Differential-Equations · Unit 3 · Video 2 · Interactive Practice

Heat, Salt, and One Equation: Building the Conduction-Diffusion Model

IKey Formulas

Formula	Name	What you need
$\dfrac{dT}{dt} = k(T_e - T)$	Conduction model (Newton's law of cooling)	The bath temperature $T_e$ and a conductivity $k > 0$
$T(0) = T_0$	Initial condition — with the equation, an IVP	The inside temperature at the starting time
$\dfrac{dC}{dt} = k_1(C_e - C)$	Diffusion model	The same equation, concentration in place of temperature
$\dfrac{dT}{dt} + k(t)\,T = k(t)\,T_e(t)$	Standard linear form, with $p = k$ and $q = kT_e$	Nothing new — the rewrite is pure algebra

Key Insight: The order inside the bracket is not a convention, it is what keeps $k$ positive. If the bath is the hotter of the two then $T_e - T > 0$ and the inside must warm, so $dT/dt > 0$ forces $k > 0$ ; written the other way round, $T - T_e$ , the very same physics would demand a negative conductivity.

IIVisualization 1 — Which Order Keeps $k$ Positive

The rate is a straight line in $T$ crossing zero at the bath temperature — one order tilts it down, the other up.

Written order

IIIVisualization 2 — One Equation, Two Stories

The gap drives the rate and the rate closes the gap, so the approach only ever slows.

Model Time t 0.0 Constant 0.35

💡 The bath is data, not a second unknown: $T_e$ is prescribed from outside the chamber, so the equation still has exactly one unknown function even when $T_e$ changes with time.

IVVisualization 3 — Standard Form, and How General $q$ Can Be

In standard form the whole bath collapses into the right-hand side $q(t) = k(t)\,T_e(t)$ .

Bath Wall

Newton's form

\dfrac{dT}{dt} = k(t)\big(T_e(t) - T\big)

Multiply out the bracket

\dfrac{dT}{dt} = k(t)T_e(t) - k(t)T

Move the $T$ -term across

\dfrac{dT}{dt} + k(t)T = k(t)T_e(t)

term under term with

y' + p(x)y = q(x)

💡 The rewrite used no physics at all, only algebra, so every model in the previous video's list arrives at the same standard form.

VQuiz Questions

Problem 1 · Reading the Model

Given: a chamber at $T = 12$ hangs in a bath held at $T_e = 20$ (degrees Celsius), and the wall has conductivity $k = 0.4$ per minute — find $\dfrac{dT}{dt}$ at this instant.

✅ Correct!

0.4(20 - 12) = 0.4 \times 8 = 3.2

, positive because the bath is the hotter of the two and heat leaks inward.

❌ The bracket is a difference, not a sum.

T_e - T = 20 - 12 = 8

, not

32

. Newton's law is driven by the gap between the two temperatures, which vanishes when they are equal.

❌ Check the order. The bracket is external minus internal,

20 - 12

; you computed

12 - 20

, which says a chamber sitting in a hotter bath gets colder.

❌ That is the temperature difference, not the rate. The difference still has to be multiplied by the conductivity:

0.4 \times 8 = 3.2

❌ Not quite. Substitute straight into

\dfrac{dT}{dt} = k(T_e - T)

with

k = 0.4

T_e = 20

and

T = 12

Show solution

The model is evaluated, not solved — everything on the right is known at this instant:

\frac{dT}{dt} = k(T_e - T) = 0.4(20 - 12) = 0.4 \times 8 = 3.2

Two checks are worth making every time. The bracket is external minus internal, so with the bath hotter it comes out positive; and $k > 0$ , so the rate comes out positive too — the chamber warms, as it must. Note also that $3.2$ is the rate right now: as $T$ climbs toward $20$ the bracket shrinks and the warming slows.

Problem 2 · The Order and the Sign of $k$

Given: an engineer models the same warming chamber, $T_e > T$ , but writes the bracket backwards as $\dfrac{dT}{dt} = k(T - T_e)$ . For that equation to describe a chamber that actually warms, what must be true of $k$ ?

✅ Correct!

T - T_e < 0

while

dT/dt > 0

, so the constant in front must be negative — and a negative conductivity is exactly what the standard order was chosen to avoid.

❌ Check the sign of the bracket. With

T_e > T

the factor

T - T_e

is negative, so a positive

k

gives

dT/dt < 0

: the chamber would cool inside a hotter bath.

❌ Then nothing happens at all.

k = 0

gives

dT/dt = 0

, a perfectly insulating wall — and the video's chamber is explicitly not perfectly insulated.

❌ The two brackets differ by a factor of $-1$ . They describe the same physics only if the constant flips sign too, so the choice of order really does decide the sign of

k

❌ Not quite. Fix the physics first — a chamber in a hotter bath has

dT/dt > 0

— then read off the sign the constant is forced to carry.

Show solution

The physics fixes the left-hand side. Heat leaks inward, so the inside temperature rises: $dT/dt > 0$ .

The written bracket has a fixed sign too, because $T_e > T$ is given: $T - T_e < 0$ .

\underbrace{\frac{dT}{dt}}_{>\,0} \; = \; k \cdot \underbrace{(T - T_e)}_{<\,0} \quad \Longrightarrow \quad k < 0

A positive quantity equal to $k$ times a negative quantity forces $k$ to be negative. Nobody wants to carry a negative conductivity through a calculation — physical parameters are arranged so they come out positive — so the difference is always written external minus internal, and $k > 0$ .

Problem 3 · Into Standard Linear Form

Given: $\dfrac{dT}{dt} = k(T_e - T)$ with constant conductivity $k = 0.25$ and a bath that rises steadily, $T_e(t) = 30 + 4t$ — rewrite it as $T' + pT = q$ and identify the two coefficients.

What is $p$ ?

What is $q$ ?

✅ Correct! Multiplying out gives

T' = 7.5 + t - 0.25T

, so

T' + 0.25T = 7.5 + t

: the coefficient

p

is the conductivity and

q

is the conductivity times the bath.

❌ Check $p$ .

p

is whatever multiplies

T

after the

T

-term has crossed the equals sign, and

-0.25T

crosses as

+0.25T

. So

p = k = 0.25

, a positive constant — never the bath, and never negative.

❌ Check $q$ . What is left on the right is

kT_e(t)

, and every term of the bath is multiplied by

k

0.25(30 + 4t) = 7.5 + t

. Leaving it as

30 + 4t

forgets the

k

;

7.5 + 4t

scales only the first term.

Show solution

Step 1 — multiply out the bracket.

\frac{dT}{dt} = 0.25\big((30 + 4t) - T\big) = 7.5 + t - 0.25\,T

Step 2 — move the $T$ -term to the left.

\frac{dT}{dt} + 0.25\,T = 7.5 + t

Step 3 — compare with $y' + py = q$ . Term under term, $p$ sits against $0.25$ and $q$ against the whole right-hand side:

p = 0.25, \qquad q(t) = 7.5 + t

The equation is visibly linear with a positive $p$ , and its right-hand side is a general function of $t$ — here a rising straight line, because the bath is rising.

Problem 4 · A Congealing Membrane

Given: salt diffuses into a chamber through a membrane that keeps setting, so its constant is really a function of time, $k_1(t) = \dfrac{2}{1+t}$ , while the brine outside is stirred to follow $C_e(t) = 5 + \sin t$ — which standard-form equation is this model?

✅ Correct!

p(t) = k_1(t)

and

q(t) = k_1(t)C_e(t)

, both now functions of

t

— which is a general first-order linear equation, not a special one.

❌ The right-hand side is $k_1C_e$ , not $C_e$ . Multiplying out

k_1(C_e - C)

scales the brine by the very same

k_1

that ends up multiplying

C

on the left.

❌ Check the sign. Moving

-k_1C

from the right to the left turns it into

+k_1C

, so

p = +k_1

— and

p

coming out positive is the whole reason the bracket is written external minus internal.

❌ $p$ and $q$ are swapped.

p

is whatever multiplies the unknown

C

, and that is the conductivity

k_1(t)

; the brine

C_e(t)

never touches the unknown and stays on the right.

❌ Not quite. Multiply out

k_1(t)\big(C_e(t) - C\big)

, then move the

C

-term across the equals sign.

Show solution

The diffusion model is word for word the conduction model, so run the same two algebraic steps:

\frac{dC}{dt} = k_1(t)\big(C_e(t) - C\big) = k_1(t)C_e(t) - k_1(t)C

\frac{dC}{dt} + \frac{2}{1+t}\,C = \frac{2}{1+t}\,(5 + \sin t) = \frac{2(5 + \sin t)}{1+t}

So $p(t) = \dfrac{2}{1+t}$ and $q(t) = \dfrac{2(5 + \sin t)}{1+t}$ .

Nothing in those two lines used the fact that $C$ is a concentration; the identical steps turn the conduction model into $T' + k(t)T = k(t)T_e(t)$ . And with both coefficients free to depend on $t$ , the last trace of specialness is gone: this is $y' + p(x)y = q(x)$ in full generality, so one method for that equation solves every one of these models.

Solved: 0 / 4

Heat, Salt, and One Equation: Building the Conduction-Diffusion Model

IKey Formulas

IIVisualization 1 — Which Order Keeps kkk Positive

IIIVisualization 2 — One Equation, Two Stories

IVVisualization 3 — Standard Form, and How General qqq Can Be

VQuiz Questions

IIVisualization 1 — Which Order Keeps $k$ Positive

IVVisualization 3 — Standard Form, and How General $q$ Can Be