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

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

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

$\cong 1.79$

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

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