If two quantities are related in such a way that when one changes, there is a corresponding change in the other, then we say there is a variation between the two quantities.

When a symbol or a letter, say ** t**, is used to denote a changeable quantity, then

**is referred to as a variable.**

*t*However, if a symbol or letter, say ** k**, is used to denote an unchangeable quantity, then we say

**is a constant.**

*k*There are four different types of variation, namely; Direct variation, indirect variation, joint variation, partial variation.

We are going to be examining direct variation in this post. we will examine the other types of variations in subsequent posts.

## Direct Variation

When two-variable $x$ and $y$ are related in such a way that when $x$ increases, there is a corresponding increase in $y$, then we say there is a direct variation between $x$ and $y$.

If $x$ varies directly as $y$, then we write $x\alpha y$. To replace the sign $\alpha$ with an equal sign, we introduce the constant k of proportionality. That is $x=ky$

Let's try some illustration:

1. If $x$ varies directly as the square of $y$, then we write $x\alpha y^2$, which is $x=ky^2$

2. If $a$ is directly proportional to the cube of $b$, then we write $a\alpha b^3$, which is, $a=kb^3$

3. If $x$ varies directly as $b$, then we write $x\alpha b$, which is, $x=kb$

__Example 1__

If $p$ varies directly as $r$ and $r=11$ when $p=55$, find

A. The relationship between $p$ and $r$

B. Hence, find $p$ when $r=7$

**Solution:**

$p\alpha r$

$p=kr$......where $k$ is the constant of the proportionality

$r=11$, $p=55$

$55=k(11)$

$55=11k$

Divide both side by $11$

$\frac{55}{11}=\frac{11k}{11}$

$5=k$

$k=5$

The relationship connecting $x$ and $y$ will be:

$p=5r$

Now let's find $p$ when $r=7$

$p=5(7)$

$p=35$

__Example 2__

If $x$ varies directly as $\sqrt{y}$ and $y=9$ when $x=9$. Find $x$ when $y=\frac{16}{9}$

**Solution: **

$x\alpha \sqrt{y}$

$x=k\sqrt{y}$

When $x=9$, $y=9$

$9=k\sqrt{9}$

$9=k(3)$

Divide both side by $3$

$\frac{9}{3}=\frac{k(3)}{3}$

$3=k$

$k=3$

$x=3\sqrt{y}$

Let's find $x$ when $y=\frac{16}{9}$

$x=3\sqrt{\frac{16}{9}}$

$x=3(\frac{4}{3})$

$x=4$

__Example 3__

$x\alpha y^3$. If $y=6$ when $x=24$. Determine $y$ when $x=192$

**Solution:**

$x=ky^3$

$y=6$ and $x=24$

$24=k6^3$

$24=k216$

Divide both side by $216$

$\frac{24}{216}=\frac{k(216)}{216}$

$\frac{1}{9}=k$

$k=\frac{1}{9}$

$x=\frac{y^3}{9}$

Let's solve for $y$ when $x=192$

$192=\frac{y^3}{9}$

Cross multiply

$192\times 9=y^3$

$1728=y^3$

To get rid of the cube, you add cube root to both side, remember that $(\sqrt[3]{x})^3=x$.....want to know why? read this post

$\sqrt[3]{1728}=(\sqrt[3]{y})^3$

$12=y$

$y=12$

__Example 4__

If $a$ varies directly as the cube roots of $b$ and $b=8$ when $a=\frac{2}{3}$. Find

1. The relationship between $x$ and $y$

2. $a$ when $b=27$

**Solution:**

$a\alpha \sqrt[3]{b}$

$a=k\sqrt[3]{b}$

$\frac{2}{3}=k\sqrt[3]{8}$

$\frac{2}{3}=k(2)$

Cross multiply

$6k=2$

Divide both sides by $6$

$\frac{6k}{6}=\frac{2}{6}$

$k=\frac{1}{3}$

The relationship connecting $a$ and $b$ is

$a=\frac{\sqrt[3]{b}}{3}$

When $b=27$

$a=\frac{\sqrt[3]{27}}{3}$

$a=\frac{3}{3}$

$a=1$

**Related post**__Example 5__

The time(t) taken by a pendulum to swing to and fro varies directly as the square root of the length(l). The length when the pendulum swing to and fro for 6 seconds is 9 meters

1. Determine the relation between $t$ and $l$

2. Find the length when the length is taken to swing to and fro is 8 meters.

**Solution:**

$t\alpha l$

$t=kl$

$t=6$, $l=9$

$6=k(9)$

$6=9k$

Divide both sides by $9$

$\frac{6}{9}=\frac{9k}{9}$

$\frac{2}{3}=k$

$k=\frac{2}{3}$

The relation between $t$ and $l$ is

$t=\frac{2l}{3}$

When $t=8$

$8=\frac{2l}{3}$

$24=2l$

Divide both sides by $2$

$\frac{24}{2}=\frac{2l}{2}$

$l=12$

So, the length when the pendulum swing to and fro for $6$ seconds is $12$.

We just started our 6-post series on variation. Indirect variation is next, so we would deal with it in this post.

Meanwhile, if you have questions, do well to ask our telegram community

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