Mathematical Review & Physical World
Build the mathematical language used throughout physics: relationships, graphs, rates and accumulation. Physical World is retained as a foundation from your coaching index; the index is not itself an official NEET syllabus.
Understand the idea → see it move → solve a question. Extra theory is there when you need it.
Models, laws and sensible approximations
Physics describes observations with quantities that can be measured. A law summarises a reproducible relationship; a model keeps the features needed for a particular question. A particle model ignores the size of an object when its dimensions do not affect the motion being studied.
The same ball can be a particle in projectile motion and an extended body when its spin matters. State the system, reference frame and assumptions before choosing an equation. After calculating, check units, limiting cases and whether the magnitude is physically sensible.
A more detailed model is useful only when the neglected effect matters to the question.
A ball travels 30 m while its diameter is 0.06 m. Is a particle model reasonable for its trajectory?
Show solution
The diameter is 0.002 times the travel distance. If rotation and air drag are neglected, tracking the centre is reasonable. For rolling down a slope, rotational energy must also be included.
An Atwood machine with masses \(m_1\gt m_2\) has acceleration \(a=\dfrac{(m_1-m_2)g}{m_1+m_2}\). Check this result using two limiting cases.
Show solution
- Let \(m_2=0\). Then \(a=m_1g/m_1=g\). With nothing on the other side, \(m_1\) falls freely. This matches.
- Let \(m_1=m_2\). Then \(a=0\). Equal masses balance. This matches.
- Let \(m_1\gg m_2\). Then \(a\to g\), never more than \(g\). A pulley system cannot make a mass fall faster than free fall. This matches.
Answer: The formula passes every limiting check, so it is likely correct.
More solved problem patterns
A rope pulls a box across the floor. Name the fundamental force behind (a) the tension in the rope, (b) the friction on the box and (c) the weight of the box.
Show solution
- Tension is the pull between neighbouring atoms of the stretched rope. Atomic bonds are electric, so tension is electromagnetic.
- Friction comes from the interaction of atoms at the two surfaces. It is also electromagnetic.
- Weight is the pull of the Earth on the mass of the box. It is gravitational.
Answer: (a) electromagnetic (b) electromagnetic (c) gravitational
Assertion: The Earth can be treated as a particle when finding its orbit around the Sun. Reason: The Earth's radius is very small compared with its distance from the Sun.
Show solution
- Earth's radius is about \(6.4\times10^{6}\) m. The Earth–Sun distance is about \(1.5\times10^{11}\) m.
- The ratio is about \(4\times10^{-5}\), so the size has almost no effect on the orbit. The assertion is true.
- The reason is true and is exactly why the particle model works.
Answer: Both A and R are true, and R correctly explains A.
Go deeper · extra explanation and derivations
The four fundamental forces
Every force you meet in mechanics comes from one of four fundamental interactions. NCERT compares their strengths relative to the strong force.
| Force | Relative strength | Range | Acts between |
|---|---|---|---|
| Strong nuclear | 1 | Short, about \(10^{-15}\) m (nuclear size) | Nucleons and heavier elementary particles |
| Electromagnetic | \(10^{-2}\) | Infinite | Charged particles |
| Weak nuclear | \(10^{-13}\) | Very short, about \(10^{-16}\) m | Some elementary particles, for example electron and neutrino (beta decay) |
| Gravitational | \(10^{-39}\) | Infinite | All objects with mass |
Contact forces such as friction, normal reaction, tension and the spring force are all electromagnetic. They come from the electric forces between the atoms of the surfaces in contact. Gravity is the weakest force, yet it rules the motion of planets because large bodies are electrically neutral overall and gravity is always attractive.
Conservation laws and symmetry
Physics rests on a few conservation laws: energy, linear momentum, angular momentum and electric charge. Each one links to a symmetry of nature.
- Laws are the same at every place (homogeneity of space) → linear momentum is conserved.
- Laws are the same in every direction (isotropy of space) → angular momentum is conserved.
- Laws are the same at every time (homogeneity of time) → energy is conserved.
A conservation law tells you what stays fixed. It does not tell you how fast a process happens.
Checking a model before trusting it
After any calculation, run three quick checks.
- Units: both sides must have the same dimensions.
- Limiting cases: set one input to zero or to a very large value. The result must match a case you already know.
- Size: the magnitude must be sensible. A walking speed of 300 m/s signals an error.
A model is good when the effects it ignores are small compared with the effect being asked about.
- "Arrange the fundamental forces in increasing order of strength." Answer: gravitational < weak < electromagnetic < strong.
- "Which force is responsible for friction / tension / normal reaction?" Answer: electromagnetic.
- "Which force has the shortest range?" Answer: weak nuclear.
- Statement-type questions linking conservation laws with symmetries.
- "Can this body be treated as a particle?" Compare its size with the distance or scale of the motion.
Friction, tension and normal reaction are electromagnetic in origin, not "mechanical" or gravitational. Gravity only supplies the weight.
Weak nuclear force is far stronger than gravity (\(10^{-13}\) against \(10^{-39}\)). Its range is tiny, so it does not act over everyday distances.
Check the idea
Proportionality, powers and scaling
A proportionality states how one quantity changes when the others are fixed. If a quantity depends on a square, doubling the input multiplies the output by four. Inverse dependence means the output decreases when that input increases.
Write the complete dependence before taking a ratio: constants cancel only if they really remain unchanged. This is the basis of many NEET comparison questions in gravitation, elasticity, heat and oscillations.
Other useful formulas
An increase of 100% means doubling; it does not mean multiplication by 100.
For a fixed spring constant, a spring period varies as the square root of mass. What happens if mass becomes nine times larger?
Show solution
Using \(T\propto\sqrt m\), the period ratio is \(\sqrt9=3\). The mass increases ninefold but the period only triples.
Two bodies of masses in the ratio 1 : 4 have equal kinetic energies. Find the ratio of their momenta.
Show solution
- At fixed \(K\), \(p=\sqrt{2mK}\propto\sqrt m\).
- \(\dfrac{p_1}{p_2}=\sqrt{\dfrac{m_1}{m_2}}=\sqrt{\dfrac14}=\dfrac12\).
- The heavier body has more momentum even though both have the same energy.
Answer: 1 : 2
More solved problem patterns
The kinetic energy of a body increases by 44%. By what percentage does its momentum increase?
Show solution
- Mass is unchanged, so \(p\propto\sqrt K\).
- \(K_2=1.44K_1\Rightarrow p_2=\sqrt{1.44}\,p_1=1.2\,p_1\).
- The momentum rises by 20%. The small-change rule (half of 44% = 22%) is not accurate for such a large change.
Answer: 20%
For planets, \(T^2\propto r^3\). A student plots \(\log T\) against \(\log r\). What are the shape and slope of the graph?
Show solution
- Write \(T=Cr^{3/2}\).
- Take logs: \(\log T=\tfrac32\log r+\log C\).
- This is \(y=mx+c\) with \(y=\log T\) and \(x=\log r\).
Answer: A straight line of slope 3/2 with intercept \(\log C\).
Go deeper · extra explanation and derivations
Small changes: the percentage rule
If \(Q\propto x^a y^b\) and each input changes by a small percentage, then
\[\%\ \text{change in } Q \approx a\,(\%\ \text{change in } x) + b\,(\%\ \text{change in } y).\]This comes from the binomial approximation \((1+\varepsilon)^a\approx1+a\varepsilon\). It works well below about 5%. For large changes, use the exact factor. For example, a 10% rise in radius raises area by exactly \(1.1^2-1=21\%\), close to the approximate 20%. A 50% rise gives exactly 125%, far from the approximate 100%.
Read what is held constant
The same two quantities can scale differently depending on what is fixed. Kinetic energy and momentum are linked by \(K=\dfrac{p^2}{2m}\).
- Same momentum: \(K\propto \dfrac1m\). The lighter body has more kinetic energy.
- Same kinetic energy: \(p=\sqrt{2mK}\propto\sqrt m\). The heavier body has more momentum.
- Same mass: \(K\propto p^2\). If \(p\) doubles, \(K\) becomes four times.
Before taking a ratio, write the formula in terms of the quantities that change, with the fixed ones gathered into the constant.
Graph shapes reveal the power
- \(y\propto x\): straight line through the origin.
- \(y\propto x^2\): parabola. Plot \(y\) against \(x^2\) to get a straight line.
- \(y\propto 1/x\): rectangular hyperbola. Plot \(y\) against \(1/x\) to get a straight line.
- \(y=Cx^n\): plot \(\log y\) against \(\log x\). The line has slope \(n\) and intercept \(\log C\).
- "If the momentum increases by 50%, the kinetic energy increases by …" (exact factor needed).
- "Two bodies of masses in the ratio 1 : 4 have equal kinetic energies. Ratio of their momenta?"
- "A quantity varies as \(x^3/y^2\). Errors / changes of 1% and 2% are made …" (small-change rule).
- Graph-match questions: which plot gives a straight line for a given law.
Percentage changes multiply, they do not add. If \(x\) rises 20% and \(y\) rises 20% in \(Q=xy\), the exact change is \(1.2\times1.2=1.44\), a 44% rise, not 40%.
When comparing kinetic energy and momentum of two bodies, check whether the question fixes \(p\), \(K\) or \(m\). The answer reverses between "same momentum" and "same energy".
Check the idea
Trigonometry and small angles
Sine, cosine and tangent connect a direction with perpendicular components. Draw the right triangle and identify the side opposite the chosen angle. Changing which axis the angle is measured from exchanges sine and cosine.
Angles in calculus and small-angle approximations must be in radians. The approximation sin θ ≈ θ comes from the near-straight behaviour of the sine curve at zero; it is useful for pendulums but not for large swings.
Other useful formulas
Do not substitute a degree value directly for θ in sin θ ≈ θ.
An incline rises 3 m for every 4 m horizontally. Find sin θ and cos θ.
Show solution
The sloping length is 5 m by Pythagoras. Thus sin θ = 3/5 and cos θ = 4/5. The slope angle is measured from the horizontal.
An incline has \(\tan\theta=3/4\). Find \(\sin2\theta\) and \(\cos2\theta\).
Show solution
- From the 3–4–5 triangle, \(\sin\theta=3/5\) and \(\cos\theta=4/5\).
- \(\sin2\theta=2\times\tfrac35\times\tfrac45=\tfrac{24}{25}\).
- \(\cos2\theta=\cos^2\theta-\sin^2\theta=\tfrac{16}{25}-\tfrac{9}{25}=\tfrac{7}{25}\).
Answer: \(\sin2\theta=24/25\), \(\cos2\theta=7/25\)
More solved problem patterns
A 10 N force acts at 120° to the positive x-axis. Find its components.
Show solution
- \(F_x=10\cos120^\circ=10\times(-\tfrac12)=-5\) N. The cosine is negative in the second quadrant.
- \(F_y=10\sin120^\circ=10\times\tfrac{\sqrt3}{2}\approx8.66\) N.
- Check: \(\sqrt{5^2+8.66^2}=10\) N.
Answer: \(F_x=-5\) N, \(F_y\approx8.66\) N
A 1.0 m pendulum swings to 0.1 rad from the vertical. By how much does the bob rise?
Show solution
- The rise is \(h=\ell(1-\cos\theta)\).
- With \(\cos\theta\approx1-\theta^2/2\), \(h\approx\ell\theta^2/2=1.0\times0.01/2=0.005\) m.
- The exact value is 0.004996 m, so the approximation is excellent.
Answer: About 5 mm
Go deeper · extra explanation and derivations
Values and identities you must know
| θ | 0° | 30° | 37° | 45° | 53° | 60° | 90° |
|---|---|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 3/5 | 1/√2 | 4/5 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 4/5 | 1/√2 | 3/5 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 3/4 | 1 | 4/3 | √3 | not defined |
The 37° and 53° values come from the 3–4–5 triangle and are approximations that coaching tests use freely. Key identities:
- \(\sin^2\theta+\cos^2\theta=1\)
- \(\sin2\theta=2\sin\theta\cos\theta\) (used in projectile range)
- \(\cos2\theta=1-2\sin^2\theta=2\cos^2\theta-1\)
- \(\sin(90^\circ-\theta)=\cos\theta\), \(\sin(180^\circ-\theta)=\sin\theta\), \(\cos(180^\circ-\theta)=-\cos\theta\)
Signs by quadrant: all positive in the first, only sine in the second, only tangent in the third, only cosine in the fourth.
Radians and small angles
An angle in radians is arc length divided by radius: \(\theta=s/r\). A full turn is \(2\pi\) rad, so \(1\ \text{rad}\approx57.3^\circ\).
For a small angle the arc and the chord are nearly equal, so \(\sin\theta\approx\tan\theta\approx\theta\). The cosine is flat at zero, so \(\cos\theta\approx1-\theta^2/2\). At 10° (0.1745 rad), \(\sin\theta=0.1736\): the error is only 0.5%. These approximations drive the simple pendulum, parallax and the angular size of distant objects.
- Direct value questions: \(\sin37^\circ\), \(\cos120^\circ\), \(\sin2\theta\) given \(\tan\theta\).
- Components of a force or velocity at an angle above 90°.
- Small-angle estimates: rise of a pendulum bob, angular size, \(\sin\theta\approx\theta\).
- Match-the-column on standard values.
\(\tan37^\circ=0.754\), not exactly 3/4. Use 3/5 and 4/5 only when the question says so or the options are clearly rounded.
\(\cos(180^\circ-\theta)=-\cos\theta\). A force at 120° to the x-axis has a negative x-component.
Check the idea
Binomial approximation and small changes
When a number is close to 1, its powers are easy to estimate. For a small dimensionless \(x\), \((1+x)^n\approx1+nx\). This works for any power \(n\): positive, negative or fractional.
Rewrite the expression as a power of one plus a small fraction. Keep the first-order change only when the fraction is much smaller than one, and check whether the question needs a more accurate result.
Other useful formulas
You cannot apply the rule to \((R+h)^{-2}\) directly. Factor out \(R\) first: \((R+h)^{-2}=R^{-2}(1+h/R)^{-2}\).
Find \(g\) at a height of 32 km above Earth. Take Earth's radius as 6400 km.
Show solution
\(g_h=g(1+h/R)^{-2}\approx g(1-2h/R)\). Here \(h/R=32/6400=0.005\), so \(g_h\approx g(1-0.01)=0.99g\). The exact value is \(0.9901g\).
The length of a simple pendulum increases by 2%. By what percentage does its period change?
Show solution
- \(T\propto\ell^{1/2}\), so \(T_2/T_1=(1.02)^{1/2}\).
- \((1+0.02)^{1/2}\approx1+\tfrac12(0.02)=1.01\).
- The period increases by about 1%.
Answer: About 1% increase
More solved problem patterns
The decrease in \(g\) at a small height \(h\) equals the decrease at depth \(d\). Find \(d\) in terms of \(h\). (At depth \(d\), \(g_d=g(1-d/R)\) exactly.)
Show solution
- At height: \(g_h\approx g(1-2h/R)\), so the decrease is \(2gh/R\).
- At depth: the decrease is \(gd/R\).
- Equate: \(gd/R=2gh/R\Rightarrow d=2h\).
Answer: \(d=2h\)
Estimate \(\dfrac{1}{\sqrt{0.98}}\).
Show solution
- Write it as \((1-0.02)^{-1/2}\).
- \(\approx1+\left(-\tfrac12\right)(-0.02)=1+0.01=1.01\).
- The exact value is 1.0102.
Answer: About 1.01
Go deeper · extra explanation and derivations
Why (1 + x)ⁿ ≈ 1 + nx works
The binomial series is
\[(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots\]When \(|x|\ll1\), each term is far smaller than the one before. Keeping only the first two gives \(1+nx\). For \(n=\tfrac12\) and \(x=0.02\), the dropped \(x^2\) term is \(-\tfrac18(0.02)^2=-0.00005\), negligible next to 0.01.
Calculus gives the same result. Near \(x=0\), any smooth \(f(x)\approx f(0)+f'(0)\,x\). With \(f=(1+x)^n\), \(f(0)=1\) and \(f'(0)=n\).
Forms used in physics
- \(\dfrac{1}{1+x}\approx1-x\), \(\dfrac{1}{1-x}\approx1+x\)
- \(\sqrt{1+x}\approx1+\dfrac x2\), \(\dfrac1{\sqrt{1-x}}\approx1+\dfrac x2\)
- \((1+x)^m(1+y)^n\approx1+mx+ny\) when both are small
- \(e^x\approx1+x\), \(\ln(1+x)\approx x\)
Uses: \(g\) at small height \(g(1-2h/R)\); a 2% longer pendulum has a period about 1% longer; percentage errors in \(x^ay^b\); small changes in volume with temperature.
When the rule fails
The rule is a linear fit near \(x=0\). As \(x\) grows, the curve and the line separate. For \(x=0.2\), \((1.2)^2=1.44\) while \(1+2(0.2)=1.40\). For \(x=0.5\), \((1.5)^{-2}=0.444\) while \(1-2(0.5)=0\), which is useless. Use the exact form when \(x\) is not small.
- "Find \(g\) at a height \(h\ll R\)" or "At what height does \(g\) fall by 1%?"
- "Decrease in \(g\) at height \(h\) equals the decrease at depth \(d\). Relation between \(d\) and \(h\)?"
- "Length of a pendulum increases by 2%. Percentage change in period?"
- Pure estimates: \(\sqrt{1.02}\), \((0.998)^{-3}\), \(1/\sqrt{0.98}\).
The rule needs \(|x|\ll1\). Using it for \(h=R/2\) gives \(g(1-1)=0\), which is absurd. The true value is \(4g/9\).
Keep the sign of \(n\). For \((1+x)^{-1/2}\), the answer is \(1-x/2\), not \(1+x/2\).
Check the idea
Graphs: slope and intercept
A graph is an equation made visible. Its horizontal and vertical axes must have quantities and units. A straight line has a constant rate of change; its intercept tells you the value of the vertical quantity when the horizontal quantity is zero.
An upward line means a positive slope, not necessarily a positive value of the plotted quantity. Always read what is on each axis: the slope of displacement versus time is velocity, while the slope of velocity versus time is acceleration.
Other useful formulas
The drawn steepness depends on axis scales; the numerical slope must be calculated from labelled values.
A velocity graph passes through (2 s, 5 m/s) and (6 s, 13 m/s). Find acceleration.
Show solution
The velocity change is 8 m/s in 4 s, so acceleration is 2 m/s². Extrapolating gives an initial velocity of 1 m/s.
In a pendulum experiment, the line of \(T^2\) against \(\ell\) passes through (0.5 m, 2.0 s²) and (1.0 m, 4.0 s²). Find \(g\).
Show solution
- Slope = \(\dfrac{4.0-2.0}{1.0-0.5}=4.0\ \text{s}^2/\text{m}\).
- Slope \(=4\pi^2/g\), so \(g=4\pi^2/4.0=\pi^2\).
- \(\pi^2\approx9.87\ \text{m/s}^2\).
Answer: \(g\approx9.87\ \text{m/s}^2\)
More solved problem patterns
Two straight x–t lines on the same axes make angles of 30° and 60° with the time axis. Find the ratio of the velocities.
Show solution
- On one graph with fixed scales, velocity = slope = \(\tan\theta\).
- \(v_1:v_2=\tan30^\circ:\tan60^\circ=\tfrac1{\sqrt3}:\sqrt3\).
- Multiply both by \(\sqrt3\): the ratio is 1 : 3.
Answer: 1 : 3
A line has slope −2 and y-intercept 4. Where does it cut the x-axis?
Show solution
- Equation: \(y=-2x+4\).
- At the x-axis, \(y=0\): \(0=-2x+4\Rightarrow x=2\).
Answer: At \(x=2\)
Go deeper · extra explanation and derivations
Standard shapes and the laws behind them
- Straight line \(y=mx+c\): uniform motion (x–t), \(v=u+at\) (v–t).
- Parabola \(y=kx^2\): \(s=\tfrac12at^2\) from rest, kinetic energy against speed.
- Rectangular hyperbola \(xy=k\): Boyle's law (P–V at fixed T), \(\lambda\) against \(f\) at fixed wave speed.
- Exponential decay \(y=y_0e^{-kx}\): radioactive decay, motion with drag proportional to speed.
Recognise the shape first, then read slope, intercept or area.
Slope, intercept and area
- Slope = \(\Delta y/\Delta x\) with units of \(y\) per unit of \(x\). On x–t it is velocity; on v–t it is acceleration; on F–x (spring) it is the spring constant.
- Intercept is the value of \(y\) when \(x=0\). On v–t it is the initial velocity.
- Area has units of \(y\times x\). Area under v–t is displacement; under F–x it is work.
Slope equals \(\tan\theta\) of the drawn angle only when both axes use the same scale. If two lines on the same graph make angles 30° and 60° with the x-axis, their slopes are in the ratio \(\tan30^\circ:\tan60^\circ=1:3\).
Linearising a law to find a constant
Experiments are analysed with straight lines. For the pendulum \(T=2\pi\sqrt{\ell/g}\), squaring gives \(T^2=\dfrac{4\pi^2}{g}\ell\). A plot of \(T^2\) against \(\ell\) is a straight line through the origin with slope \(4\pi^2/g\). Then \(g=4\pi^2/\text{slope}\).
- "Which graph represents Boyle's law / uniform acceleration from rest?" (shape recognition).
- "Two lines make angles 30° and 60° with the time axis. Ratio of velocities?"
- "A plot of \(T^2\) against \(\ell\) has slope … Find \(g\)."
- "log y against log x is a straight line of slope 2 …" (find the law).
A straight line that does not pass through the origin is not a direct proportion. \(y=mx+c\) with \(c\ne0\) is linear, but doubling \(x\) does not double \(y\).
Check the idea
Rates: differentiation
An average rate compares a finite change with the interval over which it occurs. An instantaneous rate is the limit over a very short interval. On a curve, it is the slope of the tangent at that point.
Differentiating position gives velocity and differentiating velocity gives acceleration. Constants vanish because they do not change. For a power of time, the exponent becomes a multiplier and the power decreases by one.
Other useful formulas
A tangent becomes a velocity
Bring two points together on a position–time curve. Watch an average velocity approach the instantaneous velocity at the chosen instant.
The equation walkthrough below explains this experiment. Interactive controls load when JavaScript is available.
Predict before you change an input
As the positive time gap shrinks, the chord slope…
The gap tends to zero; the rate tends to the finite derivative 2t. A small time interval does not mean a small velocity.
A question gives x(t) and asks for velocity at an instant.
- Differentiate x(t) once.
- Substitute the requested time, keeping seconds throughout.
- Read the sign as direction; speed is the magnitude.
Why the equation works · optional
- 1
Compare two positions
\[ \bar v=\frac{x(t+\Delta t)-x(t)}{\Delta t} \]The chord connects two measured positions. Its rise is a displacement; its run is a time interval.
- 2
Expand the difference
\[ \bar v=\frac{(t+\Delta t)^2-t^2}{\Delta t}=2t+\Delta t \]Cancel the common position and divide each remaining term by the interval. Values here use SI units.
- 3
Shrink the time interval
\[ v=\lim_{\Delta t\to0}\bar v=2t \]The chord approaches the tangent. Velocity is the slope at one instant, in m/s.
Model and symbols: Position x(t) = t² metres when t is measured in seconds; t ≥ 0. The coefficient of t² has units m/s². The time gap remains positive.
Instantaneous speed is |v|; differentiating position gives signed velocity, not always a positive speed.
A position is x = 2t³ − 3t + 5 in SI units. Find velocity at t = 2 s.
Show solution
Differentiating gives v = 6t² − 3. At 2 s this is 24 − 3 = 21 m/s. The position offset 5 m contributes no velocity.
A particle moves as \(x=A\sin(\omega t)\). Find its velocity and acceleration, and show \(a=-\omega^2x\).
Show solution
- Chain rule: \(v=\dfrac{dx}{dt}=A\omega\cos(\omega t)\).
- Again: \(a=\dfrac{dv}{dt}=-A\omega^2\sin(\omega t)\).
- Since \(x=A\sin(\omega t)\), this is \(a=-\omega^2x\), the defining property of SHM.
Answer: \(v=A\omega\cos\omega t\), \(a=-\omega^2x\)
More solved problem patterns
The potential energy of a particle is \(U=3x^2-2x\) J, with \(x\) in metres. Find the force at \(x=1\) m.
Show solution
- \(\dfrac{dU}{dx}=6x-2\).
- \(F=-\dfrac{dU}{dx}=-(6x-2)\).
- At \(x=1\): \(F=-(6-2)=-4\) N, directed toward \(-x\).
Answer: \(-4\) N
On an x–t curve, the tangent at point P is horizontal and the curve is shaped like a cap (bending down). Describe the motion at P.
Show solution
- A horizontal tangent means \(dx/dt=0\): the body is momentarily at rest.
- Cap shape means \(d^2x/dt^2\lt0\): acceleration is negative.
- The body was moving forward, stops at P, and starts moving backward. P is the farthest forward point.
Answer: Momentarily at rest with negative acceleration, turning back.
Go deeper · extra explanation and derivations
Derivatives you must know
| Function | Derivative | Function | Derivative |
|---|---|---|---|
| \(x^n\) | \(nx^{n-1}\) | \(\sin x\) | \(\cos x\) |
| constant | 0 | \(\cos x\) | \(-\sin x\) |
| \(e^x\) | \(e^x\) | \(\ln x\) | \(1/x\) |
- Product rule: \(\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\).
- Chain rule: \(\dfrac{d}{dt}f(g(t))=f'(g)\,\dfrac{dg}{dt}\). So \(\dfrac{d}{dt}\sin(\omega t)=\omega\cos(\omega t)\).
Every rate in physics is a derivative
- \(v=dx/dt\), \(a=dv/dt=d^2x/dt^2\)
- Force \(F=dp/dt\); power \(P=dW/dt\); current \(I=dq/dt\)
- Conservative force \(F=-dU/dx\): force points toward lower potential energy.
On a graph, the derivative is the slope of the tangent. A positive slope means the quantity is increasing; zero slope means a turning point; negative slope means decreasing.
Second derivative and curvature
On an x–t curve, the second derivative is the acceleration. A curve bending upward (concave up, like a cup) has \(a\gt0\). A curve bending downward (like a cap) has \(a\lt0\). A straight x–t line has \(a=0\).
- "Position is \(x=at^3+bt^2+c\). Find velocity / acceleration at time t."
- "When does the particle come to rest?" (set \(dx/dt=0\)).
- "\(U=\dots\). Find the force at x = …" (\(F=-dU/dx\)).
- "Charge \(q=\dots\). Find current at t = …".
The derivative of \(\sin(2x)\) is \(2\cos(2x)\), not \(\cos(2x)\). Forgetting the chain-rule factor is the most common calculus slip.
Velocity is zero where the x–t tangent is flat. That does not make the acceleration zero there.
Check the idea
Maxima and minima with derivatives
At the top of a hill or the bottom of a valley on a smooth graph, the tangent is flat. So a maximum or minimum of \(y(x)\) occurs where \(dy/dx=0\).
A maximum or minimum is a candidate where the rate of change is zero. Check whether it is actually a peak or a trough, and also compare the allowed endpoints. A stationary point outside the physical range is not an answer.
Other useful formulas
\(dy/dx=0\) alone does not prove a maximum or minimum. For \(y=x^3\) the slope is zero at \(x=0\), but it is neither: the curve just flattens and keeps rising.
A ball's height is \(y=20t-5t^2\) m. Find the maximum height.
Show solution
\(dy/dt=20-10t=0\) gives \(t=2\) s. \(d^2y/dt^2=-10\lt0\), so it is a maximum. \(y_{\max}=40-20=20\) m.
Show that the range \(R=\dfrac{u^2\sin2\theta}{g}\) is maximum at 45°.
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- \(\dfrac{dR}{d\theta}=\dfrac{2u^2\cos2\theta}{g}=0\Rightarrow\cos2\theta=0\Rightarrow\theta=45^\circ\).
- \(\dfrac{d^2R}{d\theta^2}=-\dfrac{4u^2\sin2\theta}{g}\), which is negative at 45°.
- So 45° gives a maximum, \(R_{\max}=u^2/g\).
Answer: \(\theta=45^\circ\), \(R_{\max}=u^2/g\)
More solved problem patterns
A 4 kg block rests on a floor with \(\mu=0.75\). A rope pulls it at angle \(\theta\) above the horizontal. Find the least force that just moves it, and the angle. Use \(g=10\ \text{m/s}^2\).
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- Normal force: \(N=mg-F\sin\theta\). Just sliding: \(F\cos\theta=\mu N\).
- So \(F=\dfrac{\mu mg}{\cos\theta+\mu\sin\theta}\). F is least when the denominator is largest.
- Largest value of \(\cos\theta+\mu\sin\theta\) is \(\sqrt{1+\mu^2}=\sqrt{1.5625}=1.25\), at \(\tan\theta=0.75\), so \(\theta\approx37^\circ\).
- \(F_{\min}=\dfrac{0.75\times4\times10}{1.25}=24\) N. A horizontal pull would need 30 N.
Answer: 24 N at about 37° above the horizontal
Car A is 100 m east of a crossing and drives west toward it at 8 m/s. At the same moment car B leaves the crossing heading north at 6 m/s. Find the minimum distance between them.
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- At time \(t\): A is \(100-8t\) m east, B is \(6t\) m north.
- \(s^2=(100-8t)^2+(6t)^2\).
- \(\dfrac{d(s^2)}{dt}=-16(100-8t)+72t=0\Rightarrow200t=1600\Rightarrow t=8\) s.
- Then A is 36 m east and B is 48 m north: \(s=\sqrt{36^2+48^2}=60\) m.
Answer: 60 m, at t = 8 s
Go deeper · extra explanation and derivations
The recipe and why it works
- Write the quantity to be optimised as a function of one variable.
- Differentiate and set the derivative to zero. Solve for the variable.
- Check the second derivative, or check the sign of the first derivative on each side.
- Substitute back to get the maximum or minimum value.
Just before a maximum the function rises (\(dy/dx\gt0\)); just after, it falls (\(dy/dx\lt0\)). So the slope passes through zero while decreasing. That is what \(d^2y/dx^2\lt0\) says.
Standard results worth remembering
- Range \(R=\dfrac{u^2\sin2\theta}{g}\) is maximum at \(\theta=45^\circ\).
- \(a\sin\theta+b\cos\theta\) has maximum value \(\sqrt{a^2+b^2}\).
- If \(x+y\) is fixed, the product \(xy\) is largest when \(x=y\).
- Least force to drag a block on a rough floor: \(F_{\min}=\dfrac{\mu mg}{\sqrt{1+\mu^2}}\), applied at \(\tan\theta=\mu\) above the horizontal.
Closest approach of two moving bodies
Write the separation squared, \(s^2(t)\), as a function of time. Minimising \(s^2\) is easier than minimising \(s\) and gives the same time. Set \(d(s^2)/dt=0\).
- "Find the maximum height" or "the time when velocity is maximum" (set \(a=0\)).
- "At what angle is the range maximum?"
- "Minimum force needed to move a block on a rough floor" and its angle.
- "Minimum distance between two particles moving on perpendicular roads".
- Pure maths: maximum of \(3\sin\theta+4\cos\theta\).
A maximum on a restricted range may sit at an end point, where \(dy/dx\ne0\). Check the end points too.
Find the value of the variable, then substitute back. Many students stop at "\(x=2\)" when the question asks for the minimum value of \(y\).
Check the idea
Accumulation: integration and area
Integration reverses differentiation and also measures accumulated change. If a rate varies, split the time interval into tiny pieces and add rate times duration. On a graph this becomes the signed area under the curve.
To obtain position from velocity, add the initial position after finding the displacement. Areas below the time axis are negative for displacement; take their magnitudes when calculating total distance.
Other useful formulas
Build an integral from little areas
Increase the number of rectangles under a force–position graph. The total approaches the work done by a changing force.
The equation walkthrough below explains this experiment. Interactive controls load when JavaScript is available.
Predict before you change an input
For this increasing force, more left-endpoint rectangles make the estimate…
Each refinement recovers some previously omitted area. This monotonic behaviour is specific to the increasing curve and left-endpoint rule.
A force changes with position and you need its work.
- Read the force and position axes with their units.
- Find the signed area over the stated displacement.
- Use triangles/rectangles if the graph is simple; otherwise integrate.
Why the equation works · optional
- 1
One small displacement
\[ \Delta W\approx F(x_i)\Delta x \]A thin rectangle has height force and width displacement. Its area therefore has units N m = J.
- 2
Add the rectangles
\[ W_N=\sum_{i=0}^{N-1}F(x_i)\Delta x \]Smaller widths follow the curve more closely. Left rectangles omit a little area for this increasing force.
- 3
Take the continuous limit
\[ W=\int_0^4 2x\,dx=[x^2]_0^4=16\ \mathrm{J} \]Integration accounts for every small displacement. The area is also a triangle of base 4 m and height 8 N.
Model and symbols: F(x) = 2x with F in N and x in m; integrate from 0 to 4 m. Rectangles use left endpoints, so this increasing function is underestimated.
An integral gives a change between limits; it does not supply an unknown initial value.
Velocity grows uniformly from 2 to 8 m/s over 3 s. Find displacement.
Show solution
The area is a trapezium: average velocity (2+8)/2 = 5 m/s times 3 s gives 15 m. Using just the final velocity would give an incorrect 24 m.
A particle has \(a=6t\) m/s². At \(t=0\), \(v=2\) m/s and \(x=0\). Find \(v\) and \(x\) at \(t=2\) s.
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- \(v=\int6t\,dt=3t^2+C_1\). At \(t=0\), \(v=2\), so \(C_1=2\): \(v=3t^2+2\).
- \(x=\int(3t^2+2)\,dt=t^3+2t+C_2\). At \(t=0\), \(x=0\), so \(C_2=0\).
- At \(t=2\): \(v=14\) m/s and \(x=8+4=12\) m.
Answer: \(v=14\) m/s, \(x=12\) m
More solved problem patterns
A force on a 0.5 kg ball rises linearly from 0 to 10 N and falls back to 0 over 0.2 s (a triangle on the F–t graph). The ball starts at rest. Find its final speed.
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- Impulse = area of the triangle = \(\tfrac12\times0.2\times10=1.0\) N s.
- Impulse = change in momentum: \(0.5\,v-0=1.0\).
- \(v=2.0\) m/s.
Answer: 2 m/s
Evaluate \(\displaystyle\int_0^{\pi}\sin x\,dx\).
Show solution
- \(\int\sin x\,dx=-\cos x\).
- Between the limits: \(-\cos\pi-(-\cos0)=1+1=2\).
Answer: 2
Go deeper · extra explanation and derivations
Standard integrals and the constant
- \(\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\) for \(n\ne-1\)
- \(\int\dfrac{dx}{x}=\ln x+C\)
- \(\int\sin x\,dx=-\cos x+C\), \(\int\cos x\,dx=\sin x+C\), \(\int e^x\,dx=e^x+C\)
The constant \(C\) is fixed by an initial condition. If \(a=6t\), then \(v=3t^2+C\). If \(v=2\) m/s at \(t=0\), then \(C=2\). A definite integral between limits does not need \(C\), because it cancels.
What area means on each graph
| Graph | Area under it |
|---|---|
| v–t | Displacement |
| a–t | Change in velocity |
| F–t | Impulse = change in momentum |
| F–x | Work done |
| P–V | Work done by a gas |
| I–t | Charge |
Area below the horizontal axis counts as negative. For total distance, add the magnitudes of all pieces.
Average value of a varying quantity
The time-average of a quantity is its integral divided by the interval: \(\langle f\rangle=\dfrac1T\int_0^T f\,dt\). Average velocity is total displacement over time for this reason. Over a full cycle, \(\langle\sin\rangle=0\) and \(\langle\sin^2\rangle=\tfrac12\).
- "Acceleration \(a=kt\). Starting from rest, find velocity / position at time t."
- "Area under the F–t graph gives …" (impulse).
- "From the v–t graph find displacement and distance in the first 4 s."
- Direct evaluation: \(\int_1^2 3x^2dx\), \(\int_0^\pi\sin x\,dx\).
Integration gives a change, not a value. To get position, add the initial position to the displacement found from the integral.
For the area under an x–t graph there is no standard meaning. Only areas under rate graphs (v–t, a–t, F–t, F–x) are useful.
Check the idea
Exponentials and logarithms
An exponential describes a fixed fractional change per equal interval, such as cooling toward room temperature. A logarithm answers the inverse question: what exponent gives a particular ratio? Its argument must be dimensionless.
Natural logarithms use base e and appear in integration of 1/x and in reversible isothermal gas work. Logarithms turn products into sums; logarithms of a ratio turn into differences. These operations apply only to positive real arguments in the elementary formulas here.
Other useful formulas
ln(V_f/V_i) is valid; ln of a dimensional volume alone is not a complete physical expression.
For a temperature excess θ=θ₀e^(−λt), when does the excess halve?
Show solution
½=e^(−λt), so taking ln gives −ln2=−λt and t=ln2/λ. λ must have inverse-time units.
A quantity decays exponentially with a half-life of 5 s. What fraction remains after 15 s, and when does it fall to 1/16?
Show solution
- 15 s is 3 half-lives, so the fraction is \((1/2)^3=1/8\).
- \(1/16=(1/2)^4\), which needs 4 half-lives.
- Time = \(4\times5=20\) s.
Answer: 1/8 after 15 s; 1/16 after 20 s
More solved problem patterns
A plot of \(\ln y\) against \(t\) is a straight line with slope \(-0.2\ \text{s}^{-1}\). Find the half-life.
Show solution
- \(\ln y=\ln y_0-\lambda t\), so the slope is \(-\lambda\) and \(\lambda=0.2\ \text{s}^{-1}\).
- \(t_{1/2}=0.693/0.2\approx3.47\) s.
Answer: About 3.5 s
Find \(\log_{10}2000\) and \(\ln1000\).
Show solution
- \(\log_{10}(2\times10^3)=0.301+3=3.301\).
- \(\ln1000=2.303\times\log_{10}1000=2.303\times3\approx6.91\).
Answer: 3.301 and about 6.91
Go deeper · extra explanation and derivations
Log rules and base conversion
- \(\log(ab)=\log a+\log b\), \(\log(a/b)=\log a-\log b\), \(\log a^n=n\log a\)
- \(\log1=0\) in every base; \(\ln e=1\); \(\log_{10}10=1\)
- \(\ln x=2.303\log_{10}x\)
- Useful values: \(\log_{10}2=0.3010\), \(\log_{10}3=0.4771\), \(\ln2=0.693\)
Write a number in scientific notation before taking its log: \(\log_{10}(2\times10^{3})=0.301+3=3.301\).
Exponential decay and growth
In \(y=y_0e^{-\lambda t}\) the fractional drop in each equal interval is the same. Two time scales matter:
- Time constant \(\tau=1/\lambda\): \(y\) falls to \(1/e\approx0.37\) of its start.
- Half-life \(t_{1/2}=\ln2/\lambda\approx0.693/\lambda\): \(y\) halves.
After \(n\) half-lives the fraction left is \((1/2)^n\). Taking logs gives \(\ln y=\ln y_0-\lambda t\), so a graph of \(\ln y\) against \(t\) is a straight line with slope \(-\lambda\).
Why the exponent must be dimensionless
\(e^{x}=1+x+x^2/2+\cdots\). Adding 1 to \(x\) and \(x^2\) makes sense only if \(x\) has no dimensions. So in \(e^{-\lambda t}\), \([\lambda]=T^{-1}\). The same rule fixes the dimensions of constants inside \(\sin\), \(\cos\) and \(\log\).
- "What fraction remains after 3 half-lives?" or "time to fall to 1/16".
- "The graph of ln y against t is a straight line of slope … Find the half-life."
- "Dimensions of λ in \(e^{-\lambda t}\)" (dimensions chapter link).
- Direct evaluation of logs using \(\log2=0.301\).
After two half-lives, one quarter remains, not zero. "Two halves make a whole" does not apply to decay.
\(\ln\) and \(\log_{10}\) differ by the factor 2.303. Check which one the formula uses before substituting.
Check the idea
Explain each formula’s symbols and conditions without looking. Solve the examples before revealing their steps, then use chapter practice to check what needs revision.