Inverse and Joint Variation

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

Solution:

$$\begin{gathered} y\alpha z \\ y=kz \\ y=8 \\ z=4 \end{gathered}$$

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

$8=k\times 4$

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

$$\begin{gathered} \frac{8}{4}=\frac{k\times 4}{4} \\ k=2 \end{gathered}$$

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

$$\begin{gathered} y=? \\ z=14 \\ y=kz \end{gathered}$$

Substitute the known values;

$$\begin{gathered} y=2\times 14 \\ y=28 \end{gathered}$$

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 radius of the circular cookie be r

Let the length of the watch be l

From the question, our relationship is

$$\begin{gathered} r\alpha \sqrt{l} \\ r=k\sqrt{l} \end{gathered}$$

Make k the subject of the formula

$$\begin{gathered} k=\frac{r}{\sqrt{l}} \\ k=\frac{r_1}{\sqrt{l_1}}=\frac{r_2}{\sqrt{l_2}} \end{gathered}$$

We need to state our variables:

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

Thus

$$\begin{gathered} \mathrm{Perimeter}=\mathrm{circumference}\mathrm{of}\mathrm{circle}=2\pi r=44cm \\ 2\times \frac{22}{7}\times r_1=44cm \\ \frac{44r_1}{7}=44cm \end{gathered}$$

Multiply both sides of the equation by 7

$44r_1=44cm\times 7$

Divide both sides of the equation by 44

$$\begin{gathered} r_1=7cm \\ {r}_2=? \\ {l}_1=49cm \\ {l}_2=121cm \end{gathered}$$

Remember that

$\frac{r_1}{\sqrt{l_1}}=\frac{r_2}{\sqrt{l_2}}$

Substitute the values of our variables into the equation

$$\begin{gathered} \frac{7}{\sqrt{49}}=\frac{r_2}{\sqrt{121}} \\ \frac{7}{7}=\frac{r_2}{11} \\ 1=\frac{r_2}{11} \end{gathered}$$

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

$$\begin{gathered} \mathrm{Perimeter}=\mathrm{circumference}=2\pi r \\ \mathrm{Perimeter}\mathrm{of}\mathrm{cookie}=2\times \frac{22}{7}\times 11 \\ \mathrm{Perimeter}\mathrm{of}\mathrm{cookie}=\frac{484}{7}=69\frac{1}{7} \\ \mathrm{Perimeter}\mathrm{of}\mathrm{cookie}=69\frac{1}{7}cm \end{gathered}$$

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:

$$\begin{gathered} t\alpha gv \\ t=kgv \end{gathered}$$

Make k the subject of the formula

$$\begin{gathered} k=\frac{t}{gv} \\ t=16 \\ g=2 \\ v=5 \\ k=\frac{16}{2\times 5} \\ k=1.6 \end{gathered}$$

Then

$$\begin{gathered} g=3 \\ v=8 \\ t=? \\ t=kgv \\ t=1.6\times 3\times 8 \\ t=38.4 \end{gathered}$$

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:

$$\begin{gathered} x\alpha y{z}^2 \\ x=ky{z}^2 \end{gathered}$$

Make k the subject of the formula.

$$\begin{gathered} k=\frac{x}{y{z}^2} \\ k=\frac{x_1}{y_1{z_1}^2}=\frac{x_2}{y_2{z_2}^2} \\ {x}_1=15 \\ {x}_2=18 \\ {y}_1=6 \\ {y}_2=9 \\ {z}_1=2 \\ {z}_2=? \\ \frac{x_1}{y_1{z_1}^2}=\frac{x_2}{y_2{z_2}^2} \end{gathered}$$

Substitute the values of the variables into the equation

$$\begin{gathered} \frac{15}{6\times 2^2}=\frac{18}{9\times z^2} \\ \frac{15}{6\times 4}=\frac{18}{9\times z^2} \\ \frac{15}{24}=\frac{18}{9z^2} \end{gathered}$$

Simplify both sides of the equation

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

Cross multiply

$$\begin{gathered} 5\times z^2=2\times 8 \\ 5z^2=16 \end{gathered}$$

Divide both sides by 5

$$\begin{gathered} z^2=\frac{16}{5} \\ \sqrt{z^2}=\sqrt{\frac{16}{5}} \\ z=\frac{4}{\sqrt{5}} \end{gathered}$$

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

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

or

$\cong 1.79$

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

Solution:

$$\begin{gathered} w\alpha \frac{1}{u} \\ w=\frac{k}{u} \\ k=wu \\ w=6 \\ u=2 \\ k=6\times 2 \\ k=12 \end{gathered}$$

Then

$$\begin{gathered} u=6 \\ k=12 \\ w=\frac{k}{u} \\ w=\frac{12}{6} \\ w=2 \end{gathered}$$

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.

$$\begin{gathered} S\alpha \frac{1}{t} \\ S=\frac{k}{t} \\ k=St \\ k=S_1{t}_1=S_2{t}_2 \\ {S}_1{t}_1=S_2{t}_2 \\ {S}_1=100 \\ {S}_2=150 \\ {t}_1=10 \\ {t}_2=? \end{gathered}$$

Substitute the values into the equation

$$\begin{gathered} S_1{t}_1=S_2{t}_2 \\ 100\times 10=150\times t_2 \\ 1000=150t_2 \end{gathered}$$

Divide both sides by 150

$$\begin{gathered} \mathrm{Divide}\mathrm{both}\mathrm{sides}\mathrm{by}150 \\ {t}_2=\frac{1000}{150} \\ {t}_2=6.67\mathrm{seconds} \end{gathered}$$

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

$$\begin{gathered} p\alpha \frac{q}{r} \\ p=k\left(\frac{q}{r}\right) \\ p=10 \\ q=5 \\ r=3 \\ 10=k\left(\frac{5}{3}\right) \end{gathered}$$

Cross multiply

$$\begin{gathered} 30=5k \\ \frac{30}{5}=\frac{5k}{5} \\ k=6 \end{gathered}$$

When p is 3 and q is 4

$$\begin{gathered} p=k\left(\frac{q}{r}\right) \\ 3=6\left(\frac{4}{r}\right) \\ 3=\frac{24}{r} \end{gathered}$$

Cross multiply

$$\begin{gathered} 3r=24 \\ \frac{3r}{3}=\frac{24}{3} \\ r=8 \end{gathered}$$

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

$$\begin{gathered} I\alpha \frac{H}{F} \\ I=k\left(\frac{H}{F}\right) \\ I=20 \\ H=16 \\ F=8 \\ 20=k\left(\frac{16}{8}\right) \end{gathered}$$

Cross multiply

$$\begin{gathered} 160=16k \\ \frac{16k}{16}=\frac{160}{16} \\ k=10 \end{gathered}$$

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

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

Cross multiply

$$\begin{gathered} 10F=120 \\ \frac{10F}{10}=\frac{120}{10} \\ F=12 \end{gathered}$$

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

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

$$\begin{gathered} w=\frac{k}{d} \\ 100=\frac{k}{4} \\ \xcancel{\frac{100}{1}=\frac{k}{4}} \\ k=4\times 100 \\ k=400 \end{gathered}$$

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

$$\begin{gathered} w=\frac{k}{d} \\ 250=\frac{400}{d} \\ \xcancel{\frac{250}{1}=\frac{400}{d}} \\ 250d=400 \\ \frac{250d}{250}=\frac{400}{250} \\ d=1.6m \end{gathered}$$

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