PARTIAL VARIATION

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Partial variation consists of two or more parts added together.

Here are some illustrations:

if $x$ is partly constant and partly varies directly as $y$, then it is written as $x=a+ky$, where ($a$ and $k$ are constant).

If $u$ is partly constant and partly varies inversely as $v$ then it is written as $u=a+\frac{k}{y}$, where $a$ and $k$ are constant.

If $b$ varies partly as $c$ and partly as the square of d, then it is written as $b=ac+kd^2$, Where $a$ and $k$ are constant.

If $x$ is partly constant and varies indirectly as the product of $y$ and $z$, then it is written as $x=a+\frac{k}{yz}$, where $a$ and $k$ are constant.

๐Ÿ‘‰This post extends our posts on variation ๐Ÿ‘ˆ

Example 1

$x$ is partly constant and partly varies as $y$. If $x=45$ when $y=10$ and $x=87$ when $y=24$. Find

1. the formula connecting $x$ and $y$

2. $x$ when $y=18$

Solution:

$x=a+ky$

$x=45$, and $y=10$

$45=a+10k$............eqn 1

$x=87$ and $y=24$

$87=a+24k$...........eqn 2

To solve simultaneously, substract eqn 1 from eqn 2

$87=a+24k$

$-(45=a+10k)$

$42=14k$

Divide both side by $14$

$\frac{42}{14}=\frac{14k}{14}$

$3=k$

$k=3$

To find $a$, we substitute $3$ for $k$ in eqn 1

$45=a+10(3)$

$45=a+30$

$45-30=a$

$15=a$

$a=15$

The formula connecting $x$ and $y$ is:

$x=15+3y$

Let's find $x$ when $y=18$

$x=15+3(18)$

$x=15+54$

$x=69$

Example 2

A variable ($x$) varies directly as $y$ and inversely as the square of $y$. Given that $x=11$ when $y=2$ and $x=\frac{629}{25}$ when $y=5$, find $x$ when $y=7$

Solution:

$x=ay+\frac{k}{y^2}$

When $x=11$, $y=2$ 

$11=2a+\frac{k}{2^2}$

$11=2a+\frac{k}{4}$

Taking the L.C.M

$11=\frac{8a+k}{4}$

Cross multiply

$44=8a+k$.......eqn 1

When $x=\frac{629}{25}$, $y=5$

$\frac{629}{25}=5a+\frac{k}{5^2}$

$\frac{629}{25}=5a+\frac{k}{25}$

Taking the L.C.M

$\frac{629}{25}=\frac{125a+k}{25}$

For simplicity, let's Multiply both side by $25$

$\frac{629}{25}\times25=\frac{125a+k}{25}\times 25$

$629=125a+k$.....eqn 2

Subtracting eqn 1 from eqn 2

$629=125a+k$

$-(44=8a+k$

$585=117a$

Divide both sides by $117$

$\frac{585}{117}=\frac{117a}{117}$

$a=5$

Substituting $5$ for a in eqn 1

$44=8(5)+k

$44=40+k$

$44-40=k$

$4=k$

$k=4$

$x=5y+\frac{4}{y^2}$

Now, let's find $x$ when $y=7$

$x=5(7)+\frac{4}{7^2}$

$x=35+\frac{4}{49}$

Taking the L.C.M

$x=\frac{1715+4}{49}$

$x=\frac{1719}{49}$

$x=35.1$

Example 3

$b$ is partly constant and partly varies directly as $c$ when $c=5$, $b=7$, and when $c=7$, $b=7$

1. Find the equation connecting $x$ and $y$

2. Find $b$ when $c=11$

Solution:

$b=a+kc$

$c=5$, $b=7$

$7=a+k5$.......eqn 1

$c=7$, $b=7$

$8=a+k7$.......eqn 2

Substract eqn 1 from eqn 2

$8=a+k7$

$-(7=a+k5)$

$1=2k$

Divide both side by $2$

$\frac{1}{2}=\frac{2k}{2}$

$\frac{1}{2}=k$

$k=\frac{1}{2}$

To solve for $a$, let's substitute the value of $k$ into eqn 1

$7=a+(\frac{1}{2})5$

$7=a+\frac{5}{2}$

Taking the L.C.M 

$7=\frac{2a+5}{2}$

Cross multiply

$7\times2=2a+5$

$14=2a+5$

$14-5=2a$

$9=2a$

Divide both side by $2$

$\frac{9}{2}=\frac{2a}{2}$

$\frac{9}{2}=a$

$a=\frac{9}{2}$

Therefore, the equation connecting $b$ and $c$ is:

$b=\frac{9}{2}+\frac{c}{2}$

Finding $b$ when $c=11$

$b=\frac{9}{2}+\frac{11}{2}$

Taking the L.C.M

$b=\frac{9+11}{2}$

$b=\frac{20}{2}$

$b=10$

Related posts

Example 4

$p$ is partly constant and partly varies indirectly as the cube root of $q$. When $q=125$, $p=35$ and when $q=8000$, $p=20$. Find

1. The relation between $p$ and $q$

2. Find $q$ when $p=215$

Solution:

$p=a+\frac{k}{\sqrt[3]{q}}$

$p=35$, $q=125$, 

$35=a+\frac{k}{\sqrt[3]{125}}$

$35=a+\frac{k}{5}$

Taking the L.C.M

$35=\frac{5a+k}{5}$

Cross multiply

$35\times5=5a+k$

$175=5a+k$...........eqn 1

$q=8000$, $p=20$

$2t0=a+\frac{k}{\sqrt[3]{8000}}$

$20=a+\frac{k}{20}$

Taking the L.C.M of both sides

$20=\frac{20a+k}{20}$

Cross multiply

$20\times20=20a+k$

$400=20a+k$......... eqn 2

Substract eqn 1 from eqn 2

$400=20a+k$

$-(175=5a+k$

$225=15a$

Divide both side by 15

$\frac{225}{15}=\frac{15a}{15}$

$15=a$

$a=15$

To solve for $k$, you substitute $15$ for $k$ in eqn 1

$175=5(15)+k$

$175=75+k$

$175-75=k$

$100=k$

$k=100$

The relation between $p$ and $q$ is

$p=15+\frac{100}{\sqrt[3]{q}}$

Solving for $q$ when $p=215$

$215=15+\frac{100}{\sqrt[3]{q}}$

Taking the L.C.M

$215=\frac{15\sqrt[3]{q}+100}{\sqrt[3]{q}}$

Cross multiply

$215\times\sqrt[3]{q}=15\sqrt[3]{q}+100$

$215\sqrt[3]{q}=15\sqrt[3]{q}+100$

Collect like terms

$215\sqrt[3]{q}_15\sqrt{3}{q}=100$

$200\sqrt[3]{q}=100$

Divide both side by $200$

$\frac{200\sqrt[3]{q}}{200}=\frac{100}{200}$

$\sqrt[3]{q}=\frac{1}{2}$

To get rid of cube root, we add cube both sides.

$(\sqrt[3]{q})^3=(\frac{1}{2})^3$

$q=\frac{1}{8}$

There you have it. We would continue our series on variations when we explore Word problems on direct and indirect variations.

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