is given by the formula φ = arctan(y/x) (note you have to be careful that you are not wrong by 18, since y/x = ( y)/( x)). O r x r y With a fixed choice of axes (and basis vectors along them), it is often convenient to write a vector as an array of Cartesian components, e.g. r = (x, y, z). In this sense, we have i = (1,, ), j = (, 1, ), k = (,, 1). Vector arrays are also often written as column vectors, i.e. r = ( x ) yz. In this course, we will usually write vectors in terms of i, j, k rather than using arrays or column vectors. You are encouraged to get comfortable with both ways of writing vectors! The next important notion is that of the scalar product of two vectors (also called the dot product ): it is a form of multiplying two vectors which gives a scalar quantity (rather than a vector). For two vectors a = a x i + a y j + a z k, b = b x i + b y j + b z k, the scalar product a b is the scalar a b = a x b x + a y b y + a z b z. If the vectors are only in two dimensions (the xy plane), we can proceed as if a z, b z =. We can work this out assuming a and b are both in the xy plane, with a making an angle α with the x-axis, and b making an angle β with the x-axis. From what we saw above, we 19

20 can write a = a x i + a y j = a (cos α i + sin α j), b = b x i + b y j = b (cos β i + sin β j), where we have written a for a and b for b. The scalar product is therefore a b = a x b x + a y b y = a b cos α cos β + a b sin α sin β. We can idnetify the common scalar factor of a b out at the front, and apply the trigonometric identity cos α cos β + sin α sin β = cos(α β). α β is of course the angle between the two vectors, which is often called θ. This result holds in three dimensions as well, and the general result is that a b = a b cos θ.x This is the general way of finding angles between vectors! Although we call this the scalar product, it is not a multiplication in the standard sense. However, just like standard multiplication, the order of the vectors in the scalar product does not matter, i.e. a b = b a. This can easily be justified by looking at both formulas for the scalar product (the sum of component products, or using the moduli and angle). If we take the scalar product of a vector with itself, we get the modulus squared (since θ in this case is zero, and its cos is 1), a a = a. Therefore we can easily find the modulus of any vector by taking the scalar product of the vector with itself, and taking the square root: a = a a. This is consistent with what we found before for the modulus of the position vector r = xi+yj using geometry. Since the Cartesian basis vectors i, j, k are mutually perpendicular (also called orthogonal), and each is a unit vector, they have the scalar products i i = 1, i j =, i k =, j i =, j j = 1, j k =, k i =, k j =, k k = 1. We can think of the Cartesian components of any vector a = a x i + a y j + a z k as the scalar product with the appropriate Cartesian basis vector, i.e. a x = a i, a y = a j, a z = a k. Thinking about the angle relation for scalar products, we can understand why some older books refer to a vector s Cartesian components as direction cosines.

21 Practice problems 4.1. Scalar product. (a) Find the scalar product A B for: i. A = 3i 6j; B = 4i + j; ii. A = 5i + 5j; B = i 4j; iii. A = 6i + 4j; B = 4i 6j; (b) What is the angle between the vectors A and B if A B = A B? 4.. Find the scalar product A B for: (a) A = i + 3j; B = 3i + 4j; (b) A = i + 3j k; B = 3i + j; (c) A = i + 3j k; B = 3i + j + 4k A wall clock has a minute hand that has a length of.5 m and an hour hand with a length of.5 m. Taking the centre of the clock as the origin, and choosing an appropriate coordinate system, write the position of the hour and minute hands as vectors when the time reads (a) 1 : ; (b) 3 : 3; (c) 6 : 3; (d) 7 : 15. (e) Call the position of the tip of the minute hand A and the position of the tip of the hour hand B. Find A B for the times given in the parts above Find the scalar product A B for: (a) A = i + 3j; B = 3i + 4j; (b) A = i + 3j k; B = 3i + j; (c) A = i + 3j k; B = 3i + j + 4k. 1

22 4.5. Vector manipulations transforming between different descriptions of a vector: (a) Find the Cartesian components of the following vectors which have a magnitude A and lie in the xy plane and make an angle θ with the x-axis if: i. A = 1 m, θ = 3 ; ii. A = 5 m, θ = 45 ; iii. A = 7 km, θ = 6 ; iv. A = 5 km, θ = 9 ; v. A = 15 km s 1, θ = 15 ; vi. A = 1m s 1, θ = 4 ; vii. A = 8m s, θ = 7. (b) Find the magnitude and direction of the following vectors: i. A = 5i + 3j; ii. B = 1i 7j; iii. C = i 3j + 4k.

Mathematics 1 Lecture Notes Chapter 1 Algebra Review

Mathematics 1 Lecture Notes Chapter 1 Algebra Review c Trinity College 1 A note to the students from the lecturer: This course will be moving rather quickly, and it will be in your own best interests to