M2 Algebra and graphs for physics

In this chapter

• Rearrange a formula to make any quantity its subject, keeping the units consistent, and avoid the classic algebra slips.
• Use proportion and power laws: say what doubling or tripling one quantity does to another.
• Explain the square–cube law, and why giants, and very small warm-blooded animals, run into trouble.
• Read a graph two ways: its slope is a rate, and the area under it is an accumulated total.
• Straighten a curve by choosing the right axes, and use a log–log plot to read off the power in a power law.
• Check a result with approximations, limiting cases and units.

Rearranging formulas

Rearranging formulas

• \(d\): distance (m) · \(v\): speed, or final speed (m/s) · \(u\): starting speed (m/s) · \(t\): time (s) · \(a\): acceleration (m/s²)

Units ride along and check the result: \(\dfrac{\text{m/s} - \text{m/s}}{\text{s}} = \text{m/s}^2\), an acceleration. ✓

Checkpoint: Rearranging formulas (1 of 2)

Checkpoint: Rearranging formulas (2 of 2)

Squares and roots

Squares and roots

• \(E\): kinetic energy (J) · \(m\): mass (kg) · \(T\): a pendulum's period (s) · \(L\): its length (m) · \(g\): the acceleration due to gravity (m/s²)

Squares and roots

Checkpoint: Squares and roots (1 of 2)

Checkpoint: Squares and roots (2 of 2)

Four algebra mistakes

Checkpoint: Four algebra mistakes (1 of 2)

Checkpoint: Four algebra mistakes (2 of 2)

Proportion and power laws

Proportion and power laws

Proportion and power laws

Checkpoint: Proportion and power laws (1 of 3)

Checkpoint: Proportion and power laws (2 of 3)

Checkpoint: Proportion and power laws (3 of 3)

The square–cube law

The square–cube law

The square–cube law

The square–cube law

• Strength: a bone's strength goes with its cross-section (\(L^2\)), its load with the body's weight (\(L^3\)), so

  \[ \text{stress} = \frac{\text{weight}}{\text{area}} \propto \frac{L^3}{L^2} = L. \]

  Ten times taller means ten times the stress, about where a thigh-bone breaks. Big animals need disproportionately thick bones. - Falling: air resistance goes with area, weight with volume, so small animals fall gently. - Heat: warm-blooded animals make heat in their volume and lose it through their skin: small animals lose heat fast and must eat a lot; an elephant's thin ears help it shed heat. Crushed ice melts faster than a block, and cells stay small.

The square–cube law

Checkpoint: The square–cube law (1 of 2)

Checkpoint: The square–cube law (2 of 2)

Reading graphs

Reading graphs

Reading graphs

• Slope = rate. The gradient is rise ÷ run, in vertical units per horizontal unit. On a distance–time graph it is the speed: steeper is faster, flat is stopped, downhill is heading back. For a curve, the slope of the tangent at a point is the rate at that instant: what a speedometer shows.
• Area = accumulated total. Under a speed–time graph, the area is the distance: 5 m/s for 4 s is a 20 m rectangle; speeding up steadily from 0 to 10 m/s in 4 s is a triangle, \(\tfrac12\times4\times10 = 20\) m. Power × time gives energy; flow rate × time gives volume.

Reading graphs

Checkpoint: Reading graphs (1 of 2)

Checkpoint: Reading graphs (2 of 2)

Straight lines

Straight lines

• \(m\): the slope · \(c\): the intercept, where the line crosses the vertical axis (\(x = 0\))

Straight lines

Straight lines

Checkpoint: Straight lines (1 of 2)

Checkpoint: Straight lines (2 of 2)

Straightening curves

Straightening curves

Checkpoint: Straightening curves (1 of 2)

Checkpoint: Straightening curves (2 of 2)

Log–log plots

Log–log plots

Checkpoint: Log–log plots (1 of 3)

Checkpoint: Log–log plots (2 of 3)

Checkpoint: Log–log plots (3 of 3)

Approximations

Approximations

This binomial approximation says: for \(y \propto x^n\), a small percentage change in \(x\) gives about \(n\) times that percentage change in \(y\). A cube 1% longer has about 2% more area and 3% more volume.

\(T = 2\pi\sqrt{L/g}\) is itself a small-swing approximation: a 10° swing takes about 0.2% longer, 30° about 1.7%, 90° about 18%.

Checkpoint: Approximations (1 of 2)

Checkpoint: Approximations (2 of 2)

Limiting cases

Limiting cases

Checkpoint: Limiting cases (1 of 2)

Checkpoint: Limiting cases (2 of 2)

Graph mistakes

Checkpoint: Graph mistakes

Key terms

Variable

A letter standing for a physical quantity: a number with a unit.

Subject (of a formula)

The quantity standing alone on one side.

Factor / term

Something multiplied / something added.

Kinetic energy

The energy of motion, \(\tfrac12 m v^2\).

Key terms

Proportional / inversely proportional

Double one, and the other doubles / halves.

Power law

\(y\) goes as \(x\) to some power: \(y \propto x^n\).

Inverse-square law

A quantity that goes as one over distance squared, like brightness or gravity.

Square–cube law

Areas scale as length squared, volumes as length cubed.

Key terms

Stress

Force per area.

Independent / dependent variable

The one you choose (horizontal axis) / the one you measure (vertical axis).

Gradient (slope)

Rise ÷ run: the rate at which the vertical quantity changes.

Tangent

A straight line just touching a curve at one point.

Key terms

Intercept

Where a line crosses the vertical axis.

Linearise

Choose axes that turn a curve into a straight line.

Logarithmic scale / log–log plot

An axis where equal steps are equal factors / a graph with both axes logarithmic.

Binomial approximation

\((1+x)^n \approx 1 + nx\) for small \(x\).

Key terms

Limiting case

A test of a formula at an extreme value where the answer is known.

Line of best fit (worked example)

The straight line that passes closest to all the points.

Muhammad ibn Musa al-Khwarizmi (Baghdad, about 820)

Robert Recorde (London, 1557)

Galileo Galilei (Leiden, 1638)

J. B. S. Haldane (1926)

René Descartes and Pierre de Fermat (1637)

William Playfair (London, 1786)

Nicole Oresme (Paris, about 1350)

Georges Lemaître and Edwin Hubble (1927–1929)

Johannes Kepler (Linz, 1618)

G. I. Taylor (Cambridge, 1950)

Max Kleiber (California, 1932)

Christiaan Huygens (Paris, 1673)

George Atwood (Cambridge, 1784)

Numbers used:

Watch the worked example

The problem

Problem. A ball is dropped from rest through heights \(h\) to a light gate, and the fall time \(t\) is measured, with a timer good to ±0.005 s. Find \(g\), with an uncertainty.

The problem

(Illustrative data, generated for this example from \(g = 9.81\ \text{m/s}^2\) with a few milliseconds of timing scatter.)

Solution

1. Choose the axes. \(h = \tfrac12 g t^2\) has the form \(y = m x\) with \(y = h\), \(x = t^2\), \(m = \tfrac12 g\). So plot \(h\) against \(t^2\).
2. The new column: \(t^2\) = 0.0412, 0.0807, 0.1239, 0.1608, 0.2070, 0.2430 s².
3. Error bars: ±0.005 s is about 1% of the longest time, so \(t^2\) is uncertain by about 2%: \(0.243 \pm 0.005\ \text{s}^2\) (about 0.002 s² for the shortest drop).

Solution

4. Best line: the points lie close to a straight line through the origin. From \((0, 0)\) to \((0.25\ \text{s}^2, 1.226\ \text{m})\) on the line, the slope is \(1.226/0.25 = 4.90\ \text{m/s}^2\), so \(g = 2\times4.90 \approx 9.81\ \text{m/s}^2\).
5. Uncertainty: the steepest and shallowest lines that still pass through the first and last error bars have slopes of about 5.13 and 4.79 m/s², so \(g\) lies between about 9.6 and 10.3. Half the spread: \(g = 9.8 \pm 0.3\ \text{m/s}^2\).

Solution

6. Check: units of the slope, m/s², ✓; the accepted 9.81 m/s² is inside the error bar ✓; the intercept is close to zero, so there's no sign of a systematic timing delay ✓.

★ Warm-up

1. Rearrange \(v = u + a t\) to make \(t\) the subject. A car speeds up from 5 m/s to 25 m/s at 4 m/s². How long does it take?

   Your answer sCheck

   Solution

   Subtract \(u\): \(v - u = a t\). Divide by \(a\): \(t = \dfrac{v - u}{a} = \dfrac{25 - 5}{4} = 5\ \text{s}\). Units: \(\dfrac{\text{m/s}}{\text{m/s}^2} = \text{s}\). ✓

Problem 2 (★ Warm-up)

2. Pressure is \(P = F/A\). Make \(F\) the subject, then \(A\). What force does a pressure of \(2.0\times10^5\) Pa exert on 0.03 m²?

   Your answer NCheck

   Solution

   Multiply by \(A\): \(F = PA\). Then divide by \(P\): \(A = F/P\). \(F = 2.0\times10^5 \times 0.03 = 6000\ \text{N}\) (Pa × m² = N/m² × m² = N ✓).

Problem 3 (★ Warm-up)

3. If \(y \propto x^3\), what happens to \(y\) when \(x\) is (a) doubled, (b) halved, (c) tripled?

   Your prediction or attempt, before you look

   Show solution

   Solution

   (a) \(2^3 = 8\) times bigger. (b) \((\tfrac12)^3 = \tfrac18\) as big. (c) \(3^3 = 27\) times bigger.

Problem 4 (★ Warm-up)

4. A distance–time graph is a straight line from (0 s, 0 m) to (8 s, 120 m). What is its slope, and what does it tell you?

   Your answer m/sCheck

   Solution

   Slope = rise ÷ run = 120 m ÷ 8 s = 15 m/s: the (steady) speed.

Problem 5 (★ Warm-up)

5. A speed–time graph is flat at 12 m/s from 0 to 5 s. How far does the object go?

   Your answer mCheck

   Solution

   The area is a rectangle: 12 m/s × 5 s = 60 m.

★★ Standard

6. A 1000 kg car has \(4.5\times10^5\) J of kinetic energy. (a) Find its speed. (b) If its speed rises by half again, by what factor does its kinetic energy grow?

   Your answer m/sCheck

   Solution

   (a) \(v = \sqrt{2E/m} = \sqrt{2\times4.5\times10^5/1000} = \sqrt{900} = 30\ \text{m/s}\). (b) \(E \propto v^2\), so the factor is \(1.5^2 = 2.25\). (Check: at 45 m/s, \(E = \tfrac12\times1000\times45^2 \approx 1.0\times10^6\) J, which is 2.25 times as much.)

Problem 7 (★★ Standard)

7. Rearrange \(T = 2\pi\sqrt{L/g}\) for \(L\). How long must a pendulum be to have a period of 2.00 s (a) on Earth, \(g = 9.81\ \text{m/s}^2\), and (b) on the Moon, \(g = 1.62\ \text{m/s}^2\)?

   Your answer mCheck

   Solution

   Divide by \(2\pi\), square: \(\dfrac{T^2}{4\pi^2} = \dfrac{L}{g}\). Multiply by \(g\): \(L = \dfrac{g T^2}{4\pi^2}\). (a) \(9.81\times4/39.48 \approx 0.994\ \text{m}\). (b) \(1.62\times4/39.48 \approx 0.164\ \text{m}\). Check: \(L \propto g\), and the Moon's \(g\) is about a sixth of Earth's, so is its pendulum length. ✓

Problem 8 (★★ Standard)

8. A 2 m statue weighing 300 kg is copied at 6 m in the same material. How many times more paint does it need? What is its mass? How many times bigger is the pressure on its feet?

   Your prediction or attempt, before you look

   Show solution

   Solution

   Scale factor 3. Paint goes with area: \(3^2 = 9\) times. Mass goes with volume: \(3^3 = 27\) times, so 8100 kg. Pressure = weight ÷ area \(\propto L^3/L^2 = L\): 3 times bigger.

Problem 9 (★★ Standard)

9. A lamp gives 400 lux at 1.0 m. What is the illumination at 2.5 m? At what distance is it 100 lux?

   Your answer Check

   Solution

   Inverse square: \(400\times(1.0/2.5)^2 = 400/6.25 = 64\) lux. For 100 lux, a quarter of 400, the distance must double: 2.0 m.

Problem 10 (★★ Standard)

10. A cyclist speeds up steadily from rest to 12 m/s in 4 s, then rides at 12 m/s for 6 s. Sketch the speed–time graph. Find the acceleration in the first part and the total distance.

    Your prediction or attempt, before you look

    Show solution

    Solution

    Acceleration = slope = 12 ÷ 4 = 3 m/s². Distance = area = triangle + rectangle = \(\tfrac12\times4\times12 + 12\times6 = 24 + 72 = 96\ \text{m}\).

Problem 11 (★★ Standard)

11. Two formulas are proposed for the speed of a ball after falling from rest through a height \(h\): (a) \(v = \sqrt{2gh}\) and (b) \(v = 2gh\). Which can you rule out, and how?

    Your prediction or attempt, before you look

    Show solution

    Solution

    Units: (a) \(\sqrt{\text{m/s}^2\times\text{m}} = \sqrt{\text{m}^2/\text{s}^2} = \text{m/s}\) ✓. (b) \(\text{m/s}^2\times\text{m} = \text{m}^2/\text{s}^2\), not a speed ✗. Rule out (b). (a) also passes the limiting cases: \(h = 0\) gives \(v = 0\), and stronger gravity gives a faster ball.

Problem 12 (★★ Standard)

12. The radius of a steel ball is measured with a 2% uncertainty. Roughly what is the uncertainty in its volume, and so in its calculated mass?

    Your prediction or attempt, before you look

    Show solution

    Solution

    \(V \propto r^3\), so by the binomial approximation about \(3\times2\% = 6\%\). The mass (density × volume) is uncertain by about 6% too.

★★★ Stretch

13. Fit a line by hand. A spring is loaded with forces \(F\) = 0, 1, 2, 3, 4, 5 N, and its extensions are \(x\) = 0.0, 2.1, 3.9, 6.2, 7.9, 10.1 cm, each ±0.2 cm. Plot \(x\) against \(F\), draw the best line, and find the slope with an uncertainty. The spring constant is \(k = F/x\): find it in N/m.

    Hint

    Draw the steepest and shallowest lines that still pass through the error bars at \(F = 1\) N and \(F = 5\) N.

    Your prediction or attempt, before you look

    Show solution

    Solution

    The points lie close to a straight line through the origin with slope about 2.0 cm/N (from (0, 0) to (5 N, 10.0 cm)). Steepest: from (1, 1.9) to (5, 10.3): \(8.4/4 = 2.1\) cm/N. Shallowest: from (1, 2.3) to (5, 9.9): \(7.6/4 = 1.9\) cm/N. So the slope is \(2.0 \pm 0.1\) cm/N, and \(k = 1/\text{slope} = 0.50\ \text{N/cm} = 50\ \text{N/m}\), with a range of about 48 to 53 N/m: \(k = 50 \pm 3\ \text{N/m}\). Check: the line passes through the origin (no force, no stretch) ✓.

Problem 14 (★★★ Stretch)

14. Kepler for Jupiter's moons. Io orbits Jupiter at 421,700 km in 1.769 days; Callisto at 1,882,700 km in 16.69 days. Is \(T \propto a^1\), \(a^{1.5}\) or \(a^2\)? (Europa, 671,000 km and 3.551 days, and Ganymede, 1,070,400 km and 7.155 days, let you check.)

    Hint

    Work out how many times farther Callisto is than Io, then raise that ratio to each candidate power and compare with the ratio of the periods.

    Your prediction or attempt, before you look

    Show solution

    Solution

    Distance ratio: \(1{,}882{,}700/421{,}700 \approx 4.46\). Period ratio: \(16.69/1.769 \approx 9.43\). \(4.46^1 = 4.46\) ✗; \(4.46^{1.5} = 4.46\times\sqrt{4.46} \approx 4.46\times2.11 \approx 9.43\) ✓; \(4.46^2 \approx 19.9\) ✗. So \(T \propto a^{1.5}\): Kepler's third law holds for moons too. On a log–log plot, the four moons lie on a line of slope 1.5.

Problem 15 (★★★ Stretch)

15. King Kong. A film gorilla is 18 m tall. A real gorilla is about 1.7 m tall and has a mass of about 160 kg. Estimate the giant's mass, and how many times the stress in its leg bones would be, if it had the same proportions.

    Your prediction or attempt, before you look

    Show solution

    Solution

    Scale factor \(18/1.7 \approx 10.6\). Mass \(\times 10.6^3 \approx 1190\): about \(1.9\times10^5\) kg, roughly 190 tonnes. Stress \(\propto L\): about 10.6 times as great, well beyond what bone can bear, as Haldane's argument predicts.

Problem 16 (★★★ Stretch)

16. A ball thrown straight up at speed \(u\) reaches a height \(H = \dfrac{u^2}{2g}\). Check the formula with units and limiting cases (\(u \to 0\); very weak gravity). Then say what doubling \(u\) does to \(H\).

    Your prediction or attempt, before you look

    Show solution

    Solution

    Units: \(\dfrac{(\text{m/s})^2}{\text{m/s}^2} = \text{m}\) ✓. \(u \to 0\): \(H \to 0\), no throw, no height ✓. \(g \to 0\): \(H\) becomes huge, as it should with almost no gravity to stop the ball ✓. \(H \propto u^2\), so doubling \(u\) makes the ball rise four times as high.

Answer these out loud, from memory. Then try again in a few days.

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The roadmap's other §03 questions (exponential and logarithmic relationships, telling a trend from noise, and dimensionless ratios) are picked up in M5 and M14.

Check yourself

Summary