A fictional inverse-square display explains negative-power derivatives, quartered outputs, one-eighth gradient scaling, tangent lines and admissible thresholds.
Understand the inverse-square model
- Start with the model and its limits
- A table reveals the inverse-square pattern
- Why the exponent is negative
- Prove the doubling rule algebraically
- A general scale-factor rule
- The input is not allowed to be zero
- The reading approaches zero but never equals it
- Differentiate the inverse-square expression
- Why the derivative has a negative sign
- Find the gradient at two metres
- Find the gradient at four metres
- Doubling distance changes the gradient by an eighth
- Compare value and gradient
- Draw the tangent at d = 2
- Verify the tangent at its contact point
- Why the tangent is not a global model
- Check with a nearby secant gradient
- The second derivative explains the curve’s shape
- What does concave up mean here?
- A general negative-power rule
- Solve a target reading equation
- Solve a threshold inequality
- Why the sign condition matters in the inequality
- Recover the input from the output
- Why two distances can produce one reading on the unrestricted graph
- Compare a negative-power derivative with a square-root derivative
- Units are a quick check
- A six-minute independent check
- Common mistake: a negative exponent means a negative reading
- Common mistake: doubling divides the reading by two
- Common mistake: forgetting the third power after differentiation
- Common mistake: treating a tangent as the exact function
- Common mistake: ignoring the positive domain in a quadratic step
- How parents can check understanding
- What focused A-Math tuition should diagnose
- Where this belongs in Singapore G3 Additional Mathematics
- Frequently asked questions
- Continue reading across Additional Mathematics
- Official syllabus and model limits
What is Additional Mathematics when doubling an input makes an output one-quarter as large—and makes the rate of change only one-eighth as large in magnitude?
A fictional distance-reading display provides an exact illustration. The display takes a positive distance d and reports a number that follows an inverse-square rule. It is designed to demonstrate negative powers and differentiation, not to simulate or certify a real radiation sensor, wireless receiver or measurement device.
The model is simple:
R(d) = 48/d², where d > 0.
The function describes a decreasing positive reading. Its derivative describes how rapidly that reading changes as distance increases. When written in power notation, its mathematical structure becomes especially clear.
In Singapore’s 2027 G3 Additional Mathematics syllabus, rational powers, derivatives, rates of change and tangents are part of the calculus pathway. This example makes those skills visible in one relationship.
Start with the model and its limits
Imagine a display whose input d is a positive number of metres from a reference point.
Define its fictional output R in arbitrary reading units:
R(d) = 48/d².
To make the dimensions consistent, the numerical coefficient 48 carries the corresponding reading-unit times metre-squared scale.
This function is not a measured physical law for any particular device. Some idealised physical models use inverse-square relationships under specific assumptions, but those assumptions are not established here.
Our only mathematical requirements are that d is positive and R follows the stated equation.
The purpose is to study how negative powers and derivatives behave over the real domain d > 0.
A table reveals the inverse-square pattern
| Distance d | Reading R(d) |
|---|---|
| 1 m | 48 units |
| 2 m | 12 units |
| 3 m | 16/3 units |
| 4 m | 3 units |
| 6 m | 4/3 units |
| 8 m | 3/4 units |
| 12 m | 1/3 unit |
All readings are positive.
The function falls sharply at smaller distances and becomes less steep as distance increases.
Notice:
At d = 2, the reading is 12.
At d = 4, twice that distance, the reading is 3.
At d = 8, twice the distance again, the reading is 3/4.
Each doubling divides the output by four.
This is an exact property of the inverse-square formula, not a pattern guessed from a small table.
Why the exponent is negative
The expression:
48/d²
can be written as:
48d⁻².
That is the negative-index law:
d⁻² = 1/d², provided d ≠ 0.
A negative exponent does not make the output negative.
It means the positive quantity is in the denominator.
For d > 0, d² is positive, so:
48/d² > 0.
This distinction matters when students first differentiate negative powers. The sign of an exponent and the sign of a function value are different things.
Prove the doubling rule algebraically
Evaluate the function at twice the distance:
R(2d) = 48/(2d)².
Expand the square of the entire product:
(2d)² = 4d².
Therefore:
R(2d) = 48/(4d²).
So:
R(2d) = (1/4)[48/d²].
Hence:
R(2d) = R(d)/4.
This is true for every d > 0.
The table checks individual instances; the algebra proves the relationship generally.
Doubling the input does not halve the output. The exponent two causes a fourfold change.
A general scale-factor rule
Suppose distance is multiplied by a positive factor k.
Then:
R(kd) = 48/(kd)².
Rearrange:
R(kd) = 48/(k²d²).
Thus:
R(kd) = R(d)/k².
For k = 2, the reading is quartered.
For k = 3, it becomes one-ninth.
For k = 1/2, it becomes four times as large.
These statements compare readings under the mathematical function.
They are not predictions of actual equipment performance. The scale law comes solely from the chosen exponent.
The input is not allowed to be zero
At d = 0:
R(0) would require 48/0².
That is division by zero, which is undefined.
Therefore the function’s real mathematical domain excludes zero.
Our model further restricts d to positive distances.
As d approaches zero from the positive side, R(d) grows without bound.
The vertical line d = 0 is an asymptote.
A student should not replace this limiting behaviour with a claim that the function has a finite reading at zero.
The asymptote is a property of the equation, not evidence about a real sensor near its reference point.
The reading approaches zero but never equals it
As d becomes larger:
d² increases.
So:
48/d² becomes smaller.
The reading approaches zero from above as d tends to positive infinity.
But for every finite positive d:
R(d) > 0.
Thus the horizontal axis R = 0 is a horizontal asymptote.
The function does not cross or meet it at any finite permitted input.
The same equation therefore has a vertical asymptote at d = 0 and a horizontal asymptote at R = 0 on its positive branch.
These graph facts can help students check derivatives and equations without assuming the display represents a physically unlimited system.
Differentiate the inverse-square expression
Start with:
R(d) = 48d⁻².
The power rule says:
d(dⁿ)/dd = ndⁿ⁻¹
for the relevant real domain and differentiable power.
Apply it:
R′(d) = 48(−2)d⁻³.
Therefore:
R′(d) = −96d⁻³.
Rewrite without the negative exponent:
R′(d) = −96/d³.
This derivative measures the change in the reading per metre of change in d.
The negative sign indicates the reading decreases as distance increases.
Why the derivative has a negative sign
Our original function is positive:
R(d) = 48/d² > 0.
But its derivative is negative:
R′(d) = −96/d³ < 0
for every d > 0.
There is no contradiction.
A positive function can be decreasing.
The sign of R tells us where the graph sits relative to the horizontal axis.
The sign of R′ tells us whether the graph rises or falls as d increases.
This distinction recurs throughout Additional Mathematics, especially in logarithmic, exponential and rational-power models.
Find the gradient at two metres
At d = 2:
R(2) = 48/4 = 12 units.
The derivative is:
R′(2) = −96/2³.
Since 2³ = 8:
R′(2) = −12 units per metre.
The point on the graph is:
(2, 12).
Its tangent gradient is −12.
A small increase in d near two metres produces an approximate decrease of twelve reading units per metre of additional distance.
The gradient is not the reading itself, even though the numerical magnitudes happen to coincide at this chosen input.
Find the gradient at four metres
At d = 4:
R(4) = 48/16 = 3 units.
The derivative is:
R′(4) = −96/4³.
Since 4³ = 64:
R′(4) = −3/2 units per metre.
So the graph is still decreasing, but its gradient magnitude is much smaller than at d = 2.
At two metres the slope is −12.
At four metres it is −1.5.
The curve becomes flatter with increasing distance.
This behaviour is consistent with the table of readings.
Doubling distance changes the gradient by an eighth
We already proved the reading is quartered when d doubles.
Now examine the derivative:
R′(d) = −96/d³.
At twice the distance:
R′(2d) = −96/(2d)³.
Because:
(2d)³ = 8d³,
we have:
R′(2d) = −96/(8d³).
Therefore:
R′(2d) = R′(d)/8.
The signed gradient is divided by eight, so its magnitude becomes one-eighth.
The reading follows an inverse-square scaling rule.
The rate of change follows an inverse-cube scaling rule.
This extra power of d arises naturally when differentiation lowers the exponent from −2 to −3.
Compare value and gradient
| Distance | Reading | Gradient |
|---|---|---|
| 1 m | 48 units | −96 units/m |
| 2 m | 12 units | −12 units/m |
| 3 m | 16/3 units | −32/9 units/m |
| 4 m | 3 units | −3/2 units/m |
| 6 m | 4/3 units | −4/9 units/m |
| 8 m | 3/4 units | −3/16 units/m |
The table distinguishes the output from the derivative.
Both shrink in magnitude with increasing distance, but at different rates.
At twice a given distance, the output is quartered and the derivative magnitude is reduced to one-eighth.
This is an example of how calculus reveals information that is not obvious from the original scale law alone.
Draw the tangent at d = 2
The point is:
(d, R) = (2, 12).
The gradient is:
m = −12.
Use the point–gradient equation for the tangent:
y − 12 = −12(d − 2).
Expand:
y − 12 = −12d + 24.
Therefore:
y = −12d + 36.
This is the tangent line to R(d) at d = 2.
It is valid as a local linear approximation, not as an exact replacement for the inverse-square curve at arbitrary distances.
Verify the tangent at its contact point
Substitute d = 2 into the tangent:
y = −12(2) + 36.
So:
y = 12.
That matches R(2).
Its gradient is −12, matching R′(2).
These are the two necessary checks: the line passes through the chosen point and has the curve’s local slope there.
A line with the correct point but a different gradient would not be the tangent.
The verification is straightforward and helps students avoid errors when expanding point–gradient form.
Why the tangent is not a global model
At d = 4, the tangent equation predicts:
y = −12(4) + 36 = −12.
But the inverse-square function gives:
R(4) = 3.
The tangent’s negative prediction is not a true reading of the original positive function.
It shows that a tangent line is a local approximation and can become very inaccurate far from the point of contact.
This distinction is important in modelling. A derivative provides local sensitivity, not permission to extrapolate the tangent indefinitely.
An unphysical tangent prediction does not make the original function negative.
Check with a nearby secant gradient
Use a point slightly to the right of d = 2, namely d = 2.1.
The function gives:
R(2.1) = 48/(2.1²).
Since 2.1² = 4.41:
R(2.1) ≈ 10.8844.
The secant slope from d = 2 to d = 2.1 is:
[R(2.1) − R(2)]/(2.1 − 2).
Approximately:
(10.8844 − 12)/0.1 = −11.156.
That is close to the exact tangent gradient −12.
The two values are not equal because one measures an average slope over a finite interval while the derivative measures the instantaneous slope at d = 2.
If we choose a smaller interval, the secant slope approaches the derivative.
The second derivative explains the curve’s shape
Differentiate again:
R′(d) = −96d⁻³.
Apply the power rule:
R″(d) = (−96)(−3)d⁻⁴.
Therefore:
R″(d) = 288/d⁴.
For d > 0, the denominator is positive.
So:
R″(d) > 0.
The inverse-square curve is concave up on its positive domain.
Its first derivative is negative, so the function decreases.
Its second derivative is positive, so its negative gradient becomes less negative as distance increases.
Thus the graph is decreasing and concave up.
A positive second derivative does not mean the original function is increasing.
What does concave up mean here?
At d = 1, the curve is very steeply decreasing.
At d = 2, it is still steep.
At d = 4, the gradient magnitude is smaller.
At d = 8, the gradient approaches zero from below.
The first derivative increases numerically from large negative values towards zero as distance grows.
That is why the second derivative is positive.
The graph falls but gradually flattens.
This is the opposite sign pattern from an increasing, concave-down logarithmic curve, where the first derivative is positive and decreasing.
Comparing examples helps students use derivatives as evidence rather than attach meaning to a single sign in isolation.
A general negative-power rule
Consider a function:
f(x) = Kx⁻ⁿ
with K a positive constant and n > 0, for x > 0.
Differentiate:
f′(x) = −nKx⁻ⁿ⁻¹.
The value scales under x → kx by:
f(kx) = k⁻ⁿ f(x).
The derivative scales by:
f′(kx) = k⁻ⁿ⁻¹ f′(x).
Differentiation adds one more negative power to the scaling.
Our example has n = 2:
Value scale for doubling: 2⁻² = 1/4.
Derivative scale for doubling: 2⁻³ = 1/8.
The mathematical structure is general, not specific to the coefficient 48.
Solve a target reading equation
Suppose we want to know which permitted distance produces a reading of three units.
Set:
48/d² = 3.
Multiply by d², which is positive:
48 = 3d².
Divide by three:
d² = 16.
Algebraically, the squared equation has two roots:
d = 4 or d = −4.
But our model requires d > 0.
Therefore:
d = 4 metres.
Check:
48/4² = 48/16 = 3.
The negative root is an algebraic candidate excluded by the meaning and domain of the distance variable.
Solve a threshold inequality
Suppose the task is:
Find the distances for which the modelled reading is at most two units.
Write:
48/d² ≤ 2,
with d > 0.
Multiply by d². Because d² is positive, the inequality direction remains unchanged:
48 ≤ 2d².
Divide by two:
24 ≤ d².
Since d is positive:
d ≥ √24.
Simplify:
√24 = 2√6.
Therefore:
d ≥ 2√6 metres, approximately 4.90 metres.
The exact answer is useful because it preserves the boundary precisely.
The inequality is not a safety threshold or a real sensor specification; it is an illustrative mathematical condition.
Why the sign condition matters in the inequality
Multiplying an inequality by a negative quantity reverses its direction.
Here we multiplied by d².
For every permitted distance d > 0, d² is positive.
Therefore the inequality sign did not reverse.
That is a justified algebraic step.
If a student manipulates an inequality involving a denominator whose sign is unknown, the same operation cannot be performed without considering sign cases.
Writing the domain at the beginning makes this calculation reliable.
Recover the input from the output
We can reverse the relationship.
Let y = 48/d² for d > 0.
Then:
yd² = 48.
So:
d² = 48/y.
Because d is positive:
d = √(48/y).
This formula requires y > 0.
It tells us the unique positive distance corresponding to each positive reading under the model.
The restriction comes from the original function’s range and the physical definition of d.
Without the positive-domain condition, the squared equation would permit positive and negative inputs with the same reading.
Why two distances can produce one reading on the unrestricted graph
If we considered the pure algebraic expression R(d) = 48/d² for every real d except zero, then:
R(4) = R(−4) = 3.
So the unrestricted function is not one-to-one.
The inverse formula d = √(48/y) would choose only the positive branch.
In the modelled distance context, d > 0 from the beginning, so the inverse is unambiguous.
This is an example of a broader mathematical idea: restricting a function’s domain can make an inverse relationship single-valued.
A student should not silently extend a physical variable into negative values and then interpret both roots as distances.
Compare a negative-power derivative with a square-root derivative
For our inverse-square function:
R(d) = 48d⁻².
The derivative is:
R′(d) = −96d⁻³.
For a square-root function:
h(x) = (x + 1)^(1/2),
the derivative is:
h′(x) = (1/2)(x + 1)^(−1/2).
Both use the same power rule.
But the original exponent, leading coefficient, inner expression and domain produce very different graph behaviours.
Our inverse-square function decreases and is concave up.
The square-root curve increases but becomes less steep.
The rule is shared; the interpretations are not.
Units are a quick check
The display output R is measured in arbitrary reading units.
Distance d is measured in metres.
The derivative R′(d) must therefore have units:
reading units per metre.
The second derivative has units:
reading units per square metre.
A student who reports R′(2) = −12 metres has mixed up an output sensitivity with a length.
Correct units cannot prove every algebraic coefficient, but they can expose a misinterpreted final answer.
This is particularly useful in application questions involving functions of physical quantities.
A six-minute independent check
Using R(d) = 48/d² for d > 0:
1. Rewrite R using a negative exponent. 2. Find R(2), R(4) and R(8). 3. Prove R(2d) = R(d)/4. 4. Differentiate R. 5. Calculate R′(2) and R′(4). 6. Find the tangent at d = 2. 7. Find the second derivative. 8. Solve R(d) = 3. 9. Solve R(d) ≤ 2. 10. Explain why a negative distance is excluded.
Answers: 48d⁻²; 12, 3, and 3/4; substitute 2d and simplify its square; −96/d³; −12 and −3/2 reading units/m; y = −12d + 36; 288/d⁴; d = 4 m; d ≥ 2√6 m; and because d was defined as a positive distance.
The explanations behind scaling and domain restrictions are just as important as the arithmetic.
Common mistake: a negative exponent means a negative reading
A negative exponent means reciprocal powers.
For d > 0, 48d⁻² is positive.
The derivative is negative because differentiation brings down the negative exponent.
The sign of the function and the sign of its derivative must be treated separately.
Common mistake: doubling divides the reading by two
The model uses the square of distance.
So doubling d multiplies the denominator by four.
The reading is quartered, not halved.
Differentiate and the derivative’s denominator contains d³, so its magnitude changes by a factor of eight upon doubling.
Common mistake: forgetting the third power after differentiation
Differentiate 48d⁻²:
Multiply by −2 and subtract one from the exponent.
The new exponent is −3.
The result is −96/d³, not −96/d².
A good check is to differentiate the negative-power form directly.
Common mistake: treating a tangent as the exact function
The tangent at d = 2 is y = −12d + 36.
It can become negative at a distance where the original R remains positive.
That merely shows why a tangent is a local approximation.
Never extrapolate the linear approximation as an exact global description without justification.
Common mistake: ignoring the positive domain in a quadratic step
Solving d² = 16 gives algebraic candidates ±4.
Only positive four belongs to the model.
Similarly, dividing or multiplying inequalities by quantities of unknown sign requires care.
The permitted domain is part of the solution.
How parents can check understanding
Ask your child why doubling the input quarters the output rather than halves it.
Ask why the reading itself can be positive even though its derivative is negative.
Ask why the gradient’s magnitude changes by an eighth rather than a quarter when distance doubles.
Then ask why the tangent line eventually gives a misleading prediction if extended too far.
These questions test negative powers, differentiation and mathematical interpretation rather than memorised answers.
What focused A-Math tuition should diagnose
Some students struggle to rewrite 1/d² as d⁻².
Others remember the power rule but lose the negative sign or change the exponent incorrectly.
Others calculate the derivative yet fail to distinguish units per metre from the original reading.
A fourth group may solve the target equation correctly but include a negative distance or reverse a threshold inequality incorrectly.
A good tutorial identifies the first faulty step, repairs that particular idea and tests transfer with a new exponent or coefficient.
The aim is a reusable mathematical method, not a single familiar sensor story.
Where this belongs in Singapore G3 Additional Mathematics
The official 2027 Singapore–Cambridge Secondary Education Certificate G3 Additional Mathematics K341 syllabus includes differentiation of rational powers, gradients of curves, tangents and normals, rates of change and optimisation under Calculus C1.
This article uses the negative integer exponent −2, a case of a rational exponent, and connects it to a derivative and contextual inequality.
The inverse-square reading is an original hypothetical function. It is not a claim that a real device follows this precise law or that the model defines any safety threshold.
For 2027, G3 Additional Mathematics uses code K341. SEAB identifies the earlier GCE O-Level code 4049 as the reference for 2026 and earlier. G2 Additional Mathematics K232 has a separate syllabus.
Frequently asked questions
Is a negative exponent allowed in A-Math differentiation?
Yes. Negative integers are rational numbers, and the power rule applies where the function is differentiable.
Why does 48/d² have a negative derivative?
As positive d increases, the positive denominator becomes larger, so the fraction decreases.
Why does doubling d quarter the output?
Because (2d)² = 4d².
Why does doubling d reduce derivative magnitude to one-eighth?
The derivative contains 1/d³, and (2d)³ = 8d³.
Can a positive function be decreasing?
Yes. Its values can remain positive while becoming smaller, as R(d) does.
Does positive second derivative mean R is increasing?
No. R′ is negative while R″ is positive, so the function decreases while flattening.
Is the tangent a globally accurate approximation?
No. A tangent matches the value and slope locally at the contact point but may be inaccurate elsewhere.
Why must d be positive?
The model defines d as a positive distance, and the denominator is undefined at zero.
What should a student revise first?
Index laws, fractions, the power rule, inequalities, graph gradients, tangent equations and the meaning of a derivative.
Continue reading across Additional Mathematics
The square-root ramp and rational-power tangent uses a positive fractional exponent instead of a negative one.
The magnifying scale and logarithmic derivative shows another decreasing gradient, this time for an increasing function.
For a different model of connected rates, see the garden sprinkler and chain rule.
For a general subject introduction, read What Is Additional Mathematics? and How Additional Mathematics Works.
The immutable Clementi small-group Mathematics tutor reference explains how index laws, algebraic accuracy and specific feedback build the foundations for calculus.
Official syllabus and model limits
The curriculum reference is SEAB’s 2027 G3 Additional Mathematics K341 syllabus (PDF), particularly Calculus C1, and the official G3 subject listing.
Every reading and distance in the display is invented. No actual sensor, safety distance, radiation level or engineering performance is represented.
The enduring lesson is to connect a negative power with its scale factor, then use the derivative to describe how quickly the output changes rather than only whether it increases or decreases.