Linear Growth and Linear Decay — Class 9 Mathematics Chapter 2

🖼 Image PendingA wide cinematic split image — on the left, a small green plant grows steadily upwards in a clay pot, with faint upward arrows showing its progress; on the right, a tall lit candle slowly melts down into a shorter stub, with faint downward arrows showing its decline

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio) split into two halves with a soft vertical divider. Left half: a small bright green plant in a brown clay pot, growing upward with several leaves and a tall central stem; faint glowing upward arrows trace its growth alongside a vertical scale labelled '0, 1, 2, 3' (months). Right half: a tall white candle on a brass holder, half-melted, its flame burning brightly; wax pools at its base; faint glowing downward arrows trace the descending wax line alongside a vertical scale labelled '0, 1, 2, 3' (months). Both halves share the same warm twilight ambient lighting. The image conveys: linear growth and linear decay are the same algebra running in opposite directions. Painterly cinematic illustration. Dark background. No additional text labels.

✦Think About It

A village adds 50 people every year. A mobile phone loses ₹800 in value every year. Both change by a constant per year. So why does one feel like 'growth' and the other like 'loss'?

Look at the sign of the per-year change. What's different between $+50$ and $-800$ ?

✦ब्रह्मस्फुटसिद्धान्त — ब्रह्मगुप्त (६२८ CE)

From Brahmagupta's Brahmasphuṭa-siddhānta — Dhana and Ṛṇa

धनयोः धनं भवति ऋणयोः ऋणम् एव हि

ऋण-धन-योगे शून्यं — एवं संख्याः गण्यन्ते॥

(dhanayoḥ dhanaṃ bhavati ṛṇayoḥ ṛṇam eva hi / ṛṇa-dhana-yoge śūnyaṃ — evaṃ saṃkhyāḥ gaṇyante)

'दो धन (positive) मिलाने पर धन ही मिलता है। दो ऋण (negative) मिलाने पर ऋण मिलता है। बराबर के धन-ऋण को जोड़ दो — तो शून्य बनता है। संख्याओं का यही नियम है।'

'Two positives (dhana) added give a positive. Two negatives (ṛṇa) added give a negative. Equal positive and negative cancel into zero. This is the law of numbers.'

In 628 CE, Brahmagupta did something revolutionary: he treated negative numbers as full citizens of mathematics — and named them after debt (ṛṇa, ऋण), in contrast to wealth (dhana, धन). On this page, positive leading coefficient $a > 0$ is dhana — wealth, growth. Negative $a < 0$ is ṛṇa — debt, decay. The Sanskrit terminology is exactly the modern terminology of growth and decay.

Linear growth — Example 9 (NCERT cost-of-journey)

Linear expressions model situations where a quantity increases (grows) or decreases (decays) by a constant amount per unit time.

Consider the linear function for the cost of a journey:

$C(d) = 100 + 60d$

where $C$ is the cost in rupees and $d$ is the distance in kilometres.

$d$ (km)	0	1	2	3	4	5
$C$ (₹)	100	160	220	280	340	400

For every 1 km increase in $d$ , the cost $C$ rises by exactly ₹60. This is the per-step rule of every linear polynomial — the leading coefficient is the per-step change. Because $a = +60 > 0$ , this is linear growth.

A real-world image of this: a taxi meter ticking up steadily as the car covers ground.

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Worked Example 1Cost for 15 km (NCERT Think-Reflect)

NCERT

Using $C(d) = 100 + 60d$ : (a) What is the cost for travelling 15 km? (b) For how many kilometres will the cost be ₹700?

Linear decay — Example 10 (NCERT water tank)

Now consider a linear function for the height of water in a cylindrical tank:

$h(t) = 3 - 0.5t$

where $h$ is the height in metres and $t$ is the number of months since the start of summer.

$t$ (months)	0	1	2	3	4
$h$ (m)	3	2.5	2	1.5	1

For every 1-month increase in $t$ , the height $h$ decreases by exactly 0.5 m. Because the leading coefficient is $a = -0.5 < 0$ , this is linear decay.

A real-world image of this: water evaporating from an open tank in steady weekly amounts.

Same algebra, opposite story. The cost-of-journey was linear growth ( $a = +60$ ). The water tank is linear decay ( $a = -0.5$ ). Both follow $f(t) = at + b$ — the only difference is the sign of $a$ .

Worked Example 2Water tank after 5 months (NCERT Think-Reflect)

NCERT

Using $h(t) = 3 - 0.5t$ : What will the height of water be at the end of 5 months?

Formal definitions

Linear Growth and Linear Decay — Definitions

A linear pattern $f(t) = at + b$ describes:

| Sign of $a$ | Behaviour | Verbal name | |---|---|---| | $a > 0$ | Increases by $a$ per step | Linear growth | | $a < 0$ | Decreases by $|a|$ per step | Linear decay | | $a = 0$ | Constant — doesn't change | (constant pattern, not 'linear') |

The leading coefficient $a$ is also called the growth rate (when positive) or the decay rate (when negative, with magnitude $|a|$ ).

Sanity-check trick. A linear-growth pattern eventually exceeds any number. A linear-decay pattern eventually drops below any number — including zero. So most decay models have a natural domain (e.g., the water tank can only be modelled until the water runs out at $t = 6$ ). When you set up a real-world model, always think: 'where does this break down?'

Worked Example 3Plant grows 2 cm a week

SOLVED

A plant is 15 cm tall today and grows 2 cm per week. (a) Write the linear function $h(t)$ for its height after $t$ weeks. (b) When will it reach 39 cm? (c) Is this growth or decay?

Worked Example 4A bucket leaks ₹0.20 of water per minute

SOLVED

A bucket holds 10 litres of water and leaks 0.2 L per minute. (a) Write $V(t)$ for volume after $t$ minutes. (b) When does the bucket run dry? (c) Is this growth or decay?

Logical ReasoningLevel 4 · Evaluation

Plant A grows linearly: starts at 5 cm, grows 0.5 cm per week. Plant B also grows linearly: starts at 2 cm, grows 1 cm per week. Will plant B ever catch up to plant A in height? If so, when?

Practice Yourself — Spot Growth or Decay

For each scenario: (a) write the linear function $f(t)$ ; (b) classify as growth, decay, or constant; (c) answer the asked question. Cover the answers and try.

A gym membership balance starts at ₹6{,}000, decreasing ₹500 per month. When does it run out?
A tree is 3 m tall and grows 0.4 m per year. Height after 8 years?
Daily revenue of a stall is ₹1{,}200 and increases ₹40 per day. Revenue on day 25?
An ice block weighs 2 kg and melts 0.05 kg per minute. When is it gone?
A library book is checked out and ₹2 fine is added per day overdue. Total fine after 7 days overdue?
A car's odometer reads 45{,}000 km. The car is driven 80 km per day. Reading after 30 days?

Answers: 1. $f(t) = 6000 - 500t$ ; decay; runs out at $t = 12$ months. 2. $f(t) = 3 + 0.4t$ ; growth; $f(8) = 6.2$ m. 3. $f(t) = 1200 + 40t$ ; growth; $f(25) = 2200$ ₹. 4. $f(t) = 2 - 0.05t$ ; decay; gone at $t = 40$ min. 5. $f(t) = 2t$ ; growth (from zero); $f(7) = ₹14$ . 6. $f(t) = 45000 + 80t$ ; growth; $f(30) = 47{,}400$ km.

Quick Check

Q1.The water tank height $h(t) = 3 - 0.5t$ describes:

AI Generation Prompt

✦Think About It

A village adds 50 people every year. A mobile phone loses ₹800 in value every year. Both change by a constant per year. So why does one feel like 'growth' and the other like 'loss'?

Look at the sign of the per-year change. What's different between $+50$ and $-800$ ?

✦ब्रह्मस्फुटसिद्धान्त — ब्रह्मगुप्त (६२८ CE)

From Brahmagupta's Brahmasphuṭa-siddhānta — Dhana and Ṛṇa

धनयोः धनं भवति ऋणयोः ऋणम् एव हि

ऋण-धन-योगे शून्यं — एवं संख्याः गण्यन्ते॥

(dhanayoḥ dhanaṃ bhavati ṛṇayoḥ ṛṇam eva hi / ṛṇa-dhana-yoge śūnyaṃ — evaṃ saṃkhyāḥ gaṇyante)

'Two positives (dhana) added give a positive. Two negatives (ṛṇa) added give a negative. Equal positive and negative cancel into zero. This is the law of numbers.'

Linear growth — Example 9 (NCERT cost-of-journey)

Linear expressions model situations where a quantity increases (grows) or decreases (decays) by a constant amount per unit time.

Consider the linear function for the cost of a journey:

$C(d) = 100 + 60d$

where $C$ is the cost in rupees and $d$ is the distance in kilometres.

$d$ (km)	0	1	2	3	4	5
$C$ (₹)	100	160	220	280	340	400

A real-world image of this: a taxi meter ticking up steadily as the car covers ground.

Loading simulator…

Worked Example 1Cost for 15 km (NCERT Think-Reflect)

NCERT

Using $C(d) = 100 + 60d$ : (a) What is the cost for travelling 15 km? (b) For how many kilometres will the cost be ₹700?

Linear decay — Example 10 (NCERT water tank)

Now consider a linear function for the height of water in a cylindrical tank:

$h(t) = 3 - 0.5t$

where $h$ is the height in metres and $t$ is the number of months since the start of summer.

$t$ (months)	0	1	2	3	4
$h$ (m)	3	2.5	2	1.5	1

For every 1-month increase in $t$ , the height $h$ decreases by exactly 0.5 m. Because the leading coefficient is $a = -0.5 < 0$ , this is linear decay.

A real-world image of this: water evaporating from an open tank in steady weekly amounts.

Worked Example 2Water tank after 5 months (NCERT Think-Reflect)

NCERT

Using $h(t) = 3 - 0.5t$ : What will the height of water be at the end of 5 months?

Formal definitions

Worked Example 3Plant grows 2 cm a week

SOLVED

A plant is 15 cm tall today and grows 2 cm per week. (a) Write the linear function $h(t)$ for its height after $t$ weeks. (b) When will it reach 39 cm? (c) Is this growth or decay?

Worked Example 4A bucket leaks ₹0.20 of water per minute

SOLVED

A bucket holds 10 litres of water and leaks 0.2 L per minute. (a) Write $V(t)$ for volume after $t$ minutes. (b) When does the bucket run dry? (c) Is this growth or decay?

Logical ReasoningLevel 4 · Evaluation

Quick Check

Q1.The water tank height $h(t) = 3 - 0.5t$ describes: