DIRECT VARIATION — MEANING, EXPLANATION AND EXAMPLES

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 t is referred to as a variable.

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

There are four different types of variation, namely; Direct variation, indirect variation, joint variationpartial 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.

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