Chapter 01 · Mechanics · Full book

The Language of Physics: Calculus

The question that calculus answers is not, at root, a question about physics. It is a question about mathematics: what does it mean for a quantity to change smoothly, and what structure does that smoothness impose?

The physical world need not be smooth. A particle's position might, in principle, jump discontinuously; velocity might be undefined; force might exist nowhere. Classical physics proceeds on the opposite assumption—that position, velocity, force, and every mechanical quantity are smooth enough that derivatives exist. This commitment, called differentiability, is one of the most powerful and least-examined hypotheses a physicist makes. It is not a theorem derivable from first principles. It is a working assumption about nature, testable at every scale. At the quantum scale, as we shall see much later, its limits begin to show.

Newton's twenty-year delay in publishing the theory of universal gravity is instructive here—not as biographical detail, but as epistemological necessity. He had the inverse-square force law in 1666. What he could not yet prove was that the law applied: that the Earth, composed of infinitely many particles each pulling an apple from a different distance and direction, behaves gravitationally as if all its mass were concentrated at a single central point. To prove this, he needed to sum an infinite assembly of contributions from a continuously distributed body—to integrate over a sphere. The mathematics of his time had no such tool.

So Newton invented one. He called it the method of fluxions. Leibniz, arriving at the same structure independently, called it the differential calculus. Both names point to the same discovery: that smooth functions have a precise local structure, and that this local structure accumulates globally in a way that can be computed exactly.

The entire subject organises itself around two operations:

Differentiation: Given a smooth function, extract its instantaneous rate of change at each point.
Integration: Given a rate of change, reconstruct the accumulated total—or equivalently, sum a continuously varying quantity over an interval.

That these two operations are exact inverses of each other is the content of the Fundamental Theorem of Calculus. It is a genuine mathematical theorem—not at all obvious—and Newton and Leibniz's recognition of it is what unified a scattered collection of geometric techniques into a single coherent theory. We shall see, in Section 3, precisely why it holds.

Core intuition: Chapter Roadmap

Differentiation — The microscope: zoom in on a smooth curve until it looks straight, then read the slope. This gives instantaneous velocity from position, and instantaneous acceleration from velocity.
Maxima and minima — Where the derivative is zero, the quantity is momentarily stationary. The second derivative distinguishes peaks from troughs.
Integration — The scanner: slice the area under a curve into infinitely thin strips and sum them. This gives displacement from velocity, and velocity from acceleration.
SUVAT from first principles — All four kinematic equations emerge from integrating \(a = \text{const}\) twice. There is nothing to memorise; there is one fact and two integrations.
Examiner traps — The ways examiners exploit calculus ideas to mislead: framing tricks, sign errors, graph misreading, and dimensional analysis failures.

Perplexing Questions

Perplexing questions

Before we touch a single formula, sit with these questions. Each one exposes something that algebra alone cannot answer. You will meet the explicit resolution at the point in this chapter where the relevant idea is developed.

Speed in a frozen instant. A speed camera snaps a single photograph of a car and prints 82 km/h. But a photograph freezes one instant: in that frozen moment the car covers no distance, and no time passes. Speed is distance divided by time—so the speed in the photo is \(0 \div 0\), which means nothing at all. How can a car have one exact speed at an instant in which it goes nowhere in no time?
Still in the air, but not at rest. Throw a ball straight up. At the very top it stops—for one instant its speed is exactly zero, just like a ball lying on a table. The ball on the table is perfectly balanced; nothing is trying to move it. Is the ball at the top of its flight in that same do-nothing state? If not, what is different about being momentarily still in mid-air versus being still on a table?
The odometer's impossible total. On a long drive the speedometer needle never holds still for even a second—it wanders up and down the whole way. Yet your odometer ends on exactly \(134\;\mathrm{km}\), one clean number. If you never actually held a single definite speed, how can a forever-wavering needle add up to one precise distance? What, exactly, are you adding up?
The brakes' cruel arithmetic. At \(20\;\mathrm{m/s}\) a car needs about \(40\;\mathrm{m}\) to stop. Now double the speed to \(40\;\mathrm{m/s}\) and guess the new stopping distance before reading on. Almost everyone says “twice as far—\(80\;\mathrm{m}\).” The true answer is \(160\;\mathrm{m}\): four times the distance for only twice the speed. Where does that extra factor of two come from?
The bend that means “push.” A subway train pulls away from a platform. Picture its distance from the platform plotted against time: at first the line scoops upward, curving; once the train reaches cruising speed the curve straightens into a slanted line. The straight stretch is obvious—steady speed. But what is the curved part telling you? Just from how sharply it bends, could you say how hard the train is being pushed, without measuring a single number?

Every one of these questions is answered explicitly before this chapter ends.
Look for the ↳ resolves PQ N badge as you read.

Differentiation: The Microscope

The derivative of a function \(f\) at a point \(t_0\) is, when it exists, the unique number to which the ratio \[ \frac{f(t_0 + \Delta t) - f(t_0)}{\Delta t} \] converges as \(\Delta t \to 0\). Several aspects of this definition deserve attention before we compute anything.

The limit may fail to exist. At a sharp corner—consider \(|t|\) at \(t = 0\)—the ratio approaches different values depending on the sign of \(\Delta t\), and no single derivative can be assigned. At a discontinuity the numerator does not vanish and the ratio diverges. Differentiability is a genuinely restrictive condition: most functions, in the sense of abstract mathematics, are nowhere differentiable. The smooth functions of classical mechanics are, from this perspective, extraordinarily special. Their smoothness is a hypothesis about nature, not a mathematical entitlement.

Where the derivative does exist, the geometric picture is compelling. Zoom in far enough on a smooth curve at \(t_0\), and the curve becomes locally straight—indistinguishable from its tangent line. The derivative is the slope of that tangent, and that slope is what the limit computes.

Secant to tangent. As the second point \(Q\) slides toward \(P\) (gray lines), the secant slope \(\Delta x/\Delta t\) converges to the slope of the tangent at \(P\). That limiting slope is the derivative \(dx/dt\).

Fig 0.1 Local linearity: zoom in on a smooth curve at \(t_0\) and it becomes its own tangent line — the derivative is that slope.

Core intuition ↳ resolves PQ 1 Speed at a Single Instant

The paradox dissolves the moment you stop trying to divide \(0\) by \(0\). You never evaluate the ratio at the instant—you watch it approach the instant. Take the distance the car covers over a short interval \(\Delta t\) straddling the moment in question and form \(\Delta x/\Delta t\): an honest average speed over a real stretch of road. Now shrink the interval—\(0.1\,\mathrm{s}\), then \(0.01\,\mathrm{s}\), then \(0.001\,\mathrm{s}\). Both the distance and the time march toward zero, but their ratio does not collapse; it homes in on one definite number. That number is the speed “at the instant.” The camera never computes \(0\div 0\)—it reports the value the shrinking averages converge to. Geometrically, it is the secant slope settling onto the tangent slope. The frozen instant has a speed not because anything moves during it, but because the motion immediately around it points unambiguously at a single value.

The Microscope. Zoom in on a smooth curve and it becomes locally straight. The slope of this tiny straight piece is the derivative.

Fig 0.2 As \(\Delta t \to 0\) the secant through two nearby points collapses onto the tangent, so \(\Delta x/\Delta t \to dx/dt\).

Core intuition: From \(\Delta\) to \(d\): Notation as Theorem

The symbol \(\Delta t\) denotes a finite, measurable time interval. The symbol \(dt\) denotes its limiting form as that interval shrinks toward zero—not zero itself, but the direction it approaches. The instantaneous velocity is \[ v = \frac{dx}{dt}, \] which is read as “the derivative of \(x\) with respect to \(t\),” and means precisely: the limit of \(\Delta x / \Delta t\) as \(\Delta t \to 0\). Leibniz chose this notation deliberately. It looks like a fraction because it is designed to behave like one: one writes \(dx = v\,dt\), separates variables, integrates both sides. This algebraic license is not a convenient fiction—it is a theorem. Every such manipulation is underwritten by the chain rule, which guarantees that the limit operations are mutually consistent. Know that \(dx/dt\) is strictly a limit of ratios; use it freely as a ratio; the chain rule covers every case.

Misconception alert: \(\dfrac{dx}{dt}\) Is Not a Fraction… But You Can Often Treat It Like One

Beginners ask: “Can I just cancel the \(dt\)?” Strictly, \(\dfrac{dx}{dt}\) is a single symbol—the limit of a ratio as the denominator vanishes. It is not literally a numerator divided by a denominator. However, in physics we routinely rearrange it as though it were a fraction: writing \(dx = v\,dt\) and then integrating both sides. This works because Leibniz notation was designed to behave this way—every such rearrangement is quietly justified by the chain rule. JEE/NEET rule of thumb: You are safe to “separate the \(d\)'s” in any equation where both sides are being integrated. If an examiner writes \(a = v\,\dfrac{dv}{dx}\), you are expected to rewrite it as \(a\,dx = v\,dv\) and integrate. Treat \(dx\) and \(dv\) as manipulable quantities—but know that the rigorous backing is the chain rule, not ordinary algebra.

Two notational systems divide the physics literature, and fluency in both is non-negotiable. Leibniz notation makes the variable of differentiation explicit: \[ v = \frac{dx}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2x}{dt^2}. \] Newton's dot notation records each time derivative with a superposed dot: \[ \dot{x} \equiv \frac{dx}{dt} \quad (\text{velocity}), \qquad \ddot{x} \equiv \frac{d^2x}{dt^2} \quad (\text{acceleration}). \] The mathematical content is identical; the emphasis differs. Leibniz notation keeps the independent variable visible and facilitates the chain-rule manipulations that dominate mechanics. Dot notation is compact and is standard in analytical mechanics, control theory, and every treatment that inherits the Principia's conventions. This book uses Leibniz notation in derivations, but both notations appear in examples because both will appear in every subsequent text and paper you encounter.

Discovering the Power Rule

The definition of the derivative tells us what to compute—form the ratio \((f(t + \Delta t) - f(t)) / \Delta t\) and take its limit. It says nothing about the answer. For each function, one must carry out the calculation. Newton worked through this for the power functions \(t, t^2, t^3, \ldots\) and found that the results obeyed a single pattern. We follow exactly the same path.

Case 1: \(x = At\) (uniform motion).

The position changes from \(At\) to \(A(t + \Delta t)\) in time \(\Delta t\). The ratio is: \[ \frac{\Delta x}{\Delta t} = \frac{A(t + \Delta t) - At}{\Delta t} = \frac{A\,\Delta t}{\Delta t} = A. \] The \(\Delta t\) cancels completely. As \(\Delta t \to 0\), nothing changes: \(\dfrac{dx}{dt} = A\). Constant velocity for uniform motion. No surprise.

Case 2: \(x = At^2\).

Now something interesting happens: \[\begin{aligned} \Delta x &= A(t + \Delta t)^2 - At^2 \\ &= A\bigl(t^2 + 2t\,\Delta t + \Delta t^2\bigr) - At^2 \\ &= 2At\,\Delta t + A\,\Delta t^2. \end{aligned}\] Divide by \(\Delta t\): \[ \frac{\Delta x}{\Delta t} = 2At + A\,\Delta t. \] As \(\Delta t \to 0\), the second term vanishes. What remains: \(\dfrac{dx}{dt} = 2At.\)

Case 3: \(x = At^3\).

The binomial expansion of \((t + \Delta t)^3\) gives \(t^3 + 3t^2\Delta t + 3t\,\Delta t^2 + \Delta t^3\), so: \[ \frac{\Delta x}{\Delta t} = 3At^2 + 3At\,\Delta t + A\,\Delta t^2 \;\xrightarrow{\;\Delta t \to 0\;}\; 3At^2. \]

The pattern is now unmistakable. Setting \(A = 1\) and reading down the column: \[ \begin{array}{rcl} x = t^1 & \;\xrightarrow{\;d/dt\;} & 1\cdot t^{\,0} \\[6pt] x = t^2 & \;\xrightarrow{\;d/dt\;} & 2\cdot t^{\,1} \\[6pt] x = t^3 & \;\xrightarrow{\;d/dt\;} & 3\cdot t^{\,2} \\[6pt] x = t^n & \;\xrightarrow{\;d/dt\;} & n\cdot t^{\,n-1} \end{array} \]

The pattern announces itself: multiply by the exponent, then lower it by one. Carrying the same \(\Delta t\) calculation through for negative and fractional exponents—the binomial expansion handles these equally—yields the same rule for all real \(n\).

\[ \boxed{\frac{d}{dx}(x^n) = n\,x^{n-1}} \]

This is the Power Rule. You could reconstruct it at any moment from the three cases above. Knowing where it comes from is knowing that it cannot be otherwise.

One exception is worth registering immediately. The integration analogue of the power rule—recovering \(x^n\) from its derivative—works for all \(n\) except \(n = -1\). The function \(1/x\) has no power-function antiderivative; its antiderivative is \(\ln|x|\), the natural logarithm. This is not an inconvenient accident—it reflects the deep geometric connection between the reciprocal function and the exponential, a connection that resurfaces in every chapter from oscillations onward.

Example 1 · Applying the Power Rule

A particle's position is \(x(t) = t^3\). Find its velocity and acceleration.

Solution. 1. Velocity \(v(t)\): Differentiate \(x(t)\) with respect to \(t\). \[ v = \frac{dx}{dt} = \frac{d}{dt}(t^3) = 3 t^{3-1} = 3t^2. \] 2. Acceleration \(a(t)\): Differentiate \(v(t)\) with respect to \(t\). \[ a = \frac{dv}{dt} = \frac{d}{dt}(3t^2) = 3 \cdot 2 t^{2-1} = 6t. \] We have climbed down the ladder: \[ \text{Position } x(t) \;\to\; \text{Velocity } v(t) \;\to\; \text{Acceleration } a(t). \] In Newton's notation: \(\dot{x} = 3t^2\), \(\ddot{x} = 6t\). Check: Climb back up the ladder—integration must undo differentiation: \(\int a\,dt = \int 6t\,dt = 3t^2 = v\) ✓, and \(\int v\,dt = \int 3t^2\,dt = t^3 = x\) ✓.

Core intuition: The Sign of the Derivative Tells the Story

The value of the derivative tells you the slope. But in physics, even its sign carries deep meaning:

\(\dfrac{dx}{dt} \gt 0\): position is increasing — object moves in the positive direction.
\(\dfrac{dx}{dt} \lt 0\): position is decreasing — object moves in the negative direction.
\(\dfrac{dx}{dt} = 0\): position is momentarily not changing — object is at rest at that instant.

The same logic applies one level deeper: if \(\dfrac{dv}{dt} \gt 0\), velocity is increasing (which does not always mean the object is speeding up—it depends on the sign of \(v\) itself). If \(\dfrac{dv}{dt} \lt 0\), velocity is decreasing. Reading the sign of each derivative is often more important than computing its exact value—especially in qualitative JEE questions where you must determine the direction of motion from a graph without any algebra.

Core intuition ↳ resolves PQ 2 Zero Velocity with Nonzero Acceleration

Velocity is the first derivative of position; acceleration is the second. At the top of a throw, \(v = dx/dt = 0\) (the \(x\)-\(t\) slope is horizontal at that instant), but \(a = dv/dt = -g \neq 0\) (the \(v\)-\(t\) slope is constant and nonzero throughout). Two successive derivatives of the same function are independent at any given instant. Velocity being zero says nothing about how fast velocity itself is changing. The contrast with the resting ball is the whole point: on the table the velocity is zero and staying zero, so \(a = 0\)—gravity is cancelled by the table's upward push. At the top of the flight the velocity is zero but not staying zero; it is being turned from upward to downward at the full rate \(g\), because nothing there balances gravity. Same zero speed, completely different state of motion.

The power rule covers polynomial functions. For completeness, here is the full toolkit you will need. You do not need to memorise the trigonometric and exponential entries now; they become essential in later chapters on SHM, waves, and oscillations.

Function \(f(x)\)	Derivative \(f'(x)\)
\(x^n\)	\(nx^{n-1}\)
\(k\) (constant)	\(0\)
\(kx\)	\(k\)
\(\sin x\)	\(\cos x\)
\(\cos x\)	\(-\sin x\)
\(e^x\)	\(e^x\)
\(\ln x\)	\(\dfrac{1}{x}\)

Real physical signals are rarely a single entry from that table—they are products and compositions of them. Two rules handle every such case. The product rule, \(\dfrac{d}{dt}(fg) = f'g + fg'\), and the chain rule, \(\dfrac{d}{dt}f(g(t)) = f'(g)\,g'(t)\). The next example uses both at once on the function you will meet again in the chapter on damped oscillations.

Example 2 · Differentiating a Damped Oscillation (Chain + Product Rule)

A lightly damped mass on a spring has displacement \(x(t) = x_0\,e^{-t/\tau}\cos(\omega t)\), where \(x_0\), \(\tau\), and \(\omega\) are positive constants. Find the velocity \(v(t)\), evaluate \(v(0)\), and interpret it.

Solution. Given: \(x(t) = x_0\,e^{-t/\tau}\cos(\omega t)\).
Find: \(v = dx/dt\), then \(v(0)\).
Setup: \(x\) is a product of two functions, \(e^{-t/\tau}\) and \(\cos(\omega t)\), each of which is itself a composition (an outer function of an inner linear argument). So we apply the product rule, using the chain rule on each factor: \(\dfrac{d}{dt}e^{-t/\tau} = -\tfrac{1}{\tau}e^{-t/\tau}\) and \(\dfrac{d}{dt}\cos(\omega t) = -\omega\sin(\omega t)\).
Solve: \[\begin{aligned} v &= x_0\left[\frac{d}{dt}\!\left(e^{-t/\tau}\right)\cos(\omega t) + e^{-t/\tau}\,\frac{d}{dt}\cos(\omega t)\right] \\ &= x_0\left[-\tfrac{1}{\tau}e^{-t/\tau}\cos(\omega t) - \omega\,e^{-t/\tau}\sin(\omega t)\right] \\ &= -\frac{x_0}{\tau}\,e^{-t/\tau}\big[\cos(\omega t) + \omega\tau\sin(\omega t)\big]. \end{aligned}\] At \(t = 0\): \(v(0) = -\dfrac{x_0}{\tau}\big[\cos 0 + \omega\tau\sin 0\big] = -\dfrac{x_0}{\tau}\).
Answer: \(\boxed{v(t) = -\dfrac{x_0}{\tau}\,e^{-t/\tau}\big[\cos(\omega t) + \omega\tau\sin(\omega t)\big]}\), and \(v(0) = -x_0/\tau\).
Interpretation: the mass starts at its maximum displacement \(x_0\) (where an un-damped oscillator would have \(v=0\)), but because the amplitude is already decaying, it is drifting back at speed \(x_0/\tau\).
Check: Dimensions: \([x_0/\tau] = \mathrm{m/s}\) ✓. Limiting case: with no oscillation (\(\omega \to 0\)), \(x \to x_0 e^{-t/\tau}\) and the formula collapses to \(v = -\tfrac{x_0}{\tau}e^{-t/\tau}\), exactly the derivative of a pure exponential decay ✓.

Maxima and Minima: The Peak of the Mountain

A function's derivative being zero at a point says precisely this: the function is, to first order, neither increasing nor decreasing there. The tangent is horizontal. This is a necessary condition for a local maximum or minimum—but it is not sufficient. The derivative may vanish at a point that is neither. Consider \(f(t) = t^3\) at \(t = 0\): \(f'(0) = 0\), yet the function passes straight through the origin, increasing on both sides. Such a point—where the tangent is horizontal but the function does not reverse—is called an inflection point.

The logical direction is exact: if \(f\) attains a local extremum at \(t_0\), then \(f'(t_0) = 0\). The converse fails. Solving \(f'(t_0) = 0\) produces candidates; distinguishing genuine extrema from inflection points requires further analysis—the second derivative.

Example 3 · Finding Maximum Height

A ball's height is given by \[ h(t) = 20t - 5t^2 \quad (\text{in meters, } t \text{ in seconds}). \] Find the time when it reaches maximum height.

Solution. At maximum height, the rate of change of height (i.e. the vertical velocity) is zero. \[ v(t) = \frac{dh}{dt} = \frac{d}{dt}(20t - 5t^2). \] Using the power rule: \[ v(t) = 20(1) - 5 \cdot 2t = 20 - 10t. \] Set the derivative to zero: \[ 20 - 10t = 0 \;\Rightarrow\; 10t = 20 \;\Rightarrow\; t = 2\,\mathrm{s}. \] So, the ball reaches its maximum height at \(\boxed{t = 2~\text{s}}\). Check: The vertical velocity must vanish exactly at the peak: \(v(2) = 20 - 10(2) = 0\) ✓. Reading \(v_0 = 20\,\mathrm{m/s}\) and \(g = 10\,\mathrm{m/s^2}\) off \(h(t) = 20t - 5t^2\), this matches the kinematic result \(t = v_0/g = 20/10 = 2\,\mathrm{s}\) ✓.

Second Derivative: Is it a Peak or a Valley?

Setting the first derivative to zero only tells you that the slope is horizontal. That could be:

a maximum (peak of a hill),
a minimum (bottom of a valley), or
a flat inflection point (neither max nor min).

To distinguish these, we look at the second derivative. The second derivative tells you how the slope itself is changing—is the curve bending upward or downward?

Core intuition: Second Derivative: Test and Physical Meaning

Let \(y = f(x)\) with \(f'(x_0) = 0\) (slope is flat). The second derivative tells you the shape:

\(f''(x_0) \lt 0\): curve bends downward (frown) \(\Rightarrow\) local maximum.
\(f''(x_0) \gt 0\): curve bends upward (smile) \(\Rightarrow\) local minimum.
\(f''(x_0) = 0\): test is inconclusive — examine the sign change of \(f''\) across \(x_0\), or sketch.

The physics reading: If \(x(t)\) is position, then \(x''(t) = a(t)\) is acceleration. Concavity of the \(x\)–\(t\) graph is acceleration. Concave up (\(x'' \gt 0\)) means the object is accelerating; concave down (\(x'' \lt 0\)) means it is decelerating; an inflection point (\(x'' = 0\)) marks where acceleration changes sign. On a JEE MCQ with a graph, reading concavity before writing any equation often eliminates three of four options in seconds.

Telling a peak from a trough. At both stationary points the tangent is horizontal (\(f'=0\)). The second derivative breaks the tie: concave-down (\(f''\lt 0\)) is a maximum, concave-up (\(f''\gt 0\)) is a minimum.

Fig 0.3 Concavity is acceleration: a position–time graph curving upward has \(d^{2}x/dt^{2} > 0\).

Core intuition ↳ resolves PQ 5 Concavity and Acceleration

A position–time graph that curves upward is concave up: \(d^2x/dt^2 \gt 0\), which means \(a \gt 0\) at every point on that curve. No algebra is needed; the shape of the curve encodes the sign of acceleration directly. Concave down \(\Rightarrow\) \(a \lt 0\); inflection point \(\Rightarrow\) \(a = 0\). For the departing train this answers the puzzle directly: the upward-scooping (concave) stretch is exactly where the train is being pushed—the sharper the bend, the harder the push—and the instant the curve straightens into a slanted line, the bending has stopped, so \(a = 0\) and the train has settled into steady cruising speed.

Concavity is acceleration. On a position–time graph the bending direction gives the sign of \(a\) at a glance: concave up (\(a\gt 0\)), straight (\(a=0\)), concave down (\(a\lt 0\)) — no algebra required.

Fig 0.4 At a projectile’s apex the height graph has zero slope but nonzero curvature — \(y'=0\) while \(y''=-g\).

In physics: if \(y(t)\) is the height of a projectile, then \(y''(t) = a(t)\) is the acceleration. At the apex, \(y'(t_0) = 0\) (vertical velocity momentarily zero) while \(y''(t_0) = -g \lt 0\) (gravity has not paused). The negative second derivative confirms a maximum. The concavity of the position–time graph is the acceleration, directly readable without a single computation.

Example 4 · Using Double Differentiation

Consider \(y(t) = 10t - 5t^2\). Find its critical point and use the second derivative to confirm whether it is a maximum or a minimum.

Solution. First derivative, set to zero to locate the critical point: \[ y'(t) = 10 - 10t = 0 \;\Rightarrow\; t = 1\,\mathrm{s}. \] Second derivative: \[ y''(t) = \frac{d}{dt}(10 - 10t) = -10. \] Since \(y''(t) = -10 \lt 0\) for all \(t\), the curve is concave down everywhere, so the critical point at \(t = 1\,\mathrm{s}\) is indeed a maximum. Check: Independent route—\(y(t) = t(10 - 5t)\) has roots at \(t = 0\) and \(t = 2\), and a parabola's vertex always sits at the midpoint of its roots, \(t = 1\,\mathrm{s}\) ✓. The negative leading coefficient (\(-5 \lt 0\)) means the parabola opens downward, so the vertex is a maximum—agreeing with \(y'' \lt 0\) ✓.

Misconception alert: Velocity Zero \(\neq\) Acceleration Zero {(Classic JEE/NEET Trap)

At the highest point of a ball thrown vertically upward, the velocity is zero. Many students conclude that the acceleration must also be zero at that instant. This is wrong. Acceleration is \(g = 9.8\,\mathrm{m/s^2}\) downward throughout the flight—at the bottom, the top, and everywhere in between. Gravity does not “pause” at the peak. The velocity is zero, but it is changing (from upward to downward) at the full rate \(g\). Mathematically: \(v = 0\) means \(\dfrac{dx}{dt} = 0\) at that instant, but \(\dfrac{dv}{dt} = -g \neq 0\). A function can be zero without its derivative being zero. The height function \(h(t) = 20t - 5t^2\) has \(h'(2) = 0\) but \(h''(2) = -10\), not zero. This error costs marks in almost every exam cycle. Whenever you see “\(v = 0\),” never assume “\(a = 0\).” Check \(\dfrac{dv}{dt}\) independently.

Example 5 · When the Second-Derivative Test Fails

Locate and classify every critical point of \(f(x) = x^4 - 4x^3\).

Solution. Given: \(f(x) = x^4 - 4x^3\).
Find: all critical points and their nature.
Setup: critical points solve \(f'(x)=0\); classify with \(f''\), but be ready for the inconclusive case \(f''=0\).
Solve: \[ f'(x) = 4x^3 - 12x^2 = 4x^2(x-3) = 0 \;\Rightarrow\; x = 0 \ \text{or}\ x = 3. \] \[ f''(x) = 12x^2 - 24x = 12x(x-2). \] At \(x=3\): \(f''(3) = 12(3)(1) = 36 \gt 0 \Rightarrow\) local minimum, with \(f(3) = 81 - 108 = -27\). At \(x=0\): \(f''(0) = 0\), so the test is inconclusive. Fall back to the sign of \(f'\) on either side: \(f'(x) = 4x^2(x-3)\), and since \(4x^2 \ge 0\) while \((x-3)\lt 0\) for all \(x\) near \(0\), \(f'\lt 0\) just left and just right of \(x=0\). The slope does not change sign, so \(x=0\) is not an extremum — it is a flat (horizontal) inflection point.
Answer: \(\boxed{x=3 \text{ is a local minimum } (f=-27); \ x=0 \text{ is a flat inflection, not an extremum.}}\)
Check: An independent test of the \(f''=0\) point: the lowest non-vanishing derivative at \(x=0\) is \(f'''(0) = (24x-24)\big|_{0} = -24 \neq 0\). A non-zero odd-order derivative confirms an inflection, not a turning point ✓. Trap: a student who stops at “\(f''(0)=0\), therefore it is an inflection” is right here only by luck — \(f''(c)=0\) alone proves nothing, as \(f(x)=x^4\) at \(x=0\) (a genuine minimum with \(f''=0\)) shows.

Example 6 · Optimising a Cost Built from Two Competing Terms

The energy a vehicle spends per unit distance at steady speed \(v\) is modelled as \(E(v) = \dfrac{A}{v} + Bv\), where \(A\gt 0\) collects speed-independent losses (paid over more time the slower you go) and \(B\gt 0\) collects aerodynamic drag (which grows with speed). Find the speed that minimises \(E\), the minimum value, and the relationship between the two terms at that speed.

Solution. Given: \(E(v) = A/v + Bv\), with \(A, B \gt 0\), \(v\gt 0\).
Find: the minimising speed \(v^*\), \(E(v^*)\), and the balance condition.
Setup: the two terms pull opposite ways — \(A/v\) falls as \(v\) rises, \(Bv\) grows. A minimum sits where they trade off, i.e. where \(dE/dv = 0\).
Solve: \[ \frac{dE}{dv} = -\frac{A}{v^2} + B = 0 \;\Rightarrow\; v^2 = \frac{A}{B} \;\Rightarrow\; v^* = \sqrt{\frac{A}{B}}. \] \[ E(v^*) = \frac{A}{\sqrt{A/B}} + B\sqrt{\frac{A}{B}} = \sqrt{AB} + \sqrt{AB} = 2\sqrt{AB}. \] Answer: \(\boxed{v^* = \sqrt{A/B}, \quad E_{\min} = 2\sqrt{AB}}\). At the optimum the two terms are equal: \(A/v^* = Bv^* = \sqrt{AB}\).
Check: Minimum confirmed: \(E''(v) = 2A/v^3 \gt 0\) for all \(v\gt 0\), so the curve is convex and \(v^*\) is a true minimum ✓. Dimensions: writing \([A] = \) (energy\(\cdot\)speed/distance) and \([B] = \) (energy/(distance\(\cdot\)speed)), the ratio \(A/B\) has units of speed\(^2\), so \(v^*\) is a speed ✓. General insight: for any cost of the form \(A/v + Bv\) the minimum lies where the competing terms balance — the same “equal-contribution” principle governs optimal range, impedance matching, and least-time paths.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Reading velocity and acceleration off a position function

A particle moves along a straight line with position \(x(t) = 2t^3 - 5t^2 + 3t\) (metres, \(t\) in seconds). Find its velocity and acceleration at \(t = 2\,\mathrm{s}\).

Solution. Given: \(x(t) = 2t^3 - 5t^2 + 3t\).
Find: \(v(2)\) and \(a(2)\).
Setup: Velocity is the first derivative of position, acceleration the second. Differentiate term by term with the power rule.
Solve: \[\begin{aligned} v(t) &= \frac{dx}{dt} = 6t^2 - 10t + 3, & v(2) &= 24 - 20 + 3 = 7\,\mathrm{m/s},\\ a(t) &= \frac{dv}{dt} = 12t - 10, & a(2) &= 24 - 10 = 14\,\mathrm{m/s^2}. \end{aligned}\] Answer: \(\boxed{v(2) = 7\,\mathrm{m/s},\quad a(2) = 14\,\mathrm{m/s^2}}\).
Check: Dimensions: differentiating metres once yields m/s, twice yields m/s\(^2\) ✓. Both are positive at \(t = 2\,\mathrm{s}\), so the particle is moving in the \(+x\) direction and speeding up there.

Example 2 · Displacement as the area under a \(v\)–\(t\) line

A cart starts at the origin and moves with velocity \(v(t) = 6t\) (m/s). How far does it travel in the first \(4\,\mathrm{s}\)?

Solution. Given: \(v(t) = 6t\), motion from \(t = 0\) to \(t = 4\,\mathrm{s}\).
Find: displacement \(\Delta x\).
Setup: Displacement is the integral of velocity over the interval—the area under the \(v\)–\(t\) graph.
Solve: \[ \Delta x = \int_0^4 6t\,dt = \big[\,3t^2\,\big]_0^4 = 48\,\mathrm{m}. \] Answer: \(\boxed{\Delta x = 48\,\mathrm{m}}\).
Check: The \(v\)–\(t\) graph is a straight line from \(0\) to \(v(4) = 24\,\mathrm{m/s}\), so the area is a triangle: \(\tfrac12 \times 4\,\mathrm{s} \times 24\,\mathrm{m/s} = 48\,\mathrm{m}\) ✓.

Example 3 · Differentiating a trigonometric position function

A particle oscillates with position \(x(t) = 5\sin 2t\) (metres, \(t\) in seconds). Find its velocity and acceleration at \(t = 0\), and its maximum speed.

Solution. Given: \(x(t) = 5\sin 2t\).
Find: \(v(0)\), \(a(0)\), and the maximum speed.
Setup: Differentiate using the chain rule, \(\frac{d}{dt}\sin(kt) = k\cos(kt)\). Velocity is \(x'\), acceleration is \(x''\).
Solve: \[ v(t) = 10\cos 2t, \qquad a(t) = -20\sin 2t. \] \[ v(0) = 10\,\mathrm{m/s}, \qquad a(0) = 0, \qquad v_{\max} = 10\,\mathrm{m/s}\ (\text{when } \cos 2t = \pm 1). \] Answer: \(\boxed{v(0) = 10\,\mathrm{m/s},\ a(0) = 0,\ v_{\max} = 10\,\mathrm{m/s}}\)
Check: Consistency: \(a = -20\sin 2t = -4(5\sin 2t) = -4x\), the signature \(a = -\omega^2 x\) of SHM with \(\omega = 2\), so \(v_{\max} = \omega A = 2(5) = 10\,\mathrm{m/s}\) ✓. Phase: at \(t = 0\) the particle is at the centre (\(x = 0\)), where speed is greatest and acceleration zero ✓.

Example 4 · Displacement from a definite integral

A particle moves with velocity \(v(t) = 3t^2 + 2\) (m/s). Find its displacement between \(t = 1\,\mathrm{s}\) and \(t = 3\,\mathrm{s}\).

Solution. Given: \(v(t) = 3t^2 + 2\).
Find: displacement over \([1,3]\,\mathrm{s}\).
Setup: Displacement is the definite integral of velocity, \(\Delta x = \int_{1}^{3} v\,dt\).
Solve: \[ \Delta x = \int_{1}^{3}(3t^2 + 2)\,dt = \big[\,t^3 + 2t\,\big]_{1}^{3} = (27 + 6) - (1 + 2) = 30\,\mathrm{m}. \] Answer: \(\boxed{\Delta x = 30\,\mathrm{m}}\)
Check: Differentiation undoes it: \(\frac{d}{dt}(t^3 + 2t) = 3t^2 + 2 = v(t)\) ✓. Bounding: on \([1,3]\) the speed runs from \(5\) to \(29\,\mathrm{m/s}\); a crude midpoint estimate \(\sim 17\,\mathrm{m/s}\times 2\,\mathrm{s} = 34\,\mathrm{m}\) is the right order ✓.

Example 5 · Average value of a velocity

A cart has velocity \(v(t) = 4t\) (m/s). Use the integral definition of the average to find its average velocity over the first \(5\,\mathrm{s}\).

Solution. Given: \(v(t) = 4t\), interval \([0,5]\,\mathrm{s}\).
Find: the average velocity \(\bar v\).
Setup: The average value of a function over \([a,b]\) is \(\bar v = \frac{1}{b-a}\int_a^b v\,dt\).
Solve: \[ \bar v = \frac{1}{5}\int_0^5 4t\,dt = \frac{1}{5}\big[\,2t^2\,\big]_0^5 = \frac{1}{5}(50) = 10\,\mathrm{m/s}. \] Answer: \(\boxed{\bar v = 10\,\mathrm{m/s}}\)
Check: Special case: velocity is linear, so the average equals \(\tfrac12(v_0 + v_5) = \tfrac12(0 + 20) = 10\,\mathrm{m/s}\) ✓. Dimensions: integral of \(\mathrm{m/s}\) over \(\mathrm{s}\), divided by \(\mathrm{s}\), gives \(\mathrm{m/s}\) ✓.