Chapter 2: Mathematical Methods
In the 1840s, the Irish mathematician William Rowan Hamilton was struggling with a problem that seems simple today: how do you write down a quantity that carries a direction with it, and then do algebra with it? Ordinary numbers were not enough. A number can tell you how much, but it cannot tell you which way.
The story goes that while walking along a canal in Dublin with his wife, the key idea finally arrived, and he scratched the relation onto the stone of a bridge he was crossing so that he would not lose it. His system, called quaternions, was later trimmed down by physicists into something leaner and far more usable — the vector algebra you are about to learn. The arrows, the dot product, the cross product, the unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\): all of it grew out of that search for a way to calculate with direction.
Around the same period in the history of science, another mathematical tool was reshaping physics — calculus, developed independently by Isaac Newton and G. W. Leibnitz in the 17th century, which allowed physicists to handle quantities that change continuously rather than in jumps.
These two tools — vector analysis and calculus — are the language in which the rest of this book is written. This chapter teaches you that language.
Diagram
Fig. 2.1 – An opening illustration showing a stone bridge over a canal with a walking figure, and floating above it the symbols of vector algebra: an arrow labelled \(\vec{A}\), the unit vectors \(\hat{i}, \hat{j}, \hat{k}\), and the sign \(\int\).
By the end of this chapter, you will be able to answer each of these questions yourself.
2.1 Introduction
Physics is not only about ideas; it is also about calculating with those ideas and getting numbers that can be checked against experiment. To follow the topics covered in this book, you will therefore need certain mathematical tools kept ready at hand.
Two of these tools matter most: vector analysis and elementary calculus.
Vector analysis is the algebra of quantities that carry a direction along with their size — quantities such as force, velocity and displacement. Calculus is the mathematics of quantities that change continuously, such as the position of a moving particle or the area swept under a graph.
You will study calculus in full detail in mathematics in the XIIth standard. For that reason, this chapter deals with vector analysis thoroughly, but gives only a preliminary introduction to calculus — enough for you to understand the physics that appears in the later chapters, without having to wait for the mathematics course to reach the same point.
2.2 Vector Analysis
In the previous chapter you studied several aspects of physical quantities: how they are divided into fundamental and derived quantities, and how each one is expressed through its unit and its dimensions. That gave you a way of saying how much of a quantity is present.
However, you also need to understand that not all physical quantities are fully described by their magnitudes and units alone. Some of them hide a second piece of information, and if you leave it out, the description is incomplete.
The Walking Man: Why a Number Is Not Always Enough
Consider a simple situation. You are told the time for which a man has walked and the speed with which he walked. From these two numbers you can certainly find the distance travelled by the man — speed multiplied by time.
But can you say where exactly he has reached? You cannot. Unless you also know the direction in which he walked, his final position remains unknown. He may have walked 6 km due north, or 6 km due east, and the distance would be the same 6 km in both cases while his final position would be completely different.
Diagram
Fig. 2.2 – A man starting from a point O, with two different 6 km paths shown, one due north and one due east, ending at two different points; both have equal distance travelled but different displacements.
Notice in Fig. 2.2 that the two journeys agree in “how much” and differ only in “which way”. This single observation is the reason physics needs two different kinds of quantities.
Two Families of Physical Quantities
Therefore we can say that some physical quantities, called scalars, can be described with magnitude alone. Other physical quantities, called vectors, need to be described with magnitude as well as direction.
In the example above, the distance travelled by the man is a scalar quantity. The final position of the man relative to his initial position — that is, his displacement — needs both magnitude and direction to be stated, and is therefore a vector quantity.
Keep this pairing in mind, because it repeats throughout physics: distance and displacement, speed and velocity, and so on. In each pair, the scalar member reports only the size, while the vector member also fixes the direction.
In this chapter you will study the different aspects of scalar and vector quantities: how they are represented, how they are added, subtracted, resolved into components, and multiplied.
2.2.1 Scalars
Physical quantities which can be completely described by their magnitude alone are called scalars. To specify a scalar, all you need is a number together with a unit.
Take a simple example. When we say that a given metal rod has a length of 2 m, we are stating that the rod is two times longer than a certain standard unit, the metre. Here the number 2 is the magnitude and the metre is the unit. Together, the number and the unit give a complete idea of the length of the rod — nothing further needs to be added, because length does not point anywhere.
Length, therefore, is a scalar quantity. In the same way, mass, time, temperature, density and many other quantities are examples of scalars.
Because scalars carry no direction, they behave like ordinary numbers. Scalars can be added or subtracted by the rules of simple algebra: a rod of length 2 m joined end to end with a rod of length 3 m gives a length of 5 m, and 5 s of waiting followed by 7 s of waiting is simply 12 s.
2.2.2 Vectors
Physical quantities which need magnitude as well as direction for their complete description are called vectors. Displacement, velocity and force are common examples of vectors.
How a Vector Is Drawn and Written
A vector can be represented by a directed line segment, that is, by an arrow. The length of the line segment, drawn to some chosen scale, gives the magnitude of the vector, while the way the arrow points gives its direction.
For example, the displacement of a body from a point P to a point Q can be drawn as an arrow running from P to Q, as shown in Fig. 2.3. The starting point P is called the tail of the vector and the end point Q, where the arrow head is drawn, is called the head of the vector.
Diagram
Fig. 2.3 – An arrow drawn from point P to point Q, with P labelled as the tail and Q as the head, representing the displacement vector.
Symbolically this vector is written as \(\vec{PQ}\). Vectors are also represented by a single capital letter with an arrow above it, for example \(\vec{X}\), \(\vec{A}\), and so on.
The magnitude of a vector \(\vec{X}\) is written as \(|\vec{X}|\), or simply as the plain letter \(X\). The magnitude is always a non-negative number with a unit — it tells you the “size” of the vector after its direction has been set aside.
Let us now see a few examples of different types of vectors.
Types of Vectors
(a) Zero vector (Null vector): A vector having zero magnitude, with a particular direction which is arbitrary, is called a zero vector. Symbolically it is represented as \(\vec{0}\).
Two familiar situations produce a zero vector:
The velocity vector of a stationary particle is a zero vector, since the particle is not moving in any direction at all.
The acceleration vector of an object moving with uniform velocity is a zero vector, because a uniform velocity does not change either in magnitude or in direction, and acceleration measures exactly that change.
(b) Resultant vector: The resultant of two or more vectors is that single vector which produces the same effect as that produced by all the vectors together.
This idea is the whole reason vector addition exists. If three ropes pull a boat in three different directions, we do not need to keep track of all three forces separately once we know the resultant — one force, in one direction, would drag the boat in exactly the same way.
(c) Negative vector (opposite vector): A negative vector of a given vector is a vector of the same magnitude but opposite in direction to that of the given vector.
In Fig. 2.4, \(\vec{B}\) is a negative vector to \(\vec{A}\). Notice that the two arrows in Fig. 2.4 have equal lengths, so the magnitudes are equal, and only the sense of the arrow head is reversed. We therefore write \(\vec{B} = -\vec{A}\).
Diagram
Fig. 2.4 – Two arrows of equal length drawn one above the other, \(\vec{A}\) pointing to the right and \(\vec{B}\) pointing to the left, illustrating a negative vector.
(d) Equal vector: Two vectors \(\vec{A}\) and \(\vec{B}\) representing the same physical quantity are said to be equal if and only if they have the same magnitude and the same direction. Two equal vectors are shown in Fig. 2.5.
Diagram
Fig. 2.5 – Two arrows of equal length pointing in the same direction, labelled \(\vec{A}\) and \(\vec{B}\), illustrating equal vectors.
Observe carefully what this definition does not demand: it says nothing about where the vectors are drawn. Two equal vectors need not start from the same point. As shown in Fig. 2.5, they may be drawn at different places on the page and still be equal, because a vector is fixed by magnitude and direction alone, not by its location.
(e) Position vector: A vector which gives the position of a particle at a point with respect to the origin of a chosen co-ordinate system is called the position vector of the particle.
In Fig. 2.6, \(\vec{r} = \vec{OP}\) is the position vector of the particle present at P, drawn from the origin O to the point P.
Diagram
Fig. 2.6 – An x–y coordinate system with origin O and a point P in the first quadrant, with an arrow \(\vec{r}\) drawn from O to P.
The words “of a chosen co-ordinate system” in this definition are not decoration. Shift the origin and the position vector of the same particle changes, even though the particle has not moved at all. This is why the origin must always be stated before a position vector is quoted.
(f) Unit vector: A vector having unit magnitude in a given direction is called a unit vector in that direction. A unit vector carries direction and nothing else — its whole job is to point.
If \(\vec{M}\) is a non-zero vector, that is, if its magnitude \(M = |\vec{M}|\) is not zero, then the unit vector along \(\vec{M}\) is written as \(\hat{u}_M\) and is given by
\[\vec{M} = \hat{u}_M \, M \qquad \text{--- (2.1)}\]
\[\text{or,} \quad \hat{u}_M = \frac{\vec{M}}{M} \qquad \text{--- (2.2)}\]
Here \(\vec{M}\) is the given vector, \(M\) is its magnitude (with whatever SI unit the quantity carries — metre for displacement, newton for force, and so on), and \(\hat{u}_M\) is the unit vector along \(\vec{M}\), which is a pure number of magnitude 1 with no unit.
Read Eq. (2.1) as a sentence: any vector is its magnitude multiplied by a unit vector along its own direction. Read Eq. (2.2) the other way round: dividing a vector by its own magnitude strips away the size and leaves only the direction. Hence \(\hat{u}_M\) has magnitude unity and has the same direction as that of \(\vec{M}\).
The Unit Vectors of the Coordinate Axes
The three most-used unit vectors in this book are the ones that point along the coordinate axes. We use \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\), respectively, as unit vectors along the x, y and z directions of a rectangular (three dimensional) coordinate system.
\[\hat{u}_x = \hat{i}, \quad \hat{u}_y = \hat{j} \quad \text{and} \quad \hat{u}_z = \hat{k}\]
\[\therefore \; \hat{i} = \frac{\vec{x}}{x}, \quad \hat{j} = \frac{\vec{y}}{y} \quad \text{and} \quad \hat{k} = \frac{\vec{z}}{z} \qquad \text{--- (2.3)}\]
Here \(\vec{x}\), \(\vec{y}\) and \(\vec{z}\) are vectors along the x, y and z axes respectively, and \(x\), \(y\), \(z\) are their magnitudes. Equation (2.3) is nothing but Eq. (2.2) applied three times over, once to each axis.
Diagram
Fig. 2.7 – A three-dimensional rectangular coordinate system showing the unit vectors \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) drawn along the x, y and z axes from the origin.
As shown in Fig. 2.7, these three unit vectors are mutually perpendicular. Once they are fixed, any vector in space can be built out of them, and this is the machinery you will use in almost every numerical problem in the chapters that follow.
2.3 Vector Operations
So far you have learnt what a vector is and how it is drawn and written. The next question is a practical one: what can we do with vectors?
Scalars, as you saw, obey ordinary algebra. Vectors do not, because every operation on them must respect direction as well as magnitude. A new set of rules is therefore needed.
In the following subsections you will study three basic operations. First, multiplying a vector by a scalar, which changes its size but not the line along which it points. Then, adding and subtracting vectors of the same type. Finally, two geometrical laws — the triangle law and the parallelogram law — which tell you how to combine vectors that act at an angle to each other.
2.3.1 Multiplication of a Vector by a Scalar
Multiplying a vector \(\vec{P}\) by a scalar quantity, say \(s\), yields another vector. Let us write
\[\vec{Q} = s\vec{P} \qquad \text{--- (2.4)}\]
Here \(\vec{P}\) is the original vector, \(s\) is a scalar (a plain number, possibly with a unit of its own), and \(\vec{Q}\) is the new vector produced. Then \(\vec{Q}\) will be a vector whose direction is the same as that of \(\vec{P}\) and whose magnitude is \(s\) times the magnitude of \(\vec{P}\).
In other words, this operation stretches or shrinks the arrow without turning it. If \(s = 3\), the arrow becomes three times as long and continues to point the same way; if \(s = \tfrac{1}{2}\), it becomes half as long along the same line.
Note also that the scalar may carry a unit, and if it does, the new vector is a different physical quantity from the old one. Multiplying a vector by a pure number leaves the physical nature unchanged, but multiplying it by a scalar with a unit changes it.
2.3.2 Addition and Subtraction of Vectors
The addition or subtraction of two or more vectors of the same type, that is, of vectors describing the same physical quantity, gives rise to a single vector. This single vector is built so that its effect is the same as the net effect of all the vectors which have been added or subtracted. You already have a name for it from Section 2.2.2 — it is the resultant.
Only Like Vectors May Be Added
It is important to understand that only vectors of the same type can be added or subtracted. For example, a force \(\vec{F_1}\) and a force \(\vec{F_2}\) can be added to give the resultant force
\[\vec{F} = \vec{F_1} + \vec{F_2}\]
But a force vector can not be added to a velocity vector. The sum would have no physical meaning at all, because there is no single quantity that is “part force and part velocity”.
Vectors Along the Same Line
It is easy to find the addition of vectors \(\vec{AB}\) and \(\vec{BC}\) having the same or opposite direction but different magnitudes. Two cases arise.
Parallel vectors. If the individual vectors are parallel, that is, in the same direction, the magnitude of their resultant is the addition of the individual magnitudes,
\[\left|\vec{AC}\right| = \left|\vec{AB}\right| + \left|\vec{BC}\right|\]
and the direction of the resultant is the same as that of the individual vectors, as shown in Fig. 2.8 (a). In that figure a displacement of 5 cm followed by a further displacement of 7 cm in the same direction produces a net displacement of 12 cm in that same direction.
Diagram
Fig. 2.8 (a) – A displacement of 5 cm from A to B followed by 7 cm from B to C along the same straight line, with the resultant 12 cm displacement from A to C drawn below.
Anti-parallel vectors. However, if the individual vectors are anti-parallel, that is, in the opposite direction, the magnitude of their resultant is the difference of the individual magnitudes, and the direction is that of the larger vector, i.e.,
\[\left|\vec{AC}\right| = \left|\;\left|\vec{AB}\right| - \left|\vec{BC}\right|\;\right|\]
as shown in Fig. 2.8 (b). There, a force of 11 N acting towards the right and a force of 5 N acting towards the left leave a net force of 6 N towards the right — the direction of the larger of the two forces.
Diagram
Fig. 2.8 (b) – A force of 11 N from A towards B pointing right, a force of 5 N at C pointing left, and the resultant 6 N force from A to C pointing right.
The outer pair of bars in the formula matters. They ensure that the magnitude of the resultant is written as a positive number, whichever of the two vectors happens to be the larger.
Subtraction as Addition of a Negative Vector
Subtraction needs no separate rule. To subtract \(\vec{B}\) from \(\vec{A}\), simply reverse \(\vec{B}\) to get \(-\vec{B}\), and then add:
\[\vec{A} - \vec{B} = \vec{A} + (-\vec{B})\]
This is why the anti-parallel case above looks like ordinary subtraction of numbers: turning a vector around and adding it is the same as taking it away.
2.3.3 Triangle Law for Vector Addition
The rules of the previous subsection worked only when the vectors lay along one straight line. When vectors of a given type do not act in the same or opposite directions, the resultant can be determined by using the triangle law of vector addition, which is stated as follows.
If two vectors describing the same physical quantity are represented in magnitude and direction by the two sides of a triangle taken in order, then their resultant is represented in magnitude and direction by the third side of the triangle drawn in the opposite sense — that is, from the starting point of the first vector to the end point of the second vector.
The phrase “taken in order” is the heart of the law. It means the tail of the second vector must be placed at the head of the first. Only then does the closing side of the triangle, drawn from the very first tail to the very last head, represent the resultant.
Let \(\vec{A}\) and \(\vec{B}\) be two vectors in the plane of the paper as shown in Fig. 2.9 (a). The sum of these two vectors can be obtained by using the triangle law described above, as shown in Fig. 2.9 (b). The resultant vector is indicated by \(\vec{C}\).
Diagram
Fig. 2.9 (a) – Two separate vectors \(\vec{A}\) and \(\vec{B}\) drawn in a plane, not yet joined.
Diagram
Fig. 2.9 (b) – Vector \(\vec{B}\) shifted so that its tail sits at the head of \(\vec{A}\), with the resultant \(\vec{C} = \vec{A} + \vec{B}\) drawn from the tail of \(\vec{A}\) to the head of \(\vec{B}\).
Notice in Fig. 2.9 (b) that \(\vec{B}\) has been shifted from its position in Fig. 2.9 (a) without changing its length or its direction. This is allowed, and you already know why: from Section 2.2.2, a vector is fixed by magnitude and direction alone, so sliding it about the page does not alter it.
\[\vec{C} = \vec{A} + \vec{B}\]
We can use the triangle law for showing that vector addition obeys two important properties.
Vector Addition Is Commutative
For any two vectors \(\vec{P}\) and \(\vec{Q}\),
\[\vec{P} + \vec{Q} = \vec{Q} + \vec{P} \qquad \text{--- (2.5)}\]
Figure 2.10 (a) shows the addition of the two vectors \(\vec{P}\) and \(\vec{Q}\) in two different ways. Triangle OAB shows \(\vec{P} + \vec{Q} = \vec{R} = \vec{OB}\), while triangle OCB shows \(\vec{Q} + \vec{P} = \vec{R} = \vec{OB}\).
Diagram
Fig. 2.10 (a) – A parallelogram OABC in which triangle OAB gives \(\vec{P}+\vec{Q}=\vec{R}=\vec{OB}\) and triangle OCB gives \(\vec{Q}+\vec{P}=\vec{R}=\vec{OB}\), the same diagonal \(\vec{R}\) serving both.
Both routes end at the same point B, so the same arrow \(\vec{OB}\) is obtained in both cases.
\[\therefore \; \vec{P} + \vec{Q} = \vec{Q} + \vec{P}\]
Vector Addition Is Associative
If \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\) are three vectors, then
\[(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})\]
Figure 2.10 (b) shows the addition of the three vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\) in two different ways to give the resultant \(\vec{R}\).
Diagram
Fig. 2.10 (b) – Points O, P, Q, R with \(\vec{A}\), \(\vec{B}\), \(\vec{C}\) placed head to tail, showing \(\vec{A}+\vec{B}\) and \(\vec{B}+\vec{C}\) as the two crossing intermediate resultants and \(\vec{R}\) as the common final resultant from O to R.
\[\vec{R} = (\vec{A} + \vec{B}) + \vec{C} \qquad \text{--- from triangle OQR}\]
\[\vec{R} = \vec{A} + (\vec{B} + \vec{C}) \qquad \text{--- from triangle OPR}\]
i.e.,
\[(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C}) \qquad \text{--- (2.6)}\]
Thus the associative law is proved. In words: when three or more vectors are to be added, it does not matter which pair you choose to combine first. This is what allows us, in the next example, to add four vectors one after another without worrying about the grouping.
2.3.4 Law of Parallelogram of Vectors
Another geometrical method of adding two vectors is called the parallelogram law of vector addition, which is stated as follows.
If two vectors of the same type, originating from the same point (that is, with their tails at the same point), are represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant vector is given in magnitude and direction by the diagonal of the parallelogram starting from the same point, as shown in Fig. 2.15.
Diagram
Fig. 2.15 – Parallelogram OACD with \(\vec{OA} = \vec{P}\) along the base and \(\vec{OB} = \vec{Q}\) inclined at angle θ, the diagonal \(\vec{OC} = \vec{R} = \vec{P} + \vec{Q}\), the angle α between \(\vec{R}\) and \(\vec{P}\), the angle β between \(\vec{R}\) and \(\vec{Q}\), and the perpendicular CD dropped on OA extended.
In Fig. 2.15, vector \(\vec{OA} = \vec{P}\) and vector \(\vec{OB} = \vec{Q}\) represent two vectors originating from the point O, inclined to each other at an angle θ. If we complete the parallelogram, then according to this law, the diagonal \(\vec{OC} = \vec{R}\) represents the resultant vector.
Magnitude of the Resultant
To find the magnitude of \(\vec{R}\), drop a perpendicular from C to reach OA (extended) at D. This perpendicular is what allows us to use right-angled triangles, and hence Pythagoras’ theorem.
In the right angled triangle ODC, by application of Pythagoras’ theorem,
\[\text{OC}^2 = \text{OD}^2 + \text{DC}^2\]
\[= (\text{OA} + \text{AD})^2 + \text{DC}^2\]
\[\text{OC}^2 = \text{OA}^2 + 2\,\text{OA}\cdot\text{AD} + \text{AD}^2 + \text{DC}^2\]
In the right angled triangle ADC, by application of Pythagoras’ theorem,
\[\text{AD}^2 + \text{DC}^2 = \text{AC}^2\]
Substituting this in the previous line, the two terms \(\text{AD}^2 + \text{DC}^2\) collapse into the single term \(\text{AC}^2\):
\[\therefore \; \text{OC}^2 = \text{OA}^2 + 2\,\text{OA}\cdot\text{AD} + \text{AC}^2 \qquad \text{--- (2.7)}\]
Also, from the figure,
\[\vec{OA} = \vec{P}, \quad \vec{AC} = \vec{OB} = \vec{Q} \quad \text{and} \quad \vec{OC} = \vec{R}\]
In \(\Delta\) ADC, \(\cos\theta = \dfrac{\text{AD}}{\text{AC}}\)
\[\therefore \; \text{AD} = \text{AC}\cos\theta = Q\cos\theta\]
Substituting in Eq. (2.7),
\[R^2 = P^2 + Q^2 + 2\,P\,Q\cos\theta\]
\[\therefore \; R = \sqrt{P^2 + Q^2 + 2\,P\,Q\cos\theta} \qquad \text{--- (2.8)}\]
Equation (2.8) gives us the magnitude of the resultant vector \(\vec{R}\). Here \(P\) and \(Q\) are the magnitudes of the two vectors (in whatever SI unit the quantity carries — newton for forces, metre per second for velocities), θ is the angle between them, and \(R\) is the magnitude of the resultant in the same unit.
Look at what Eq. (2.8) is telling you physically. The first two terms, \(P^2 + Q^2\), would be the whole answer if the vectors were perpendicular. The extra term \(2PQ\cos\theta\) is the “cooperation term”: it is largest and positive when θ = 0° and the vectors help each other fully, it vanishes at θ = 90°, and it becomes negative when the vectors begin to oppose each other.
Direction of the Resultant
To find the direction of the resultant vector \(\vec{R}\), we will have to find the angle (α) made by \(\vec{R}\) with \(\vec{P}\).
In \(\Delta\) ODC,
\[\tan\alpha = \frac{\text{DC}}{\text{OD}} = \frac{\text{DC}}{\text{OA} + \text{AD}} \qquad \text{--- (2.9)}\]
From the figure, \(\sin\theta = \dfrac{\text{DC}}{\text{AC}}\)
\[\therefore \; \text{DC} = \text{AC}\sin\theta = Q\sin\theta\]
Also,
\[\text{AD} = \text{AC}\cos\theta = Q\cos\theta \quad \text{and} \quad \text{OA} = P\]
Substituting these in Eq. (2.9), we get
\[\tan\alpha = \frac{Q\sin\theta}{P + Q\cos\theta}\]
\[\therefore \; \alpha = \tan^{-1}\left(\frac{Q\sin\theta}{P + Q\cos\theta}\right) \qquad \text{--- (2.10)}\]
Equation (2.10) gives us the direction of the resultant vector \(\vec{R}\), measured as the angle α that \(\vec{R}\) makes with \(\vec{P}\).
If β is the angle between \(\vec{R}\) and \(\vec{Q}\), it can be similarly derived that
\[\beta = \tan^{-1}\left(\frac{P\sin\theta}{Q + P\cos\theta}\right)\]
The symmetry between the two formulae is worth noticing: interchanging \(P\) and \(Q\) turns one into the other, exactly as it should, because α and β are the same angle measured from the two different sides.
2.4 Resolution of Vectors
Adding vectors is one half of the story. The other half is the reverse operation: taking a single vector apart into pieces that are easier to handle.
A vector can be written as a sum of two or more vectors along certain fixed directions. Thus a vector \(\vec{V}\) can be written as
\[\vec{V} = V_1\hat{\alpha} + V_2\hat{\beta} + V_3\hat{\gamma} \qquad \text{--- (2.11)}\]
where \(\hat{\alpha}\), \(\hat{\beta}\), \(\hat{\gamma}\) are unit vectors along chosen directions. \(V_1\), \(V_2\) and \(V_3\) are known as the components of \(\vec{V}\) along the three directions \(\hat{\alpha}\), \(\hat{\beta}\) and \(\hat{\gamma}\).
The process of splitting a given vector into its components is called resolution of the vector.
The components can be found along directions at any required angles. But if these components are found along the directions which are mutually perpendicular, they are called rectangular components.
Rectangular Components in Two Dimensions
Let us see how to find rectangular components in two dimensions.
Consider a vector \(\vec{R} = \vec{OC}\), originating from the origin of a rectangular co-ordinate system as shown in Fig. 2.17.
Diagram
Fig. 2.17 – A vector \(\vec{R} = \vec{OC}\) drawn from the origin O in the x–y plane, with perpendiculars dropped from C to the x-axis at A and to the y-axis at B, showing components \(R_x\) along OA and \(R_y\) along OB, and the angle θ between \(\vec{R}\) and the x-axis.
Drop perpendiculars from C that meet the x-axis at A and the y-axis at B.
\[\vec{OA} = \vec{R_x} \quad \text{and} \quad \vec{OB} = \vec{R_y}\]
\(\vec{R_x}\) and \(\vec{R_y}\) being the components of \(\vec{OC}\) along the x and y axes, respectively.
Notice in Fig. 2.17 that OACB is a rectangle, which is a parallelogram with right angles. Then by the law of parallelogram of vectors — the very law you proved in the previous section, now read backwards — the diagonal \(\vec{R}\) is the resultant of the two sides:
\[\vec{R} = \vec{R_x} + \vec{R_y}\]
\[\vec{R} = R_x\,\hat{i} + R_y\,\hat{j}\]
where \(\hat{i}\) and \(\hat{j}\) are unit vectors along the x and y axes respectively, and \(R_x\) and \(R_y\) are the magnitudes of the two components of \(\vec{R}\). Each of \(R_x\) and \(R_y\) carries the same SI unit as \(\vec{R}\) itself.
Finding the Components from the Angle
Let θ be the angle made by \(\vec{R}\) with the x-axis. Then, from the right-angled triangle OAC in Fig. 2.17,
\[\cos\theta = \frac{R_x}{R}\]
\[\therefore \; R_x = R\cos\theta \qquad \text{--- (2.12)}\]
\[\sin\theta = \frac{R_y}{R}\]
\[\therefore \; R_y = R\sin\theta \qquad \text{--- (2.13)}\]
These two results are worth memorising, but they must be used with care. The cosine belongs to the axis from which the angle θ is measured, and the sine to the axis perpendicular to it. If in a problem the angle is measured from the y-axis instead, the roles of sine and cosine are exchanged. Blindly writing “\(R\cos\theta\) for the horizontal component” is one of the commonest sources of lost marks in numerical problems.
Recovering the Vector from Its Components
Squaring and adding Eqs. (2.12) and (2.13), we get
\[R^2\cos^2\theta + R^2\sin^2\theta = R_x^{\,2} + R_y^{\,2}\]
Since \(\cos^2\theta + \sin^2\theta = 1\), the left side is simply \(R^2\):
\[\therefore \; R^2 = R_x^{\,2} + R_y^{\,2}\]
\[\text{or,} \quad R = \sqrt{R_x^{\,2} + R_y^{\,2}} \qquad \text{--- (2.14)}\]
Equation (2.14) gives the magnitude of \(\vec{R}\).
To find the direction of \(\vec{R}\), from Fig. 2.17,
\[\tan\theta = \frac{R_y}{R_x}\]
\[\therefore \; \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \qquad \text{--- (2.15)}\]
Equations (2.14) and (2.15) are the exact inverse of Eqs. (2.12) and (2.13). Given the magnitude and direction, you can find the components; given the components, you can recover the magnitude and direction. A vector may therefore be quoted in either language, and you must be fluent in both.
Rectangular Components in Three Dimensions
Similarly, if \(\vec{R_x}\), \(\vec{R_y}\) and \(\vec{R_z}\) are the rectangular components of \(\vec{R}\) along the x, y and z axes of the rectangular Cartesian co-ordinate system in three dimensions, then
\[\vec{R} = \vec{R_x} + \vec{R_y} + \vec{R_z} = R_x\hat{i} + R_y\hat{j} + R_z\hat{k}\]
\[\text{or,} \quad \left|\vec{R}\right| = \sqrt{R_x^{\,2} + R_y^{\,2} + R_z^{\,2}} \qquad \text{--- (2.16)}\]
Equation (2.16) is Pythagoras’ theorem carried into three dimensions, and it is the form in which almost every numerical answer of this chapter will finally be expressed.
Equal Vectors Have Equal Components
If two vectors are equal, it means that their corresponding components are also equal, and vice versa.
If \(\vec{A} = \vec{B}\), i.e., if
\[A_x\hat{i} + A_y\hat{j} + A_z\hat{k} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}, \quad \text{then}\]
\[A_x = B_x, \quad A_y = B_y \quad \text{and} \quad A_z = B_z\]
This single statement is extremely powerful. It converts one vector equation into three ordinary algebraic equations, one for each axis — which is exactly how vector equations are actually solved in practice.
2.5 Multiplication of Vectors
We saw that we can add or subtract vectors of the same type to get resultant vectors of the same type. Two forces add to give a force; two velocities add to give a velocity. Addition never changes what kind of quantity you are dealing with.
Multiplication behaves quite differently. When we multiply vectors of the same or different types, we get a new physical quantity, which may either be a scalar (scalar product) or a vector (vector product). Force multiplied by displacement, for instance, gives work — which is neither a force nor a displacement.
Also note the two simpler cases you already know. The multiplication of a scalar with a scalar is always a scalar. The multiplication of a scalar with a vector is always a vector — this was Eq. (2.4) of Section 2.3.1.
Let us now study the characteristics of a scalar product and a vector product of two vectors.
2.5.1 Scalar Product (Dot Product)
The scalar product or dot product of two nonzero vectors \(\vec{P}\) and \(\vec{Q}\) is defined as the product of the magnitudes of the two vectors and the cosine of the angle θ between the two vectors.
The scalar product of \(\vec{P}\) and \(\vec{Q}\) is written as
\[\vec{P}\cdot\vec{Q} = PQ\cos\theta \qquad \text{--- (2.17)}\]
where θ is the angle between \(\vec{P}\) and \(\vec{Q}\), and \(P\) and \(Q\) are the magnitudes of the two vectors in their respective SI units.
Observe that the right-hand side of Eq. (2.17) contains no direction whatsoever — only magnitudes and a cosine. That is why the answer is a scalar, and this is the whole point of the name.
Characteristics of Scalar Product
(1) The scalar product of two vectors is equivalent to the product of the magnitude of one vector with the magnitude of the component of the other vector in the direction of the first.
This is best seen in Fig. 2.18, where the perpendicular dropped from the head of \(\vec{Q}\) onto the line of \(\vec{P}\) cuts off a length \(Q\cos\theta\) — the projection of \(\vec{Q}\) on \(\vec{P}\).
Diagram
Fig. 2.18 – Vectors \(\vec{P}\) and \(\vec{Q}\) drawn from a common point O with angle θ between them, with a perpendicular dropped from the head of \(\vec{Q}\) onto \(\vec{P}\) marking the projection \(Q\cos\theta\), and a second perpendicular marking \(P\cos\theta\) on \(\vec{Q}\).
From Fig. 2.18,
\[\vec{P}\cdot\vec{Q} = PQ\cos\theta\]
\[= P\,(Q\cos\theta)\]
\[= P\,(\text{component of } \vec{Q} \text{ in the direction of } \vec{P})\]
Similarly,
\[\vec{P}\cdot\vec{Q} = Q\,(P\cos\theta)\]
\[= Q\,(\text{component of } \vec{P} \text{ in the direction of } \vec{Q})\]
So the dot product asks a single physical question: how much of one vector lies along the other?
(2) Scalar product obeys the commutative law of vector multiplication.
\[\vec{P}\cdot\vec{Q} = P\,Q\cos\theta = Q\,P\cos\theta = \vec{Q}\cdot\vec{P}\]
This follows immediately because \(P\) and \(Q\) are ordinary numbers, and the angle between the two vectors is the same whichever of the two you start measuring from.
(3) Scalar product obeys the distributive law of multiplication.
\[\vec{P}\cdot(\vec{Q} + \vec{R}) = \vec{P}\cdot\vec{Q} + \vec{P}\cdot\vec{R}\]
(4) Special cases of the scalar product \(\vec{P}\cdot\vec{Q} = PQ\cos\theta\):
(i) If θ = 0, i.e., the two vectors \(\vec{P}\) and \(\vec{Q}\) are parallel to each other, then
\[\vec{P}\cdot\vec{Q} = P\,Q\cos\theta = P\,Q\]
Thus,
\[\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1\]
because each unit vector is parallel to itself and has magnitude 1.
(ii) If θ = 180°, i.e., the two vectors \(\vec{P}\) and \(\vec{Q}\) are anti-parallel, then
\[\vec{P}\cdot\vec{Q} = P\,Q\cos 180^\circ = -P\,Q\]
The negative sign is physically meaningful: it signals that the two vectors oppose each other. A braking force opposing the motion of a car does negative work for exactly this reason.
(iii) If θ = 90°, i.e., the two vectors are perpendicular to each other, then
\[\vec{P}\cdot\vec{Q} = P\,Q\cos 90^\circ = 0\]
Thus,
\[\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0\]
(5) If \(\vec{P} = \vec{Q}\), then
\[\vec{P}\cdot\vec{Q} = P^2 = Q^2\]
In other words, the dot product of a vector with itself gives the square of its magnitude — a compact way of writing \(|\vec{P}|^2\).
(6) Scalar product of vectors expressed in terms of rectangular components:
Let
\[\vec{P} = P_x\hat{i} + P_y\hat{j} + P_z\hat{k}\]
and
\[\vec{Q} = Q_x\hat{i} + Q_y\hat{j} + Q_z\hat{k}\]
Then
\[\vec{P}\cdot\vec{Q} = P_x Q_x + P_y Q_y + P_z Q_z\]
Proof:
\[\vec{P}\cdot\vec{Q} = (P_x\hat{i} + P_y\hat{j} + P_z\hat{k})\cdot(Q_x\hat{i} + Q_y\hat{j} + Q_z\hat{k})\]
Expanding by the distributive law of characteristic (3), term by term,
\[= P_x\hat{i}\cdot(Q_x\hat{i} + Q_y\hat{j} + Q_z\hat{k})\]
\[+ P_y\hat{j}\cdot(Q_x\hat{i} + Q_y\hat{j} + Q_z\hat{k})\]
\[+ P_z\hat{k}\cdot(Q_x\hat{i} + Q_y\hat{j} + Q_z\hat{k})\]
\[= (\hat{i}\cdot\hat{i})P_xQ_x + (\hat{i}\cdot\hat{j})P_xQ_y + (\hat{i}\cdot\hat{k})P_xQ_z\]
\[+ (\hat{j}\cdot\hat{i})P_yQ_x + (\hat{j}\cdot\hat{j})P_yQ_y + (\hat{j}\cdot\hat{k})P_yQ_z\]
\[+ (\hat{k}\cdot\hat{i})P_zQ_x + (\hat{k}\cdot\hat{j})P_zQ_y + (\hat{k}\cdot\hat{k})P_zQ_z\]
Since,
\[\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1\]
and
\[\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = \hat{i}\cdot\hat{k} = \hat{j}\cdot\hat{i} = \hat{k}\cdot\hat{j} = 0\]
every “mixed” term dies and only the three “matching” terms survive:
\[\therefore \; \vec{P}\cdot\vec{Q} = P_xQ_x + 0 + 0\]
\[+\; 0 + P_yQ_y + 0\]
\[+\; 0 + 0 + P_zQ_z\]
\[\therefore \; \vec{P}\cdot\vec{Q} = P_xQ_x + P_yQ_y + P_zQ_z\]
This component form is the one you will actually use in calculations, because it needs no angle at all — only the components. Conversely, comparing it with Eq. (2.17) gives a standard method for finding the angle between two vectors, which you will meet shortly in a solved example.
(7) If \(\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}\), where \(\vec{a} \neq 0\), it is not necessary that \(\vec{b} = \vec{c}\). Using the distributive law, we can write
\[\vec{a}\cdot\left(\vec{b} - \vec{c}\right) = 0\]
It implies that either \(\vec{b} - \vec{c} = 0\), or \(\vec{a}\) is perpendicular to \(\vec{b} - \vec{c}\). It does not necessarily imply that \(\vec{b} - \vec{c} = 0\).
This is a warning against a habit carried over from ordinary algebra. With plain numbers you may cancel a common non-zero factor from both sides. With dot products you may not, because a vanishing dot product can also mean perpendicularity rather than equality.
2.5.2 Vector Product (Cross Product)
The vector product or cross product of two vectors (\(\vec{P}\) and \(\vec{Q}\)) is a vector whose magnitude is equal to the product of the magnitudes of the two vectors and the sine of the smaller angle (θ) between the two vectors. The direction of the product vector is given by \(\hat{u}_r\), which is a unit vector perpendicular to the plane containing the two vectors, and is given by the right hand screw rule. This is shown in Fig. 2.20 (a) and (b).
\[\text{a)} \quad \vec{R} = \vec{P} \times \vec{Q} = PQ\sin\theta \; \hat{u}_r \qquad \text{--- (2.18)}\]
\[\text{b)} \quad \vec{S} = \vec{Q} \times \vec{P} = PQ\sin\theta \; \hat{u}_s \qquad \text{--- (2.19)}\]
Diagram
Fig. 2.20 (a) – Vectors \(\vec{P}\) and \(\vec{Q}\) drawn from a common point O with angle θ between them, and the vector product \(\vec{R} = \vec{P}\times\vec{Q}\) drawn upward along \(\hat{u}_r\), perpendicular to their plane, with a curved arrow showing the rotation from \(\vec{P}\) to \(\vec{Q}\).
Diagram
Fig. 2.20 (b) – The same two vectors with the rotation taken from \(\vec{Q}\) to \(\vec{P}\), so that \(\vec{S} = \vec{Q}\times\vec{P}\) points downward along \(\hat{u}_s\), opposite to \(\vec{R}\).
Notice the essential difference from the dot product. There, the answer was a single number. Here the answer is a vector, and it does not lie in the plane of \(\vec{P}\) and \(\vec{Q}\) at all — it sticks straight out of that plane.
The Right Hand Screw Rule
According to the right hand screw rule, if the screw is rotated in a direction from \(\vec{P}\) to \(\vec{Q}\) through the smaller angle, then the direction in which the tip of the screw advances is the direction of \(\vec{R}\), perpendicular to the plane containing \(\vec{P}\) and \(\vec{Q}\).
Compare Fig. 2.20 (a) with Fig. 2.20 (b). In the first, the screw is turned from \(\vec{P}\) towards \(\vec{Q}\) and advances upward. In the second, it is turned from \(\vec{Q}\) towards \(\vec{P}\) and advances downward. The order of multiplication decides which way the product points.
One example of a vector or cross product is the force \(\vec{F}\) experienced by a charge \(q\) moving with velocity \(\vec{v}\) through a uniform magnetic field of magnetic induction \(\vec{B}\). It is an empirical law (experimentally determined) given by
\[\vec{F} = q\,\vec{v} \times \vec{B}\]
Characteristics of Vector Product
(1) Vector product does not obey the commutative law of multiplication.
\[\vec{P} \times \vec{Q} \neq \vec{Q} \times \vec{P} \qquad \text{--- (2.20)}\]
However, \(|\vec{P} \times \vec{Q}| = |\vec{Q} \times \vec{P}|\), i.e., the magnitudes are the same but the directions are opposite to each other. In symbols, \(\vec{Q} \times \vec{P} = -\,(\vec{P} \times \vec{Q})\).
This is a genuine break from ordinary algebra, and it must be respected in every calculation. Writing \(\vec{r} \times \vec{F}\) where the problem demands \(\vec{F} \times \vec{r}\) will reverse the direction of the torque and change the physics of the answer.
(2) The vector product obeys the distributive law of multiplication.
\[\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C} \qquad \text{--- (2.21)}\]
Here the order within each product has been preserved, and that is what makes the law safe to use despite characteristic (1).
(3) Special cases of cross product:
\[|\vec{P} \times \vec{Q}| = P\,Q\sin\theta \qquad \text{--- (2.22)}\]
(i) If θ = 0, i.e., if the two nonzero vectors are parallel to each other, their vector product is a zero vector.
\[|\vec{P} \times \vec{Q}| = P\,Q \cdot 0 = 0\]
(ii) If θ = 180°, i.e., if the two nonzero vectors are anti-parallel, their vector product is a zero vector.
\[|\vec{P} \times \vec{Q}| = P\,Q\sin 180^\circ = P\,Q\sin\pi = 0\]
(iii) If θ = 90°, i.e., if the two nonzero vectors are perpendicular to each other, the magnitude of their vector product is equal to the product of the magnitudes of the two vectors.
\[|\vec{P} \times \vec{Q}| = P\,Q\sin 90^\circ = P\,Q\]
Thus
\[\hat{i} \times \hat{j} = \hat{k}, \quad \hat{j} \times \hat{k} = \hat{i} \quad \text{and} \quad \hat{k} \times \hat{i} = \hat{j}\]
These three results follow the cyclic order \(x \to y \to z \to x\). Taken in that order the answer is positive; taken against it, the sign reverses.
(4) If \(\vec{P} = \vec{Q}\), then
\[|\vec{P} \times \vec{Q}| = |\vec{P} \times \vec{P}| = |\vec{Q} \times \vec{Q}| = 0\]
Thus
\[\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0\]
(5) Vector product of vectors expressed in terms of rectangular components:
Let
\[\vec{P} = P_x\hat{i} + P_y\hat{j} + P_z\hat{k}\]
and
\[\vec{Q} = Q_x\hat{i} + Q_y\hat{j} + Q_z\hat{k}\]
\[\vec{P} \times \vec{Q} = \left(P_x\hat{i} + P_y\hat{j} + P_z\hat{k}\right) \times \left(Q_x\hat{i} + Q_y\hat{j} + Q_z\hat{k}\right)\]
Expanding using the distributive law,
\[= P_xQ_x\left(\hat{i}\times\hat{i}\right) + P_xQ_y\left(\hat{i}\times\hat{j}\right) + P_xQ_z\left(\hat{i}\times\hat{k}\right)\]
\[+ P_yQ_x\left(\hat{j}\times\hat{i}\right) + P_yQ_y\left(\hat{j}\times\hat{j}\right) + P_yQ_z\left(\hat{j}\times\hat{k}\right)\]
\[+ P_zQ_x\left(\hat{k}\times\hat{i}\right) + P_zQ_y\left(\hat{k}\times\hat{j}\right) + P_zQ_z\left(\hat{k}\times\hat{k}\right)\]
Now
\[\hat{i}\times\hat{i} = \hat{j}\times\hat{j} = \hat{k}\times\hat{k} = 0, \; \text{and}\]
\[\hat{i}\times\hat{k} = -\hat{j}, \quad \hat{j}\times\hat{i} = -\hat{k}, \quad \hat{k}\times\hat{j} = -\hat{i}\]
\[\hat{i}\times\hat{j} = \hat{k}, \quad \hat{j}\times\hat{k} = \hat{i}, \quad \hat{k}\times\hat{i} = \hat{j}\]
Substituting these, the three “self” terms vanish and the six remaining terms combine in pairs:
\[\therefore \; \vec{P} \times \vec{Q} = 0 + P_xQ_y\hat{k} - P_xQ_z\hat{j}\]
\[-\; P_yQ_x\hat{k} + 0 + P_yQ_z\hat{i}\]
\[+\; P_zQ_x\hat{j} - P_zQ_y\hat{i} + 0\]
\[= (P_yQ_z - P_zQ_y)\,\hat{i}\]
\[+ (P_zQ_x - P_xQ_z)\,\hat{j}\]
\[+ (P_xQ_y - P_yQ_x)\,\hat{k}\]
This can be written in a determinant form as
\[\vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ P_x & P_y & P_z \\ Q_x & Q_y & Q_z \end{vmatrix} \qquad \text{--- (2.23)}\]
The determinant is only a memory device — it packs the nine products above into one compact box. Note the order of the rows: the first vector of the product occupies the middle row. Interchanging the two rows changes the sign of the determinant, which is exactly the non-commutativity of characteristic (1) showing itself again.
(6) The magnitude of the cross product of two vectors is numerically equal to the area of a parallelogram whose adjacent sides represent the two vectors.
As shown in Fig. 2.21, \(\vec{P} = \vec{OA}\), \(\vec{Q} = \vec{OB}\), and \(\vec{P}\) and \(\vec{Q}\) are inclined at an angle θ.
Diagram
Fig. 2.21 – Parallelogram OACB with \(\vec{P} = \vec{OA}\) as base and \(\vec{Q} = \vec{OB}\) as the adjacent side inclined at θ, and the perpendicular BD of height \(h\) dropped from B onto OA.
The perpendicular BD, of length \(h\) drawn on OA, gives the height of the parallelogram with OA as base.
Area of parallelogram
\[= \text{base} \times \text{height}\]
\[= \text{OA} \times \text{BD}, \quad \text{as } \sin\theta = \frac{\text{BD}}{\text{OB}}\]
\[= P\,Q\sin\theta\]
\[= \left|\vec{P} \times \vec{Q}\right|\]
\[= \text{magnitude of the vector product} \qquad \text{--- (2.24)}\]
So the cross product has a neat geometrical meaning: its magnitude measures the area swept out between the two vectors, and this area is largest when they are perpendicular and zero when they lie along the same line.
2.6 Introduction to Calculus
Vectors gave us a way of handling direction. Calculus gives us a way of handling change.
Calculus is the study of continuous (not discrete) changes in mathematical quantities. The word “continuous” is the key. A quantity that jumps from one value to the next in steps can be handled by ordinary algebra; a quantity that slides smoothly from one value to another, without ever skipping a value in between, cannot.
This branch of mathematics was first developed by G. W. Leibnitz and Sir Isaac Newton in the 17th century, and is extensively used in several branches of science.
You will study calculus in mathematics in the XIIth standard. Here we will learn the basics of the two branches of calculus, namely differential calculus and integral calculus. These are necessary to understand the topics covered in this book.
In short, differential calculus answers the question “how fast is this quantity changing?”, while integral calculus answers the reverse question “given how fast it is changing, how much has accumulated?” The two subsections that follow take up these questions one at a time.
2.6.1 Differential Calculus
Let us consider a function \(y = f(x)\). Here \(x\) is called an independent variable, and \(f(x)\) gives the value of \(y\) for different values of \(x\); \(y\) is the dependent variable. For example, \(x\) could be the position of a particle moving along the x-axis, and \(y = f(x)\) could be its velocity at that position.
We can thus draw a graph of \(y\) against \(x\), as shown in Fig. 2.22 (a).
Diagram
Fig. 2.22 (a) – A curve \(y = f(x)\) with two points A at \(x = x_0\) and B at \(x = x_0 + \Delta x\) marked on it, the increments \(\Delta x\) and \(\Delta y\) shown as the sides of a small right triangle, and the chord AB drawn with the angle θ it makes with the x-axis.
From a Chord to a Tangent
Let A and B be two points on the curve giving values of \(y\) at \(x = x_0\) and \(x = x_0 + \Delta x\), where \(\Delta x\) is a small increment in \(x\). As you move from A to B, the value of \(y\) changes by a corresponding amount \(\Delta y\).
The slope of the straight line joining A and B is given by
\[\tan\theta = \frac{\Delta y}{\Delta x}\]
This ratio is the average rate of change of \(y\) over the stretch from A to B. It is an honest number, but it is a blunt one: it tells you nothing about what is happening at the single point A, only about the average behaviour over a whole interval.
If we make \(\Delta x\) smaller, the point B will come closer to A. And if we keep making \(\Delta x\) smaller and smaller, we will ultimately reach a stage when B will coincide with A. This process is called taking the limit of \(\Delta x\) going to zero, and is written as \(\displaystyle\lim_{\Delta x \to 0}\).
In this limit, the line AB extended on both sides to P and Q will become the tangent to the curve at A, i.e., at \(x = x_0\), as shown in Fig. 2.22 (b).
Diagram
Fig. 2.22 (b) – The same curve \(y = f(x)\) with the chord AB replaced by the tangent line PQ touching the curve at the single point A at \(x = x_0\).
In this limit both \(\Delta x\) and \(\Delta y\) will go to zero. However, when two quantities tend to zero, their ratio need not go to zero. In fact
\[\lim_{\Delta x \to 0}\left(\frac{\Delta y}{\Delta x}\right)\]
becomes the slope of the tangent shown by PQ in Fig. 2.22 (b). This is written as \(dy/dx\) at \(x = x_0\).
The Derivative
Thus,
\[\left.\frac{dy}{dx}\right|_{x_0} = \lim_{\Delta x \to 0} \frac{(y + \Delta y) - y}{\Delta x}\]
\[\left.\frac{df(x)}{dx}\right|_{x_0} = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}\]
We can drop the subscript zero and write a general formula which will be valid for all values of \(x\) as
\[\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{df(x)}{dx} \qquad \text{--- (2.25)}\]
\(dy/dx\) is called the derivative of \(y\) with respect to \(x\) — which is the rate of change of \(y\) with respect to a change in \(x\) — and the process of finding the derivative is called differentiation.
In the XIIth standard you will learn about the properties of derivatives and how to find derivatives of different functions. Here we will just list the properties, as we will need them in later chapters.
Properties of Differentiation
Let \(f_1(x)\) and \(f_2(x)\) be two different functions of \(x\), and let \(s\) be a constant. Some of the properties of differentiation are:
1. \[\frac{d(s\,f(x))}{dx} = s\,\frac{df(x)}{dx} \qquad \text{--- (2.26)}\]
A constant multiplier simply rides along, unaffected by the differentiation.
2. \[\frac{d}{dx}\big(f_1(x) + f_2(x)\big) = \frac{df_1(x)}{dx} + \frac{df_2(x)}{dx} \qquad \text{--- (2.27)}\]
3. \[\frac{d}{dx}\big(f_1(x) \times f_2(x)\big) = f_1(x)\,\frac{df_2(x)}{dx} + f_2(x)\,\frac{df_1(x)}{dx} \qquad \text{--- (2.28)}\]
This is the product rule. Note that the derivative of a product is not the product of the derivatives — a mistake that costs marks regularly.
4. \[\frac{d}{dx}\left(\frac{f_1(x)}{f_2(x)}\right) = \frac{1}{f_2(x)}\,\frac{df_1(x)}{dx} - \frac{f_1(x)}{f_2^{\,2}(x)}\,\frac{df_2(x)}{dx} \qquad \text{--- (2.29)}\]
This is the quotient rule.
5. If \(x\) depends on another variable \(t\) (time), then
\[\frac{df(x)}{dt} = \frac{df(x)}{dx}\,\frac{dx}{dt} \qquad \text{--- (2.30)}\]
6.
\[\frac{d}{dx} f\big(g[x]\big) = f'\big(g(x)\big) \times g'(x)\]
\[\text{where } f'\big(g(x)\big) = \frac{df}{dg}\]
\[\text{or } \frac{dy}{dx} = \frac{dy}{dv}\,\frac{dv}{dx}\]
Properties 5 and 6 are two faces of the chain rule, and they are the workhorses of physics. They let you connect one rate of change to another — for instance, how fast a quantity changes with position, and how fast the position itself changes with time.
Derivatives of Some Simple Functions
The derivatives of some simple functions of \(x\) are given below.
1. \[\frac{d}{dx}(x^n) = n\,x^{n-1} \qquad \text{--- (2.31)}\]
2. \[\frac{d(e^x)}{dx} = e^x \quad \text{and} \quad \frac{d(e^{ax})}{dx} = a\,e^{ax} \qquad \text{--- (2.32)}\]
3. \[\frac{d}{dx}(\ln x) = \frac{1}{x} \qquad \text{--- (2.33)}\]
4. \[\frac{d}{dx}(\sin x) = \cos x \qquad \text{--- (2.34)}\]
5. \[\frac{d}{dx}(\cos x) = -\sin x \qquad \text{--- (2.35)}\]
6. \[\frac{d}{dx}(\tan x) = \sec^2 x \qquad \text{--- (2.36)}\]
7. \[\frac{d}{dx}(\cot x) = -\,\mathrm{cosec}^2 x \qquad \text{--- (2.37)}\]
8. \[\frac{d}{dx}(\sec x) = \tan x \,\sec x \qquad \text{--- (2.38)}\]
9. \[\frac{d}{dx}(\mathrm{cosec}\, x) = -\,\mathrm{cosec}\, x \,\cot x \qquad \text{--- (2.39)}\]
Two of these carry a minus sign — the derivatives of \(\cos x\) and \(\cot x\), and of \(\mathrm{cosec}\,x\) — and dropping that sign is one of the most frequent slips in examination scripts. Whenever a trigonometric function decreases as \(x\) increases, its derivative must come out negative.
2.6.2 Integral Calculus
Integral calculus is the branch of mathematics dealing with the properties of integrals and their applications.
The physical interpretation of the integral of a function \(f(x)\), i.e., of \(\displaystyle\int f(x)\,dx\), is the area under the curve of \(f(x)\) versus \(x\). It is the reverse process of differentiation, as we will see below.
Keep both meanings of the word “integral” in view as you read on. Geometrically it is an area. Operationally it is an undoing of the derivative. These two descriptions look unrelated, and it is one of the pleasant surprises of mathematics that they turn out to be the same thing.
Area Under a Straight Line
We know how to find the area of a rectangle, a triangle, and so on. In Fig. 2.23 (a) we have shown \(y\), which is a function of \(x\), with A and B being two points on it.
Diagram
Fig. 2.23 (a) – A straight line graph of \(y\) against \(x\), with the region under it between \(x = a\) and \(x = b\) shaded and split into a rectangle ADEC and a triangle ABC.
The area under the curve (here a straight line) from \(x = a\) to \(x = b\) is shown by the shaded area. This can be obtained as the sum of the area of the rectangle ADEC \(= f(a)\,(b-a)\) and the area of the triangle ABC \(= \tfrac{1}{2}\,(b-a)\,(f(b) - f(a))\).
No new mathematics was needed here, because the boundary was a straight line and the region split neatly into shapes we already know.
Area Under a Curve
Figure 2.23 (b) shows another function of \(x\). We do not have a simple formula to calculate the area under this curve, because its upper boundary bends.
Diagram
Fig. 2.23 (b) – A curved graph \(y = f(x)\) with the region under it between \(x_0 = a\) and \(x_n = b\) divided into a large number of narrow vertical strips, each treated as a thin rectangle.
For this calculation, we use a simple trick. We divide the area into a large number of vertical strips, as shown in the figure. We assume the thickness (width) of each strip to be so small that it can be assumed to be a rectangle, as shown in the figure, and we add the areas of these rectangles.
Thus the area under the curve is given by
\[\text{Area under the curve} = \sum_{i=1}^{n} \Delta A_i = \sum_{i=1}^{n} (x_i - x_{i-1})\, f(x_i)\]
where \(n\) is the number of strips and \(\Delta A_i\) is the area of the \(i^{\text{th}}\) strip. In each term, \((x_i - x_{i-1})\) is the width of the strip and \(f(x_i)\) is its height.
As the strips are not really rectangles — each one has a slightly slanting top — the area calculated above is not exactly equal to the area under the curve. However, as we increase \(n\), the sum of the areas of the rectangles gets closer to the actual area under the curve, and becomes equal to it in the limit \(n \to \infty\). Thus we can write,
\[\text{Area under the curve} = \lim_{n \to \infty} \sum_{i=1}^{n} (x_i - x_{i-1})\, f(x_i) \qquad \text{--- (2.40)}\]
Integration helps us in getting the exact area if the change is really continuous, i.e., if \(n\) is really infinite. It is represented as
\[\int_{x=a}^{x=b} f(x)\,dx\]
and is called the definite integral of \(f(x)\) from \(x = a\) to \(x = b\). The numbers \(a\) and \(b\) are called the limits of integration.
Thus,
\[\int_{x=a}^{x=b} f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} (x_i - x_{i-1})\, f(x_i) \qquad \text{--- (2.41)}\]
Notice how the notation itself remembers where it came from: the stretched-out S of \(\int\) stands for the sum, and \(dx\) is what the strip-width \((x_i - x_{i-1})\) becomes when it shrinks towards zero.
Integration as the Reverse of Differentiation
The process of obtaining the integral is called integration. We can also write
\[F(x) = \int f(x)\,dx \qquad \text{--- (2.42)}\]
\(F(x)\) is called the indefinite (without any limits on \(x\)) integral of \(f(x)\).
Differentiation is the reverse process to that of integration. Therefore,
\[f(x) = \frac{d}{dx}\big(F(x)\big) \qquad \text{--- (2.43)}\]
\[\therefore \; \left. F(x)\right|_a^b = F(b) - F(a) = \int_a^b f(x)\,dx \qquad \text{--- (2.44)}\]
Equation (2.44) is the practical rule you will use again and again: to evaluate a definite integral, first find the indefinite integral \(F(x)\), then substitute the upper limit, then the lower limit, and subtract.
Properties of Integration
1. \[\int \big(f_1(x) + f_2(x)\big)\,dx = \int f_1(x)\,dx + \int f_2(x)\,dx \qquad \text{--- (2.45)}\]
2. \[\int K\, f(x)\,dx = K \int f(x)\,dx \quad \text{for } K = \text{constant} \qquad \text{--- (2.46)}\]
These two properties mirror Eqs. (2.26) and (2.27) of differentiation, as you would expect of an inverse operation.
Indefinite Integrals of Some Basic Functions
Indefinite integrals of some basic functions are given below. Their definite integrals can be obtained by using Eq. (2.44).
1. \[\int x^n\,dx = \frac{x^{n+1}}{n+1} \qquad \text{--- (2.47)}\]
2. \[\int \frac{1}{x}\,dx = \ln x \qquad \text{--- (2.48)}\]
3. \[\int \sin x\,dx = -\cos x \qquad \text{--- (2.49)}\]
4. \[\int \cos x\,dx = \sin x \qquad \text{--- (2.50)}\]
5. \[\int e^x\,dx = e^x \qquad \text{--- (2.51)}\]
Read each of these backwards and you will recover a derivative from the previous subsection. For instance, differentiating \(-\cos x\) gives \(+\sin x\), which is precisely why Eq. (2.49) carries its minus sign. Checking an integral by differentiating the answer is the quickest way to catch a sign error.
2.7 Summary
Scalars and Vectors
Scalars are physical quantities completely described by magnitude alone — a number with a unit. Examples: length, mass, time, temperature, density. They are added and subtracted by the rules of simple algebra.
Vectors are physical quantities that need magnitude as well as direction for a complete description. Examples: displacement, velocity, force. A vector is drawn as an arrow whose length (to scale) gives the magnitude, with tail and head fixing the direction. It is written as \(\vec{PQ}\) or \(\vec{A}\), and its magnitude as \(|\vec{A}|\).
Types of Vectors
- Zero (null) vector \(\vec{0}\): magnitude zero, direction arbitrary. The velocity of a stationary particle, and the acceleration of a body moving with uniform velocity, are zero vectors.
- Resultant vector: the single vector producing the same effect as all the given vectors together.
- Negative vector: same magnitude, opposite direction.
- Equal vectors: same magnitude and same direction (their location does not matter).
- Position vector: drawn from the origin of a chosen co-ordinate system to the point where the particle lies.
- Unit vector: a vector of unit magnitude in a given direction.
\[\vec{M} = \hat{u}_M\,M \qquad \text{--- (2.1)} \qquad\qquad \hat{u}_M = \frac{\vec{M}}{M} \qquad \text{--- (2.2)}\]
- \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) are the mutually perpendicular unit vectors along the x, y and z axes:
\[\hat{i} = \frac{\vec{x}}{x}, \quad \hat{j} = \frac{\vec{y}}{y}, \quad \hat{k} = \frac{\vec{z}}{z} \qquad \text{--- (2.3)}\]
Vector Operations
- Multiplying a vector by a scalar \(s\) gives a vector along the same direction with \(s\) times the magnitude:
\[\vec{Q} = s\vec{P} \qquad \text{--- (2.4)}\]
- Only vectors of the same type (same physical quantity) may be added or subtracted. A force cannot be added to a velocity.
- For parallel vectors: \(\left|\vec{AC}\right| = \left|\vec{AB}\right| + \left|\vec{BC}\right|\), along the same direction.
- For anti-parallel vectors: \(\left|\vec{AC}\right| = \left|\,\left|\vec{AB}\right| - \left|\vec{BC}\right|\,\right|\), along the direction of the larger vector.
Triangle Law
- If two vectors of the same type are represented in magnitude and direction by two sides of a triangle taken in order, their resultant is the third side, drawn in the opposite sense — from the tail of the first to the head of the second.
- Vector addition is commutative:
\[\vec{P} + \vec{Q} = \vec{Q} + \vec{P} \qquad \text{--- (2.5)}\]
- Vector addition is associative:
\[(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C}) \qquad \text{--- (2.6)}\]
Parallelogram Law
- If two vectors drawn from a common point are represented by two adjacent sides of a parallelogram, the diagonal from that same point is their resultant. For \(\vec{P}\) and \(\vec{Q}\) inclined at angle θ:
\[R = \sqrt{P^2 + Q^2 + 2\,P\,Q\cos\theta} \qquad \text{--- (2.8)}\]
\[\alpha = \tan^{-1}\left(\frac{Q\sin\theta}{P + Q\cos\theta}\right) \qquad \text{--- (2.10)}\]
where α is the angle made by \(\vec{R}\) with \(\vec{P}\). If β is the angle between \(\vec{R}\) and \(\vec{Q}\),
\[\beta = \tan^{-1}\left(\frac{P\sin\theta}{Q + P\cos\theta}\right)\]
- The resultant is maximum (\(P + Q\)) at θ = 0° and minimum (\(|P - Q|\)) at θ = 180°.
Resolution of Vectors
- A vector may be split into components along chosen directions:
\[\vec{V} = V_1\hat{\alpha} + V_2\hat{\beta} + V_3\hat{\gamma} \qquad \text{--- (2.11)}\]
- Components along mutually perpendicular directions are called rectangular components. In two dimensions, with θ measured from the x-axis:
\[R_x = R\cos\theta \qquad \text{--- (2.12)} \qquad\qquad R_y = R\sin\theta \qquad \text{--- (2.13)}\]
\[R = \sqrt{R_x^{\,2} + R_y^{\,2}} \qquad \text{--- (2.14)} \qquad\qquad \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \qquad \text{--- (2.15)}\]
- In three dimensions:
\[\vec{R} = R_x\hat{i} + R_y\hat{j} + R_z\hat{k}, \qquad \left|\vec{R}\right| = \sqrt{R_x^{\,2} + R_y^{\,2} + R_z^{\,2}} \qquad \text{--- (2.16)}\]
- Two vectors are equal if and only if all their corresponding components are equal: \(A_x = B_x\), \(A_y = B_y\), \(A_z = B_z\). Equal magnitude alone does not make two vectors equal.
Scalar (Dot) Product
\[\vec{P}\cdot\vec{Q} = PQ\cos\theta \qquad \text{--- (2.17)}\]
- The result is a scalar. It equals the magnitude of one vector times the component of the other along it.
- Commutative: \(\vec{P}\cdot\vec{Q} = \vec{Q}\cdot\vec{P}\). Distributive: \(\vec{P}\cdot(\vec{Q} + \vec{R}) = \vec{P}\cdot\vec{Q} + \vec{P}\cdot\vec{R}\).
- Special cases: at θ = 0°, \(\vec{P}\cdot\vec{Q} = PQ\); at θ = 180°, \(\vec{P}\cdot\vec{Q} = -PQ\); at θ = 90°, \(\vec{P}\cdot\vec{Q} = 0\).
\[\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1, \qquad \hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0\]
- If \(\vec{P} = \vec{Q}\), then \(\vec{P}\cdot\vec{Q} = P^2 = Q^2\).
- In component form:
\[\vec{P}\cdot\vec{Q} = P_xQ_x + P_yQ_y + P_zQ_z\]
- If \(\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}\) with \(\vec{a} \neq 0\), it does not follow that \(\vec{b} = \vec{c}\); \(\vec{a}\) may instead be perpendicular to \(\vec{b} - \vec{c}\).
- Physical uses: \(W = \vec{F}\cdot\vec{S}\) and \(P = \vec{F}\cdot\vec{v}\).
Vector (Cross) Product
\[\vec{R} = \vec{P} \times \vec{Q} = PQ\sin\theta\,\hat{u}_r \qquad \text{--- (2.18)} \qquad\qquad \vec{S} = \vec{Q} \times \vec{P} = PQ\sin\theta\,\hat{u}_s \qquad \text{--- (2.19)}\]
- The result is a vector, perpendicular to the plane containing \(\vec{P}\) and \(\vec{Q}\), its sense given by the right hand screw rule, θ being the smaller angle between them.
- Not commutative: \(\vec{P} \times \vec{Q} \neq \vec{Q} \times \vec{P}\) — (2.20), though \(|\vec{P} \times \vec{Q}| = |\vec{Q} \times \vec{P}|\); the directions are opposite.
- Distributive: \(\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}\) — (2.21)
- Magnitude: \(|\vec{P} \times \vec{Q}| = PQ\sin\theta\) — (2.22). It is a zero vector at θ = 0° and θ = 180°, and equals \(PQ\) at θ = 90°.
\[\hat{i} \times \hat{j} = \hat{k}, \quad \hat{j} \times \hat{k} = \hat{i}, \quad \hat{k} \times \hat{i} = \hat{j}, \qquad \hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0\]
- Determinant form:
\[\vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ P_x & P_y & P_z \\ Q_x & Q_y & Q_z \end{vmatrix} \qquad \text{--- (2.23)}\]
\[= (P_yQ_z - P_zQ_y)\hat{i} + (P_zQ_x - P_xQ_z)\hat{j} + (P_xQ_y - P_yQ_x)\hat{k}\]
- The magnitude of the cross product equals the area of the parallelogram whose adjacent sides are the two vectors — (2.24).
- Physical uses: \(\vec{F} = q\,\vec{v} \times \vec{B}\), \(\vec{v} = \vec{\omega} \times \vec{r}\), and \(\vec{\tau} = \vec{r} \times \vec{F}\).
Differential Calculus
- The derivative is the limiting value of the ratio of small changes; geometrically, the slope of the tangent to the \(y\)–\(x\) curve:
\[\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{df(x)}{dx} \qquad \text{--- (2.25)}\]
- Properties of differentiation:
\[\frac{d(s\,f(x))}{dx} = s\,\frac{df(x)}{dx} \qquad \text{--- (2.26)}\]
\[\frac{d}{dx}\big(f_1(x) + f_2(x)\big) = \frac{df_1(x)}{dx} + \frac{df_2(x)}{dx} \qquad \text{--- (2.27)}\]
\[\frac{d}{dx}\big(f_1(x) \times f_2(x)\big) = f_1(x)\frac{df_2(x)}{dx} + f_2(x)\frac{df_1(x)}{dx} \qquad \text{--- (2.28)}\]
\[\frac{d}{dx}\left(\frac{f_1(x)}{f_2(x)}\right) = \frac{1}{f_2(x)}\frac{df_1(x)}{dx} - \frac{f_1(x)}{f_2^{\,2}(x)}\frac{df_2(x)}{dx} \qquad \text{--- (2.29)}\]
\[\frac{df(x)}{dt} = \frac{df(x)}{dx}\,\frac{dx}{dt} \qquad \text{--- (2.30)} \qquad\qquad \frac{d}{dx}f\big(g[x]\big) = f'\big(g(x)\big) \times g'(x)\]
- Standard derivatives:
\[\frac{d}{dx}(x^n) = n\,x^{n-1} \; \text{--- (2.31)}; \quad \frac{d(e^x)}{dx} = e^x, \; \frac{d(e^{ax})}{dx} = a\,e^{ax} \; \text{--- (2.32)}; \quad \frac{d}{dx}(\ln x) = \frac{1}{x} \; \text{--- (2.33)}\]
\[\frac{d}{dx}(\sin x) = \cos x \; \text{--- (2.34)}; \quad \frac{d}{dx}(\cos x) = -\sin x \; \text{--- (2.35)}; \quad \frac{d}{dx}(\tan x) = \sec^2 x \; \text{--- (2.36)}\]
\[\frac{d}{dx}(\cot x) = -\,\mathrm{cosec}^2 x \; \text{--- (2.37)}; \quad \frac{d}{dx}(\sec x) = \tan x \sec x \; \text{--- (2.38)}; \quad \frac{d}{dx}(\mathrm{cosec}\,x) = -\,\mathrm{cosec}\,x \cot x \; \text{--- (2.39)}\]
Integral Calculus
- The integral of \(f(x)\) is the area under the curve of \(f(x)\) versus \(x\), obtained as the limit of a sum of strip areas:
\[\text{Area} = \lim_{n \to \infty} \sum_{i=1}^{n} (x_i - x_{i-1})\,f(x_i) \qquad \text{--- (2.40)}\]
\[\int_{x=a}^{x=b} f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} (x_i - x_{i-1})\,f(x_i) \qquad \text{--- (2.41)}\]
- Indefinite integral: \(F(x) = \displaystyle\int f(x)\,dx\) — (2.42), and integration is the reverse of differentiation:
\[f(x) = \frac{d}{dx}\big(F(x)\big) \qquad \text{--- (2.43)}\]
\[\left.F(x)\right|_a^b = F(b) - F(a) = \int_a^b f(x)\,dx \qquad \text{--- (2.44)}\]
- Properties of integration:
\[\int \big(f_1(x) + f_2(x)\big)dx = \int f_1(x)\,dx + \int f_2(x)\,dx \; \text{--- (2.45)}; \qquad \int K f(x)\,dx = K\int f(x)\,dx \; \text{--- (2.46)}\]
- Standard integrals:
\[\int x^n dx = \frac{x^{n+1}}{n+1} \; \text{--- (2.47)}; \quad \int \frac{1}{x}dx = \ln x \; \text{--- (2.48)}; \quad \int \sin x\,dx = -\cos x \; \text{--- (2.49)}\]
\[\int \cos x\,dx = \sin x \; \text{--- (2.50)}; \quad \int e^x dx = e^x \; \text{--- (2.51)}\]
2.8 Check Your Concepts
A student says, “A quantity that can be positive or negative must be a vector, because the sign shows its direction.” Using temperature and displacement as examples, explain why this statement is wrong.
The velocity of a stationary particle and the acceleration of a body moving with uniform velocity are both zero vectors. Explain, in each case, why the vector vanishes.
Two vectors have equal magnitudes but are not equal vectors. Is this possible? Justify your answer with a suitable example written in component form.
Why can a force vector never be added to a velocity vector, even though both are vectors and both can be represented by arrows?
A boat is always steered due north across a river flowing due east, yet it never lands at the point directly opposite. Explain the physics behind this, and state what the boatman must change if he wants to land exactly opposite.
Why is the resultant of two vectors maximum when the angle between them is 0° and minimum when it is 180°? Answer using the term \(2PQ\cos\theta\) in Eq. (2.8).
While resolving a vector, we almost always choose two mutually perpendicular directions rather than any two arbitrary directions. What advantage does this choice give?
If \(\vec{P}\cdot\vec{Q} = 0\) for two non-zero vectors, what can you conclude about them? If instead \(|\vec{P}\times\vec{Q}| = 0\) for two non-zero vectors, what can you conclude?
Work done involves \(\cos\theta\) while torque involves \(\sin\theta\). Explain why the two quantities demand different trigonometric factors, and why one is a scalar while the other is a vector.
In ordinary algebra, if \(ab = ac\) and \(a \neq 0\), we may cancel \(a\) and conclude \(b = c\). Show, with reasoning, why the same cancellation is not permitted when \(\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}\).
State the right hand screw rule. Using it, explain why \(\vec{P}\times\vec{Q}\) and \(\vec{Q}\times\vec{P}\) have equal magnitudes but opposite directions.
A door is hinged along one edge. Explain, using \(\vec{\tau} = \vec{r}\times\vec{F}\), why pushing the door near its hinge, or pushing along the plane of the door, hardly turns it at all.
When \(\Delta x \to 0\), both \(\Delta y\) and \(\Delta x\) tend to zero. Why then does the ratio \(\Delta y/\Delta x\) tend to a definite non-zero value in general? What is this limiting value called, and what does it represent geometrically?
The area under a curve is found by dividing it into a very large number of thin rectangular strips. Since the top of each strip is flat while the curve is not, explain how this method can still give an exact answer.
Why does an indefinite integral give a function while a definite integral gives a number? Relate your answer to Eq. (2.44).
A student integrates \(\sin x\) and writes \(\int \sin x\,dx = \cos x\). Without looking up any table, describe a quick check that would immediately reveal the error, and state the correct result.
2.9 Practice with Numericals
A man walks 4 km due east and then 3 km due north. Find the total distance travelled by him and the magnitude and direction of his displacement.
Find the magnitude of the vector \(\vec{A} = 6\hat{i} - 3\hat{j} + 2\hat{k}\), and obtain the unit vector along its direction. Verify that the unit vector so obtained has magnitude 1.
Two forces of magnitude 12 N and 5 N act at a point at right angles to each other. Determine the magnitude of their resultant and the angle it makes with the 12 N force.
Two forces of 8 N and 6 N act at a point with an angle of 60° between them. Using Eq. (2.8), find the magnitude of the resultant, and using Eq. (2.10), find the angle it makes with the 8 N force.
An aeroplane flies with a speed of 200 km/hr relative to still air, heading due north. A wind blows from the west at 40 km/hr. Determine the magnitude and direction of the resultant velocity of the aeroplane relative to the ground.
A force of 100 N acts at an angle of 60° to the horizontal. Resolve it into its horizontal and vertical rectangular components, and write the force in the form \(F_x\hat{i} + F_y\hat{j}\).
Given \(\vec{A} = 3\hat{i} + 4\hat{j} - 2\hat{k}\) and \(\vec{B} = \hat{i} - \hat{j} + 2\hat{k}\), determine (i) \(\vec{A} + \vec{B}\), (ii) \(\vec{A} - \vec{B}\), and (iii) the magnitude of \(\vec{A} + \vec{B}\).
Find the scalar product of \(\vec{P} = 2\hat{i} - \hat{j} + 3\hat{k}\) and \(\vec{Q} = \hat{i} + 4\hat{j} - 2\hat{k}\), and hence determine the angle between them.
A constant force \(\vec{F} = 5\hat{i} + 3\hat{j} - 2\hat{k}\) N displaces a body through \(\vec{s} = 2\hat{i} - \hat{j} + 4\hat{k}\) m. Calculate the work done by the force, with its SI unit.
Determine the value of \(\lambda\) for which the vectors \(\vec{A} = 2\hat{i} + \lambda\hat{j} + \hat{k}\) and \(\vec{B} = 4\hat{i} - 2\hat{j} + 2\hat{k}\) are perpendicular to each other.
Find \(\vec{A} \times \vec{B}\) for \(\vec{A} = 2\hat{i} + \hat{j} - \hat{k}\) and \(\vec{B} = \hat{i} - 2\hat{j} + 3\hat{k}\), using the determinant form of Eq. (2.23).
Two adjacent sides of a parallelogram are represented by \(\vec{P} = 4\hat{i} + \hat{j}\) and \(\vec{Q} = \hat{i} + 3\hat{j}\). Using Eq. (2.24), find the area of the parallelogram.
A force \(\vec{F} = 3\hat{i} + 2\hat{j}\) N acts at a point whose position vector is \(\vec{r} = \hat{i} + 4\hat{j}\) m with respect to the axis of rotation. Determine the torque \(\vec{\tau} = \vec{r} \times \vec{F}\) and state its direction.
For \(\vec{P} = 6\hat{i} - 2\hat{j} + 4\hat{k}\) and \(\vec{Q} = 3\hat{i} - \hat{j} + m\hat{k}\), find \(m\) if the two vectors have the same direction.
Obtain the derivatives of the following functions with respect to \(x\):
- \(x^5 + 3x^2 - 7\)
- \(x^2\cos x\)
- \(e^{3x}\)
- \(\dfrac{\sin x}{x}\)
The displacement of a particle moving along a straight line is \(x = 4t^3 - 3t^2 + 2t\), where \(x\) is in metre and \(t\) in second. Find its velocity and acceleration at \(t = 1\) s, with correct SI units.
Evaluate the following integrals:
- \(\displaystyle\int x^5\,dx\)
- \(\displaystyle\int_{1}^{3} (2x + 1)\,dx\)
- \(\displaystyle\int_{0}^{\pi/2} \cos x\,dx\)
The velocity of a particle varies with time as \(v = 6t + 2\) m s\(^{-1}\). Using integration, find the distance travelled by the particle between \(t = 0\) and \(t = 4\) s.
Show that \(\vec{a} = \dfrac{\hat{i} - \hat{j}}{\sqrt{2}}\) is a unit vector.
If \(\vec{v_1} = 3\hat{i} + 4\hat{j} + \hat{k}\) and \(\vec{v_2} = \hat{i} - \hat{j} - \hat{k}\), determine the magnitude of \(\vec{v_1} + \vec{v_2}\).
For \(\vec{v_1} = 2\hat{i} - 3\hat{j}\) and \(\vec{v_2} = -6\hat{i} + 5\hat{j}\), determine the magnitude and direction of \(\vec{v_1} + \vec{v_2}\).
Find a vector which is parallel to \(\vec{v} = \hat{i} - 2\hat{j}\) and has a magnitude 10.
Show that the vectors \(\vec{a} = 2\hat{i} + 5\hat{j} - 6\hat{k}\) and \(\vec{b} = \hat{i} + \dfrac{5}{2}\hat{j} - 3\hat{k}\) are parallel.
Determine \(\vec{a} \times \vec{b}\), given \(\vec{a} = 2\hat{i} + 3\hat{j}\) and \(\vec{b} = 3\hat{i} + 5\hat{j}\).
Show that the vectors \(\vec{a} = 2\hat{i} + 3\hat{j} + 6\hat{k}\), \(\vec{b} = 3\hat{i} - 6\hat{j} + 2\hat{k}\) and \(\vec{c} = 6\hat{i} + 2\hat{j} - 3\hat{k}\) are mutually perpendicular.
Determine the vector product of \(\vec{v_1} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\vec{v_2} = \hat{i} + 2\hat{j} - 3\hat{k}\).
Given \(\vec{v_1} = 5\hat{i} + 2\hat{j}\) and \(\vec{v_2} = a\hat{i} - 6\hat{j}\) are perpendicular to each other, determine the value of \(a\).
Obtain the derivatives of the following functions:
- \(x\sin x\)
- \(x^4 + \cos x\)
- \(x/\sin x\)
Using the rule for differentiation for the quotient of two functions, prove that \[\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \sec^2 x\]
Evaluate the following integrals:
- \(\displaystyle\int_{0}^{\pi/2} \sin x\,dx\)
- \(\displaystyle\int_{1}^{5} x\,dx\)
2.10 Objective Questions
- The resultant of two forces 10 N and 15 N acting along \(+x\) and \(-x\) axes respectively, is
- 25 N along \(+x\)-axis
- 25 N along \(-x\)-axis
- 5 N along \(+x\)-axis
- 5 N along \(-x\)-axis
- For two vectors to be equal, they should have the
- same magnitude
- same direction
- same magnitude and direction
- same magnitude but opposite direction
- The magnitude of the scalar product of two unit vectors perpendicular to each other is
- zero
- 1
- \(-1\)
- 2
- The magnitude of the vector product of two unit vectors making an angle of 60° with each other is
- 1
- 2
- 3/2
- \(\sqrt{3}/2\)
- If \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\) are three vectors, then which of the following is not correct?
- \(\vec{A}\cdot\left(\vec{B} + \vec{C}\right) = \vec{A}\cdot\vec{B} + \vec{A}\cdot\vec{C}\)
- \(\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}\)
- \(\vec{A}\times\vec{B} = \vec{B}\times\vec{A}\)
- \(\vec{A}\times\left(\vec{B} + \vec{C}\right) = \vec{A}\times\vec{B} + \vec{B}\times\vec{C}\)