# Inverse proportionalities

*inverse relation*between variable quantities \(\normalsize{x}\) and \(\normalsize{y}\) given by the equation \({y=\dfrac{a}{x}}\) for some fixed number \(\normalsize a\gt 0\). This connects directly with the geometry of the hyperbola that we have been studying. In this step, we introduce inverse relations and discuss basic examples.

## Inverse relations and symmetry

An*inverse relation*, or

*inverse proportionality*is a relation of the form \[\Large{y=\frac{a}{x}}\]for some fixed constant of proportionality \(\normalsize a\gt 0\). In this kind of situation, the bigger \(\normalsize{x}\) gets the smaller \(\normalsize{y}\) gets, and conversely, the smaller \(\normalsize{x}\) gets the larger \(\normalsize{y}\) gets. In fact by thinking about the particular inverse relation \(\normalsize{y=1/x}\) we see that the two intervals \(\normalsize{(0,1]}\) and \(\normalsize{[1,\infty)}\) can be completely matched up: for every number in the first interval, its reciprocal is a number in the second interval, and vice versa. Of course this matching severely distorts measurements.If we rewrite the basic inverse relation in the form \[\Large{xy=a}\]we see that it is completely symmetric between \(\normalsize{x}\) and \(\normalsize{y}\): the product of \(\normalsize{x}\) and \(\normalsize{y}\) is constant.Also we see that if one of \(\normalsize{x}\) and \(\normalsize{y}\) is positive, so is the other one. Conversely if one is negative, then so is the other one. We can also state the relation in the form \[\Large y \propto \frac{1}{x}.\]

## Speed and time

The familiar equation \[\Large{\operatorname{speed}=\frac{\operatorname{distance}}{\operatorname{time}}}\]expresses the fact that if \(\normalsize{\operatorname{distance}}\) is constant, then \(\normalsize{\operatorname{speed}}\) and \(\normalsize{\operatorname{time}}\) are inversely proportional. In this discussion the terms*velocity*and

*speed*are essentially synonymous.For example suppose we want to travel \(\normalsize{10}\;\text{km}\). If we have a speed of \(\normalsize{50\;\text{km}/\text{hr}}\), then the time taken will be \(\normalsize{10/50=1/5}\) hours, or \(\normalsize{12}\) minutes. If we have a speed of \(\normalsize{100\;\text{km}/\text{hr}}\), then the time taken will be only \(\normalsize{6}\) minutes. If we double one quantity, the other halves. If we triple one quantity, the other is divided by three, and so on.

Q1(E): If an Olympic walker takes \(\normalsize{3}\) hours to walk a race, and I walk half as fast as the Olympic walker, then how long will it take me to walk the same race?Q2(E): If a marathon runner runs \(\normalsize{42}\) kilometers in three hours, and you run \(\normalsize{5}\) kilometers in \(\normalsize{45}\) minutes, then what is the ratio between your average speed and the marathon runner’s average speed?

## Many hands make light work

^{Concrete blocks By Justanother GFDL or CC BY-SA 3.0, via Wikimedia Commons}

*person-hours*.

Q3(M): If \(\normalsize{8}\) people take \(\normalsize{5}\) hours to paint a house, how long would it take \(\normalsize{3}\) people to paint the same house?

^{egg, By Hans, Free for commercial use / No attribution required}

Q4(C): If \(\normalsize{6}\) children take \(\normalsize{5}\) hours to paint \(\normalsize{4}\) Easter eggs, then how long would it take \(\normalsize{5}\) children to paint \(\normalsize{3}\) Easter eggs?

## Answers

A1.Since we halve the speed, the time doubles, and so it will take \(\normalsize{6}\) hours to walk the same race.A2.The average speed of the marathon runner is \(\normalsize{\frac{42\;\text{km}}{3\;\text{hr}}=14\;\text{km}/\text{hr}}\). My average speed is \(\normalsize{\frac{5\;\text{km}}{45\;\text{min}}}\), or equivalently, \(\normalsize{\frac{5\;\text{km}}{3/4\;\text{hr}}=20/3 \;\text{km}/\text{hr}}.\) Therefore, the ratio of average speeds is \(\normalsize{\frac{20/3}{14}}=10/21.\)A3.A single person will paint the house in \(\normalsize{8\times 5\;\text{hr}=40\;\text{hr}}.\) Therefore, three people will paint the same house in \(\normalsize{40/3}\approx 13.3\) hours.A4.One child will paint \(\normalsize{4}\) eggs in \(\normalsize{6\times 5\;\text{hr}=30\;\text{hr}}\). Therefore, one child will paint one egg in \(\normalsize{30/4\;\text{hr}=15/2\;\text{hr}}\). Thus, \(\normalsize{5}\) children will paint one egg in \(\normalsize{\frac{15/2}{5}\;\text{hr}=3/2\;\text{hr}}\) (this is an inverse relation), and so \(\normalsize{5}\) children will paint \(\normalsize{3}\) eggs in \(\normalsize{3\times 3/2\;\text{hr}=9/2\;\text{hr}}\) (this is a direct relation).

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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