Direct Variation: Power Functions $y = kx^n$
Two variables exhibit direct variation when an increase in one causes a proportional increase in a power of the other:
Common real-world models:
- Circle Area: $A = \pi r^2$ ($n = 2$, $k = \pi$). Doubling the radius quadruples ($2^2 = 4$) the area.
- Sphere Volume: $V = \frac{4}{3}\pi r^3$ ($n = 3$). Doubling radius multiplies volume by $8$ ($2^3 = 8$).
- Pendulum Period: $T = k\sqrt{L} = kL^{0.5}$ ($n = 0.5$).
Inverse Variation: $y = \frac{k}{x^n}$ ($y = kx^{-n}$)
In inverse variation, an increase in $x$ causes a proportional decrease in $y$:
Examples:
- Travel Time: $t = \frac{d}{v}$ (Time is inversely proportional to speed).
- Inverse Square Law: Light intensity, gravitational force, or radiation: $I = \frac{k}{d^2}$. Doubling the distance reduces intensity to $\frac{1}{4}$ of its original value.
Asymptotes & Real-World Limitations
For rational models of the form $f(x) = \frac{a}{x - b} + c$:
Physical Model Breakdown: In $y = \frac{k}{x}$, as $x \to 0$, $y \to \infty$. In reality, physical quantities (e.g. speed, pressure, density) cannot become infinite. A valid IB response comments that the model ceases to be realistic near zero.
Identify the exponent and validate the model
- For positive data fitting $y=kx^n$, divide two observations: $y_2/y_1=(x_2/x_1)^n$, so $n=\ln(y_2/y_1)/\ln(x_2/x_1)$ for distinct x-values.
- The units of k are [y]/[x]ⁿ. Check $y/x^n$ on further measurements; a constant from two points alone does not validate the model.
- For positive x and k, n>0 gives increasing power variation and n<0 decreasing inverse variation. Broader signs/domains need separate analysis.
A variable $y$ is directly proportional to $x^2$. When $x = 3$, $y = 72$. Find the equation connecting $y$ and $x$, and calculate $y$ when $x = 5$.
Solution:
$y = kx^2 \implies 72 = k(3^2) = 9k \implies k = 8$ [M1][A1]
Equation: $y = 8x^2$
When $x = 5$: $y = 8(5^2) = 8(25) = 200$. [A1]
The time $t$ (hours) taken to complete a paving project is inversely proportional to the number of workers $n$. If 4 workers take 18 hours, find the time taken by 6 workers.
Solution:
$t = \frac{k}{n} \implies 18 = \frac{k}{4} \implies k = 72\text{ worker-hours}$ [M1][A1]
For $n = 6$: $t = \frac{72}{6} = 12\text{ hours}$. [A1]
The intensity of light $I$ (lux) varies inversely as the square of the distance $d$ (meters) from a lamp. At a distance of 2 meters, the intensity is 300 lux.
- Find the constant of proportionality $k$.
- Calculate the light intensity at a distance of 5 meters.
Solution:
(a) $I = \frac{k}{d^2} \implies 300 = \frac{k}{2^2} = \frac{k}{4} \implies k = 1200$. [M1][A1]
(b) When $d = 5$: $I = \frac{1200}{5^2} = \frac{1200}{25} = 48\text{ lux}$. [M1][A1]
If distance $d$ triples in an inverse square law relationship $I = \frac{k}{d^2}$, state the factor by which the intensity $I$ changes.
Solution:
$I_{\text{new}} = \frac{k}{(3d)^2} = \frac{k}{9d^2} = \frac{1}{9} I_{\text{old}}$ [M1][R1]
The intensity is divided by 9 (multiplied by a factor of $\frac{1}{9}$ or reduced to $11.1\%$). [A1]
Consider the rational function $f(x) = \frac{6}{x - 4} + 3$.
- Write down the equation of the vertical asymptote.
- Write down the equation of the horizontal asymptote.
- Find the coordinates of the $y$-intercept.
Solution:
(a) Vertical asymptote occurs where denominator is zero: $x - 4 = 0 \implies x = 4$. [A1]
(b) Horizontal asymptote: $y = 3$. [A1]
(c) When $x = 0$: $f(0) = \frac{6}{0 - 4} + 3 = -1.5 + 3 = 1.5 \implies (0, 1.5)$. [M1][A1]
The cost per person $C$ (in dollars) of organizing a club excursion is modeled by $C(n) = \frac{1200}{n} + 25$, where $n$ is the number of participants ($n \ge 10$).
- Find the cost per person if 40 people participate.
- Write down the horizontal asymptote of $C(n)$, and interpret what it represents.
Solution:
(a) $C(40) = \frac{1200}{40} + 25 = 30 + 25 = 55$ dollars per person. [M1][A1]
(b) Horizontal asymptote: $y = 25$ (or $C = 25$). [A1]
Interpretation: As the number of participants becomes very large ($n \to \infty$), the fixed bus cost $\frac{1200}{n} \to 0$, so the cost per person approaches the individual variable cost of $25 per ticket. [R1]
Explain why the model $C(n) = \frac{1200}{n} + 25$ is restricted to a domain such as $10 \le n \le 100$ in a realistic context.
Solution:
Lower bound: A minimum group size is needed to run the trip; $n = 0$ is mathematically undefined (division by zero). [R1]
Upper bound: Physical capacity constraints (bus seating capacity, venue capacity) prevent $n$ from increasing indefinitely. [A2]
The braking distance $d$ (meters) of a car is directly proportional to the square of its speed $v$ (km/h). A car travelling at $60\text{ km/h}$ requires $27\text{ meters}$ to stop.
- Find the formula for $d$ in terms of $v$.
- Calculate the braking distance at $100\text{ km/h}$.
Solution:
(a) $d = kv^2 \implies 27 = k(60^2) = 3600k \implies k = \frac{27}{3600} = 0.0075$. [M1][A1]
Formula: $d = 0.0075v^2$.
(b) At $v = 100$: $d = 0.0075(100^2) = 0.0075(10000) = 75\text{ meters}$. [M1][A1]
A power model $y = a x^b$ passes through $(2, 12)$ and $(6, 108)$. Determine the exact values of $a$ and $b$.
Solution:
Equation 1: $12 = a(2^b)$. Equation 2: $108 = a(6^b)$. [M1]
Divide Eq 2 by Eq 1: $\frac{108}{12} = \frac{a(6^b)}{a(2^b)} \implies 9 = \left(\frac{6}{2}\right)^b = 3^b$. [M1]
$3^b = 9 = 3^2 \implies b = 2$. [A1]
Substitute $b = 2$: $12 = a(2^2) = 4a \implies a = 3$. [A1]
Find the roots (horizontal intercepts) of the rational function $g(x) = \frac{2x - 10}{x + 3}$.
Solution:
Set $g(x) = 0 \implies \frac{2x - 10}{x + 3} = 0$. [M1]
The fraction is zero when the numerator is zero (and denominator $\ne 0$): $2x - 10 = 0 \implies 2x = 10 \implies x = 5$. [M1]
The root is at $(5, 0)$. [A1]
Data(2,12),(4,48),(6,108)follow a power model. Find n,k and check the third observation.
Ratio4=2ⁿ givesn2;k12/4=3. Prediction at6is108, agreeing with the third value.
For light intensity $I=1200/d^2$ with Iinlux and dinmetres, give units of1200and predict atd10. Explain one limitation.
Constant unitslux·m²; intensity12lux. A point-source inverse-square model ignores reflected light and background illumination.
Core Formulas (Topic 2.3 & 2.6)
| Direct Power Variation | $$y = kx^n \iff \frac{y}{x^n} = k$$ |
| Inverse Variation | $$y = \frac{k}{x^n} \iff yx^n = k$$ |
| Rational Function Asymptotes | $$f(x) = \frac{a}{x - b} + c \implies \text{VA: } x = b, \quad \text{HA: } y = c$$ |
| Inverse Square Law | $$I_1 d_1^2 = I_2 d_2^2 = k$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Confusing direct and inverse proportionality: In direct variation, $\frac{y}{x^n}$ is constant; in inverse variation, $y \cdot x^n$ is constant.
- Trap 2: Forgetting power exponents: If $y \propto x^2$, doubling $x$ multiplies $y$ by 4, not 2!
- Trap 3: Writing asymptotes as numbers instead of line equations: Writing "asymptote is 4" loses the mark. You must write the equation of the line: "$x = 4$" or "$y = 3$".
- Trap 4: Missing the units of $k$: When finding constant $k$, deduce its dimensional units (e.g. worker-hours, $\text{lux}\cdot\text{m}^2$).
- Trap 5: Ignoring vertical asymptote in practical domains: A real model cannot have an input equal to the vertical asymptote where division by zero occurs.
5-Question Concept Check
Score: - / 51. If $y$ is inversely proportional to $x$, what happens to $y$ when $x$ is doubled?
2. What is the vertical asymptote of $f(x) = \frac{8}{2x - 6} + 5$?
3. In the formula $y = kx^3$, if $x = 2$ gives $y = 40$, what is the value of $k$?
4. What is the horizontal asymptote of $g(x) = \frac{15}{x} - 4$?
5. If intensity obeys an inverse square law $I \propto \frac{1}{d^2}$, and distance is doubled ($2d$), what is the new intensity?