Closes the chapter with power functions met through the situations that produce them. Translates the phrases direct, inverse and joint variation into formulas with an unknown constant, finds that constant from a single data point, and works out how the output responds to scaling the input — where the exponent, not the constant, does all the work.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 3 — Polynomial and Rational Functions
§3.9 Modeling Using Variation, pp. 438-447
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 438-447 — the pages these objectives are drawn from
Warm-up
The word is used loosely in ordinary speech and precisely here.
Discussion prompt
A newspaper says crime is proportional to unemployment. What exact claim would that be, mathematically, and is it likely to be what was meant?
Hint: What would the formula have to be, and what would it force at zero?
Answer:
Mathematically it claims crime equals a constant times unemployment — a straight line through the origin. In particular, zero unemployment would mean zero crime.
That is almost certainly not what was meant. The looser everyday sense is just 'they rise and fall together', which is a statement about correlation from §2.4 rather than about proportionality.
So this section's vocabulary is narrower and sharper than ordinary usage. 'Varies directly as' means a specific formula, and knowing which formula the words demand is most of the work here.
Concept
A variation statement specifies a formula completely except for one unknown constant. Substituting a single pair of known values determines that constant, and the model is then complete.
constant of variation — The unknown multiplier in a variation formula, usually written k. It is not given in the problem and is always found by substituting one pair of corresponding values.
\[ y = kx^n \quad \text{(direct)}, \qquad y = \frac{k}{x^n} \quad \text{(inverse)} \]
This is why variation problems are shorter than the general modelling of §2.3. There the rate and the initial value both had to be found, needing two facts; here the words fix the entire shape and only the scale is unknown, so one fact suffices.
Figure (svg): The three-step variation procedure shown as a flow: translate the words into a formula with an unknown constant, substitute the given data to find it, then use the completed formula
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 438-441
Section
Section 1
Concept
One quantity varies directly as a power of another when it equals a constant times that power. The two rise and fall together, and the ratio of output to that power is constant.
\[ y=kx^n \]
The last point is the sharpest test of whether a situation really is direct variation. If the relationship has a nonzero value when the input is zero — a fixed fee, a starting amount — then it is linear but not proportional, and the variation language does not apply.
Figure (svg): The three kinds of variation shown side by side with their formulas and the shape of the relationship each produces
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 438-442
Picture it
Each phrase corresponds to one formula shape.
Figure (svg): The three kinds of variation shown side by side with their formulas and the shape of the relationship each produces
In every case the constant is unknown and one data point finds it. The words decide the shape and nothing else.
Worked example
Translate, substitute, then use.
\[ \text{y varies directly as the cube of x, and } y=54 \text{ when } x=3. \text{ Find } y \text{ when } x=5. \]
Translate the words
Why: Directly as the cube.
\[ y = k x ^{3} \]
Substitute the given pair
Why: Fifty four and three.
\[ 54 = k(27) \]
Solve for the constant
Why: Divide by 27.
\[ k = 2 \]
Use the completed model
Why: At x equal to 5.
\[ y = 2(125) = 250 \]
Figure (svg): The three-step variation procedure shown as a flow: translate the words into a formula with an unknown constant, substitute the given data to find it, then use the completed formula
\[ y=2x^3, \quad y(5)=250 \]
Verify: check the ratio is constant
Why: At x equal to 3 the ratio of y to x cubed is 54 over 27, which is 2. At x equal to 5 it is 250 over 125, also 2. The ratio to the relevant power staying constant is exactly what direct variation means, so this check tests the model rather than just the arithmetic.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 439-441
Sorting
Direct variation means a constant times a power, with nothing added.
Sort into buckets
Sort each relationship.
Worked example
A nonzero value at zero rules it out.
\[ \text{Is } y=3x+7 \text{ a direct variation?} \]
Test the value at zero
Why: Substitute x equal to zero.
\[ y = 7 \]
Compare with the requirement
Why: Direct variation gives zero at zero.
\[ \text{should be } 0 \]
Check the ratio
Why: At x equal to 1 it is 10; at 2 it is 6.5.
Conclude
Why: It is linear but not proportional.
Figure (svg): The solution to Worked example recognise what is not direct variation shown as a ladder of expressions, one row per legal move
\[ \text{Linear but not a direct variation: } y(0)=7\ne 0. \]
Verify: say what the distinction costs
Why: The scaling rule fails as a result. Doubling the input from 1 to 2 takes the output from 10 to 13, which is nowhere near doubling — because the constant 7 does not scale. Direct variation's convenient scaling behaviour depends entirely on there being no constant term.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 441-442
Trap
\[ y=5x+2 \;\Longrightarrow\; y \text{ varies directly as } x, \text{ with } k=5 \]
Identify the coefficient of x
Why: The slope is 5, which is taken as the constant of variation.
The relationship is described as a direct variation.
Direct variation has no constant term. This rule gives 2 when x is zero, so it is not proportional and no single k describes it.
The ratio of y to x is 7 at x equal to 1 and 6 at x equal to 2 — not constant, which is the definitive test.
Every direct variation is linear, but not every linear rule is a direct variation. The extra requirement is passing through the origin, and it is exactly what makes the scaling rules work.
Faded example
y varies directly as the square of x, and y is 45 when x is 3.
Fill in the blanks
45 = k(9) \;\Longrightarrow\; k = 5, \text___ y = ___x^2
Why: Substituting the pair gives 45 equal to 9k, so k is 5. The completed model is 5 times x squared. Note that the given pair is used only to find k; once k is known, the model works for every input.
Prediction
A quantity varies directly as another.
Predict first
What is its value when the input is zero?
Correct: Zero, always.
Why: A direct variation is a constant times a positive power of the input, and any positive power of zero is zero. So the graph passes through the origin without exception, which is the sharpest test for distinguishing direct variation from a general linear relationship.
Translation
Each phrase names one formula shape.
Match the pairs
Why: The phrase after 'varies directly as' names the power, and 'proportional to' means the same thing. The last two are the same statement in different words, which is worth noticing since problems use both interchangeably.
Section
Section 2
Concept
One quantity varies inversely as a power of another when it equals a constant divided by that power. As one grows the other shrinks, and their product is constant.
\[ y=\frac{k}{x^n} \]
The excluded input is worth noting rather than skipping. An inverse variation is undefined at zero, so the model always carries a domain restriction — and physically that usually corresponds to something impossible, like measuring the brightness of a light from zero distance away.
Figure (svg): An inverse square relationship graphed against a direct proportion, showing how much faster the inverse square falls away as the input grows
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 442-445
Picture it
Compare the two curves as the input grows.
Figure (svg): An inverse square relationship graphed against a direct proportion, showing how much faster the inverse square falls away as the input grows
Doubling the input quarters the output and tripling it takes the output to a ninth. That steepness is why inverse square laws produce such dramatic falls with distance.
Worked example
Same three steps, different shape.
\[ \text{y varies inversely as x, and } y=12 \text{ when } x=5. \text{ Find } y \text{ when } x=15. \]
Translate the words
Why: Inversely as x.
\[ y = \frac{k}{x} \]
Substitute the given pair
Why: Twelve and five.
\[ 12 = \frac{k}{5} \]
Solve for the constant
Why: Multiply by 5.
\[ k = 60 \]
Use the model
Why: At x equal to 15.
\[ y = \frac{60}{15} = 4 \]
Figure (svg): The solution to Worked example build an inverse variation model shown as a ladder of expressions, one row per legal move
\[ y=\frac{60}{x}, \quad y(15)=4 \]
Verify: check the product is constant
Why: At the given pair the product is 12 times 5, which is 60. At the new pair it is 4 times 15, also 60. A constant product is exactly what inverse variation means, and it is a stronger check than merely confirming the arithmetic. Note the input tripled and the output fell to a third, as it must.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 443-444
Faded example
y varies inversely as the square of x, and y is 8 when x is 3.
Fill in the blanks
8 = \frac72___ \;\Longrightarrow\; k = ___, \text___ y = \frac___}___
Why: Multiplying both sides by 9 gives k equal to 72. The completed model is 72 over x squared. Checking: the product of y with x squared is 8 times 9, which is 72 — constant, as inverse square variation requires.
Worked example
The exponent changes the behaviour dramatically.
\[ \text{Light intensity varies inversely as the square of distance. At } 2 \text{ m it is } 100. \text{ At } 6 \text{ m?} \]
Translate
Why: Inversely as the square.
\[ I = k / d ^{2} \]
Substitute the given pair
Why: One hundred at two metres.
\[ 100 = \frac{k}{4} \]
Solve for the constant
Why: Multiply by 4.
\[ k = 400 \]
Use the model
Why: At six metres.
\[ I = \frac{400}{36} = 11.1 \]
Figure (svg): An inverse square relationship graphed against a direct proportion, showing how much faster the inverse square falls away as the input grows
\[ I=\frac{400}{d^2}, \quad I(6)\approx 11.1 \]
Verify: check with the scaling rule
Why: The distance tripled, from 2 to 6, so the intensity should fall to one ninth. One ninth of 100 is 11.1 — matching. The scaling rule gives the answer without computing k at all, which is often the faster route and is a good independent check when k has been computed.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 444-445
Error analysis
A student models an inverse relationship.
Annotate
On: \( y \text{ varies inversely as } x \;\Longrightarrow\; y = k - x \)
Decreasing together is not enough — many formulas decrease. Inverse variation is the specific claim that the product is constant, and only division produces that.
Prediction
A quantity varies inversely as the square of another.
Predict first
What happens to it when the input doubles?
Correct: It falls to a quarter.
Why: Doubling the input multiplies its square by 4, and the output is divided by that square, so the output is divided by 4. This is the defining behaviour of an inverse square law and the reason light and gravity fade so quickly with distance — twice as far away is four times as weak.
Discrimination
The words decide, and so does what stays constant.
Sort into buckets
Sort each described relationship.
Edge cases
An inverse variation is undefined at input zero.
Discussion prompt
What does that mean physically for an inverse square law, and is it a defect of the model?
Hint: What would it mean to be zero distance from a light source?
Answer:
It predicts an infinite intensity at zero distance, which is physically impossible. The model says nothing sensible there.
It is not really a defect, because zero distance is outside the situation's domain anyway. A light source has physical size, so you cannot be at zero distance from it, and the model was only ever intended for distances comfortably larger than the source.
This is §2.3's domain point in a new setting. The formula's domain and the model's domain differ, and the model's is decided by the physics. An inverse square law is excellent at ordinary distances and meaningless very close in, which is a limitation worth stating rather than a reason to distrust it.
Section
Section 3
Concept
A quantity varies jointly as several others when it equals a constant times their product. Combined variation mixes direct and inverse dependence in one formula.
The formula can be read straight off the sentence once the convention is known: everything named after 'varies jointly as' or 'directly as' goes on top, and everything after 'inversely as' goes underneath, with the powers as stated. Translating carefully is the whole difficulty.
Figure (svg): The three kinds of variation shown side by side with their formulas and the shape of the relationship each produces
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 445-447
Picture it
Joint variation is the same idea with more than one variable in the product.
Figure (svg): The three kinds of variation shown side by side with their formulas and the shape of the relationship each produces
Combined variation puts some variables on top and some underneath in one formula, and the sentence says which is which.
Worked example
Two variables in the numerator.
\[ \text{y varies jointly as x and z. When } x=2 \text{ and } z=3, \; y=24. \text{ Find } y \text{ when } x=5, z=4. \]
Translate
Why: Both variables in the product.
\[ y = k x z \]
Substitute the given values
Why: Two and three give six.
\[ 24 = k(6) \]
Solve for the constant
Why: Divide by 6.
\[ k = 4 \]
Use the model
Why: Five times four is twenty.
\[ y = 4(20) = 80 \]
Figure (svg): The solution to Worked example joint variation shown as a ladder of expressions, one row per legal move
\[ y=4xz, \quad y(5,4)=80 \]
Verify: check the ratio to the product
Why: At the first pair, y over xz is 24 over 6, which is 4. At the second it is 80 over 20, also 4. A constant ratio to the product of the variables is what joint variation means, and the check confirms the model rather than only the arithmetic.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 445-446
Translation
Each phrase places one factor.
Match the pairs
Why: Everything after 'directly as' or 'jointly as' goes on top and everything after 'inversely as' goes underneath, with the stated powers. The third puts an entire product underneath, which is different from putting each factor underneath separately — though in this case they happen to agree.
Worked example
Read the sentence carefully and put each factor on the right side of the line.
\[ \text{y varies directly as x and inversely as the square of z. When } x=6, z=2, \; y=9. \]
Put the direct factor on top
Why: After 'directly as'.
Put the inverse factor underneath
Why: After 'inversely as'.
\[ z ^{2}\text{ underneath} \]
Write the model
Why: With the unknown constant.
\[ y = k x / z ^{2} \]
Substitute and solve
Why: Nine equals k times 6 over 4.
\[ k = 6 \]
Figure (svg): The solution to Worked example combined variation shown as a ladder of expressions, one row per legal move
\[ y=\frac{6x}{z^2} \]
Verify: test the model on the given data
Why: At x equal to 6 and z equal to 2: 6 times 6 is 36, over 4, which is 9 — the given value. Note the two variables behave oppositely: increasing x raises y while increasing z lowers it sharply, and the sentence's two phrases are what put them on opposite sides of the fraction bar.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 446-447
Trap
\[ \text{directly as } x, \text{ inversely as } z \;\Longrightarrow\; y = kxz \]
Collect both variables into the formula
Why: Both are named in the sentence, so both are included as factors.
Both variables are placed in the numerator.
'Inversely as z' puts z in the denominator. The correct model is k times x, over z.
The wrong version makes y increase when z increases, which is the opposite of what 'inversely' means. Testing the direction catches it immediately.
Read each phrase and place its variable accordingly: 'directly as' and 'jointly as' put a factor on top, 'inversely as' puts one underneath. Then check each variable moves the way the words say.
Faded example
y varies jointly as x and the square of z, with y equal to 60 when x is 3 and z is 2.
Fill in the blanks
60 = k(3)(4) \;\Longrightarrow\; 60 = 12k \;\Longrightarrow\; k = 5
Why: The square of 2 is 4, and 3 times 4 is 12, so 60 equals 12k and k is 5. The completed model is 5 times x times z squared. Squaring z before multiplying is the step to watch, since substituting 2 rather than 4 would give the wrong constant.
Prediction
A quantity varies directly as x and inversely as z.
Predict first
What happens if both x and z double?
Correct: It is unchanged, since the two effects cancel.
Why: Doubling x doubles the output and doubling z halves it, so the two changes exactly cancel. This is worth noticing because it shows the two variables genuinely pull in opposite directions, and it is a good check that a combined model was assembled with the factors on the correct sides.
Explain it to yourself
A combined variation may involve three or four variables.
Discussion prompt
Explain why a single data point still determines the model, however many variables it has.
Hint: How many unknowns are there in the formula?
Answer:
However many variables appear, there is still exactly one unknown: the constant k. The words fixed every exponent and every position, so nothing else is left to determine.
One equation determines one unknown, so one data point suffices — and the data point must supply values for every variable, since they all appear in the equation.
This is why variation problems are so much shorter than the general modelling of §2.3, where both the rate and the initial value had to be found. Here the sentence has done most of the work, and the measurement supplies only the scale.
Section
Section 4
Concept
Multiplying an input by a factor multiplies the output by that factor raised to the exponent. The constant of variation never enters, so it need not be known.
The cancellation is the point. Taking the ratio of the output at the new input to the output at the old one, the constant appears in both and divides out, leaving only the ratio of the powers. So the scaling behaviour is a property of the exponent alone.
Figure (svg): A table showing how the output responds to doubling the input for several different exponents, from inverse square through direct proportion to a cube
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 441-446
Picture it
One table covers every case in the section.
Figure (svg): A table showing how the output responds to doubling the input for several different exponents, from inverse square through direct proportion to a cube
The exponent's size decides the factor and its sign decides the direction. Nothing else in the model has any effect on scaling.
Worked example
Take the ratio and watch the constant cancel.
\[ \text{y varies directly as the square of x. If x triples, what happens to y?} \]
Write both outputs
Why: At x and at 3x.
\[ k x ^{2}\text{ and } k(3 x) ^{2} \]
Expand the second
Why: The 3 is squared too.
\[ 9 k x ^{2} \]
Take the ratio
Why: New over old.
\[ 9 k x ^{2} / k x ^{2} \]
Cancel
Why: The constant and the power both go.
\[ 9 \]
Figure (svg): A table showing how the output responds to doubling the input for several different exponents, from inverse square through direct proportion to a cube
\[ y \text{ is multiplied by } 3^2 = 9 \]
Verify: test with actual numbers
Why: Take k equal to 2. At x equal to 1, y is 2; at x equal to 3, y is 18 — nine times as large. Any other value of k gives the same factor of 9, which is what the cancellation showed. Being able to answer without k is what makes this the quicker route.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 442-443
Matching
For an input that doubles.
Match the pairs
Why: The doubling factor is raised to the exponent's size, giving 2, 8, and for the inverse cases the reciprocal of 2 and of 4. Cubing amplifies most dramatically among these, which is why volume grows so much faster than length.
Worked example
The same method, with the power in the denominator.
\[ \text{Intensity varies inversely as the square of distance. If distance is halved, what happens?} \]
Write both outputs
Why: At d and at half d.
\[ k / d ^{2}\text{ and } k / (\frac{d}{2}) ^{2} \]
Simplify the second denominator
Why: Half squared is a quarter.
\[ \frac{k}{d ^{2} / 4} \]
Rewrite as a multiplication
Why: Dividing by a quarter multiplies by 4.
\[ 4 k / d ^{2} \]
Compare
Why: Four times the original.
Figure (svg): The solution to Worked example scaling an inverse square shown as a ladder of expressions, one row per legal move
\[ \text{intensity } \times 4 \]
Verify: check against the earlier example
Why: Earlier, tripling the distance took the intensity to a ninth. Halving it here quadruples the intensity — the reciprocal relationship, with the factor squared in both directions. Moving twice as close makes a light four times as bright, which is a fact worth having a feel for.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 444-445
Error analysis
A student scales a direct square variation.
Annotate
On: \( y=kx^2: \quad x \text{ doubles} \;\Longrightarrow\; y \text{ doubles} \)
The scaling factor is raised to the same power the variable is. That is the entire content of the scaling rule, and applying the factor unchanged is the standard error.
Faded example
A quantity varies directly as the cube of x, and x is doubled.
Fill in the blanks
\text3 2^8} = ___
Why: The exponent is 3, so the doubling factor is cubed, giving 8. The constant of variation plays no part and need not be known. This is why doubling a cube's edge multiplies its volume by 8, which is the same computation in a geometric setting.
Prediction
y varies jointly as x and z, and both are tripled.
Predict first
What happens to y?
Correct: It is multiplied by 9.
Why: Each variable contributes its own factor of 3, and the two multiply to give 9. Adding the factors to get 6 is the standard error: the variables appear as a product in the formula, so their scaling factors multiply rather than add.
Socratic
Scaling questions can be answered without ever finding k.
Discussion prompt
Explain why the constant of variation never affects a scaling question.
Hint: What happens when you divide the new output by the old one?
Answer:
Take the ratio of the output at the new input to the output at the old one. The constant k appears as a factor in both, so it divides out completely.
What remains is the ratio of the two powers, which depends only on the scaling factor and the exponent. So the answer is the same whatever k happens to be.
That is why a problem can ask 'what happens if the input doubles' without supplying any data at all. The scaling behaviour is a property of the exponent, and the constant only ever sets the overall scale — which cancels the moment two outputs are compared.
Section
Section 5
Concept
Most of the named laws in physics and chemistry are variation statements, and the exponent is usually the interesting part of the claim.
The recurring theme is that the exponent carries the content. Whether a force falls off as the reciprocal of distance or its square is a deep physical fact, testable by measurement; the constant is comparatively uninteresting and often just reflects a choice of units.
Figure (svg): An inverse square relationship graphed against a direct proportion, showing how much faster the inverse square falls away as the input grows
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 444-447
Picture it
The shape of the most important variation law in physics.
Figure (svg): An inverse square relationship graphed against a direct proportion, showing how much faster the inverse square falls away as the input grows
The steep fall near the origin and the long flat tail are what make inverse square laws distinctive: enormous close up, and negligible far away.
Worked example
An inverse square law, applied by scaling.
\[ \text{Gravity varies inversely as the square of distance from a planet's centre. At } 4 \text{ times the radius?} \]
Identify the relationship
Why: Inverse square in the distance.
\[ F = k / d ^{2} \]
Note the scaling factor
Why: Distance multiplied by 4.
\[ \text{factor } 4 \]
Raise it to the power
Why: Four squared.
\[ 16 \]
Apply it inversely
Why: Divided rather than multiplied.
Figure (svg): An inverse square relationship graphed against a direct proportion, showing how much faster the inverse square falls away as the input grows
\[ F \text{ is } \tfrac{1}{16} \text{ of the surface value} \]
Verify: sanity-check the size
Why: One sixteenth is about 6 percent, so an object four planet-radii out feels only a small fraction of surface gravity. That matches the physical fact that gravity fades quickly but never vanishes — an inverse square approaches zero without ever reaching it, which is why orbits at any distance are still bound.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 444-445
Sorting
The exponent is the content of the law.
Sort into buckets
Sort each relationship by whether it is an inverse square.
Worked example
Scaling a length scales a volume by the cube.
\[ \text{Mass varies as the cube of length for a scaled object. Double the length: what happens?} \]
Identify the relationship
Why: Direct, with exponent 3.
\[ m = k L ^{3} \]
Note the scaling factor
Why: Length doubled.
\[ \text{factor } 2 \]
Cube it
Why: Two cubed.
\[ 8 \]
Apply it
Why: Directly, so multiply.
\[ \text{mass } \times 8 \]
Figure (svg): The solution to Worked example a cube relationship shown as a ladder of expressions, one row per legal move
\[ m \times 8 \]
Verify: notice the consequence
Why: The surface area only quadruples, since area varies as the square. So doubling an animal's height multiplies its weight by 8 while its bone cross-sections only quadruple — which is why large animals need proportionally thicker legs. The mismatch between a square and a cube law is one of the most consequential facts in biology, and it is this section's arithmetic.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 446-447
Trap
\[ \text{temperature falls as time passes} \;\Longrightarrow\; T = \frac{k}{t} \]
Note that one quantity decreases as the other increases
Why: That is what inverse variation describes, so the model is chosen.
The cooling is modelled as an inverse variation in time.
Decreasing is not the same as inverse variation. Inverse variation makes a specific claim: the product of the two quantities is constant.
Cooling does not behave that way — it approaches the surrounding temperature rather than zero, and its rate depends on the current difference. It is exponential decay, which is Chapter 4's subject.
Test the product before choosing the model. If halving the time does not double the quantity, it is not inverse variation, however clearly the two move in opposite directions.
Prediction
Mass varies as the cube of length and bone strength as the square.
Predict first
What happens as an animal is scaled up?
Correct: Mass outgrows strength, so proportions must change.
Why: A cube grows faster than a square, so doubling the length multiplies mass by 8 and strength by only 4. A scaled-up animal would be relatively half as strong, which is why large animals have proportionally thicker limbs than small ones and why there is an upper limit to size on land.
Faded example
Sound intensity varies inversely as the square of distance. The distance is multiplied by 5.
Fill in the blanks
\text25 5^2 = ___, \text___ ___\text___
Why: Five squared is 25, and the inverse relationship divides rather than multiplies, so the intensity falls to one twenty-fifth. This is why moving a modest distance from a sound source reduces its loudness so dramatically, and why sound level is usually measured on a logarithmic scale.
Real world
A variation model has two parts, and only one of them is interesting.
Discussion prompt
Why do physicists care so much more about whether a law is inverse square than about the value of its constant?
Hint: What would change the constant, and what would change the exponent?
Answer:
The constant depends on the units. Measuring in metres rather than feet changes it, so it carries no physical content beyond a choice of scale.
The exponent is a claim about how the world works. That gravity falls off as the square of distance rather than the cube reflects the geometry of three-dimensional space, and it is testable independently of any units.
This is why deviations from an expected exponent are historically so important. Measurements suggesting an exponent other than 2 for gravity would indicate new physics, whereas an unexpected constant would just mean somebody used the wrong units.
Comparison
Fill the blanks from memory. Three shapes, one procedure.
Comparison matrix
| direct | inverse | joint | |
|---|---|---|---|
| formula | y = kx^n | y = k/x^n | y = kxz |
| what stays constant | the ratio y over x^n | the product y times x^n | the ratio y over xz |
| as the input grows | the output grows | the output shrinks | depends on which variable |
| value at zero input | zero | undefined | zero |
| data points needed | one | one | one |
The last row is the same in every column, because every one of these formulas has exactly one unknown. That is what makes the whole section short.
Pattern
Five steps, and the first is the only one that takes thought.
Step 5's shortcut is worth knowing: a question asking only what happens when an input doubles can be answered from the exponent alone, with no data and no constant.
OpenStax Algebra and Trigonometry 2e, §5.8 Modeling Using Variation §5.8
Check
Which side of the fraction bar?
Check your understanding
If y varies inversely as the square of x, which formula is correct?
Answer: A
Why: 'Inversely as' puts the variable in the denominator, and 'the square of x' says the power is 2. The product of y with x squared is then the constant k, which is the defining property.
Check
Raise the factor to the power.
Check your understanding
y varies directly as the square of x. If x is multiplied by 3, y is multiplied by what?
Answer: A
Why: The scaling factor is raised to the exponent, so 3 squared gives 9. The constant of variation cancels when the two outputs are compared, so it need not be known.
Check
Direct variation passes through the origin.
Check your understanding
Which of these is a direct variation?
Answer: A
Why: A direct variation is a constant times a power with nothing added, so it gives zero when the input is zero and the ratio of output to input is constant. Only the first satisfies both.
Real world
Scaling arguments settle a surprising number of practical questions without any data at all.
Discussion prompt
Why can a mouse fall from a height unharmed while a horse cannot? Answer using this section's exponents.
Hint: How do mass and surface area scale with length?
Answer:
Mass varies as the cube of the animal's length, while the cross-sectional area of its bones and its air resistance vary as the square. So as size increases, mass outgrows both.
A mouse has a large surface area relative to its mass, so air resistance slows it substantially and the impact force is spread over bones that are relatively strong for its weight. A horse has the reverse on both counts.
The striking part is that this argument needs no measurements at all — only the exponents. That is the scaling shortcut from this section, applied to a question that looks biological and is really about the difference between a square and a cube.
Commit first
State your confidence along with your answer.
Predict first
y varies inversely as x. If x is multiplied by 4, what happens to y?
Correct: It falls to a quarter.
Why: Inverse variation divides by the scaling factor, and the exponent is 1, so the factor is used unchanged: the output is divided by 4. Falling to a sixteenth would be the inverse SQUARE answer, and multiplying by 4 would be direct variation — the two distinctions that decide this question are the sign and the size of the exponent.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate the difference between 'linear' and 'directly proportional', and why the distinction matters.
Hint: What does each one do at input zero?
Answer:
Linear means a straight line: a constant rate of change, with any starting value. Directly proportional means a straight line through the origin, so the output is zero when the input is.
Every proportional relationship is linear, and most linear ones are not proportional. The difference is the constant term, and it is exactly what the variation language forbids.
It matters because the scaling rule depends on it. Doubling the input doubles the output for a proportional relationship, and does not for a general linear one — because the constant term does not double. A gym charging a joining fee plus a monthly rate is linear, and doubling the months does not double the bill.
Exit ticket
One honest answer, so the next chapter can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The first is where the errors are, since a misplaced factor produces a model that is wrong in direction rather than merely in scale. The third is the most useful beyond this section, because scaling arguments settle real questions with no data at all.
Connect it up
One page, drawn from memory, closes the chapter.
Draw it
Write the three variation formulas with their names, and beside each note what stays constant. Then make a small table of what happens to the output when the input doubles, for exponents from inverse square up to a cube. Finally, write one sentence describing a combined variation and translate it into a formula, checking each variable is on the side its phrase demands.
If your table shows the factor being raised to the exponent in every row, you have the one rule that answers every scaling question in the section.
Recap
Five things, closing a chapter that began with a new number system and ends with a vocabulary.
| if you remember one thing | it should be this |
|---|---|
| about translating | directly and jointly go on top, inversely goes underneath |
| about the constant | one unknown means one data point, always |
| about scaling | raise the factor to the exponent; the constant cancels |
| about proportion | direct variation passes through the origin, and linear need not |
Chapter 4 introduces the one family of functions that outgrows every power: exponentials, where the variable moves from the base to the exponent and the growth changes character entirely.
OpenStax, Precalculus, §3.9 Modeling Using Variation §3.9, pp. 438-447 — everything on these slides traces back here
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