Single-Variable-Calculus · Unit 7 · Video 3 · Interactive Practice

Combining Linear Approximations: An Algebraic Shortcut and GPS Time Dilation

IKey Formulas

Formula	Name	When it applies
$e^{u} \approx 1 + u$	Exponential entry	Any small $u$ , e.g. $u = -3x$
$(1+x)^{r} \approx 1 + rx$	Power entry	Small $x$ ; roots and quotients via $r = -\tfrac{1}{2}$ , $r = -1$ , …
$(1 + ax)(1 + bx) \approx 1 + (a + b)\,x$	Multiply, drop $x^2$	Linear coefficients add
$T' \approx T\left(1 + \tfrac{1}{2}\dfrac{v^2}{c^2}\right)$	Time dilation, linearized	$u = v^2/c^2$ tiny

Key Insight: The result is $f(0) + f'(0)\,x$ , so $\dfrac{e^{-3x}}{\sqrt{1+x}} \approx 1 - \tfrac{7}{2}x$ hands over $f'(0) = -\tfrac{7}{2}$ with no product rule.

IIVisualization 1 — Three Moves to $1 - \tfrac{7}{2}x$

Two catalogue entries, multiplied and cut off at degree one, give the tangent line of $e^{-3x}/\sqrt{1+x}$ at $0$ .

Step 1

Step 1 — Rewrite as a product of powers

f(x) = \frac{e^{-3x}}{\sqrt{1+x}} = e^{-3x}\,(1+x)^{-1/2}

Step 2 — Substitute into the catalogue

e^{-3x} \approx 1 - 3x \qquad (u = -3x)

(1+x)^{-1/2} \approx 1 - \tfrac{1}{2}x \qquad (r = -\tfrac{1}{2})

Step 3 — Multiply out

(1 - 3x)(1 - \tfrac{1}{2}x) = 1 - 3x - \tfrac{1}{2}x + \tfrac{3}{2}x^2

Step 4 — Keep degree $\le 1$

f(x) \approx 1 - \tfrac{7}{2}x \quad\Longrightarrow\quad f(0) = 1,\ \ f'(0) = -\tfrac{7}{2}

Step 5 — Why $\tfrac{3}{2}x^2$ had to go

f(x) = 1 - \tfrac{7}{2}x + \tfrac{51}{8}x^2 + \cdots

Each factor's own

x^2

term was already discarded in Step 2, so

\tfrac{3}{2}x^2

is only one of several quadratic pieces.

IIIVisualization 2 — How Small Are the Dropped Terms?

The error of $1 - \tfrac{7}{2}x$ is of size $x^2$ : a tenfold smaller $x$ gives a roughly hundredfold smaller error.

x 0.100

IVVisualization 3 — The GPS Clock Correction

With $u = v^2/c^2$ , the kept term $\tfrac{1}{2}u$ sets how fast the two clocks drift apart each day.

Satellite speed v (km/s) 4.0

💡 Speed is only half the story: gravity (general relativity) pushes the other way and wins, so overall the satellite clock gains about $38\,\mu\text{s}$ per day.

VQuiz Questions

Problem 1 · Two Catalogue Entries

Given: $f(x) = e^{2x}(1+x)^{3}$ — find its linear approximation near $x = 0$ without differentiating.

✅ Correct!

(1 + 2x)(1 + 3x) = 1 + 5x + 6x^2

, and only

1 + 5x

survives: the linear coefficients add.

❌ Close, but… the linear coefficients add, they do not multiply:

2x \cdot 3x = 6x^2

is a degree-two term, and the

x

terms are

2x + 3x

❌ Not quite.

6x^2

is only one of several

x^2

pieces — each factor's own

x^2

term was already thrown away — so it must be dropped too.

❌ Not quite. Approximate both factors:

e^{2x} \approx 1 + 2x

and

(1+x)^3 \approx 1 + 3x

, then multiply.

Show solution

Catalogue entries: $e^{u} \approx 1 + u$ with $u = 2x$ , and $(1+x)^r \approx 1 + rx$ with $r = 3$ :

e^{2x} \approx 1 + 2x, \qquad (1+x)^3 \approx 1 + 3x

Multiply and keep degree $\le 1$ :

(1 + 2x)(1 + 3x) = 1 + 2x + 3x + 6x^2 \approx 1 + 5x

So $f(x) \approx 1 + 5x$ , which also says $f'(0) = 5$ . (Product rule check: $f'(0) = 2\cdot 1 + 1 \cdot 3 = 5$ .)

Problem 2 · Dividing Means a Negative Power

Given: $g(x) = \dfrac{e^{x}}{(1+x)^{2}}$ — find its linear approximation near $x = 0$ .

✅ Correct!

g(x) = e^{x}(1+x)^{-2} \approx (1 + x)(1 - 2x) \approx 1 - x

❌ Close, but… dividing by

(1+x)^2

is multiplying by

(1+x)^{-2}

, so

r = -2

, not

+2

: that factor contributes

-2x

❌ Not quite. The

-2x^2

from multiplying the two lines is just one of several quadratic pieces (the true coefficient is

\tfrac{3}{2}

), so it cannot be kept — drop every term of degree two or more.

❌ Not quite. Both factors contribute a linear term:

e^{x} \approx 1 + x

as well as

(1+x)^{-2} \approx 1 - 2x

Show solution

Rewrite as a product of powers: $g(x) = e^{x}\,(1+x)^{-2}$ .

Substitute: $e^{x} \approx 1 + x$ and, with $r = -2$ , $(1+x)^{-2} \approx 1 - 2x$ .

Multiply, keep degree $\le 1$ :

(1 + x)(1 - 2x) = 1 + x - 2x - 2x^2 \approx 1 - x

So $g(x) \approx 1 - x$ and $g'(0) = -1$ . The discarded $-2x^2$ is not the true quadratic term: the actual expansion is $g(x) = 1 - x + \tfrac{3}{2}x^2 + \cdots$ .

Problem 3 · A Derivative for Free

Given: $h(x) = \dfrac{e^{-2x}(1+x)^{3}}{\sqrt{1-x}}$ — find $h'(0)$ by building the linear approximation near $x = 0$ .

✅ Correct!

h(x) \approx 1 + (-2 + 3 + \tfrac{1}{2})x = 1 + \tfrac{3}{2}x

, so

h'(0) = \tfrac{3}{2}

❌ Close, but check the root factor.

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

has

r = -\tfrac{1}{2}

and small quantity

-x

, so

rx

becomes

(-\tfrac{1}{2})(-x) = +\tfrac{1}{2}x

— the two minus signs cancel.

❌ Not quite.

(-2)(3)(\tfrac{1}{2})

multiplies the linear coefficients; products of linear terms have degree two or more. The linear coefficients add.

❌ Not quite.

-2 + 3 = 1

leaves out the denominator: dividing by

\sqrt{1-x}

is multiplying by

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

, which contributes its own linear term.

❌ Not quite. Write

h

as three powers, approximate each by

1 + (\text{coefficient})\,x

, and add the coefficients.

Show solution

Rewrite as a product of powers:

h(x) = e^{-2x}\,(1+x)^{3}\,(1-x)^{-1/2}

Substitute into the catalogue:

e^{-2x} \approx 1 - 2x, \qquad (1+x)^3 \approx 1 + 3x, \qquad (1-x)^{-1/2} \approx 1 + \left(-\tfrac{1}{2}\right)(-x) = 1 + \tfrac{1}{2}x

Multiply, keep degree $\le 1$ : every cross term has degree at least two, so only the constant and the sum of the linear coefficients remain:

h(x) \approx 1 + \left(-2 + 3 + \tfrac{1}{2}\right)x = 1 + \tfrac{3}{2}x

Matching with $h(0) + h'(0)\,x$ : $h(0) = 1$ and $h'(0) = \tfrac{3}{2}$ .

Problem 4 · Same Move, Different Physics

Given: the energy of a mass $m$ moving at speed $v$ is $E = \dfrac{mc^2}{\sqrt{1 - v^2/c^2}}$ — find its linear approximation in the small quantity $u = v^2/c^2$ , written in terms of $m$ , $v$ , $c$ .

✅ Correct!

mc^2(1 - u)^{-1/2} \approx mc^2(1 + \tfrac{1}{2}u) = mc^2 + \tfrac{1}{2}mv^2

— rest energy plus the Newtonian kinetic energy.

❌ Close, but check the sign. The small quantity is

x = -u

and

r = -\tfrac{1}{2}

, so

rx = (-\tfrac{1}{2})(-u) = +\tfrac{1}{2}u

— exactly as for the GPS clock, where

T'

came out larger than

T

❌ Not quite. A square root in the denominator means

r = -\tfrac{1}{2}

, not

r = -1

, so the coefficient is

\tfrac{1}{2}

❌ Not quite. Rewrite as

E = mc^2(1 - u)^{-1/2}

and use

(1+x)^r \approx 1 + rx

with

r = -\tfrac{1}{2}

x = -u

Show solution

Rewrite as a power: with $u = v^2/c^2$ ,

E = mc^2\,(1 - u)^{-1/2}

Substitute into $(1+x)^r \approx 1 + rx$ with $r = -\tfrac{1}{2}$ and $x = -u$ :

(1 - u)^{-1/2} \approx 1 + \left(-\tfrac{1}{2}\right)(-u) = 1 + \tfrac{1}{2}u

Multiply by $mc^2$ :

E \approx mc^2\left(1 + \tfrac{1}{2}\frac{v^2}{c^2}\right) = mc^2 + \tfrac{1}{2}mv^2

This is the same computation as $T' \approx T\left(1 + \tfrac{1}{2}\frac{v^2}{c^2}\right)$ ; the dropped term is of size $u^2$ , negligible at everyday speeds.

Solved: 0 / 4

Combining Linear Approximations: An Algebraic Shortcut and GPS Time Dilation

IKey Formulas

IIVisualization 1 — Three Moves to 1−72x1 - \tfrac{7}{2}x1−27​x

IIIVisualization 2 — How Small Are the Dropped Terms?

IVVisualization 3 — The GPS Clock Correction

VQuiz Questions

IIVisualization 1 — Three Moves to $1 - \tfrac{7}{2}x$