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Proportional Reasoning and Functional Dependence

Functional dependence says how one quantity responds when another changes.

Practice 2: Mathematical Routines

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Question set for this topic

Part 1 · Hook

Why this matters

Drive twice as fast and your braking distance does not double; it becomes four times as long. Questions like "if the speed triples, what happens to..." appear all over the exam, and you can answer them in seconds without a single number, once you know how to read the equation.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Solve d = vt for t.

  1. t = d/v
  2. t = v/d
  3. t = dv
  4. t = d − v
Show the answer

Divide both sides by v.

  • Correct: t = d/v:
  • t = v/d:
  • t = dv:
  • t = d − v:

2. What is 3²?

  1. 9
  2. 6
  3. 3
  4. 1.5
Show the answer

3² = 3 × 3 = 9.

  • Correct: 9:
  • 6:
  • 3:
  • 1.5:

Part 4 · See it

See it first

Three panels. Directly proportional, d = v t at steady speed: a car covers distance d in time t and 2d in time 2t, so doubling t doubles d. Inversely proportional, t = d over v for a fixed trip: a car at speed v needs time t, and at speed 2v needs time t over 2, so doubling v halves the time. Square relationship, A = s squared: a square of side s has area A; a square of side 2s holds four copies of the small square, so doubling s multiplies the area by 4.
Doubling one quantity: in a direct proportion the other doubles, in an inverse one it halves, and in a square relationship it becomes 4 times as large. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. An equation links two quantities while the others stay fixedthe power of each quantity in the equation sets how the other responds
  2. y = kx (first power)multiplying x by a factor multiplies y by the same factor
  3. y = k/x (x in the denominator)multiplying x by a factor divides y by that factor
  4. y = kx² (x squared)multiplying x by a factor multiplies y by the factor squared

Part 6 · Key ideas

Key ideas

  • To find a functional dependence, solve for the quantity asked, hold everything else fixed, and look at the power of the quantity that changes.
  • Directly proportional (y ∝ x): factor in, same factor out. Inversely proportional (y ∝ 1/x): factor in, divide by it.
  • Square (y ∝ x²): factor in, factor squared out. Triple x → 9y. A square root (y ∝ √x): quadruple x → 2y.
  • Several changes at once: write a factor for each and multiply them. In y = kx²/z, doubling x and z gives 2²/2 = 2.

Part 7 · Misconception

A common mistake

The wrong idea: If the speed doubles, the braking distance doubles, because a bigger speed means a proportionally bigger distance.

What actually happens: Read the equation: d = v²/(2a), so d ∝ v². Doubling v multiplies d by 2² = 4.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. A cyclist rides at a steady speed, so the distance covered is d = vt. If she rides for 3 times as long, how does the distance change?

  1. It is 9 times as large
  2. It is 3 times as large
  3. It is one third as large
  4. It is unchanged
Show the answer

d = vt with v fixed: d ∝ t. Multiply t by 3 and d is multiplied by 3.

  • It is 9 times as large: That would be true if d depended on t², but at a steady speed d = vt has t to the first power.
  • Correct: It is 3 times as large: Right: d is directly proportional to t, so tripling t triples d.
  • It is one third as large: Riding longer at the same speed covers more distance, not less.
  • It is unchanged: Distance depends on time here; only the speed is held steady.

2. A bus route has a fixed length d. The time for the trip is t = d/v. If the bus drives 1.5 times as fast, how does the trip time change?

  1. It becomes about 0.67 times as long
  2. It becomes about 1.5 times as long
  3. It becomes about 2.25 times as long
  4. It becomes about 0.50 times as long
Show the answer

t = d/v: t ∝ 1/v. New t = t/1.5 ≈ 0.67t.

  • Correct: It becomes about 0.67 times as long: Right: t is inversely proportional to v, so multiplying v by 1.5 divides t by 1.5.
  • It becomes about 1.5 times as long: Faster means less time; v is in the denominator.
  • It becomes about 2.25 times as long: v is not squared in t = d/v, and the time should shrink, not grow.
  • It becomes about 0.50 times as long: Halving the time would need twice the speed, not 1.5 times.

3. A cart starts from rest with a steady acceleration, so its displacement is Δx = ½at². After 2.0 s it has moved 0.50 m. How far has it moved after 6.0 s?

Type a number in m.

Show the answer

Δx ∝ t². The time is 3 times as long, so Δx is 3² = 9 times as large: 9 × 0.50 m = 4.5 m.

  • Answer: 4.5 m

4. For a car braking with the same steady deceleration, the stopping distance is d = v²/(2a). Driving at 3 times the speed multiplies the stopping distance by

  1. 3
  2. 6
  3. 9
  4. 1/9
Show the answer

d = v²/(2a), a fixed: d ∝ v². Factor = 3² = 9. This is why speed matters so much for braking.

  • 3: That treats d as proportional to v, but the speed is squared.
  • 6: Squaring 3 gives 9, not 2 × 3. A square multiplies the factor by itself.
  • Correct: 9: Right: d ∝ v², so 3 times the speed gives 3² = 9 times the distance.
  • 1/9: The speed is on top of the fraction, so a larger speed gives a larger distance.

5. In y = kx²/z, x is doubled and z is doubled at the same time. By what factor does y change?

  1. 1
  2. 2
  3. 4
  4. 8
Show the answer

Write the factor for each change and multiply: (2)²/(2) = 4/2 = 2.

  • 1: The two changes do not cancel: x is squared but z is not.
  • Correct: 2: Right: doubling x multiplies y by 2² = 4, and doubling z divides it by 2. 4 ÷ 2 = 2.
  • 4: This counts the effect of x but forgets that z also doubled, in the denominator.
  • 8: z is in the denominator, so doubling it divides y by 2 rather than multiplying.

6. A table of measurements shows that each time t doubles, a quantity Q becomes 4 times as large. Which relationship fits?

  1. Q ∝ t
  2. Q ∝ 1/t
  3. Q ∝ √t
  4. Q ∝ t²
Show the answer

Test with a factor: if t × 2 gives Q × 4 = Q × 2², then Q ∝ t².

  • Q ∝ t: A direct proportion would only double Q when t doubles.
  • Q ∝ 1/t: An inverse relationship would halve Q when t doubles.
  • Q ∝ √t: A square root would multiply Q by about 1.41 when t doubles.
  • Correct: Q ∝ t²: Right: a factor of 2 in t giving a factor of 4 = 2² in Q is the mark of a square relationship.

Part 9 · Summary

Summary

Functional dependence says how one quantity responds when another changes. Directly proportional: the same factor. Inversely proportional: divide by the factor. Square: the factor squared. For several changes, multiply their factors.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections