Why is A equal to Q naught minus P?

At t equals zero, Q equals P plus A, so A equals Q naught minus P.

Why is A equal to Q naught minus P?

At t equals zero, Q equals P plus A, so A equals Q naught minus P.

Year 12 Maths Extension 1 (2027) Further applications of calculus

Rate ∝ (Q−P); Newton’s Law of Cooling; Q = P + Ae^{kt}

Q: What is Newton's Law of Cooling?

A model where the rate of temperature change is proportional to the gap from the surroundings: dQ by dt equals k times Q minus P, giving Q equals P plus A e to the k t.

Q: How do you find A and k?

Use the initial value for A equals Q naught minus P, then a second data point to solve for k, keeping e to the k t exact.

Q: What is the limiting value?

When k is negative, Q approaches P as t grows; Q equals P is the horizontal asymptote.

Q: Does this model growth too?

Yes, the same equation models any quantity approaching a limit P, such as a population nearing a carrying capacity.

20 practice questions 2 video lessons Theory + worked examples

Explore Newton's Law of Cooling in NSW Year 12 Mathematics Extension 1, where a quantity changes at a rate proportional to its difference from the surroundings. Solving this gives the exponential model that describes how objects cool and warm towards a limiting value.

You will learn to set up and verify the differential equation and its solution, find the constants from given conditions, and interpret the limiting value and asymptote — a practical application of calculus that recurs in the HSC Extension 1 exam.

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Theory

When a quantity changes at a rate proportional to its gap from a fixed value $P$ , it follows $\frac{d Q}{d t} = k (Q - P)$ , solved by $Q = P + A e^{k t}$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-05) models cooling and limiting values, with $A = Q_{0} - P$ .

When a quantity changes at a rate proportional to how far it is from a fixed value $P$ , it follows $\frac{d Q}{d t} = k (Q - P)$ , whose solution is $Q = P + A e^{k t}$ . Newton's Law of Cooling is this model with $P$ the surrounding temperature.

Differentiating $Q = P + A e^{k t}$ gives $\frac{d Q}{d t} = A k e^{k t} = k (Q - P)$ , which verifies the solution. The constant $A = Q_{0} - P$ is fixed by the initial value.

If $k < 0$ then $Q \to P$ as $t \to \infty$ , so $Q = P$ is a horizontal asymptote — the limiting value.

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01. The syllabus uses $\frac{d Q}{d t} = k (Q - P)$ , $Q = P + A e^{k t}$ to model cooling and ecosystems with a carrying capacity.

Cooling

(k < 0, A > 0)

Q

falls toward the asymptote

Q = P

Approaching a limit from below

(A < 0)

, as with a carrying capacity.

\frac{d Q}{d t} = k (Q - P) ⟺ Q = P + A e^{k t} .

dQ/dt = k(Q - P) has solution Q = P + A e^(kt)

The initial value fixes $A$ , and the long-run value is $P$ when $k < 0$ :

A = Q_{0} - P, Q \to P as t \to \infty (k < 0) .

A = Q0 - P; Q approaches P as t grows when k < 0

Keep it exact. Carry $e^{k t}$ exactly (e.g. $e^{4 k} = \frac{4}{7}$ ) and use index laws — $e^{8 k} = (e^{4 k})^{2}$ — rather than rounding $k$ early.

How to solve a cooling / limiting-value problem

Set the model: $Q = P + A e^{k t}$ with $P$ the limiting value.
Find $A$ from the initial value: $A = Q_{0} - P$ .
Find $k$ from a second data point, keeping $e^{k t}$ exact.
Answer the question — evaluate $Q$ at a time (use index laws) or find the limit $P$ .

Example 1 — Verify the solution

Show

Q = P + A e^{k t}

satisfies

\frac{d Q}{d t} = k (Q - P)

Solution

\frac{d Q}{d t}

=

A k e^{k t} = k (A e^{k t}) = k (Q - P)

dQ/dt = A k e^(kt) = k(Q - P)

So the equation is satisfied.

Example 2 — Find A and k

A body cools as

T = 20 + A e^{k t}

. Given

T = 90

t = 0

and

T = 60

t = 5

, find

A

and

k

Solution

$90 = 20 + A$	$\Rightarrow$	$A = 70$
$60 = 20 + 70 e^{5 k}$	$\Rightarrow$	$e^{5 k} = \frac{4}{7}$
$k$	$=$	$\frac{1}{5} \ln \frac{4}{7}$

A = 70, k = (1/5) ln(4/7)

$A = 70$ , $k = \frac{1}{5} \ln \frac{4}{7}$ .

Example 3 — Cooling coffee

Coffee at

95^{\circ}

C in a

25^{\circ}

C room cools to

65^{\circ}

C after

4

min. Find the temperature after

8

min.

Solution

$T = 25 + 70 e^{k t}$ ; $e^{4 k} = \frac{40}{70} = \frac{4}{7}$ .

$T (8)$	$=$	$25 + 70 {(e^{4 k})}^{2}$
	$=$	$25 + 70 {(\frac{4}{7})}^{2} \approx {47.9}^{\circ}$ C

T(8) = 25 + 70 (4/7)^2 approx 47.9 C

About ${47.9}^{\circ}$ C.

Example 4 — Population to a limit

A population is

P = 800 - 500 e^{- 0.1 t}

. State the initial and limiting populations, and find

P

t = 10

Solution

$P (0)$	$=$	$800 - 500 = 300$
$P (\infty)$	$=$	$800$
$P (10)$	$=$	$800 - 500 e^{- 1} \approx 616$

initial 300, limit 800, P(10) approx 616

Initial $300$ , limit $800$ , $P (10) \approx 616$ .

Common pitfalls

Wrong $A$ .

A = Q_{0} - P

— the initial value minus the limiting value

P

, not just

Q_{0}

Sign of $k$ . For cooling,

k < 0

and

Q

approaches

P

from above.

Rounding early. Keep

e^{k t}

exact and use index laws (

e^{8 k} = (e^{4 k})^{2}

); round only at the end.

Ignoring the asymptote. The limiting value is

P

, the horizontal asymptote of

Q = P + A e^{k t}

Frequently asked questions

What is Newton's Law of Cooling?

A model where the rate of temperature change is proportional to the gap from the surroundings: $\frac{d Q}{d t} = k (Q - P)$ , giving $Q = P + A e^{k t}$ .

How do you find A and k?

Use the initial value for $A = Q_{0} - P$ , then a second data point to solve for $k$ , keeping $e^{k t}$ exact.

What is the limiting value?

When $k < 0$ , $Q \to P$ as $t \to \infty$ ; $Q = P$ is the horizontal asymptote.

Why is A equal to Q0 minus P?

At $t = 0$ , $Q = P + A$ , so $A = Q_{0} - P$ .

Does this model growth too?

Yes — the same equation models any quantity approaching a limit $P$ , such as a population nearing a carrying capacity.

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Rate ∝ (Q−P); Newton’s Law of Cooling; Q = P + Ae^{kt}

📖 Theory

How to solve a cooling / limiting-value problem

Common pitfalls

Frequently asked questions

Theory