# Inverse and Joint Variation

Imagine if your height is determined by how much food you consume daily. It means you can project a height you would wish to attain by the consumption of a certain amount of food. Hereafter, you would learn how this occurs and is calculated through the concept of variation.

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## What is variation?

In Mathematics, variation tells us the relationship between variables or quantities. Such relationship could be direct, inverse, joint or partial.

## Direct variation

This occurs when the relationship between two variables is affected in the same manner. This means that an increase in one variable leads to an increase in the other variable. Also, a decrease in one variable would mean a decrease in the other. The ratio between these variables is often defined by a constant k.

When a person says p varies directly as q, it is expressed as:

$p\alpha q\phantom{\rule{0ex}{0ex}}p=kq$

Recall that k is the relationship constant also known as the constant of proportionality.

Knowing that for all possible values of p and q, k never changes (because it is constant) then:

${p}_{1}=k{q}_{1}\phantom{\rule{0ex}{0ex}}{p}_{2}=k{q}_{2}\phantom{\rule{0ex}{0ex}}k=\frac{{p}_{1}}{{q}_{1}}\phantom{\rule{0ex}{0ex}}k=\frac{{p}_{2}}{{q}_{2}}\phantom{\rule{0ex}{0ex}}k=\frac{{p}_{1}}{{q}_{1}}=\frac{{p}_{2}}{{q}_{2}}\phantom{\rule{0ex}{0ex}}\frac{{p}_{1}}{{q}_{1}}=\frac{{p}_{2}}{{q}_{2}}\phantom{\rule{0ex}{0ex}}{p}_{1}{q}_{2}={p}_{2}{q}_{1}$

There are other ways to express direct variation such as;

"p varies proportionately as q" or "p varies by the same proportion as q".

Given that y varies directly as z, and when y is 8, z is 4. Find y when z is 14.

Solution:

$y\alpha z\phantom{\rule{0ex}{0ex}}y=kz\phantom{\rule{0ex}{0ex}}y=8\phantom{\rule{0ex}{0ex}}z=4$

To find k, substitute the values of y and z in the equation

$8=k×4$

Make k the subject of the formula by dividing both sides of the equation by 4. Thus

$\frac{8}{4}=\frac{k×4}{4}\phantom{\rule{0ex}{0ex}}k=2$

Now, k has been solved and equals to 2, we can now apply its value in the second case. Thus

$y=?\phantom{\rule{0ex}{0ex}}z=14\phantom{\rule{0ex}{0ex}}y=kz$

Substitute the known values;

$y=2×14\phantom{\rule{0ex}{0ex}}y=28$

The radius of a circular cookie varies directly as the square root of the length of a watch. When the perimeter of the cookie is 44 cm, the watch has a length of 49 cm. What is the circumference of the cookie when the length of the watch is 121 cm?

Solution:

Let the length of the watch be l

From the question, our relationship is

$r\alpha \sqrt{l}\phantom{\rule{0ex}{0ex}}r=k\sqrt{l}$

Make k the subject of the formula

$k=\frac{r}{\sqrt{l}}\phantom{\rule{0ex}{0ex}}k=\frac{{r}_{1}}{\sqrt{{l}_{1}}}=\frac{{r}_{2}}{\sqrt{{l}_{2}}}\phantom{\rule{0ex}{0ex}}$

We need to state our variables:

r1 - note that we have the perimeter of the cookie given. We can calculate the radius.

Thus

$\mathrm{Perimeter}=\mathrm{circumference}\mathrm{of}\mathrm{circle}=2\pi r=44cm\phantom{\rule{0ex}{0ex}}2×\frac{22}{7}×{r}_{1}=44cm\phantom{\rule{0ex}{0ex}}\frac{44{r}_{1}}{7}=44cm$

Multiply both sides of the equation by 7

$44{r}_{1}=44cm×7$

Divide both sides of the equation by 44

${r}_{1}=7cm\phantom{\rule{0ex}{0ex}}{r}_{2}=?\phantom{\rule{0ex}{0ex}}{l}_{1}=49cm\phantom{\rule{0ex}{0ex}}{l}_{2}=121cm$

Remember that

$\frac{{r}_{1}}{\sqrt{{l}_{1}}}=\frac{{r}_{2}}{\sqrt{{l}_{2}}}$

Substitute the values of our variables into the equation

$\frac{7}{\sqrt{49}}=\frac{{r}_{2}}{\sqrt{121}}\phantom{\rule{0ex}{0ex}}\frac{7}{7}=\frac{{r}_{2}}{11}\phantom{\rule{0ex}{0ex}}1=\frac{{r}_{2}}{11}$

Multiply both sides of the equation by 11

${r}_{2}=11$

Now, we have the radius so we can calculate the circumference (perimeter) of the cookie. Thus

$\mathrm{Perimeter}=\mathrm{circumference}=2\pi r\phantom{\rule{0ex}{0ex}}\mathrm{Perimeter}\mathrm{of}\mathrm{cookie}=2×\frac{22}{7}×11\phantom{\rule{0ex}{0ex}}\mathrm{Perimeter}\mathrm{of}\mathrm{cookie}=\frac{484}{7}=69\frac{1}{7}\phantom{\rule{0ex}{0ex}}\mathrm{Perimeter}\mathrm{of}\mathrm{cookie}=69\frac{1}{7}cm$

## Joint variation

This occurs when one variable is related to the product of two or more variables in the same manner. In other words, this takes place when a quantity varies directly as the product of two or more quantities. With the constant of variation as k, if b varies jointly as c and d then:

$b\alpha cd\phantom{\rule{0ex}{0ex}}b=kcd\phantom{\rule{0ex}{0ex}}k=\frac{b}{cd}\phantom{\rule{0ex}{0ex}}k=\frac{{b}_{1}}{{c}_{1}{d}_{1}}=\frac{{b}_{2}}{{c}_{2}{d}_{2}}$

t varies jointly as g and v. When t is 16, g is 2 and v is 5. Find t when g is 3 and v is 8.

Solution:

$t\alpha gv\phantom{\rule{0ex}{0ex}}t=kgv$

Make k the subject of the formula

$k=\frac{t}{gv}\phantom{\rule{0ex}{0ex}}t=16\phantom{\rule{0ex}{0ex}}g=2\phantom{\rule{0ex}{0ex}}v=5\phantom{\rule{0ex}{0ex}}k=\frac{16}{2×5}\phantom{\rule{0ex}{0ex}}k=1.6$

Then

$g=3\phantom{\rule{0ex}{0ex}}v=8\phantom{\rule{0ex}{0ex}}t=?\phantom{\rule{0ex}{0ex}}t=kgv\phantom{\rule{0ex}{0ex}}t=1.6×3×8\phantom{\rule{0ex}{0ex}}t=38.4$

x varies jointly as y and the square of z. When x is 15, y is 6 and z is 2. Find the z when x is 18 and y is 9.

Solution:

$x\alpha y{z}^{2}\phantom{\rule{0ex}{0ex}}x=ky{z}^{2}$

Make k the subject of the formula.

$k=\frac{x}{y{z}^{2}}\phantom{\rule{0ex}{0ex}}k=\frac{{x}_{1}}{{y}_{1}{{z}_{1}}^{2}}=\frac{{x}_{2}}{{y}_{2}{{z}_{2}}^{2}}\phantom{\rule{0ex}{0ex}}{x}_{1}=15\phantom{\rule{0ex}{0ex}}{x}_{2}=18\phantom{\rule{0ex}{0ex}}{y}_{1}=6\phantom{\rule{0ex}{0ex}}{y}_{2}=9\phantom{\rule{0ex}{0ex}}{z}_{1}=2\phantom{\rule{0ex}{0ex}}{z}_{2}=?\phantom{\rule{0ex}{0ex}}\frac{{x}_{1}}{{y}_{1}{{z}_{1}}^{2}}=\frac{{x}_{2}}{{y}_{2}{{z}_{2}}^{2}}$

Substitute the values of the variables into the equation

$\frac{15}{6×{2}^{2}}=\frac{18}{9×{z}^{2}}\phantom{\rule{0ex}{0ex}}\frac{15}{6×4}=\frac{18}{9×{z}^{2}}\phantom{\rule{0ex}{0ex}}\frac{15}{24}=\frac{18}{9{z}^{2}}$

Simplify both sides of the equation

$\frac{5}{8}=\frac{2}{{z}^{2}}$

Cross multiply

$5×{z}^{2}=2×8\phantom{\rule{0ex}{0ex}}5{z}^{2}=16$

Divide both sides by 5

${z}^{2}=\frac{16}{5}\phantom{\rule{0ex}{0ex}}\sqrt{{z}^{2}}=\sqrt{\frac{16}{5}}\phantom{\rule{0ex}{0ex}}z=\frac{4}{\sqrt{5}}$

Rationalize by multiplying the denominator and numerator by $\sqrt{5}$

$z=\frac{4\sqrt{5}}{5}$

or

$\cong 1.79$

## Inverse variation

Inverse variation is a relationship between two variables with changes in them moving in an opposite direction. This means that when one variable rises, the other variable falls and vice versa.

When b varies inversely as a

$b\alpha \frac{1}{a}\phantom{\rule{0ex}{0ex}}b=k×\frac{1}{a}\phantom{\rule{0ex}{0ex}}b=\frac{k}{a}\phantom{\rule{0ex}{0ex}}k=ba\phantom{\rule{0ex}{0ex}}k={b}_{1}{a}_{1}={b}_{2}{a}_{2}\phantom{\rule{0ex}{0ex}}{b}_{1}{a}_{1}={b}_{2}{a}_{2}$

If w is inversely proportional to u and w = 6 when u = 2. Find w when u = 6.

Solution:

$w\alpha \frac{1}{u}\phantom{\rule{0ex}{0ex}}w=\frac{k}{u}\phantom{\rule{0ex}{0ex}}k=wu\phantom{\rule{0ex}{0ex}}w=6\phantom{\rule{0ex}{0ex}}u=2\phantom{\rule{0ex}{0ex}}k=6×2\phantom{\rule{0ex}{0ex}}k=12$

Then

$u=6\phantom{\rule{0ex}{0ex}}k=12\phantom{\rule{0ex}{0ex}}w=\frac{k}{u}\phantom{\rule{0ex}{0ex}}w=\frac{12}{6}\phantom{\rule{0ex}{0ex}}w=2$

The speed of a train varies inversely to the time it takes. When it moves at 100 m/s it takes 10 seconds to cover a certain distance, how long would it take for the train to cover the same distance if it moves at 150 m/s?

Solution:

Let S represent the speed of the train and t represent the time spent by the train.

$S\alpha \frac{1}{t}\phantom{\rule{0ex}{0ex}}S=\frac{k}{t}\phantom{\rule{0ex}{0ex}}k=St\phantom{\rule{0ex}{0ex}}k={S}_{1}{t}_{1}={S}_{2}{t}_{2}\phantom{\rule{0ex}{0ex}}{S}_{1}{t}_{1}={S}_{2}{t}_{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{S}_{1}=100\phantom{\rule{0ex}{0ex}}{S}_{2}=150\phantom{\rule{0ex}{0ex}}{t}_{1}=10\phantom{\rule{0ex}{0ex}}{t}_{2}=?$

Substitute the values into the equation

${S}_{1}{t}_{1}={S}_{2}{t}_{2}\phantom{\rule{0ex}{0ex}}100×10=150×{t}_{2}\phantom{\rule{0ex}{0ex}}1000=150{t}_{2}$

Divide both sides by 150

$\mathrm{Divide}\mathrm{both}\mathrm{sides}\mathrm{by}150\phantom{\rule{0ex}{0ex}}{t}_{2}=\frac{1000}{150}\phantom{\rule{0ex}{0ex}}{t}_{2}=6.67\mathrm{seconds}$

## What is the difference between inverse and joint variation?

There are several differences between joint variation and inverse variation.

1. Inverse variation shows relationships between two variables while joint variation shows the relationship between more than two variables.

2. In inverse variation an increase in one variable would bring a decrease to the other variable. However, in joint variation, an increase in the first variable would lead to an increase in the product of the remaining variables.

## Combined variation

Combined variation occurs when a variable varies directly to one or more variables and is inversely related to the rest.

This means that when this variable increases another variable increases and at the same time others decrease. For example, speed varies directly with distance but inversely with time. This means that with an increase in speed, more distance is covered but the time is reduced to get to that distance. Thus

$s\alpha \frac{d}{t}$

A quantity p varies directly as q and inversely as r. When p is 10, q is 5 and r is 3. Find r when p is 3 and q is 4.

Solution:

Write out the relationship

$p\alpha \frac{q}{r}\phantom{\rule{0ex}{0ex}}p=k\left(\frac{q}{r}\right)\phantom{\rule{0ex}{0ex}}p=10\phantom{\rule{0ex}{0ex}}q=5\phantom{\rule{0ex}{0ex}}r=3\phantom{\rule{0ex}{0ex}}10=k\left(\frac{5}{3}\right)$

Cross multiply

$30=5k\phantom{\rule{0ex}{0ex}}\frac{30}{5}=\frac{5k}{5}\phantom{\rule{0ex}{0ex}}k=6$

When p is 3 and q is 4

$p=k\left(\frac{q}{r}\right)\phantom{\rule{0ex}{0ex}}3=6\left(\frac{4}{r}\right)\phantom{\rule{0ex}{0ex}}3=\frac{24}{r}$

Cross multiply

$3r=24\phantom{\rule{0ex}{0ex}}\frac{3r}{3}=\frac{24}{3}\phantom{\rule{0ex}{0ex}}r=8$

Ireti, Kohe and Finicky run a family business based on a combined relationship with respect to the proportion of their individual earnings. Ireti's contribution is directly proportional to that of Kohe but inversely to that of Finicky; when Ireti contributes 20%, Kohe contributes 16% while Finicky contributes 8%. What percentage of Finicky's income would be contributed if Ireti and Kohe contribute 10% and 12% of their earnings respectively?

Solution:

Let us represent I for Ireti, H for Kohe and F for Finicky. So

$I\alpha \frac{H}{F}\phantom{\rule{0ex}{0ex}}I=k\left(\frac{H}{F}\right)\phantom{\rule{0ex}{0ex}}I=20\phantom{\rule{0ex}{0ex}}H=16\phantom{\rule{0ex}{0ex}}F=8\phantom{\rule{0ex}{0ex}}20=k\left(\frac{16}{8}\right)$

Cross multiply

$160=16k\phantom{\rule{0ex}{0ex}}\frac{16k}{16}=\frac{160}{16}\phantom{\rule{0ex}{0ex}}k=10$

Therefore, when I is 10, H is 12, F would be

$10=10\left(\frac{12}{F}\right)$

Cross multiply

$10F=120\phantom{\rule{0ex}{0ex}}\frac{10F}{10}=\frac{120}{10}\phantom{\rule{0ex}{0ex}}F=12$

Thus, when Ireti and Kohe contribute 10% and 12% of their income respectively, Finicky would contribute 12% of his income.

## Real-life examples of inverse and joint variation

The concept of variation is indeed much understandable when related to our day-to-day activities.

The weight of Tom's bag varies inversely with the distance he covers per second. When his bag weighs 100 N he covers a distance of 4 m per second. What distance will he cover every second if he was to carry a luggage of 250 N.

Solution:

Let w represent weight and d represent the distance covered every second. From the question, the relationship is

$w\alpha \frac{1}{d}$

With our values substituted we have

$w=\frac{k}{d}\phantom{\rule{0ex}{0ex}}100=\frac{k}{4}\phantom{\rule{0ex}{0ex}}\overline{)\frac{100}{1}=\frac{k}{4}}\phantom{\rule{0ex}{0ex}}k=4×100\phantom{\rule{0ex}{0ex}}k=400$

Now, that the constant, k, has been gotten we can re-substitute it into the equation to find d when w is 250 N and k is 400

$w=\frac{k}{d}\phantom{\rule{0ex}{0ex}}250=\frac{400}{d}\phantom{\rule{0ex}{0ex}}\overline{)\frac{250}{1}=\frac{400}{d}}\phantom{\rule{0ex}{0ex}}250d=400\phantom{\rule{0ex}{0ex}}\frac{250d}{250}=\frac{400}{250}\phantom{\rule{0ex}{0ex}}d=1.6m$

Therefore, Tom will cover 1.6 m every second with his 250 N bag.

This explains why you can walk or run much faster when carrying nothing as compared to when you are carrying a bag or any kind of load.

## Inverse and Joint Variation - Key takeaways

• Variation tells us the relationship between variables or quantities.

• Direct variation occurs when the relationship between two variables is affected in the same manner.

• Joint variation occurs when one variable is related to the product of two or more variables in the same manner.

• Inverse variation is a relationship between two variables with changes in them moving in an opposite direction.

• Combined variation occurs when a variable varies directly to one or more variables and is inversely related to the rest.

#### Flashcards in Inverse and Joint Variation 10

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What is inverse and joint variation?

Inverse and joint variation are variation types where inverse means an opposite relationship on the quantities involved while joint variation has a direct relationship with with more than two quantities.

How to solve inverse and joint variation?

To solve inverse and joint variation questions, you need to represent the relationship with an equation where k is constant.

How do you calculate Z in joint variation?

Z is a variable in joint variation where z varies jointly as x and y. To calculate z, the variation equation must be written to express the relationship between the three variables.

What is combined variation?

Combined variation occurs when a variable varies directly to one or more variables and is inversely related to the rest.

What is the difference between joint and combined variation?

In joint variation the relation between a variable and the other variables are the same but in a combined variation, the relationship between a variable and other variables are not the same because for some the variable would be directly proportional and for the rest it would be inversely proportional.

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