This is the book itself, not a sample chapter written for a website. It runs from the opening to the derivative as a slope and as a rate — 39 pages. Nothing is asked of you to read it: no account, no card.
The wall falls where calculus turns into technique. What is behind it is listed at the bottom of this page.
WHY THIS CHAPTER EXISTS
No syllabus lists this chapter. Every other chapter assumes it.
Physics is written in the language of mathematics. If you are still translating that language while you read, you cannot follow the physics. That is the only reason this chapter is here. It is neither a mathematics chapter nor a physics chapter. It is the place where the two meet: every tool in it is here because a later chapter picks it up and uses it without stopping to explain.
It is not the JEE Mathematics syllabus and should not be read as one. A topic you know from your mathematics course and do not find here was left out because the physics in these thirty-six chapters does not need it, not because it is unimportant.
The standard is that you do these things without noticing that you have done them, not that you can do them given time and a formula sheet.
It is a toolkit, not a chapter to be read straight through. Work it in the six passes overleaf, with the physics chapters in between.
IN THIS CHAPTER
0 How do I choose the tool? · 1 Algebra that must be automatic · 2 Approximations, and the magnitude of the error · 3 Trigonometry · 4 Differentiation · 5 Integration · 6 Averages and root mean square · 7 Differential equations you will meet · 8 The graphs physics reuses · 9 Vectors · 10 Where you will use all this
BEFORE YOU START
School algebra and trigonometry, to the point where you are fluent. No calculus is assumed — it is built here, starting from the idea of a slope.
A few examples borrow a situation from a later chapter — a cell with internal resistance, a rolling wheel, a turning frame. There the physics is handed to you in the question, because what is being tested is the mathematics.
HOW TO USE THIS CHAPTER
Six passes, in this order. The chapter is built to be worked this way. The standard for leaving it is not “I have read it” — six testable conditions are printed on the last page, and they are worth reading now.
1 Tool map → 2 Core sections and their examples → 3 Beyond the Paper → 4 Diagnostic → 5 Levels 1, 2, 3 → 6 Revision, traps, Diagnostic again
Weeks apart, not hours: passes 2 to 6 are meant to have physics chapters standing between them.
By the end of this chapter you will hold about thirty tools. The paper does not tell you which one to use. It gives you a situation in words, and the real difficulty — in this chapter and in every chapter after it — is the step between reading those words and writing the first line.
That step has a structure. It can be trained, and it is trained by reading questions rather than by solving them. This section is the map. Read it now, once more after each section as the tools become real, and again before the paper.
COREA large share of the selections in this chapter are settled by three questions, asked in this order. They will not settle all of them — §0.3 is about what to do when they do not — but they get the first line written far more often than any amount of staring at the question does.
1. What is changing, and with respect to what? This picks the variable. “Speed changes with time” and “speed changes with position” are different sentences and send you to different tools — dv ⁄ dt in the first, v dv ⁄ dx in the second. Reading this wrongly is the most expensive slip in kinematics, because the working that follows is correct and still worthless.
2. Am I asked for a rate, for a total, or for a special value? This picks the operation. A rate is a derivative. A total accumulated over a stretch of time, length or angle is an integral. A special value — largest, smallest, the instant a body turns back — usually comes from a stationary condition, which most often means setting a derivative to zero. Then check the ends of the allowed range, because the largest value can sit at an end where no derivative vanishes (§4.4).
3. Does the quantity carry a direction? This picks the algebra. If the quantity carries a direction, decide next which of four things the question wants: the two added or subtracted, one of them resolved into components, a scalar product when the answer is a plain number that depends on the angle between them, or a vector product when the answer itself has a direction. A quantity without a direction is handled by ordinary arithmetic.
The order matters. Question 1 fixes what you are differentiating or integrating with respect to; until that is fixed, question 2 has no answer. Students who start at question 2 write ∫ a dt for an acceleration that was given as a function of x, and lose the question in the first line.
Take any question you have already solved. Read the stem alone, answer the three questions in under ten seconds, and only then look at what you wrote last time. If you cannot answer them in ten seconds, the trouble was the selection and not the algebra.
What follows is the same three questions, written out as the words a paper actually uses. The left column is what you read. The right column is what should come to mind, with the section that builds it. Nothing here is a rule you obey without thinking — it is a first guess, made in two seconds, which the three questions then confirm or overturn.
| IF THE QUESTION SAYS … | REACH FOR |
|---|---|
| A quadratic comes out of the physics | Find both roots, then ask of each whether the physics allows it — often both — §1.1 |
| Successive terms differ by a constant — displacements in equal successive intervals, readings taken at fixed steps | Arithmetic progression — §1.2 |
| A sum of many terms with a constant ratio between them | Geometric sum — §1.2 |
| The unknown sits in the exponent | Take logarithms of both sides — §1.3 |
| “Least possible” or “greatest possible”, with no function in front of you to differentiate | AM ≥ GM: x + c ⁄ x ≥ 2√c — §1.4 |
| “Is this possible at all?”, or a quadratic that comes out of the physics | The discriminant read as a condition — §1.4 |
| “Distance” and not displacement; an answer that comes out negative or impossible | Modulus, and the physical domain — §1.5 |
| A small percentage change; “by what per cent does … change” | The index rule Δy ⁄ y = n (Δx ⁄ x) — §2.1 |
| One quantity is much smaller than another, and the answer may be approximate | Binomial to first order — §2.1 |
| An angle is described as small, or a distance as far away | sin θ ≈ tan θ ≈ θ, cos θ ≈ 1 − θ² ⁄ 2 — §2.2 |
| “Component along”, “at an angle to”, a force on a slope | Resolution — §3.1 |
| A triangle of three quantities with no right angle in it | Sine and cosine rules — §3.3 |
| “Rate of change”, “how fast is … changing”, “per second” | Differentiate — §4.1 |
| Acceleration is given as a function of position | a = v dv ⁄ dx — §4.3 |
| Acceleration is given as a function of time | Integrate once for v, again for x — §5.1 |
| “Maximum”, “minimum”, “least”, “greatest”, “just enough” | Find the stationary points, then check the ends of the allowed range — §4.4 |
| “Total”, “net”, “over the whole length”, “in the first t seconds” | Integrate — §5.1, §5.3 |
| The body is not a point — a rod, a ring, a disc, a shell | Choose the element, then integrate — §5.5 |
| The shape has an axis of symmetry | Test for symmetry before integrating — §5.4 |
| “Average value over one cycle” | Mean by integration — §6.1 |
| “Root mean square”, or a squared quantity to be averaged | RMS, and the mean of sin² is ½ — §6.2 |
| The rate of a quantity is proportional to the quantity itself | dy ⁄ dt = ky, so y = y₀ekt — §7.1 |
| “Approaches a steady value”, “terminal”, “fully charged” | dy ⁄ dt = k(A − y) — §7.2 |
| Acceleration is proportional to displacement and opposite in sign | a = −ω²y, so the motion is simple harmonic — §7.3 |
| A graph of a rate against a variable is given, and a total over that variable is asked for | Area under the graph — §8.1 |
| A graph is given, and a rate is asked for | Slope of the graph — §8.1 |
| Readings are given and a power law is to be found from them | Plot log against log; the slope is the power — §8.2 |
| An experimental straight line does not pass through the origin | Read the intercept, then decide what it represents — a real effect, or an offset in the apparatus — §8.3 |
| A graph drawn from readings, with axes somebody else chose | The eight questions before you read a value off it — §8.4 |
| Two vectors and the angle between them; the answer is a number that is largest when they are parallel | Scalar product — §9.3 (a resultant magnitude is a number too, and comes from the cosine rule — §9.1) |
| Two vectors and the angle between them; the answer itself has a direction, and a sense of rotation | Vector product — §9.4 |
| Three vectors, and a volume, a coplanarity test, or a product of three | Triple product — §9.5 |
| “Relative to”, or two velocities to be compared | Vector subtraction — §9.1 (completed in Chapter 5) |
Two entries are worth reading twice. Acceleration as a function of position and acceleration as a function of time look almost the same on the page and need different first lines. And for most graph questions the first two things to ask are what the slope means and what the area means — a rate for the first, a total for the second. §8 adds three more readings that a graph can be asked for: its shape, its intercept, and what it becomes when you plot logarithms instead.
DEEPERIt will happen, and it is meant to — a paper made entirely of recognisable questions would not separate anybody. Three moves are available, in this order.
Write the definition of the quantity asked for. Not a formula for it — the definition. Power is dW ⁄ dt; current is dq ⁄ dt; average velocity is total displacement over total time. Writing the definition converts an unfamiliar question into a familiar one more often than any other single move.
Check the dimensions of what you are being asked for. A quantity with dimensions of length × something has very often come from an integral over length, and a quantity per second from a derivative with respect to time. Dimensions do not prove which operation was used — a product has dimensions too — but they narrow it to one or two, which is enough to start.
Take the smallest case you can work out completely. Two particles instead of n. One element instead of a rod. A quarter cycle instead of a cycle. The small case shows the pattern, and the pattern shows the tool.
None of these three is a trick. They are what a physicist does when a problem is genuinely new, which is the situation the paper is trying to put you in.
One test of this section exists, and solved examples cannot be it. On the page before Level 1 there is a Tool Selection Diagnostic: twenty questions for which you write only the tool and the first line, and not the answer. Work it there, and work it again after Level 3.
A large share of the marks lost in a physics paper are lost in the algebra, not in the physics. A quadratic is solved correctly and then the wrong root is kept. A ratio is inverted. A logarithm is handled as though it were a bracket. None of this is difficult. All of it must be automatic, because in the examination hall your attention is needed for the physics.
COREWHEN THIS COMES UP Any time the physics itself produces a term in t² or x² beside a term in t or x — a body thrown up and asked when it passes a given height, two bodies asked when they meet. The tool is deciding which of the two roots survives; the formula is the easy part.
Quadratic equation. An equation in which the highest power of the unknown is two. It is written ax² + bx + c = 0, with a not equal to zero. Counted with multiplicity it has two roots — but whether those are two different real roots, one repeated real root, or a pair of complex roots is decided by the discriminant below.
The two roots are given by
The quantity b² − 4ac under the root sign is called the discriminant, and it decides the nature of the roots. If it is positive, the roots are real and different. If it is zero, the roots are real and equal. If it is negative there is no real root, and in a physics problem that almost always means the event you asked about never happens.
Two further results save time. The sum of the roots is −b ⁄ a and the product of the roots is c ⁄ a. If a question asks only for the sum or only for the product, you need not find the roots at all.
Now the part that is physics rather than algebra. In a physics problem the quadratic usually comes from a displacement, so its two roots are two instants of time, or two positions. Both are correct answers to the equation. Only one may be the answer to the question. A negative time means an instant before you started the clock; a negative length may mean a point on the other side of the wall. Read the physical situation and decide.
Do not throw away the negative root out of habit. Sometimes both roots are physical. A stone thrown upward passes a given height twice — once going up, once coming down — so a question asking when it is at that height has two correct answers.
Sometimes the negative root is the useful one. If you started the clock at a convenient instant rather than at the beginning of the motion, a negative root simply means the event happened before that instant, and it is a perfectly good answer.
A stone is thrown vertically upward from the ground with a speed of 20 m s⁻¹. Taking g = 10 m s⁻², find the times at which it is 15 m above the ground.
Modelling decision. The height is a quadratic in time, so expect two answers. Check whether both are physical before rejecting either; do not stop at the first root.
Taking upward as positive and the ground as the origin, h = ut − ½gt², so 15 = 20t − 5t².
Rearranging: 5t² − 20t + 15 = 0, that is t² − 4t + 3 = 0.
Factorising, (t − 1)(t − 3) = 0, so t = 1 s and t = 3 s.
Both answers are physical. At t = 1 s the stone is at 15 m on the way up; at t = 3 s it is at the same height coming down.
Remark. The stone reaches its highest point midway between the two, at t = 2 s. That follows from the symmetry of the motion about the top, so the two times are equally spaced about it.
A ball is thrown vertically upward with a speed of 10 m s⁻¹ from the top of a tower 25 m high. With g = 10 m s⁻², find the time it takes to reach the ground.
Modelling decision. Fix the origin and the positive direction first and keep them for the whole problem. Taking upward as positive with the top of the tower as origin puts the ground at −25 m. The sign of that displacement is the whole difficulty in this question.
With s = −25 m, u = +10 m s⁻¹ and a = −10 m s⁻²,
−25 = 10t − 5t², so 5t² − 10t − 25 = 0, that is t² − 2t − 5 = 0.
t = [2 ± √(4 + 20)] ⁄ 2 = 1 ± √6.
√6 = 2.449, so the roots are t = 3.449 s and t = −1.449 s.
Time cannot be negative here, so t = 3.45 s.
Remark. The rejected root still means something. It is the instant at which a stone thrown from the ground, passing the top of the tower at 10 m s⁻¹ upward, would have left the ground.
A particle is projected vertically upward with speed u. It is found to be at a height H at only one instant during its flight. Find u in terms of H and g.
Modelling decision. “At only one instant” is a statement about the number of roots, not about the motion. Two equal roots means the discriminant is zero. Turning that sentence into an algebraic condition is where the physics of this question lies.
The height at time t is H = ut − ½gt², so ½gt² − ut + H = 0.
For the two roots to coincide the discriminant must vanish:
u² − 4(½g)(H) = 0, so u² = 2gH and u = √(2gH).
The single instant is t = u ⁄ g, the time to reach the highest point.
Remark. The condition for a double root is exactly the condition that H is the maximum height. The algebra and the physics say the same thing: a height reached twice lies below the top, a height reached once is the top, and a height reached never lies above it — which is the case of a negative discriminant.
WHEN THIS COMES UP A sum of many terms whose ratio to each other is fixed: a ball rebounding to the same fraction of its height each time, a chain of charges each half the last, a body losing a fixed share of its speed at every stage.
Arithmetic progression. A list of numbers in which each term is obtained by adding a fixed number, the common difference d, to the previous term.
Geometric progression. A list of numbers in which each term is obtained by multiplying the previous term by a fixed number, the common ratio r.
For an arithmetic progression with first term a,
and for a geometric progression with first term a,
When the magnitude of r is less than 1 the terms shrink, and the sum of infinitely many of them is finite:
That result is worth having by heart. It is the reason a ball that bounces forever still travels a finite distance and still comes to rest in a finite time. Both facts come out of a geometric series and have nothing to do with the details of the collisions.
Where these appear in physics: the displacements in successive equal intervals under constant acceleration form an arithmetic progression — signed displacements, so if the body turns round the distances no longer follow it and have to be handled by splitting at the turn (§1.5, §5.3). Successive amplitudes of a damped oscillator form a geometric progression, and so do the successive heights of a bouncing ball and the charges left after each stage of a chain of capacitors.
A body starts from rest and moves with a constant acceleration of 2 m s⁻². Show that the distances covered in the first, second and third seconds form an arithmetic progression, and find the distance covered in the 5th second.
Modelling decision. Distance in the nth second means the distance between t = n − 1 and t = n, not the distance up to t = n. Compute the totals and subtract; the pattern then shows itself.
With u = 0 and a = 2, the distance in time t is s = ½at² = t².
s(1) = 1 m, s(2) = 4 m, s(3) = 9 m, s(4) = 16 m, s(5) = 25 m.
Distances in successive seconds: 1, 3, 5, 7, 9 m.
These differ by a constant 2 m, an arithmetic progression with a = 1 and d = 2.
Distance in the 5th second = 9 m.
Remark. The common difference is numerically equal to the acceleration. That is why Chapter 3 can find the acceleration from two such distances without ever writing down a time.
A ball is dropped from a height of 10 m. After each bounce it rises to 0.8 times the height from which it fell. Find the total distance it travels before coming to rest.
Modelling decision. The first fall happens once; every later height is covered twice, going up and coming down. Write the first fall separately and put everything after it into one geometric series, or the first rise gets counted twice.
First fall: 10 m.
After the first bounce the ball rises to 10(0.8) and falls the same distance; after the second it rises to 10(0.8)² and falls the same, and so on.
Total = 10 + 2[10(0.8) + 10(0.8)² + 10(0.8)³ + …]
The bracket is a geometric progression with a = 8 and r = 0.8, so its sum is 8 ⁄ (1 − 0.8) = 40 m.
Total distance = 10 + 2(40) = 90 m.
Remark. The answer is finite although the number of bounces is not. A student who adds the bounces one by one never finishes; a student who recognises the series is done in two lines.
For the ball of the previous example, find the total time it takes to come to rest. Take g = 10 m s⁻².
Modelling decision. Time goes as the square root of the height, so the times form a geometric progression with ratio √r, not r. Getting that ratio right settles the question; after it the series is the same as before.
Time for the first fall: t₀ = √(2h ⁄ g) = √(2 × 10 ⁄ 10) = √2 s.
After the first bounce the ball rises to 0.8h, so that up-and-down journey takes 2√(2 × 0.8h ⁄ g) = 2√0.8 × t₀.
The journeys after that take 2(√0.8)² t₀, 2(√0.8)³ t₀, and so on.
Total = t₀[ 1 + 2(√0.8 + (√0.8)² + …) ] = t₀[ 1 + 2√0.8 ⁄ (1 − √0.8) ].
√0.8 = 0.8944, so the bracket is 1 + 2(0.8944) ⁄ 0.1056 = 17.94.
Total time = 1.414 × 17.94 = 25.4 s.
Round √0.8 to three places here and the answer comes out 25.3 s: the denominator 1 − √0.8 is small, so it magnifies any rounding above it.
Remark. The model predicts infinitely many bounces in a finite time, and that is not a trick of the mathematics — the geometric series converges, so the times really do add to something finite. What a real ball does near the end is another question: a fixed rebound ratio stops describing it once the bounces are small, because deformation, sound and contact time no longer scale with the drop.
WHEN THIS COMES UP The unknown sits in an exponent, or a quantity is described as halving, doubling or falling to 1 ⁄ e in a fixed time. The words “half-life” and “time constant” both mean that a logarithm has to be taken.
Logarithm. The power to which the base must be raised to give the number. If ax = N then logaN = x. In physics the base is almost always e ≈ 2.718, and we write ln in place of loge.
Three rules do all the work:
together with ln 1 = 0 and ln e = 1. To move between the two bases, ln x = 2.303 log₁₀ x.
The exponential is the reverse of the logarithm, and its defining property is that it multiplies when the exponents add:
This single line explains why decay problems become easy once you see them properly. If a quantity falls to half its value in one interval, then over two such intervals it falls to a quarter and over three to an eighth. The fractions multiply. So instead of using the formula you can often just count the intervals.
There is no rule for the logarithm of a sum. ln(a + b) is not ln a + ln b, and no manipulation will make it two separate logarithms. It can be rewritten: ln(a + b) = ln a + ln(1 + b ⁄ a), and when b ≪ a that second term is the small correction §2.1 expands. Take the larger of the two outside and that rewrite is the only move available. If your working has produced ln a + ln b, either you have made a mistake or the expression was meant to stay as it was.
Equally, ea + eb is not ea+b. Adding exponentials and adding exponents are two different operations, and on a hurried page the two look alike.
A radioactive sample reduces to one eighth of its original amount in 15 days. Find its half-life.
Modelling decision. Count halvings instead of using the decay formula. One eighth is one half applied three times, so the interval contains exactly three half-lives and no logarithm is needed.
1 ⁄ 8 = (1 ⁄ 2)³, so three half-lives have passed.
3 T½ = 15 days, so T½ = 5 days.
Remark. Use this counting method whenever the fraction is a power of one half. Reach for the formula only when it is not.
A radioactive sample decays as N = N₀e−λt with λ = 0.10 per day. Find the time in which 90 per cent of the sample decays.
Modelling decision. Ninety per cent decayed means ten per cent remaining. Read which of the two the formula needs — this is where the mark is usually lost, not in the logarithm.
90 per cent decayed leaves N ⁄ N₀ = 0.10.
0.10 = e−0.10t.
Taking natural logarithms of both sides: ln(0.10) = −0.10t.
−2.303 = −0.10t, so t = 23.0 days.
Remark. As a check, the half-life here is ln2 ⁄ λ = 6.93 days, so 23.0 days is about 3.3 half-lives. Three half-lives leave one eighth, which is 12.5 per cent, so a little over three is right for 10 per cent.
The potential difference across a discharging capacitor falls from 10 V to 5 V in 2 s. Find the potential difference after a further 4 s.
Modelling decision. Do not look for the time constant. The discharge is exponential, so equal intervals of time multiply the voltage by the same factor. Seeing that turns the problem into a geometric progression and removes every logarithm from the working.
In the first 2 s the voltage is multiplied by 5 ⁄ 10 = 0.5.
A further 4 s is two more such intervals, so the voltage is multiplied by 0.5 twice more.
V = 5 × 0.5 × 0.5 = 1.25 V.
Remark. The formula route gives the same answer: V = V₀e−t/RC with RC = 2 ⁄ ln 2, then t = 6 s. It takes four lines instead of two, and every extra line is a place to make a mistake.
WHEN THIS COMES UP “Least possible”, “greatest possible”, “at most”, “at least”, or a minimum you are asked for without being given a function to differentiate. Also any question that asks whether a situation is possible at all.
Arithmetic and geometric mean. For two positive numbers a and b the arithmetic mean is (a + b) ⁄ 2 and the geometric mean is √(ab). The first is never smaller than the second, and they are equal only when a = b.
This one line settles a whole family of minimum problems without a derivative. Put a = x and b = c ⁄ x, both positive. Their product is c whatever x is, so the geometric mean is fixed at √c while the arithmetic mean can be anything above it:
Physics is full of that shape. The period of a compound pendulum, which carries h + k² ⁄ h and is least when the pivot sits one radius of gyration from the centre of mass; the power a source delivers into a load, whose denominator is R + r² ⁄ R + 2r; the distance from source to screen when a lens sits between them — each of them is a sum of a quantity and a constant over that quantity, and each has its least value where the two parts are equal. That is worth saying plainly: the sum is least where the two terms are equal, and you can write the answer down before you have found the derivative.
AM ≥ GM holds for positive numbers only. If x is allowed to be negative, x + c ⁄ x has no least value at all — it runs down without limit as x approaches zero from below. Check the sign of every quantity before you use the inequality, and say why it is positive.
The second bound is one you already have and may not have read as a bound. The quadratic formula contains √(b² − 4ac), and a real answer needs what is under that root to be positive or zero. In algebra that is a statement about roots. In physics it is a statement about whether the situation can happen.
Three bounds worth having.
1. x + c ⁄ x ≥ 2√c for x > 0, with the least value at x = √c.
2. b² ≥ 4ac is the condition for ax² + bx + c = 0 to have a real solution. When the physics produces the quadratic, this is the condition for the event to be possible.
3. |sin θ| ≤ 1 and |cos θ| ≤ 1, always. An answer of sin θ = 1.3 means one of two things, of unequal likelihood: an arithmetic slip, which is the commoner by far, or — once the arithmetic has been checked and survives — the mathematics reporting that no such angle exists and therefore that the situation cannot happen. Check the working first. §1.5 says what to do when it holds.
Two resistors R₁ and R₂ are joined in parallel, so that the pair behaves as a single resistance R = R₁R₂ ⁄ (R₁ + R₂). Show, without putting numbers in, that R can never reach a quarter of R₁ + R₂, except in one case.
Modelling decision. Nothing here needs a derivative. Twice the parallel resistance is the harmonic mean of the two, so the question is really AM ≥ HM with resistances in place of numbers — and that one is proved by a square that cannot be negative.
2R = 2R₁R₂ ⁄ (R₁ + R₂), which is the harmonic mean of the two resistances.
Compare it with the arithmetic mean (R₁ + R₂) ⁄ 2. Their ratio is 4R₁R₂ ⁄ (R₁ + R₂)².
Now (R₁ + R₂)² − 4R₁R₂ = (R₁ − R₂)², which is never negative.
So the ratio is at most 1, giving 2R ≤ (R₁ + R₂) ⁄ 2, that is R ≤ (R₁ + R₂) ⁄ 4 — with equality only when the two resistances are equal.
With numbers: 40 Ω and 60 Ω in parallel give 24 Ω, against a quarter-sum of 25 Ω.
Remark. The quantity 2R₁R₂ ⁄ (R₁ + R₂) is the harmonic mean, and the result you have just proved is the outer pair of AM ≥ GM ≥ HM: that the harmonic mean never passes the arithmetic mean. You will meet the same three means wherever two equal shares are combined — two springs in series, two lenses in contact, a journey done in two equal halves, which §6.1 takes up — and the ordering does not change.
A small source and a screen are a fixed distance D apart. A converging lens of focal length f is placed between them. Show that no sharp image can be formed unless D is at least 4f.
Modelling decision. Write the lens equation with the two distances forced to add to D. That constraint turns two unknowns into one and produces a quadratic, and the question “is an image possible” becomes the question “is the discriminant positive”.
Let the object distance be u and the image distance v, with u + v = D.
The lens equation gives 1 ⁄ u + 1 ⁄ v = 1 ⁄ f.
Putting v = D − u: 1 ⁄ u + 1 ⁄ (D − u) = 1 ⁄ f, so f D = u(D − u).
That is u² − Du + fD = 0.
For a real u the discriminant must not be negative: D² − 4fD ≥ 0, so D(D − 4f) ≥ 0.
D is positive, so D ≥ 4f, with a single position at D = 4f and two positions whenever D is larger.
Remark. The algebra has answered a physical question that no amount of moving the lens would settle: below four focal lengths there is nowhere to put it. Notice also what D = 4f gives — one root instead of two — which is the lens exactly halfway, u = v = 2f. Chapter 31 uses this as the displacement method for measuring f.
A cell of emf E and internal resistance r drives a resistance R. Show, without differentiating, that the power delivered to R is greatest when R = r.
Modelling decision. Write the power and look for the shape x + c ⁄ x in the denominator. A fraction is largest when its denominator is smallest, so a minimum in the denominator is a maximum in the power.
The current is I = E ⁄ (R + r), so the power in R is P = I²R = E²R ⁄ (R + r)².
Divide top and bottom by R: P = E² ⁄ [ (R + r)² ⁄ R ].
Now (R + r)² ⁄ R = R + 2r + r² ⁄ R = (R + r² ⁄ R) + 2r.
Both R and r² ⁄ R are positive, so their sum is at least 2√(r²) = 2r, with equality at R = r.
The denominator is therefore at least 4r, and P is at most E² ⁄ (4r), reached when R = r.
Remark. Example 1.39 solves this same problem again with a derivative, in about the same number of lines. That is on purpose: neither route is a shortcut, and holding both is worth more than holding either, because one of them will be the one you see under pressure. Note also the efficiency at that point: half the power is lost inside the cell, which is why matching is generally not the operating condition wanted where bulk power has to be delivered. It is the condition wanted where the aim is to get the largest possible signal out of a weak source. Chapter 24 returns to this.
WHEN THIS COMES UP Every time an answer comes out negative, or a square root has to be taken, or a question asks for distance rather than displacement. Also whenever an answer looks impossible — sin θ above one, a negative time, a mass below zero.
Modulus. |x| is the magnitude of x with its sign removed: |−5| = 5 and |5| = 5. In physics it is the difference between speed and velocity, between distance and displacement, and between the magnitude of a force and its component along an axis.
A sign in physics is never decoration. It carries the direction you chose at the start, and three separate steps can destroy it without any warning on the page.
√(x²) is |x|, not x. This is why v = √(u² + 2as) gives the speed and says nothing about the direction of travel. The direction has to come back from the physics.
A square root written in a formula is the positive root unless the physics asks for both. Writing ± everywhere postpones a decision you are being asked to make.
Squaring an equation can create a solution the original did not have. Every answer obtained after squaring must be put back into the equation you started from.
The second half of this is the physical domain — the set of values a quantity may take before any physics is done. A time measured from the start runs forward. A mass is positive. A speed is positive or zero, though a velocity may be negative. For the usual absolute refractive index of a transparent medium, n ≥ 1. A probability lies between zero and one. Checking an answer against its domain costs five seconds and catches a whole class of error that no amount of rechecking the algebra will find.
An answer outside the domain may be telling you something. Sometimes it is the physics reporting that the situation you described cannot happen: sin θ = 1.3 at a glass surface means there is no refracted ray, which is total internal reflection and is the answer. Before you hunt for the arithmetic slip, ask whether the impossible answer is telling you something.
Solve √(2x + 3) = x, and say which solution survives if x is the length of a rod.
Modelling decision. Squaring is the only route, and squaring is exactly the step that can manufacture a root. Every answer therefore goes back into the original equation before it is written down.
Squaring: 2x + 3 = x², so x² − 2x − 3 = 0.
Factorising: (x − 3)(x + 1) = 0, giving x = 3 or x = −1.
Check x = 3 in the original equation: √9 = 3. It holds.
Check x = −1: √1 = 1, which is not −1. It fails.
So x = 3 is the only solution of the equation, and being positive it is also the only one a length could be: x = 3.
Remark. The root x = −1 came from the squared equation, which is a different equation, and not from a mistake in the working. Squaring replaced “equals” with “equals in magnitude”, and the discarded root is the one that satisfied the second and not the first.
Light inside glass of refractive index 1.51 meets the surface at 60° to the normal. Find the angle at which it leaves into the air.
Modelling decision. Snell's law is handed to you here; what is being tested is what you do when the answer leaves the domain. Write the law, get the number, and then read it.
n₁ sin θ₁ = n₂ sin θ₂, with n₁ = 1.51, θ₁ = 60° and n₂ = 1.00.
sin θ₂ = 1.51 × sin 60° = 1.51 × 0.866 = 1.31.
No angle has a sine above 1.
So no ray leaves the glass: the light is totally internally reflected at the surface.
Remark. Nothing went wrong in the algebra. The domain |sin θ| ≤ 1 is the physics, and running into it is the result. The same reading saves you elsewhere: a negative time means the event was before your chosen origin, and a negative mass means an assumption is wrong. Chapter 31 builds the critical angle out of this line.
A particle starts with a speed of 5 m s⁻¹ and moves with a constant acceleration of −2 m s⁻² along the same straight line. Using v² = u² + 2as, find its speed after it has covered 6 m, and say why that equation on its own cannot tell you which way it is then moving.
Modelling decision. The equation gives v², and taking a square root of v² gives |v|. The magnitude comes out of the algebra; the direction has to come from somewhere else, and knowing where is the point of the question.
v² = 5² + 2(−2)(6) = 25 − 24 = 1.
√(v²) = |v|, so the speed is 1 m s⁻¹. Both v = +1 and v = −1 satisfy the equation.
To settle the direction, use an equation that carries the sign of v: v = u + at.
First find the time. s = ut + ½at² gives 6 = 5t − t², so t² − 5t + 6 = 0 and t = 2 s or t = 3 s.
Both are physical: the particle passes s = 6 m going forward at t = 2 s, turns round at t = 2.5 s, and passes it again coming back at t = 3 s.
v = 5 − 2t gives v = +1 m s⁻¹ at t = 2 s and −1 m s⁻¹ at t = 3 s.
Remark. The formula v² = u² + 2as gives you v², so its square root is the magnitude and the sign is yours to supply; and having no t in it, it cannot tell the two visits apart. Every equation you use throws something away, and knowing what each one throws away is worth more than knowing the equation. This one keeps the magnitude and drops the sign.
An approximation is a decision that some term is small enough to drop, and like every decision in this book it rests on a condition that can be checked. Two approximations carry almost all of the load in JEE physics.
COREWHEN THIS COMES UP Anything described as slight, small, or a one per cent change; a height small beside the radius of the earth; a square root or a power with a small correction inside it.
Binomial approximation. When x is very small compared with 1, the quantity (1 + x)ⁿ may be replaced by 1 + nx. The index n may be positive or negative, whole or fractional.
Two conditions are running together here and they differ. If n is a whole number and not negative, the expansion stops on its own after n + 1 terms and holds for every x. For any other n it is an endless series, and for the series used here the relevant convergence condition is |x| < 1. Keeping only the first two terms needs more again: the first term dropped is n(n − 1)x² ⁄ 2, so for a fixed n the error falls away as x². If x is one per cent, that error is about one part in ten thousand, which no examination will ask you to resolve.
A first check in a physics problem is |x| ≪ 1, and when |n| is large you must look at the magnitude of the term you are throwing away as well. That term is n(n − 1)x² ⁄ 2, so what actually has to be small is that, not x on its own. The short test |nx| ≪ 1 catches most cases, and it is a test rather than a definition — what it really checks is that the second term nx is itself small, so that the third is smaller again. A large n can ruin the approximation even when x is tiny, which §2.3 returns to.
The skill being tested is the ability to force an expression into the shape (1 + something small)ⁿ. Take the value of g at a height h above the surface of the earth:
and now the approximation applies directly, with n = −2:
See what this has gained. The exact expression is an inverse square and tells you little at a glance. The approximate one is a straight line, and it says at once that g falls off twice as fast with height as the bare fraction h ⁄ R would suggest.
For a small fractional change, use the index. If y = k xn, then Δy ⁄ y = n (Δx ⁄ x). A one per cent increase in length changes the period T = 2π√(l ⁄ g) by half a per cent, because the index is ½.
This is usually the quickest route through a percentage-change question, and it avoids writing out the expansion at all.
Two more first-order approximations belong beside the binomial, and physics uses them just as often. For small x,
Both come from the same idea as the binomial: over a small step a curve and its tangent agree. The first is needed whenever a logarithm has to be undone at the end of a decay problem, the second whenever an exponential has only just begun. Like the binomial, each drops a term of order x².
Find the value of √(1.02) correct to four decimal places, without using a calculator.
Modelling decision. Write the number as 1 plus a small quantity, so that the approximation applies with n = ½. The whole art is in the writing.
√(1.02) = (1 + 0.02)1/2 ≈ 1 + ½(0.02) = 1.0100.
The true value is 1.00995, so the error is 5 × 10⁻⁵.
√(1.02) ≈ 1.0100.
Remark. The dropped term is n(n − 1)x² ⁄ 2 = (½)(−½)(0.0004) ⁄ 2 = −5 × 10⁻⁵, which is exactly the error found. The estimate of the error is as reliable as the approximation itself.
The radius of the earth is 6400 km. Find the percentage decrease in the acceleration due to gravity at a height of 32 km above the surface.
Modelling decision. Use g′ = g(1 − 2h ⁄ R) rather than the exact inverse square. Here h ⁄ R is 1 ⁄ 200, far inside the condition, so the working is one line.
h ⁄ R = 32 ⁄ 6400 = 0.005.
g′ = g(1 − 2 × 0.005) = g(1 − 0.01) = 0.99 g.
Percentage decrease = 1 per cent.
Remark. Doing it exactly gives 0.993 per cent. The approximation is wrong in the third significant figure, which is well below the precision of any data you will be given.
A wire is stretched so that its length increases by 0.5 per cent, its volume staying the same. Its resistance is R = ρL ⁄ (πr²). Find the percentage change in its resistance.
Modelling decision. Two things change at once, L and r, and each depends on the other — constant volume ties them together. Eliminate r first so that R depends on one variable only; then the binomial is applied once instead of twice, and no cross term can be lost.
Constant volume means πr²L is fixed, so r² ∝ 1 ⁄ L.
Substituting into R = ρL ⁄ (πr²) gives R ∝ L ⁄ (1 ⁄ L) = L².
So R = kL², and a small change gives R′ = k L²(1 + 0.005)².
By the binomial, (1 + 0.005)² ≈ 1 + 2(0.005) = 1.01.
The resistance rises by 1 per cent.
Remark. Answering 0.5 per cent means treating R as proportional to L and forgetting that the wire also thins. Answering 1.5 per cent means letting the radius shrink by the whole 0.5 per cent: constant volume gives r² ∝ 1 ⁄ L, so the radius falls by only half of that, 0.25 per cent. Reducing to one variable before expanding removes both traps at once.
The length of a simple pendulum is increased by 2 per cent, and the value of g at the place is 1 per cent less than was assumed. Find the percentage change in the time period T = 2π√(l ⁄ g).
Modelling decision. Two small changes act at once, so work with fractional changes rather than with the quantities themselves. Each factor contributes its own index times its own fractional change, and the contributions add.
Writing T ∝ l1/2 g−1/2,
ΔT ⁄ T = ½(Δl ⁄ l) − ½(Δg ⁄ g).
Here Δl ⁄ l = +0.02 and Δg ⁄ g = −0.01.
ΔT ⁄ T = ½(0.02) − ½(−0.01) = 0.01 + 0.005 = 0.015.
The period increases by 1.5 per cent.
Remark. Watch the sign. A smaller g makes the pendulum slower, so it lengthens the period and its contribution adds to that of the longer string instead of cancelling it. Do not carry this straight into error analysis. Here the direction of each change is given, so the signs are kept and they decide whether the two contributions add or cancel. In an error calculation no direction is known, so every contribution is taken at its magnitude and added, which gives the largest the error could be — Level 3 Question 5 does exactly that with the same formula. Same rule, opposite treatment of the signs.
WHEN THIS COMES UP A small angle, a thin lens, a distant object, a pendulum through a few degrees, an arc treated as a straight line. Once θ is in radians and small, sin θ, tan θ and θ are the same number.
Radian. The angle subtended at the centre of a circle by an arc whose length equals the radius. A full circle is 2π radian, so 180° = π radian and 1 radian = 57.3°.
Angles in physics are measured in radian, and not as a matter of taste. Every approximation below, and every derivative of a trigonometric function in this book, is written for an angle in radian. Feed a figure in degrees into one of them without converting and the answer is out by π ⁄ 180, or by a power of it.
Look at the figure. For a small angle θ the arc, the sine and the tangent all become nearly equal, and all three become nearly equal to Rθ. That is the content of the small-angle results:
Keep the cosine to its second term; do not set it equal to 1. In the simple pendulum the restoring force needs sin θ ≈ θ, while the height risen needs 1 − cos θ ≈ θ² ⁄ 2. A student who puts cos θ = 1 finds the pendulum rises through no height at all and therefore has no potential energy, which is plainly wrong.
Radian is not a choice. Every result in §2.2, and every trigonometric derivative in §4.2, is written for an angle in radian. Convert a degree figure to radian before it enters any formula, and think in radian thereafter, wherever an angle meets an approximation or a derivative.
Keep one order more than you think you need whenever the leading term cancels. The θ² in the cosine survives precisely because the 1 cancelled against something else.
A simple pendulum of length 1.0 m swings so that its bob moves 5.0 cm horizontally from the lowest point. Find the angle of the string with the vertical at that extreme, and the fraction by which the tension in the string there falls short of mg.
Modelling decision. Two approximations in one problem, wanted to two different orders. The sine settles the angle. The cosine settles the tension — and here the leading term of the cosine survives, so keeping it is enough. §2.3 has the case where it is not.
Horizontal displacement ≈ lθ, so θ = 0.050 ⁄ 1.0 = 0.050 rad, about 2.9°.
At the extreme of the swing the bob is momentarily at rest, so there is no centripetal term and T = mg cos θ.
cos θ ≈ 1 − θ² ⁄ 2 = 1 − 0.00125.
The tension falls short of mg by 0.125 per cent, about one part in eight hundred.
Remark. Here cos θ ≈ 1 would have given T = mg, wrong by one part in eight hundred — which no measurement in this experiment could detect. Example 1.23 takes the same pendulum and asks instead for the height the bob has risen, where that same approximation gives zero and destroys the answer. The difference is whether the leading terms cancel: in the tension they do not, in the height they do.
The angular diameter of the moon seen from the earth is about 0.50°, and its distance from the earth is 3.84 × 10⁵ km. Estimate the diameter of the moon.
Modelling decision. The moon is far away, so the angle is small and its diameter may be treated as an arc of a circle centred on the observer. Convert the angle to radian first; D ≈ Rθ needs θ in radian.
θ = 0.50° = 0.50 × π ⁄ 180 = 8.73 × 10⁻³ rad.
For a small angle, diameter D ≈ Rθ, where R is the distance.
D = 3.84 × 10⁵ × 8.73 × 10⁻³ = 3.35 × 10³ km.
Remark. The accepted diameter is 3475 km, so the estimate is within four per cent. That error comes from the rounded angular diameter, not from the approximation: at half a degree the small-angle step is good to a few parts in a million, far finer than the data it is applied to.
A ray of light strikes the surface of water at a small angle of incidence i. Using the small-angle approximation, show that the angle of deviation is i(1 − 1 ⁄ μ), where μ is the refractive index of water.
Modelling decision. Snell's law contains sines, and sines are awkward to subtract. For small angles replace each sine by the angle itself; the law becomes a simple proportion and the deviation follows in one step.
Snell's law: sin i = μ sin r.
For small angles, i ≈ μ r, so r ≈ i ⁄ μ.
The deviation is the angle turned through: d = i − r.
d = i − i ⁄ μ = i(1 − 1 ⁄ μ).
Remark. The deviation is proportional to the angle of incidence, so a narrow bundle of rays arriving nearly straight on stays a bundle after refraction. Chapter 31 uses this to derive the apparent depth of a pool.
WHEN THIS COMES UP The moment after you approximate, and before you trust the result. “Is the next term small enough to drop?” is a question to answer, not to assume.
Order of the error. The magnitude of the first term you threw away. The order of the approximation is a different thing — it is the highest power you kept, so (1 + x)ⁿ ≈ 1 + nx is first order and cos θ ≈ 1 − θ² ⁄ 2 is second. An approximation is only as good as that discarded term is small, so estimating it is part of using the approximation and not an optional extra.
Every approximation in this section carries a condition. It is worth knowing how much room that condition leaves, so that you can use the approximation with confidence in the paper and put it down when it does not apply.
For the sine, the first dropped term is θ³ ⁄ 6. At θ = 0.1 rad the error is about one part in six hundred. At θ = 0.5 rad, a little under 30°, it is already 4 per cent. The approximation is excellent for a few degrees and poor by thirty.
|x| ≪ 1 is not by itself enough. The series needs |x| < 1 to converge at all, but converging is a weaker thing than being well approximated by its first two terms. The first term dropped is ½ n(n − 1)x², so what has to be small is that term — and with a large index it is not. At x = 0.1 and n = 10 the approximation gives 1 + 10(0.1) = 2, against a true (1.1)¹⁰ = 2.594: an error of 23 per cent. As a quick check in the examination hall, ask whether |nx| is small; as a statement of the condition, measure the dropped term itself, which is what §2.3 is for.
An approximation that is accurate in a quantity can be useless in a difference of two such quantities, because the large parts cancel and leave the errors behind. Whenever a derivation subtracts two nearly equal approximated quantities, go back and keep one more term in each.
A pendulum bob on a string of length L is pulled aside through a small angle θ. Find how far it rises above its lowest point. Show that cos θ ≈ 1 is useless here, while cos θ ≈ 1 − θ² ⁄ 2 is not.
Modelling decision. The height is a difference of two nearly equal lengths, L and L cos θ. In such a difference the leading terms cancel, so the answer lives entirely in the term you would normally discard. Recognising that before you approximate is what is taught in this section.
The bob rises h = L − L cos θ = L(1 − cos θ).
With cos θ ≈ 1: h = L(1 − 1) = 0.
That says the bob never rises, so its potential energy never changes and a pendulum cannot swing. The approximation has destroyed the answer, not approximated it.
With cos θ ≈ 1 − θ² ⁄ 2: h = L(1 − 1 + θ² ⁄ 2) = Lθ² ⁄ 2.
Check it: for L = 1.0 m and θ = 0.10 rad this gives h = 5.0 mm, against an exact 4.996 mm.
Remark. The 1 in the cosine cancelled against the other 1, so the θ² term is not a correction to the answer — it is the answer. Whenever a quantity is the difference of two nearly equal things, keep one order more than you think you need, then check whether the leading term survived. Chapter 19 takes this same h straight into the energy of a pendulum.
Find the percentage error made in writing sin θ ≈ θ at θ = 5°, 15° and 30°. At which of these would you refuse to use the approximation in a pendulum problem?
Modelling decision. Convert to radian first — sin θ ≈ θ sets an angle equal to a pure number, and only the radian measure can be set equal to one. Then compare the error with the accuracy the problem actually needs, which for a pendulum timed with a stopwatch is about one per cent.
At 5°: θ = 0.08727 rad, sin θ = 0.08716. Error = 0.13 per cent.
At 15°: θ = 0.26180, sin θ = 0.25882. Error = 1.15 per cent.
At 30°: θ = 0.52360, sin θ = 0.50000. Error = 4.72 per cent.
The error grows as θ² ⁄ 6, so it quadruples when the angle doubles.
At 5° the approximation is safe. At 15° the error is 1.15 per cent, already past the one per cent this experiment can resolve, so the approximation is no longer good enough there either. At 30° it is hopeless: refuse it.
Remark. The simple pendulum formula T = 2π√(L ⁄ g) is built on this very approximation, so it too is approximate. At an amplitude of 30° the true period is about 1.7 per cent longer. That is why a pendulum experiment is done at small swing, and the instruction is physics, not over-carefulness.
Swing a pendulum at 60° instead of 5°. Does the standard formula T = 2π√(L ⁄ g) predict a period that is too long or too short? The formula replaces the restoring force mg sin θ by mgθ, and sin θ is the smaller of the two, so the formula assumes a stronger pull towards the lowest point than there really is. A stronger pull means a quicker return, so the predicted period is too short: the real pendulum is slower. About 1.7 per cent slower at 30°, and some 7 per cent at 60°.
You have just dropped a term. Before you write the answer down: how big was it, and how big is the accuracy the question asked for? Two numbers, and the second is the one students never look up.
In physics, trigonometry is the machinery of resolution. A vector — a quantity with a magnitude and a direction, whose rules §9 sets out in full — pointing at an angle is replaced by two vectors along convenient axes, and the sine and the cosine decide which is which. Everything in this section exists to serve that one operation. You need nothing from §9 to follow it: a component is a length, and lengths are what trigonometry is for.
One habit will save you many marks: read the angle off the diagram every time. Do not memorise that the horizontal component takes the cosine. The moment the axes are tilted, as they are on an inclined plane, a memorised pairing is wrong and a read angle is still right.
COREWHEN THIS COMES UP A force, velocity or field that does not lie along the direction you want to work in — an incline, a string at an angle, a projectile. Resolve before writing any equation, every time.
Resolution of a vector. Replacing one vector by two perpendicular vectors which, added together, produce exactly the same effect. The component along a direction making angle θ with the vector is A cos θ, and the component perpendicular to it is A sin θ.
There are six ratios. Three do most of the work. Of the reciprocal three, sec is the one that returns: it appears as sec² in the identity below, in the derivative of the tangent in §4.2 and in the integral that undoes it in §5.2. All six must be recognised on sight:
In calculus and in every small-angle result, angles must be in radian, for the reason given in §2.2. The standard values are quoted at the familiar degree marks because that is how a question states them, and they should come without hesitation: sin 0° = 0, sin 30° = ½, sin 45° = 1 ⁄ √2, sin 60° = √3 ⁄ 2, sin 90° = 1, with the cosines reading the same list backwards.
Two more are worth knowing because papers use them to keep the arithmetic clean: sin 37° ≈ 3 ⁄ 5 and sin 53° ≈ 4 ⁄ 5, with the cosines the other way round. The true values are 0.602 and 0.799, so these are rounded; the angles of the 3–4–5 triangle are 36.87° and 53.13°. But a paper that quotes 37° or 53° intends the fractions, and working with them is what the setter has costed the arithmetic for.
Beyond 90° the ratios change sign, and in physics that sign is usually the answer. A phase angle in an alternating-current problem, or the position of an oscillator a little past its extreme, both turn on it. Read the signs off the two curves rather than memorising a mnemonic:
Two facts from the picture are worth remembering. The two curves are the same shape, shifted by a quarter turn — which is why a velocity in simple harmonic motion leads the displacement by 90°. And each repeats after 2π, so an angle and that angle plus 2π describe the same physical state.
A block of mass 10 kg rests on a plane inclined at 30° to the horizontal. Resolve its weight along and perpendicular to the plane. Take g = 10 m s⁻².
Modelling decision. Take the axes along and perpendicular to the plane, not horizontal and vertical. The angle between the weight and the perpendicular to the plane equals the angle of the incline — read that from the diagram rather than recalling a pairing.
Weight = mg = 10 × 10 = 100 N, acting vertically downward.
Component along the plane, down the slope = mg sin 30° = 100 × 0.5 = 50 N.
Component perpendicular to the plane = mg cos 30° = 100 × 0.866 = 86.6 N.
Remark. Check the parts against the whole: √(50² + 86.6²) = 100 N, as it must be. The check takes five seconds and catches a swapped sine and cosine at once.
The displacement of a particle in simple harmonic motion is y = 4 sin θ centimetre, where θ is the phase angle. Find y at θ = 150°, 210° and 300°, and say at which of these the particle is on the negative side of its mean position.
Modelling decision. Every angle here is beyond 90°, so the sign of the sine is the whole question. Read each off the curve of §3.1: sine is positive in the first two quadrants and negative in the last two. The magnitude comes from the related acute angle.
At θ = 150° (second quadrant): sin 150° = +sin 30° = +0.5,
so y = +2 cm.
At θ = 210° (third quadrant): sin 210° = −sin 30° = −0.5,
so y = −2 cm.
At θ = 300° (fourth quadrant): sin 300° = −sin 60° = −0.866,
so y = −3.46 cm.
The particle is on the negative side at 210° and 300°.
Remark. Learn to take the sign from the quadrant rather than from a table of values. In a phase question the angle is usually a symbol and not a number, so there is nothing to look up: the sign of sin(π + φ) has to be read off the quadrant, and Chapter 19 asks for that constantly.
WHEN THIS COMES UP Two sines or two cosines to be added, a double angle, or a product of two oscillating terms in a wave or an alternating circuit. These identities are what turn a product into a sum, and a sum can be averaged or integrated term by term, which a product cannot.
Identity. An equation between two expressions that is true for every value of the angle, not merely for some. An identity may therefore be used to replace one expression by the other anywhere, at any time, without solving anything first.
sin C + sin D looks useless until two waves of nearly equal frequency are added together, when it produces beats in a single line. The double-angle form of the cosine is what turns the average of sin²θ into ½, and that is the basis of root mean square values in alternating current.
The two product forms run that one backwards. A product of two oscillating terms becomes a sum, and a sum can be integrated or averaged term by term while a product cannot. That is the only reason they are here — and the addition formulae above them are what the product forms are built from, which is why they are printed too.
The last two are the half-angle forms, and they are the ones a student is most often missing when a chord or a rise is wanted. The straight line joining the two ends of an arc of angle θ on a circle of radius R is 2R sin(θ ⁄ 2) — not R sin θ — and the height a pendulum bob rises is L(1 − cos θ) = 2L sin²(θ ⁄ 2). Both come straight from these two lines.
Two forces act on a particle along the same line and vary with time as F₁ = 3 sin ωt and F₂ = 4 cos ωt newton. Show that their sum is a single sinusoid of the same frequency, and find its amplitude and phase.
Modelling decision. A sine and a cosine of the same frequency combine into one sinusoid. Write the target form first, expand it, and match coefficients — that is quicker and safer than hunting for an identity.
Let F = R sin(ωt + φ) = R cos φ sin ωt + R sin φ cos ωt.
Matching: R cos φ = 3 and R sin φ = 4.
Squaring and adding: R² = 9 + 16 = 25, so R = 5 N.
Dividing: tan φ = 4 ⁄ 3, so φ = 53.1° = 0.927 rad.
Hence F = 5 sin(ωt + 0.927) newton.
Remark. The amplitude is 5 N, not 7 N. Two quantities of the same frequency add their peaks only when they peak together; these two are a quarter cycle apart, so they do not. Note also that the phase is written in radian. ωt is an angle in radian, so anything added to it must be too; sin(ωt + 53.1°) adds a degree to a radian and is meaningless. This is the heart of the superposition of two simple harmonic motions of the same frequency, and Chapter 19 uses it without re-deriving it.
Two sound waves of nearly equal angular frequencies ω₁ and ω₂ and of the same amplitude A arrive together. Show that the result may be read as a single wave whose amplitude varies slowly, and find the frequency with which the loudness rises and falls.
Modelling decision. Add the two displacements and convert the sum of sines into a product. The product form separates a fast oscillation from a slow one, which is exactly what the ear hears as beats.
y = A sin ω₁t + A sin ω₂t.
Using sin C + sin D = 2 sin[(C + D) ⁄ 2] cos[(C − D) ⁄ 2],
y = 2A cos[(ω₁ − ω₂)t ⁄ 2] · sin[(ω₁ + ω₂)t ⁄ 2].
The sine term is a wave at the average frequency. The cosine in front is a slowly changing amplitude, because ω₁ − ω₂ is small.
Loudness is greatest whenever the cosine is +1 or −1, which happens twice in each of its cycles.
The cosine has angular frequency (ω₁ − ω₂) ⁄ 2, so by ω = 2πf its own frequency is (ω₁ − ω₂) ⁄ (4π). Loudness peaks at twice that:
beat frequency = (ω₁ − ω₂) ⁄ (2π) = f₁ − f₂.
Remark. The factor of two is the part that is missed. The amplitude envelope has frequency (f₁ − f₂) ⁄ 2, but the ear responds to loudness, which peaks twice in every envelope cycle. Chapter 21 derives the same result in its own language.
A particle has x = a sin ωt along one axis and y = a sin(ωt + φ) along the other, with the same amplitude and the same frequency. Show that the path is in general an ellipse, and find the values of φ for which it becomes a straight line and for which it becomes a circle.
Modelling decision. The path is what is left when the time is eliminated between the two coordinates, so eliminate it. Expand the second with the addition formula, replace sin ωt by x ⁄ a, and the only awkward term left is cos ωt — which goes when the equation is squared.
Expanding, y = a sin ωt cos φ + a cos ωt sin φ — §3.2.
With sin ωt = x ⁄ a: y = x cos φ + a sin φ cos ωt.
Move the first term across and square, using a²cos²ωt = a² − x²:
(y − x cos φ)² = sin²φ (a² − x²).
Multiplying out and collecting:
x² − 2xy cos φ + y² = a² sin²φ, an ellipse.
φ = 0: the right side vanishes and the left is (y − x)², so y = x — a straight line. Likewise φ = π gives y = −x.
φ = π ⁄ 2: cos φ = 0 and sin²φ = 1, leaving x² + y² = a² — a circle.
Remark. One phase angle carries the geometry, and the two extreme cases are the ones worth holding: in step, the particle runs along a line; a quarter cycle apart, it runs round a circle. Everything between is an ellipse leaning one way or the other. Chapter 19 calls these the superposition of two perpendicular simple harmonic motions and uses the result without deriving it again; Chapter 32 calls the same three cases plane, circular and elliptical polarisation.
WHEN THIS COMES UP A closed triangle of three quantities with no right angle anywhere in it — three forces in equilibrium, two velocities and their resultant, a vector triangle read as plain geometry.
Sine and cosine rules. Two relations that hold in any triangle, not only a right-angled one. The sine rule links each side to the sine of the angle opposite it. The cosine rule gives one side when the other two and the angle between them are known.
These two are vector addition written another way. The cosine rule is the formula for the magnitude of a resultant, with the sign of the cosine term settled by which angle you choose to name. Because you meet them here, §9 has nothing new to prove when it comes to add two vectors.
Two forces of 3 N and 4 N act at a point with an angle of 60° between them. Find the magnitude of the resultant and the angle it makes with the 3 N force.
Modelling decision. Place the two forces head to tail; the resultant closes the triangle. The angle inside that triangle is 180° − 60° = 120°, which is where the sign in the cosine rule comes from. Getting that angle wrong is the usual error.
With θ = 60° between the forces, R² = P² + Q² + 2PQ cos θ.
R² = 9 + 16 + 2(3)(4)(0.5) = 25 + 12 = 37.
R = √37 = 6.08 N.
For the direction, tan α = Q sin θ ⁄ (P + Q cos θ)
= 4(0.866) ⁄ (3 + 4 × 0.5) = 3.464 ⁄ 5 = 0.693.
α = 34.7° from the 3 N force.
Remark. Check the answer for sense. The resultant must lie between 1 N and 7 N and must lean towards the larger force. 6.08 N at 34.7° from the 3 N force, and so 25.3° from the 4 N force, satisfies both.
Three forces acting at a point hold it in equilibrium. Two of them are 8 N and 6 N with an angle of 90° between them. Find the third force and the angles it makes with the other two.
Modelling decision. Equilibrium means the three close a triangle, so the third force is equal and opposite to the resultant of the other two. Find that resultant, then reverse it.
The resultant of the perpendicular pair is √(8² + 6²) = 10 N.
It makes an angle tan α = 6 ⁄ 8 with the 8 N force, so α = 36.9°.
The third force is 10 N, opposite to that resultant.
So it makes 180° − 36.9° = 143.1° with the 8 N force,
and 180° − 53.1° = 126.9° with the 6 N force.
Remark. The three angles add to 360°, as they must. Check the answer against Lami's theorem, which is the sine rule applied to this triangle: each force divided by the sine of the angle between the other two is the same number. Here 8 ⁄ sin 126.9° = 6 ⁄ sin 143.1° = 10 ⁄ sin 90° = 10.
Everything up to here you have met in school. Calculus is where physics begins to be written properly. A law of physics almost never says what a quantity is; it says how fast that quantity changes. Velocity is how fast position changes. Acceleration is how fast velocity changes. Force is how fast momentum changes. Current is how fast charge flows. Power is how fast work is done. Induced emf is how fast flux changes. Six laws, five different chapters, one idea. That idea is the derivative.
COREWHEN THIS COMES UP “How fast”, “rate of”, “per second”, “at the instant”, or a graph whose steepness is what the question wants.
Derivative. The rate at which one quantity changes when another changes. If y depends on x, take a small change Δx, see how much y changes, and form the ratio Δy ⁄ Δx. The value this ratio settles down to as Δx is made smaller and smaller is the derivative, written dy ⁄ dx.
Look at the graph of y against x. Pick two points P and Q on the curve and join them by a straight line. That line is a chord, and its slope is Δy ⁄ Δx — the average rate of change between the two points. Now slide Q towards P. The chord turns, and as Q arrives at P the chord becomes the tangent at P. The slope of that tangent is the derivative: the rate of change at the point P itself.
So one symbol carries two pictures, and you should be able to switch between them without effort. Geometrically, dy ⁄ dx is a slope. Physically, dy ⁄ dx is a rate. When you are reading a graph you want the first picture. When you are reading a law you want the second.
The two names physics uses most are
and the notation d²x ⁄ dt² is only shorthand for differentiating twice. It is not a square of anything.
Average rate and instantaneous rate are different quantities, and a question that says “average” is not asking for a derivative. Average velocity is total displacement divided by total time; the velocity at an instant is dx ⁄ dt. The two carry the same number at every instant only when the velocity is constant. In any smooth motion there is at least one instant inside the interval where they happen to be equal — Example 1.32 has one at t = 1 s — but you cannot find that instant without working out the average first, so it is no help.
At any one instant, a quantity and its rate of change say nothing about each other. At the top of its flight a stone has v = 0 while a = g, and at the far end of a swing a pendulum bob has its displacement greatest while its velocity is zero. Over a whole interval it is different — a velocity that stays at zero does force the acceleration to be zero — but a single instant carries no such information, and a question about one instant is the usual case.
The position of a particle moving along a straight line is x = 5t² − 3t + 2, where x is in metres and t in seconds. Find its velocity at t = 2 s, and its average velocity over the first two seconds. Why do the two differ?
Modelling decision. Read which one the question wants. The velocity “at” an instant is the derivative; the “average” velocity over an interval is displacement divided by time. Compute both and compare.
v = dx ⁄ dt = 10t − 3.
At t = 2 s: v = 20 − 3 = 17 m s⁻¹.
For the average, x(0) = 2 m and x(2) = 20 − 6 + 2 = 16 m.
Average velocity = (16 − 2) ⁄ 2 = 7 m s⁻¹.
They differ because the velocity varies: it rises from −3 m s⁻¹ at t = 0 to 17 m s⁻¹ at t = 2 s.
Remark. Notice that the average, 7 m s⁻¹, is exactly the mean of the two end velocities, (−3 + 17) ⁄ 2. That is guaranteed only when the acceleration is constant, as it is here. Do not carry the shortcut into a problem where the acceleration varies: it may happen to hold there, but nothing makes it hold.
A particle moves so that x = t³ − 6t² + 9t metres. Find the instants at which it is momentarily at rest, and its acceleration at each of them.
Modelling decision. “At rest” is a statement about the derivative, not about the position. Set v = 0 and solve. The acceleration at those instants tells you what the particle does next.
v = dx ⁄ dt = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3).
v = 0 at t = 1 s and t = 3 s.
a = dv ⁄ dt = 6t − 12.
At t = 1 s, a = −6 m s⁻²; at t = 3 s, a = +6 m s⁻².
Remark. The two signs tell the story. At t = 1 s the particle is at rest and being pushed backwards, so it turns round and moves in the negative direction. At t = 3 s it is at rest and being pushed forwards, so it turns round again. Between them it travels backwards. A question asking for distance rather than displacement over 0 to 4 s must be split at these two instants.
END OF THE FREE PART
The free part stops where calculus turns into technique. Still in this one chapter:
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