Inverse variation: find the constant and missing values

A 180 km journey is modelled at constant speed with no stops. At 60 km/h it takes three hours. Predict the time at 90 km/h.

Vary the constant speed for a fixed 180 km journey. Compare 40 with 80 km/h and 60 with 90 km/h. Time is shown to two decimal places where necessary

Which relationship stays true throughout the investigation?

Use the fixed product

An inverse model makes one positive quantity smaller when the other grows, with their product fixed

1. 1Find the constant product Multiply a known input and output. Attach the product of their units \(\displaystyle k=xy\)
2. 2Find either unknown Divide the constant by the known quantity \(\displaystyle \begin{aligned}y&=k/x\\x&=k/y\end{aligned}\)
3. 3Check positivity and assumptions Positive physical quantities exclude zero. Fixed distance and constant speed are assumptions, not guarantees about a real journey
4. 4Test and revise Compare a new observation with the prediction. Investigate fixed delays or changed conditions before choosing a revised model

Data or rule

\(\displaystyle v=72\text{ km/h},\quad t=2.5\text{ h}\)

A journey takes 2.5 hours at a constant 72 km/h with no stops. Find the distance, which is the inverse constant for speed and time

Data or rule

\(\displaystyle vt=180\)

For this 180 km journey, find the time at 80 km/h.

Data or rule

\(\displaystyle vt=180\)

At what constant speed is the 180 km journey completed in 1.5 hours?

Data or rule

\(\displaystyle t=2.25\text{ h}\)

Express 2.25 hours as a number of minutes

Data or rule

\(\displaystyle xy=k\)

The positive variables satisfy xy=k. A known pair is x=2.4, y=15. Find k.

Data or rule

\(\displaystyle xy=36,\quad x=8\)

With xy=36, find y when x=8.

Data or rule

\(\displaystyle xy=36,\quad y=7.5\)

With xy=36, find x when y=7.5

A pupil uses the same product model but adds two to y whenever x increases by two. What is the error?

Could zero speed give a finite travel time for the nonzero 180 km distance?

Data or rule

\(\displaystyle \begin{gathered}t_{\mathrm{predicted}}=3\text{ h}\\t_{\mathrm{observed}}=3.5\text{ h}\end{gathered}\)

At 60 km/h the no-stop model predicts three hours. The observed total journey takes 3.5 hours. Find the extra time in minutes

A 180 km journey takes 3.5 hours at 60 km/h and 2.5 hours at 90 km/h. Which model fits both observations if each journey includes the same fixed delay?

Data or rule

\(\displaystyle t=\frac{180}{v}+0.5\)

Use the revised fixed-delay model to predict total time at 80 km/h.

A 180 km journey takes 3.5 hours at 60 km/h and 2.5 hours at 90 km/h. A proposed model gives 180 divided by speed, plus a fixed half-hour delay, for total time. Which conclusion is justified?

Inverse variation with a stated square

Use the fixed-product method, but multiply by the square when the relationship specifies inverse-square variation

1. 1Multiply by the powered input Ordinary inverse variation fixes xy; inverse-square variation fixes x²y. \(\displaystyle y=k/x^n\quad\Rightarrow\quad k=x^ny\)
2. 2Reverse carefully For positive x in an inverse-square model, first find x²=k/y, then take the positive square root \(\displaystyle x=\sqrt{k/y},\quad x>0\)
3. 3Keep domains and units Zero input is excluded. The units of k are the output units multiplied by the nth power of the input units

Data or rule

\(\displaystyle y=k/x^2,\quad(x,y)=(4,9)\)

Find the inverse-square constant

Data or rule

\(\displaystyle y=144/x^2,\quad x=6\)

Predict y when x increases to six

Data or rule

\(\displaystyle 16=144/x^2,\quad x>0\)

Recover the positive input when y=16.

For the inverse-square pairs (2,36), (3,16), (6,4), which check is appropriate?

Write and check the full inverse rule

On paper, model y inversely proportional to x², given y=12 at positive x=3. Write the completed rule, find y at x=6, and solve y=27. Then compare your working with the model response; the writing is self-reviewed

Model response: Write and check the full inverse rule

Given y inversely proportional to x² and y=12 at positive x=3, write the rule, find y at x=6, and solve y=27. Compare your working with this model

1. 1Model response Find the constant, write the rule, evaluate it at six, then solve for the positive input. Substitution verifies both answers \(\displaystyle \begin{aligned}k&=3^2\times12=108\\y&=\frac{108}{x^2},\quad x>0\\x=6:quad y&=\frac{108}{6^2}=3\\y=27:quad x^2&=\frac{108}{27}=4,\quad x=2\\x=2:quad y&=\frac{108}{2^2}=27\end{aligned}\)