Single-Variable-Calculus · Unit 2 · Video 2 · Interactive Practice

Derivatives Without a Clock: From Weather Fronts to GPS

IKey Formulas

Formula	Name	What you need
$\dfrac{dT}{dx} = \lim\limits_{\Delta x \to 0} \dfrac{\Delta T}{\Delta x}$	Temperature gradient	$T$ as a function of position $x$ ; units $^\circ\mathrm{C/km}$
$L^2 = h^2 - s^2$	Flat-earth GPS relation	Fixed satellite height $s$ , measured distance $h$
$\dfrac{dL}{dh} = \dfrac{2h}{2L} = \dfrac{h}{L}$	Sensitivity of the deduced position	The difference quotient of the relation above
$\Delta L \approx \dfrac{dL}{dh}\,\Delta h$	Error propagation	The derivative and the measurement error $\Delta h$

Key Insight: Nothing in $\lim_{\Delta x \to 0} \Delta y/\Delta x$ mentions time. The bottom variable is whatever the top one depends on: a distance underneath makes the derivative a gradient in $^\circ\mathrm{C/km}$ , a measured length underneath makes it a dimensionless sensitivity.

IIThe Temperature Gradient

Two towns $15$ km apart give an average of $-0.4\ ^\circ\mathrm{C/km}$ ; shrinking the interval finds the gradient at a point.

Interval width Δx (km) 15.00

IIISensitivity of a Deduced Position

The satellite height $s$ is fixed, so $L$ is deduced from the measured $h$ — and every error in $h$ is magnified.

Receiver position L 5.0

💡 An aircraft on approach needs its position to within a few feet — a specification on $\Delta L$ , which the sensitivity $h/L$ turns into a much tighter specification on the measured $\Delta h$ .

IVWhere $dL/dh = h/L$ Comes From

Pythagoras plus a nudge: the difference quotient $\Delta L/\Delta h$ settles at $h/L$ as $\Delta h \to 0$ .

Step 1

Step 1 — Pythagoras, with $s$ fixed

L^2 = h^2 - s^2

Step 2 — Nudge the measurement

(L + \Delta L)^2 = (h + \Delta h)^2 - s^2

Step 3 — Subtract step 1

2L\,\Delta L + (\Delta L)^2 = 2h\,\Delta h + (\Delta h)^2

Step 4 — Factor each side

\Delta L\,(2L + \Delta L) = \Delta h\,(2h + \Delta h)

Step 5 — The difference quotient

\frac{\Delta L}{\Delta h} = \frac{2h + \Delta h}{2L + \Delta L}

Step 6 — Let $\Delta h \to 0$ , and $\Delta L \to 0$ with it

\frac{dL}{dh} = \frac{2h}{2L} = \frac{h}{L} \qquad\Longrightarrow\qquad \Delta L \approx \frac{h}{L}\,\Delta h

VQuiz Questions

Problem 1 · A Rate With a Distance Underneath

Given: town A at $x = 0$ reads $24\ ^\circ\mathrm{C}$ , and town B, $15$ km east at $x = 15$ , reads $18\ ^\circ\mathrm{C}$ — find the average rate of change $\Delta T/\Delta x$ .

✅ Correct! The denominator is a distance, so the unit is degrees per kilometre — a rate of change with no clock anywhere in it.

❌ Not quite. That is

\Delta x/\Delta T

. The gradient keeps the temperature change on top and the distance underneath.

❌ Close, but check the sign. Going east,

18 - 24 = -6

: the temperature falls, so

\Delta T/\Delta x

is negative.

❌ Not quite. That is

\Delta T

on its own. A rate still has to be divided by the

15

km you travelled.

❌ Not quite. Use

\Delta T = 18 - 24

and

\Delta x = 15 - 0

, then divide.

Show solution

The average rate of change is the difference quotient with position in the denominator:

\frac{\Delta T}{\Delta x} = \frac{T_B - T_A}{x_B - x_A} = \frac{18 - 24}{15 - 0} = \frac{-6}{15} = -0.4\ ^\circ\mathrm{C/km}

Two things to read off. The unit is degrees per kilometre, because the bottom variable is a distance — nothing forced it to be a time. The sign is negative because the temperature drops as you move east.

Shrinking the interval turns this average into $\dfrac{dT}{dx}$ , the temperature gradient at a point.

Problem 2 · From a Gradient Back to a Change

Given: at $x = 7.5$ km the gradient is $\dfrac{dT}{dx} = -1.5\ ^\circ\mathrm{C/km}$ — estimate the temperature change $\Delta T$ over the next $0.4$ km east.

✅ Correct!

\Delta T \approx \dfrac{dT}{dx}\,\Delta x = (-1.5)(0.4) = -0.6\ ^\circ\mathrm{C}

— the same rearrangement that turns

dL/dh

into an error estimate.

❌ You divided.

-1.5 / 0.4 = -3.75

carries the units

^\circ\mathrm{C/km^2}

, which is not a temperature. Multiply the rate by the interval.

❌ That is the rate itself.

-1.5\ ^\circ\mathrm{C/km}

only becomes a temperature change after it is multiplied by a distance.

❌ Not quite.

-1.5 - 0.4

subtracts a distance from a rate; the gradient multiplies the interval.

❌ Not quite. Start from

\Delta T \approx (dT/dx)\,\Delta x

Show solution

The derivative is the limit of $\Delta T/\Delta x$ , so for a small step it is the ratio itself:

\frac{\Delta T}{\Delta x} \approx \frac{dT}{dx} \qquad\Longrightarrow\qquad \Delta T \approx \frac{dT}{dx}\,\Delta x

\Delta T \approx (-1.5\ ^\circ\mathrm{C/km})(0.4\ \mathrm{km}) = -0.6\ ^\circ\mathrm{C}

The units settle the question by themselves: $\dfrac{^\circ\mathrm{C}}{\mathrm{km}} \times \mathrm{km} = {}^\circ\mathrm{C}$ .

This is exactly the move used for GPS, with different letters: $\Delta L \approx \dfrac{dL}{dh}\,\Delta h$ .

Problem 3 · Flat-Earth GPS in Two Steps

Given: the satellite sits at height $s = 12$ and the receiver stands at $L = 9$ along the ground. The radio measurement of $h$ is off by $\Delta h = 3$ ft.

What is $h$ ?

Estimate the resulting error $\Delta L$

✅ Correct! With

h = 15

and

L = 9

the sensitivity is

\tfrac{5}{3}

, so a

3

ft measurement error becomes a

5

ft position error.

❌ Legs do not add.

9 + 12 = 21

would be the path along the two legs; the hypotenuse satisfies

h^2 = 9^2 + 12^2

❌ Wrong leg.

L^2 = h^2 - s^2

solves for the horizontal leg. Here

h

is the hypotenuse, so

h^2 = L^2 + s^2

❌ Not quite.

h

is the hypotenuse of a right triangle with legs

9

and

12

❌ The ratio is upside down.

\tfrac{L}{h}\Delta h = 1.8

; the derivative is

\dfrac{dL}{dh} = \dfrac{h}{L}

, which is bigger than

1

❌ Not quite.

\Delta L = \Delta h

would need

h = L

, impossible when the satellite is

12

units up.

❌ Not quite.

15 \times 3

uses

h

as the sensitivity; the sensitivity is the ratio

h/L

, which is dimensionless.

❌ Not quite. Use

\Delta L \approx \dfrac{h}{L}\,\Delta h

with the

h

you just found.

Show solution

Step 1 — the hypotenuse. The vertical leg is $s = 12$ and the horizontal leg is $L = 9$ :

h^2 = L^2 + s^2 = 81 + 144 = 225 \qquad\Longrightarrow\qquad h = 15

Step 2 — the sensitivity. Differentiating $L^2 = h^2 - s^2$ with $s$ fixed gives

\frac{dL}{dh} = \frac{h}{L} = \frac{15}{9} = \frac{5}{3} \approx 1.67

Step 3 — propagate the error.

\Delta L \approx \frac{dL}{dh}\,\Delta h = \frac{5}{3}(3) = 5\ \mathrm{ft}

The ratio is dimensionless, so feet go in and feet come out. And since $h$ is a hypotenuse while $L$ is a leg, $h/L > 1$ always: the deduced position is always less accurate than the measurement it came from.

Problem 4 · Designing to a Specification

Given: $s = 12$ and the receiver stands at $L = 5$ , so $h = 13$ . The landing specification demands $|\Delta L| \le 2\ \mathrm{ft}$ .

How accurate must the measurement of $h$ be?

Sliding the receiver toward the point directly beneath the satellite makes the same $\Delta h$ produce

✅ Correct! A sensitivity of

2.6

turns a

2

ft allowance on

\Delta L

into a

0.77

ft allowance on

\Delta h

— and the allowance only gets tighter as the receiver approaches the satellite's foot.

❌ You multiplied.

2 \times 2.6 = 5.2

runs the sensitivity the wrong way; solve

2.6\,|\Delta h| \le 2

instead.

❌ Not quite.

2/5

divides by

L

alone. The sensitivity is the ratio

h/L = 13/5

, not the leg

L

❌ Not quite. That would need

dL/dh = 1

. Because

h > L

, the error in

L

is always bigger than the error in

h

❌ Not quite. Start from

|\Delta L| \approx \dfrac{h}{L}|\Delta h| \le 2

and solve for

|\Delta h|

❌ The opposite happens. As

L

shrinks,

h/L

grows — the sensitivity is worst right under the satellite.

❌ Not quite.

s

is fixed, but

h/L

is not: with

s = 12

and

L = 1

the sensitivity is already

\sqrt{145} \approx 12.04

❌ There is no ceiling. As

L \to 0

with

s

fixed,

h \to 12

while

L \to 0

, so

h/L

passes every bound.

❌ Not quite. Compare

h/L

L = 5

with

h/L

L = 1

Show solution

Part 1 — invert the error estimate. With the $5$ – $12$ – $13$ triangle,

\frac{dL}{dh} = \frac{h}{L} = \frac{13}{5} = 2.6

|\Delta L| \approx 2.6\,|\Delta h| \le 2 \qquad\Longrightarrow\qquad |\Delta h| \le \frac{2}{2.6} \approx 0.77\ \mathrm{ft}

The specification on what you infer is a tighter specification on what you measure, tightened by exactly the derivative.

Part 2 — the blow-up. Hold $s = 12$ and let $L \to 0$ . Then $h = \sqrt{L^2 + 144} \to 12$ , a finite number, while the denominator collapses:

\frac{h}{L} = \frac{\sqrt{L^2 + 144}}{L} \longrightarrow \infty

At $L = 1$ it is already $\sqrt{145} \approx 12.04$ , so the same $1$ ft measurement error costs about $12$ ft on the ground. Directly beneath the satellite the flat-earth model gives no usable horizontal fix at all.

Solved: 0 / 4

Derivatives Without a Clock: From Weather Fronts to GPS

IKey Formulas

IIThe Temperature Gradient

IIISensitivity of a Deduced Position

IVWhere dL/dh=h/LdL/dh = h/LdL/dh=h/L Comes From

VQuiz Questions

IVWhere $dL/dh = h/L$ Comes From