Difficulty: Introductory | Prerequisites: Basic algebra, familiarity with logarithms and exponents
Physics 114 at UW opens with measurement, units, and the quantitative reasoning that underpins every later topic. Scaling questions test whether you can predict how a physical quantity (volume, mass, force) changes when lengths or other dimensions change by a fixed factor. Log-log plots and power-law fitting show up whenever biological or physical data span many orders of magnitude, and the exam expects you to move fluently between a best-fit line on a log-log graph and the underlying power-law equation. If you are comfortable with exponent rules and can take a natural log of both sides of an equation, you are ready for this material.
When every linear dimension of a container scales by a factor k, its volume scales by k³. On a log-log plot, a power-law relationship y = A·xⁿ appears as a straight line whose slope is the exponent n and whose y-intercept gives ln(A). Read the slope and intercept off the best-fit line, then reconstruct the original equation.
Dimensional (proportional) scaling
The principle that when every linear dimension of an object is multiplied by a factor k, areas scale by k² and volumes scale by k³. In simple terms, making something ¾ the size in every direction does not give you ¾ the volume; it gives you (¾)³ = 27/64 of the volume.
Power law
A relationship of the form y = Axⁿ, where A is a constant and n is the exponent (the "power"). Think of it as: one quantity grows as a fixed power of another, rather than in direct proportion.
Log-log plot (ln-ln plot)
A graph where both axes display the natural logarithm (or log base 10) of the variable rather than the variable itself. In simple terms, this is how you turn a curved power-law relationship into a straight line so you can read off the slope and intercept.
Linear regression (best-fit line)
The straight line y = mx + b that minimises the sum of squared distances from the data points. On a log-log plot, m is the power-law exponent and b is the log of the coefficient.
Coefficient of determination (R or R²)
A measure of how well the best-fit line matches the data. R = 1 (or R² = 1) means a perfect fit. Values close to 1 indicate the power-law model describes the data well.
A cylinder's volume is V = π(d/2)²h, where d is the diameter and h is the height.
If both d and h are scaled by a factor k, the new volume is V' = π(kd/2)²(kh) = k³ · V.
Two such scaled containers together hold 2k³ · V.
Worked example (Exam Q1): Container A is full with 2 litres. Containers B and C each have diameter and height that are ¾ of A's. Each has volume (3/4)³ = 27/64 of A. Two of them hold 2 × (27/64) = 54/64 = 27/32 of A's volume. Since 27/32 ≈ 0.844, the combined volume of B and C is about 84.4% of A. That is less than 100%, so you cannot transfer all 2 litres. Answer: A.
Start with the power law: fc = A · mⁿ.
Take the natural log of both sides: ln(fc) = n · ln(m) + ln(A).
This has the form y = mx + b, so on a plot of ln(fc) vs ln(m), the slope is n and the y-intercept is ln(A).
Reading the graph (Exam Q2): The best-fit line on the ln-ln plot gives y = 2.7387x − 1.0891, with R = 0.99231.
Slope = 2.7387, so the exponent n ≈ 2.7.
Intercept = −1.0891, so ln(A) = −1.0891, which means A = e^(−1.0891) ≈ 0.34.
The relationship is fc = 0.34 · m^(2.7). Answer: C.
Method tip: Always check the axis labels. If the axes are ln vs ln, you use e^(intercept). If they are log₁₀ vs log₁₀, you use 10^(intercept).
V_{\text{cylinder}} = \pi \left(\frac{d}{2}\right)^2 hV' = k^3 \cdot V \quad \text{(when all linear dimensions scale by } k\text{)}\ln(y) = n \cdot \ln(x) + \ln(A) \quad \Leftrightarrow \quad y = A \cdot x^nA = e^{\text{intercept}} \quad \text{(for ln-ln plots)} \qquad A = 10^{\text{intercept}} \quad \text{(for log}_{10}\text{-log}_{10}\text{ plots)}Volume scaling is why you cannot simply "shrink" a storage tank and expect the same capacity: halving every dimension cuts the volume to one-eighth. Biologists use power-law relationships between bone dimensions and body mass to estimate the weight of extinct animals from fossil measurements, which is exactly what the femur-circumference question models.
Students often assume that scaling every dimension by ¾ scales the volume by ¾. It does not. Volume scales as the cube of the linear factor: (¾)³ = 27/64, which is roughly 0.42, not 0.75.
Students sometimes confuse the slope of a log-log best-fit line with the coefficient A. The slope is the exponent n. The coefficient comes from the intercept.
Forgetting to exponentiate the intercept is a common error. On an ln-ln plot, ln(A) = intercept, so A = e^(intercept). Writing A = intercept gives a wrong answer.
Mixing up ln and log₁₀ axes leads to using the wrong base when recovering A. Always check the axis labels on the graph.
⚠️ Volume-scaling problems appear frequently. Expect a question where dimensions change by a simple fraction and you must determine whether a container can hold the contents of another.
⚠️ Log-log plot interpretation is a near-certainty on the midterm. You will be given a best-fit equation on a log-transformed graph and asked to identify the correct power-law equation from a list of options.
⚠️ Pay close attention to the axis labels: ln vs log₁₀ changes how you recover the coefficient.
True or false: If you double every dimension of a box, its volume doubles. (False: it increases by a factor of 8.)
Fill in the blank: On a ln-ln plot, a power law y = Axⁿ appears as a straight line with slope ____ and y-intercept ____. (n; ln(A))
True or false: Two containers, each with dimensions ¾ of the original, together hold more than the original. (False: they hold about 84% of the original.)
Fill in the blank: If the y-intercept of a best-fit line on a ln-ln plot is −1.09, then A = ____. (e^(−1.09) ≈ 0.34)
True or false: R = 0.99 means the power-law model is a poor fit for the data. (False: R close to 1 indicates an excellent fit.)
Q: A sphere has radius R. A second sphere has radius 0.5R. What fraction of the first sphere's volume does the second sphere contain?
A: (0.5)³ = 1/8 of the original volume.
Q: A best-fit line on a log₁₀-log₁₀ plot of y vs x has equation log₁₀(y) = 3.2 · log₁₀(x) + 0.5. What is the power-law relationship between y and x?
A: The exponent is 3.2 and A = 10^(0.5) ≈ 3.16, so y = 3.16 · x^(3.2).
Q: Container A has volume V. Container B has all linear dimensions scaled by factor k = 2/3 relative to A. How many copies of container B would you need to hold all of A's contents?
A: Each B has volume (2/3)³ · V = 8/27 · V ≈ 0.296V. You need at least 1/0.296 ≈ 3.38, so you need 4 containers.
Q: On a ln(mass) vs ln(femur circumference) plot, the slope is 2.74 and the intercept is −1.09. Write the power-law equation for mass as a function of femur circumference.
A: ln(m) = 2.74 · ln(fc) − 1.09, so m = e^(−1.09) · fc^(2.74) ≈ 0.34 · fc^(2.74). Note this gives mass as a function of fc. If the question asks for fc as a function of m, you must invert.
Scaling and dimensional analysis reappear whenever you work with forces and pressure (force scales with area, not volume) and in fluid mechanics. Power-law fitting is the same mathematical technique used in analysing exponential decay and growth, which comes up in later units on energy and oscillations. The vector-addition skills tested elsewhere on this midterm also rely on careful handling of components and trigonometry, so the algebra here is good preparation.
volume scaling, proportional reasoning, dimensional analysis, power law, log-log plot, ln-ln plot, best-fit line, linear regression, exponent, coefficient, R-squared, R value, femur circumference, allometry, cylinder volume, scaling factor, cube law, Physics 114, UW, midterm 1