C. Polynomial and Rational Functions · Rational Functions and Equations

Modelling with Rational Functions and Inequalities

C3.7 C4.2About 75 minutes

Learning goals

I can build and use rational models for intensity, concentration, average cost and growth that levels off.
I can interpret intercepts and asymptotes in context, and state a sensible domain for a model.
I can solve rational equations and inequalities that answer real questions, by algebra and by reading graphs.
I can use the discriminant to find the maximum value of a model such as C(t) = 5t/(t² + 1).

Words to know

mathematical model
A function that describes a real situation approximately, usually only for a limited domain.

inverse-square model
A model of the form y = k/x², the reciprocal of a quadratic. Light, sound intensity and gravity follow inverse-square laws.

domain of a model
The input values that make sense in the situation, for example t ≥ 0 for time, or t ≤ 40 if a tank is full after 40 minutes.

levelling off
Approaching a horizontal asymptote: the output changes less and less as the input grows.

Rational functions in the real world

Throughout this unit, rational functions have appeared in real situations: travel times, batting averages, average costs and concentrations. In this last lesson you will put everything together: build a model, read its asymptotes in context, choose a sensible domain, and answer 'when is it at least...?' questions with inequalities.

Remember that a model is a simplification. Every answer must be checked against the situation: negative times, or concentrations beyond what the tank can hold, are signs that you have gone outside the model's domain.

Model 1: inverse-square light

Light from a small lamp spreads out over the surface of a growing sphere, whose area is proportional to the square of the distance. So the brightness on a surface (its illuminance, measured in lux) is modelled by I(d) = k/d², the reciprocal of a quadratic from the first lessons of this unit. In context only d > 0 makes sense, so you see only the right branch.

Worked example

Example 1: How far can you read?

A lamp gives I(d) = 900/d² lux at a distance of d metres. (a) Find I at 5 m. (b) Where is I = 100 lux? (c) For which distances is I ≥ 4 lux?

(a) I(5) = 900/25 = 36 lux.
(b) 900/d² = 100 gives d² = 9, so d = 3 m (d > 0).
(c) 900/d² ≥ 4. Since d² > 0, multiplying by d² is safe: 900 ≥ 4d², so d² ≤ 225.
With d > 0: 0 < d ≤ 15.

(a) 36 lux (b) 3 m (c) 0 < d ≤ 15 m.

The right branch of y = 900/d². The graph is on or above the line I = 4 for every distance up to 15 m (shaded). Doubling the distance from 5 m to 10 m cuts the intensity from 36 to 9 lux, a quarter.

Model 2: concentration in a filling tank

A 500 L tank starts with 100 L of fresh water. Brine (salt water) with 30 g of salt per litre flows in at 10 L/min, and nothing flows out. After t minutes:

salt = 30 × 10t = 300t grams;
volume = 100 + 10t litres;
concentration C(t) = 300t/(100 + 10t) = 30t/(10 + t) g/L.

The tank is full when 100 + 10t = 500, so the model's domain is 0 ≤ t ≤ 40. The horizontal asymptote C = 30 is the concentration of the incoming brine: the mixture can get closer and closer to it but never exceed it. In this tank it never gets very close, because the tank fills first.

Worked example

Example 2: When is the water salty enough?

Using C(t) = 30t/(10 + t), find when the concentration is at least 20 g/L, and the concentration when the tank is full.

For t ≥ 0, 10 + t > 0, so multiplying is safe: 30t ≥ 20(10 + t).
30t ≥ 200 + 20t, so 10t ≥ 200 and t ≥ 20.
Combine with the domain: 20 ≤ t ≤ 40.
When full: C(40) = 1200/50 = 24 g/L.

From 20 min until the tank is full at 40 min; the final concentration is 24 g/L.

The solid part is the model's domain, 0 ≤ t ≤ 40. The dashed continuation shows what the formula would do if the tank were bigger: level off toward C = 30.

Model 3: a rise and a fall

Some quantities rise and then fall, like the concentration of a medication in the bloodstream after one dose. A simple model is C(t) = 5t/(t² + 1) (mg/L, t in hours). It is 0 at t = 0, rises, then falls back toward the horizontal asymptote C = 0. (This is a textbook model, not medical advice.)

Worked example

Example 3: How long is the level at least 2 mg/L?

Solve 5t/(t² + 1) ≥ 2 for t ≥ 0.

t² + 1 is always positive, so multiply: 5t ≥ 2t² + 2.
2t² − 5t + 2 ≤ 0, so (2t − 1)(t − 2) ≤ 0.
The parabola y = 2t² − 5t + 2 opens up and has zeros 1/2 and 2, so it is ≤ 0 between them.

0.5 h ≤ t ≤ 2 h: the level stays at or above 2 mg/L for 1.5 hours.

Worked example

Example 4: The maximum level, without calculus

Find the highest concentration of C(t) = 5t/(t² + 1).

Ask: for which values k does 5t/(t² + 1) = k have a solution?
Rearrange: kt² − 5t + k = 0. For k > 0 this quadratic in t has a real root when its discriminant is ≥ 0: 25 − 4k² ≥ 0, so k ≤ 2.5.
So the largest possible value is k = 2.5. Then 2.5t² − 5t + 2.5 = 0, so t² − 2t + 1 = 0 and t = 1.

The maximum is 2.5 mg/L at t = 1 h. (In Calculus and Vectors, MCV4U, you will find maximums like this with derivatives.)

The medication model rises to its maximum (1, 2.5) and falls back toward C = 0. It is at or above 2 mg/L for 0.5 ≤ t ≤ 2 (shaded).

Worked example

Example 5: Growth that levels off

A stocked lake's trout population is modelled by P(t) = (800t + 200)/(t + 2), t in years. (a) Find the initial population. (b) What does the horizontal asymptote mean? (c) When will the population be at least 600?

(a) P(0) = 200/2 = 100 trout.
(b) HA P = 800/1 = 800: the model predicts the population levels off near 800, the most the lake can support in this model. The VA t = −2 has no meaning (before stocking).
(c) For t ≥ 0, t + 2 > 0, so multiply: 800t + 200 ≥ 600t + 1200, so 200t ≥ 1000 and t ≥ 5 years.
Check: P(5) = 4200/7 = 600 ✓.

(a) 100 (b) the population approaches 800 (c) after 5 years or more.

Just for fun

Stand twice as far from a lamp and each square centimetre of your page gets only about a quarter of the light. That is the inverse-square law at work, and the reason the far corner of the room always seems to be where you left your book.

Hands-on activity

Flashlight inverse-square check

You need: a flashlight (a phone light works), a measuring tape, a ruler, a sheet of paper taped to a wall, and a dim room. Do not shine the light into anyone's eyes.

Hold the flashlight pointing straight at the wall at d = 50 cm. Mark the edge of the bright circle on the paper and measure its diameter D.
Repeat at d = 100 cm and d = 150 cm.
Make a table of d, D and the spot area A = π(D/2)².
The same light spreads over the area A, so the brightness on the paper is roughly proportional to 1/A. Calculate 1/A for each distance and compare the ratios with 1/d².
Does doubling the distance (50 cm to 100 cm) make the spot area about 4 times larger?

What to look for: the spot diameter grows roughly in proportion to the distance, so the area grows roughly like d² and the light per square centimetre falls roughly like 1/d². Real flashlights have lenses and reflectors, so your ratios will be close to 4 and 9 but probably not exact. That gap between model and reality is normal and worth describing.

Your turn

For I(d) = 900/d², find I at d = 5 m.

For I(d) = 900/d², at what distance is I = 100 lux?

For I(d) = 900/d², solve I ≥ 36.

For the salt tank, C(t) = 30t/(10 + t). Find C(10).

For the salt tank, when is C(t) ≥ 24 g/L? Use the domain 0 ≤ t ≤ 40.

For P(t) = (800t + 200)/(t + 2), what is the initial population and what does it approach?

For P(t) = (800t + 200)/(t + 2), when is P ≥ 700?

More practice: warm-up

Find the average cost C(60) for C(n) = (120 + 1.5n)/n.

State the horizontal asymptote of C(t) = 30t/(10 + t).

Find P(0) for P(t) = (800t + 200)/(t + 2).

Find I(30) for I(d) = 900/d².

What is a sensible domain for I(d) = 900/d² in the lamp situation?

More practice: on your own

How many pins must the club make so the average cost C(n) = (120 + 1.5n)/n is at most $2?

For C(t) = 5t/(t² + 1), when is C ≥ 1 mg/L? Give exact values and decimals.

For I(d) = 900/d², at what distance is the intensity one quarter of its value at 2 m?

A tank holds 50 L of fresh water. Brine with 20 g/L of salt flows in at 5 L/min. Write C(t) and find when C = 15 g/L.

Explain why the vertical asymptote of P(t) = (800t + 200)/(t + 2) has no meaning in the lake model.

Challenge

Show that P(t) = (800t + 200)/(t + 2) is less than 800 for every t ≥ 0.

Pipe A fills a pool in t hours, and pipe B takes 2 hours longer. Together they must fill it in at most 2.4 h. What is the largest possible t?

Use the discriminant method to find the maximum value of f(x) = 4x/(x² + 9) for x > 0, and where it occurs.

Check your understanding

Quiz

For I(d) = 900/d² with d > 0, where is I ≥ 4?

In the salt-tank model C(t) = 30t/(10 + t), what does the horizontal asymptote C = 30 represent?

How many pins make the average cost (120 + 1.5n)/n at most $2?

When solving 5t/(t² + 1) ≥ 2, why may you multiply both sides by t² + 1 without reversing the sign?

For t ≥ 0, what is the solution of 5t/(t² + 1) ≥ 2?

What does the vertical asymptote t = −2 of P(t) = (800t + 200)/(t + 2) mean in the lake model?

For parents and teachers

Big idea (C3.7, C4.2): rational functions model real quantities: inverse-square intensity, concentrations in a filling tank, average costs and growth that levels off. The asymptotes have real meanings (a limit that is approached, or a value outside the situation), and every model has a limited domain.

How to help: In the flashlight activity, ask your teen why the measured ratios are not exactly 4 and 9. Discussing the gap between a model and reality is exactly what university science courses expect. The medication example is a made-up textbook model, not medical information.

Common difficulties:

Ignoring the domain (giving negative times, or times after a tank is full).
Multiplying an inequality by an expression without first checking that it is positive.
Misreading what an asymptote means in context.
Rounding too early in multi-step problems.