Differential-Equations · Unit 4 · Video 1 · Interactive Practice

Scale Before You Solve: The Fourth-Power Cooling Law Made Dimensionless

IKey Formulas

Formula	Name	What it says
$x_1 = \dfrac{x}{a}, \quad y_1 = \dfrac{y}{b}$	Scaling	Measure $x$ in units of size $a$ and $y$ in units of size $b$ ( $a$ , $b$ constants)
$\dfrac{dT}{dt} = k\,(M^4 - T^4)$	Fourth-power cooling law (very high temperatures)	Body $T$ , constant surroundings $M$ , both absolute; $k > 0$
$T = M\,T_1: \quad \dfrac{dT}{dt} = M\,\dfrac{dT_1}{dt}, \quad T^4 = M^4\,T_1^4$	Substitution	Define $T_1 = T/M$ , substitute for $T$ ; the derivative step needs $M$ constant
$\dfrac{dT_1}{dt} = k_1\,(1 - T_1^4), \quad k_1 = kM^3$	Scaled cooling law	$T_1$ has no units; $k_1$ is in per second

Key Insight: Define $T_1 = T/M$ , but substitute $T = MT_1$ , because $T$ is what appears in the equation. The scaled equation is no easier to solve; it has a dimensionless temperature and one constant, $k_1 = kM^3$ , in place of two.

IIVisualization 1 — Choosing the Unit of Temperature

Measuring $T$ in units of $a$ relabels the axis and leaves the curve alone; only $a = M$ turns the bracket's constant into $1$ .

Unit a (kelvin) 100

IIIVisualization 2 — Why the Temperatures Must Be Absolute

A difference of two temperatures ignores where zero is; a difference of their fourth powers does not.

Body temperature T (kelvin) 1200

IVVisualization 3 — Two Constants Lumped into One

After scaling, $k$ and $M$ enter only through $k_1 = kM^3$ : equal $k_1$ and equal $T_1(0)$ give the same $T_1$ curve.

Surroundings M (kelvin) 400 Hold

💡 The substitution relies on $M$ being constant: if $M$ varied with $t$ , the product rule would give $\dfrac{dT}{dt} = M'\,T_1 + M\,\dfrac{dT_1}{dt}$ , and the scaled equation would gain the extra term $-\dfrac{M'}{M}\,T_1$ .

VQuiz Questions

Problem 1 · Lump the Constants

Given: $\dfrac{dT}{dt} = k\,(M^4 - T^4)$ with $k = 2\times10^{-12}$ per second per kelvin cubed and surroundings at $M = 500$ K. The substitution $T = MT_1$ turns it into $\dfrac{dT_1}{dt} = k_1\,(1 - T_1^4)$ . Find $k_1$ in per second.

✅ Correct!

k_1 = kM^3 = (2\times10^{-12})(1.25\times10^{8}) = 2.5\times10^{-4}

per second: the kelvin cubed in the units of

k

cancels against

M^3

❌ That is $kM^4$ . The left side carries a factor of

M

too,

\dfrac{dT}{dt} = M\,\dfrac{dT_1}{dt}

, and dividing by it leaves

kM^3

❌ Scaling does change the constant. Substituting

T = MT_1

pulls

M^4

out of the bracket and

M

out of the derivative, so the coefficient becomes the lump

kM^3

❌ The powers of $M$ multiply $k$ . After substitution,

M\,\dfrac{dT_1}{dt} = kM^4(1 - T_1^4)

; dividing by

M

gives

kM^3

, not

k/M^3

Show solution

Since $M$ is constant, $T = MT_1$ gives $\dfrac{dT}{dt} = M\,\dfrac{dT_1}{dt}$ and $T^4 = M^4T_1^4$ . Substituting:

M\,\frac{dT_1}{dt} = k\,(M^4 - M^4T_1^4) = kM^4\,(1 - T_1^4)

Divide both sides by $M$ :

\frac{dT_1}{dt} = kM^3\,(1 - T_1^4), \qquad k_1 = kM^3

With the numbers, $M^3 = 500^3 = 1.25\times10^{8}$ , so

k_1 = (2\times10^{-12})(1.25\times10^{8}) = 2.5\times10^{-4}

Units: per second per kelvin cubed, times kelvin cubed, leaves per second.

Problem 2 · Which Quantity Ignores the Zero?

Given: a body at $T = 1200$ K in surroundings at $M = 300$ K, that is $926.85\,\degree\text{C}$ and $26.85\,\degree\text{C}$ . Which quantity has the same value whether both temperatures are written in kelvin or in degrees Celsius?

✅ Correct! Celsius is kelvin shifted by

273.15

, and a shift cancels in a difference:

926.85 - 26.85 = 1200 - 300 = 900

❌ A ratio depends on where zero is.

1200/300 = 4

, but

926.85/26.85 \approx 34.5

T_1

is the same on every absolute scale, not on Celsius.

❌ A fourth power depends on where zero is.

300^4 - 1200^4 \approx -2.07\times10^{12}

, but

26.85^4 - 926.85^4 \approx -7.38\times10^{11}

❌ Only one of them survives. Shifting the zero by

273.15

cancels in a difference, but it changes a ratio and it changes fourth powers.

Show solution

Celsius readings are kelvin readings shifted: $T_C = T - 273.15$ , $M_C = M - 273.15$ .

Difference: $T_C - M_C = (T - 273.15) - (M - 273.15) = T - M$ . Here $926.85 - 26.85 = 900 = 1200 - 300$ . Same value.

Ratio: $1200/300 = 4$ , while $926.85/26.85 \approx 34.5$ . Different: a ratio is unchanged by a change of unit size (kelvin to Rankine), not by a shift of zero.

Fourth powers: $300^4 - 1200^4 \approx -2.07\times10^{12}$ , while $26.85^4 - 926.85^4 \approx -7.38\times10^{11}$ . Different.

So the answer is $T - M$ . This is why Newton's law, which uses only $M - T$ , works in Celsius, while the fourth-power law needs absolute temperatures.

Problem 3 · The Initial Cooling Rate

Given: the scaled law $\dfrac{dT_1}{dt} = k_1\,(1 - T_1^4)$ with $k_1 = 0.0027$ per second, surroundings $M = 300$ K and $T(0) = 1200$ K. Find the rates at $t = 0$ .

What is $dT_1/dt$ at $t = 0$ , in per second?

What is $dT/dt$ at $t = 0$ , in kelvin per second?

✅ Correct!

dT_1/dt = 0.0027 \times (-255) = -0.6885

per second, and

dT/dt = 300 \times (-0.6885) = -206.55

kelvin per second.

❌ Check the fourth power.

T_1(0) = 1200/300 = 4

and

T_1^4 = 256

, so the bracket is

1 - 256 = -255

, not

1 - 4 = -3

❌ Check the sign. The body is hotter than its surroundings,

T_1 = 4 > 1

, so

1 - T_1^4 \lt 0

and the temperature falls.

❌ The $1$ is still in the bracket.

1 - T_1^4 = 1 - 256 = -255

, not

-256

: the

1

is what the surroundings' term

M^4

becomes once

M^4

is factored out.

❌ That is $dT_1/dt$ . Convert back with

T = MT_1

\dfrac{dT}{dt} = M\,\dfrac{dT_1}{dt}

❌ Multiply by $M$ , do not divide.

T = MT_1

, so

\dfrac{dT}{dt} = M\,\dfrac{dT_1}{dt} = 300 \times (-0.6885)

❌ That uses $1 - 4 = -3$ . The bracket is

1 - 4^4 = -255

Show solution

Step 1: scaled starting value. $T_1(0) = T(0)/M = 1200/300 = 4$ .

Step 2: scaled rate.

\frac{dT_1}{dt} = 0.0027\,(1 - 4^4) = 0.0027 \times (-255) = -0.6885 \text{ per second}

Step 3: back to kelvin. $T = MT_1$ with $M$ constant, so

\frac{dT}{dt} = M\,\frac{dT_1}{dt} = 300 \times (-0.6885) = -206.55 \text{ kelvin per second}

Check with the original law. $k = k_1/M^3 = 0.0027/300^3 = 1\times10^{-10}$ , and $k\,(M^4 - T^4) = 10^{-10}\,(300^4 - 1200^4) = 10^{-10} \times (-2.0655\times10^{12}) = -206.55$ ✓

Problem 4 · Scale a Falling Body

Given: a body falling against air resistance, $\dfrac{dv}{dt} = g - c\,v^2$ , with $g$ and $c$ positive constants. Scale by the terminal speed: $v_1 = v/V$ with $V = \sqrt{g/c}$ . The equation becomes $\dfrac{dv_1}{dt} = k_1\,(1 - v_1^2)$ . Find $k_1$ .

✅ Correct!

k_1 = g/V = \sqrt{gc}

. With

g

in metres per second squared and

c

in per metre,

\sqrt{gc}

is in per second: inverse time only, like

k_1 = kM^3

❌ That is $V$ itself.

\sqrt{g/c}

is a speed (metres per second), but

k_1

must be in per second because

v_1

has no units.

❌ The left side carries a factor of $V$ .

v = Vv_1

gives

\dfrac{dv}{dt} = V\,\dfrac{dv_1}{dt}

, so you must divide

g

V

❌ Divide by $V$ , do not multiply.

V\,\dfrac{dv_1}{dt} = g\,(1 - v_1^2)

, so

k_1 = g/V

Show solution

Define $v_1 = v/V$ , but substitute $v = Vv_1$ ( $V$ is constant):

V\,\frac{dv_1}{dt} = g - cV^2v_1^2

Since $V^2 = g/c$ , the term $cV^2 = g$ , and the right side factors:

V\,\frac{dv_1}{dt} = g\,(1 - v_1^2)

Divide by $V$ :

\frac{dv_1}{dt} = \frac{g}{V}\,(1 - v_1^2), \qquad k_1 = \frac{g}{\sqrt{g/c}} = \sqrt{gc}

Two constants, $g$ and $c$ , have been lumped into one, exactly as $k$ and $M$ were lumped into $k_1 = kM^3$ .

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