Single-Variable-Calculus · Unit 8 · Video 1 · Interactive Practice

A Linear Approximation Is a Proportionality: Clocks, Pitches and Rules of Thumb

IKey Formulas

Formula	Name	What it says
$T' \approx T\left(1 + \tfrac{1}{2}\frac{v^2}{c^2}\right)$	Time dilation, linearized	Earth watch $T'$ against satellite clock $T$ (last lecture)
$\frac{\Delta T}{T} \approx \tfrac{1}{2}\cdot\frac{v^2}{c^2}$	Relative change	Proportional to $v^2/c^2$ with factor $\tfrac{1}{2}$ ; both sides dimensionless
$\left.\frac{d}{du}(1-u)^{-1/2}\right\|_{u=0} = \tfrac{1}{2}$	Factor = rate of change	With $u = \frac{v^2}{c^2}$ , $\ \frac{\Delta T}{T} = (1-u)^{-1/2} - 1$ exactly
$\Delta v \approx a\,\Delta h$	Rule of thumb	$a$ = tangent slope at the base point; left out: order $(\Delta h)^2$

Key Insight: A linear approximation says small change in, proportional change out. The constant of proportionality is the derivative at the base point, and everything dropped — the curvature terms, a slightly wrong constant — is a small quantity times another small quantity.

IIVisualization 1 — The Factor $\tfrac{1}{2}$ Is a Slope

Near $u = 0$ the exact curve $\frac{\Delta T}{T} = (1-u)^{-1/2} - 1$ merges with its tangent line of slope $\tfrac{1}{2}$ .

window 0 ≤ u ≤ 0.60

IIIVisualization 2 — What the Rule Leaves Out

The tangent misses the curve by an amount of order $(\Delta h)^2$ , negligible beside $a\,\Delta h$ for small $\Delta h$ .

Δh 1.50

IVVisualization 3 — The Constant Depends on the Base Point

Pitchers at 80, 90 and 100 mph have slightly different constants; borrowing the wrong one costs only $(\Delta a)(\Delta h)$ .

pitcher Δh 1.00

VQuiz Questions

Problem 1 · The GPS Proportionality

Given: $T' \approx T\left(1 + \tfrac{1}{2}\frac{v^2}{c^2}\right)$ and, for a GPS satellite, $\frac{v^2}{c^2} \approx 1.8\times10^{-10}$ — find the relative change $\frac{\Delta T}{T}$ , where $\Delta T = T' - T$ .

✅ Correct! The relative change is the factor

\tfrac{1}{2}

times

\frac{v^2}{c^2}

\tfrac{1}{2}(1.8\times10^{-10}) = 9\times10^{-11}

❌ Close, but… that is

\frac{v^2}{c^2}

itself. The proportionality factor

\tfrac{1}{2}

still multiplies it.

❌ Not quite.

3.6\times10^{-10}

2\frac{v^2}{c^2}

. The factor is

\tfrac{1}{2}

, so

\frac{v^2}{c^2}

is halved, not doubled.

❌ Not quite.

1 + 9\times10^{-11}

is the ratio

\frac{T'}{T}

. Subtracting

T

first leaves only the change

\Delta T

❌ Not quite. Multiply out, subtract

T

, divide by

T

\frac{\Delta T}{T} \approx \tfrac{1}{2}\frac{v^2}{c^2}

Show solution

Multiply out the right side and subtract $T$ :

T' \approx T + \tfrac{1}{2}T\frac{v^2}{c^2} \quad\Longrightarrow\quad \Delta T = T' - T \approx \tfrac{1}{2}T\frac{v^2}{c^2}.

Divide by $T$ :

\frac{\Delta T}{T} \approx \tfrac{1}{2}\cdot\frac{v^2}{c^2} = \tfrac{1}{2}\left(1.8\times10^{-10}\right) = 9\times10^{-11}.

Both sides are dimensionless (time over time, speed squared over speed squared), so this number is the same in any system of units. The clocks agree to about ten significant digits.

Problem 2 · Where the Factor Comes From

Given: with $u = \frac{v^2}{c^2}$ , the exact relative change is $\frac{\Delta T}{T} = (1-u)^{-1/2} - 1$ — find its rate of change with respect to $u$ at $u = 0$ .

✅ Correct! The slope of the tangent at the origin is

\tfrac{1}{2}

— exactly the proportionality factor in

\frac{\Delta T}{T} \approx \tfrac{1}{2}u

❌ Close, but check the chain rule. The inner function

1 - u

has derivative

-1

, which cancels the minus sign from the power rule.

❌ Not quite.

-\tfrac{3}{2}

is the new exponent, not the derivative. The power rule brings down the old exponent

-\tfrac{1}{2}

❌ Not quite.

0

is the value of

\frac{\Delta T}{T}

u = 0

(the two clocks agree), not its rate of change.

❌ Not quite. Differentiate

(1-u)^{-1/2}

with the power rule and the chain rule, then set

u = 0

Show solution

The constant $-1$ has derivative $0$ . Power rule with the chain rule:

\frac{d}{du}(1-u)^{-1/2} = -\tfrac{1}{2}(1-u)^{-3/2}\cdot(-1) = \tfrac{1}{2}(1-u)^{-3/2}.

At $u = 0$ :

\tfrac{1}{2}(1)^{-3/2} = \tfrac{1}{2}.

So the factor $\tfrac{1}{2}$ in $\frac{\Delta T}{T} \approx \tfrac{1}{2}\cdot\frac{v^2}{c^2}$ is a rate of change: the slope of the tangent line to the curve $\frac{\Delta T}{T} = (1-u)^{-1/2} - 1$ at the origin.

Problem 3 · A Rule of Thumb for the Mound

Given: suppose that near the standard mound a pitcher's rule of thumb is $\Delta v \approx a\,\Delta h$ with $a = 0.4$ mph per inch, and the mound is raised by $\Delta h = 2.5$ inches — find the predicted change in pitch speed. Then a second pitcher's own constant turns out to be $0.44$ mph per inch — find how much the prediction for that pitcher changes when this constant replaces $a$ .

Predicted change $\Delta v$ with $a = 0.4$ :

Change in the prediction when $0.44$ replaces $a$ :

✅ Correct!

\Delta v \approx 0.4 \times 2.5 = 1.0

mph, and the wrong constant costs

(\Delta a)(\Delta h) = 0.04 \times 2.5 = 0.1

mph: a small correction to an already small change.

❌ Not quite. A proportionality multiplies: the change out is the constant times the change in, not a quotient.

❌ Check the prediction. The rule is

\Delta v \approx a\,\Delta h

— a product, not a sum.

❌ Close, but…

0.44 \times 2.5 = 1.1

mph is the whole new prediction. The question asks how much it differs from the old one.

❌ Close, but…

0.04

is the change in the constant,

\Delta a

. It still multiplies

\Delta h

❌ Not quite. That is

(\Delta a)(\Delta h)^2

. The error is first order in each small quantity:

(\Delta a)(\Delta h)

❌ Check the second part. Subtract the two predictions:

(0.44 - 0.4)\,\Delta h

Show solution

Prediction:

\Delta v \approx a\,\Delta h = 0.4 \times 2.5 = 1.0 \text{ mph}.

Using the pitcher's own constant $a + \Delta a = 0.44$ , so $\Delta a = 0.04$ :

(a + \Delta a)\,\Delta h - a\,\Delta h = (\Delta a)(\Delta h) = 0.04 \times 2.5 = 0.1 \text{ mph}.

The shift in the constant is a second-order effect — a small change in $a$ times a small change in $h$ — so the rule $\Delta v \approx a\,\Delta h$ barely depends on which nearby base point it was worked out at.

Problem 4 · Two Effects at Once

Given: suppose two effects change a satellite clock's rate by the factors $1 + x$ and $1 + y$ , with relative changes $x = -8\times10^{-11}$ and $y = 5\times10^{-10}$ — find the combined relative change, to the accuracy the rules of thumb give.

✅ Correct!

(1+x)(1+y) = 1 + x + y + xy

, and

xy = -4\times10^{-20}

is negligible, so the relative changes simply add:

x + y = 4.2\times10^{-10}

❌ Close, but… the two effects push in opposite directions:

x

is negative, so it partly cancels

y

❌ Close, but… check the sign. The larger effect,

y = 5\times10^{-10}

, is positive, so

x + y

is positive: the negative

x

only partly cancels it.

❌ Not quite.

xy

is the one term the rule of thumb drops — a product of two small numbers. The sum

x + y

is what survives.

❌ Not quite. Expand

(1+x)(1+y)

, drop the product of the two small numbers, and keep the signs.

Show solution

The combined factor is

(1+x)(1+y) = 1 + x + y + xy.

The cross term is tiny compared with $x$ and $y$ :

xy = \left(-8\times10^{-11}\right)\left(5\times10^{-10}\right) = -4\times10^{-20}.

So the combined relative change is

x + y = -0.8\times10^{-10} + 5\times10^{-10} = 4.2\times10^{-10}.

Small relative changes combine by addition because their products are negligible — the reason engineers can reduce each effect to its own rule of thumb and add them.

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A Linear Approximation Is a Proportionality: Clocks, Pitches and Rules of Thumb

IKey Formulas

IIVisualization 1 — The Factor 12\tfrac{1}{2}21​ Is a Slope

IIIVisualization 2 — What the Rule Leaves Out

IVVisualization 3 — The Constant Depends on the Base Point

VQuiz Questions

IIVisualization 1 — The Factor $\tfrac{1}{2}$ Is a Slope