Single-Variable-Calculus · Unit 18 · Video 3 · Interactive Practice

Units Matter: Why y = x² Is Not Scale-Invariant

IKey Formulas

Formula	Meaning	Reading
$V = \dfrac{\pi}{2}a^2$	Cauldron from $y = x^2$ filled to depth $a$ (shells)	$a$ is a pure number; $V$ is in cubic units of whatever unit $x$ and $y$ use
$a = 100\ (\text{cm}): \ V = \dfrac{\pi}{2}\,10^4\ \text{cm}^3 \approx 16\ \text{liters}$	The skinny cauldron	Rim radius $\sqrt{100} = 10$ cm, $20$ cm across
$a = 1\ (\text{m}): \ V = \dfrac{\pi}{2}\ \text{m}^3 \approx 1600\ \text{liters}$	The wide vat	Rim radius $\sqrt{1} = 1$ m, $2$ m across
$1\ \text{liter} = 1000\ \text{cm}^3, \quad 1\ \text{m}^3 = (100\ \text{cm})^3 = 10^6\ \text{cm}^3$	Conversions	Cube the unit along with the number

Key Insight: In $y = x^2$ , $y$ is a length but $x^2$ is a length squared, so the equation draws a different shape in each unit. Compute with unit-less numbers and attach the units at the end, knowing which cauldron the equation describes.

IIVisualization 1 — One Depth, Many Cauldrons

Reading $y = x^2$ in a different unit draws a different cauldron, though the depth stays $1$ m.

Units of x and y

IIIVisualization 2 — Ten Times the Radius at Every Level

At every height the vat's radius is ten times the skinny cauldron's, so its cross-sections have a hundred times the area.

Height y (m) 0.25

IVQuiz Questions

Problem 1 · Liters in the Skinny Cauldron

Given: $V = \tfrac{\pi}{2}a^2$ with $a = 100$ , measuring in centimeters — find the volume in liters, to three significant figures.

✅ Correct!

\tfrac{\pi}{2}\cdot 10^4\ \text{cm}^3 = \tfrac{\pi}{2}\cdot 10

liters

\approx 15.7

liters.

❌ That is the meter reading, $a = 1$ . With

a = 100

in centimeters the formula gives

\tfrac{\pi}{2}\cdot 10^4

cubic centimeters, the skinny cauldron.

❌ Check the conversion. A liter is

1000\ \text{cm}^3

, not

100\ \text{cm}^3

10^4\ \text{cm}^3 = 10

liters.

❌ Check the conversion.

10^4\ \text{cm}^3

10

liters, since one liter is

1000\ \text{cm}^3

❌ Not quite. Compute

\tfrac{\pi}{2}(100)^2

cubic centimeters and divide by

1000

Show solution

V = \frac{\pi}{2}(100)^2 = \frac{\pi}{2}\cdot 10^4\ \text{cm}^3.

Since $1000\ \text{cm}^3 = 1$ liter, $10^4\ \text{cm}^3 = 10$ liters, so

V = \frac{\pi}{2}\cdot 10\ \text{liters} = 5\pi\ \text{liters} \approx 15.7\ \text{liters}.

Problem 2 · How Many Times Bigger?

Given: the vat ( $y = x^2$ in meters) and the skinny cauldron ( $y = x^2$ in centimeters) are both $1$ m deep, and at every height the vat's radius is $10$ times the skinny one's — find the ratio of their volumes.

✅ Correct! Each horizontal cross-section has

10^2 = 100

times the area and the depths match, so the volume is

100

times larger:

1570

liters against

15.7

liters.

❌ Only two of the three dimensions grow.

1000

would be right if the vat were the skinny cauldron scaled up by

10

in every direction, but the two have the same depth.

❌ The radius is only one length of a cross-section. A horizontal disk has area

\pi r^2

, so ten times the radius gives a hundred times the area.

❌ These are different shapes.

1

m and

100

cm are the same depth, but

y = x^2

draws a different cauldron in each unit, so the volumes differ.

❌ Not quite. Compare the areas of the cross-sections at the same height, then the depths.

Show solution

At height $y$ the vat has radius $r$ and the skinny cauldron radius $\tfrac{r}{10}$ . The cross-section areas are $\pi r^2$ and $\pi \tfrac{r^2}{100}$ , a ratio of $100$ at every level.

Slicing horizontally, each volume is the integral of its cross-section area over the same depth, so the ratio of volumes is also $100$ :

\frac{\tfrac{\pi}{2}\ \text{m}^3}{\tfrac{\pi}{2}\cdot 10^4\ \text{cm}^3} = \frac{\tfrac{\pi}{2}\cdot 10^6\ \text{cm}^3}{\tfrac{\pi}{2}\cdot 10^4\ \text{cm}^3} = 100.

Problem 3 · The Millimeter Cauldron

Given: the same depth of $1$ m, but $y = x^2$ read in millimeters, so $a = 1000$ — find the width of the rim and the volume.

How wide is the rim?

How much does it hold?

✅ Correct! The rim radius is

\sqrt{1000} \approx 31.6

mm, so the rim is about

63

mm across, and

V = \tfrac{\pi}{2}\cdot 10^6\ \text{mm}^3 = \tfrac{\pi}{2}

liters: an even skinnier cauldron.

❌ Check the rim. At the rim

1000 = x^2

, so

x = \sqrt{1000} \approx 31.6

mm is the radius; the width across is twice that.

❌ Check the volume.

\tfrac{\pi}{2}(1000)^2 = \tfrac{\pi}{2}\cdot 10^6

cubic millimeters, and one liter is

(100\ \text{mm})^3 = 10^6\ \text{mm}^3

Show solution

Rim: $y = a = 1000$ gives $x^2 = 1000$ , $x = \sqrt{1000} \approx 31.6$ mm, so the rim is about $63$ mm across.

Volume: $V = \tfrac{\pi}{2}(1000)^2 = \tfrac{\pi}{2}\cdot 10^6\ \text{mm}^3$ . A liter is $1000\ \text{cm}^3 = 1000 \cdot 10^3\ \text{mm}^3 = 10^6\ \text{mm}^3$ , so

V = \frac{\pi}{2}\ \text{liters} \approx 1.57\ \text{liters}.

Centimeters gave $15.7$ liters and meters $1570$ : each reading of $y = x^2$ is a different cauldron.

Problem 4 · Which Equation Ignores Units?

Given: a curve through the origin is described by an equation in $x$ and $y$ , both lengths — which equation draws the same shape whatever unit of length is used?

✅ Correct! Changing units multiplies both

x

and

y

by the same factor

k

ky = 3kx

is again

y = 3x

. Both sides are lengths, so the shape is scale-invariant.

❌ That is the cauldron's equation.

ky = (kx)^2

becomes

y = kx^2

, a different parabola in each unit.

❌ The two sides have different units.

ky = (kx)^3

becomes

y = k^2x^3

; a length cannot equal a length cubed in every unit.

❌ The $x^2$ term breaks it.

ky = k^2x^2 + kx

becomes

y = kx^2 + x

, which changes with

k

❌ Not quite. Replace

x

kx

and

y

ky

and see which equation is unchanged.

Show solution

Switching from one unit to a unit $k$ times smaller multiplies every length by $k$ : $x \to kx$ , $y \to ky$ . An equation draws the same shape in both units exactly when it is unchanged by this substitution.

$y = 3x$ : $ky = 3kx \iff y = 3x$ . Unchanged.
$y = x^2$ : $ky = k^2x^2 \iff y = kx^2$ . Changed.
$y = x^3$ : $ky = k^3x^3 \iff y = k^2x^3$ . Changed.
$y = x^2 + x$ : $ky = k^2x^2 + kx \iff y = kx^2 + x$ . Changed.

Only the equation whose two sides are both plain lengths survives every change of units.

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