We can see from the above proportion that, when the value of $x$ is $2$, the value of $y$ is $3$ and when the value of $x$ is $4$, the value of $y$ is $3$ and so on.
So, whenever the value of $x$ increases, the value of $y$ also increases to a value such that their ratio does not change.
Here, we say that $y$ varies directly with $x$ and vice versa.
II. Proportionality Constant
When we have a series of ratio that are in proportion, as we saw before, instead of writing all possible values as being proportional we can simply bring in a proportionality constant. Let us take an example to understand this better.
Let us say we have a car moving at a constant speed. If its travels $45$ kms in $1$ hour, it means that it will travel $90$ kms in $2$ hours and $135$ kms in $3$ hours.
Using the concept of proportionality we can write,
$1:45::2:90::3:135$
But it doesn't end there. If you consider all the possible values this ratio can take, you will see that there are infinite possible pairs that can be reduced to any given ratio.
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To avoid all these things, we do a couple of things. First of all we say that distance travelled is proportional to the time travelled, that is:
$distance \propto time$ or, using shorter symbols,
$d \propto t$
Next, we bring in one more value, to change the proportional symbol to equality symbol, which is called the proportionality constant, and write this as:
$d = k \times t$ where $k$ is the proportionality constant.
In our example, $k$ is equal to $\dfrac{distance}{time}$, and is called the speed, which had to be constant for our proportionality relationship to be valid.
III. Direct Proportion
Whenever a quantity, $A$, increases or decreases with another related quantity, $B$, such that their ratio is fixed, we say $\unicode{0x201C}A\ varies\ directly\ with\ B\unicode{0x201D}$ or $\unicode{0x201C}A\ is\ directly\ proportional\ to\ B\unicode{0x201D}$
Mathematically we write it as:
$A \propto B$
Since the ratio $A:B$ is fixed, we can write $\dfrac{A}{B} = k$, or $A = kB$ where $k$ is called the proportionality constant.
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Similarly if we know that a quantity $y$ is directly proportional to the square of another quantity $x$, we can write:
$y \propto x^2$
$\texttip{\Rightarrow}{follows that} y = kx^2$, where, $k$ is the proportionality constant.
IV. Inverse Proportion
There are many cases where two quantities that are related to each other, vary in opposite directions, that is, when one increases the other decreases.
For example, if it takes $1\ hr$ to drive a certain distance at a speed of $20\ km/h$, it will take $\dfrac{1}{2}\ hr$ to cover the same distance if you drive at $40\ km/h$. When we increased the speed to twice, the time became half.
Similarly if you reduce the speed to half, that is $10\ km/hr$ it will take twice the time to cover the same distance.
When a quantity $A$ increases with the decrease of another quantity $B$ and decreases with an increase in $B$ we say that $\unicode{0x201C}A\ varies\ indirectly\ with\ B\unicode{0x201D}$ or $\unicode{0x201C}A\ is\ inversely\ proportional\ to\ B\unicode{0x201D}$
Mathematically we write it as:
$A \propto \dfrac{1}{B}$
Here, the product of the two quantities is constant, that is, $A\times B = k$
$\texttip{\therefore}{therefore} A = \dfrac{k}{B}$
where, $k$ is the proportionality constant.
As stated in the case of direct variation, indirect variation can also be with respect to a higher power of another quantity.
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For example, if we know that $y$ varies inversely to the square of $x$, we can write:
$y \propto \dfrac{1}{x^2}$
$\texttip{\Rightarrow}{follows that} y = \dfrac{k}{x^2}$
Let us look at some examples.
$\underline{Example\ 1:}$
Given that the distance covered is proportional to the square of time, under constant acceleration, with initial velocity equal to zero. The distance covered by a particle under a fixed acceleration in $3$ seconds is $18\ m$.
How much distance the particle will cover in $7$ seconds?
$d \propto t^2$
$\texttip{\Rightarrow}{follows that} d = kt^2$
We can find out $k$ from the given condition that the particle covers $18\ m$ in $3$ seconds.
$\texttip{\therefore}{therefore} 18 = k3^2$
$\texttip{\Rightarrow}{follows that} k = \dfrac{18}{9}$
$\texttip{\Rightarrow}{follows that} k = 2$
In $7$ seconds if the particle travels $d$ meters, then
$d = k7^2$
$\texttip{\Rightarrow}{follows that} d = 2 \times 7^2$
$\texttip{\Rightarrow}{follows that} d = 2 \times 49 = 98$
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$\underline{Example\ 2:}$
Given that the weight of a solid sphere is directly proportional to the cube of its radius.
If a solid sphere of some material of radius $2\ cm$ is $40$ gms, what will be the weight of a sphere with radius $5\ cm$?
From the given constant we can write:
$w \propto r^3$, where $w$ is the weight of the sphere and $r$ is the radius.
$\texttip{\Rightarrow}{follows that} w = kr^3$
Using the given input conditions,
$40 = k \times 2^3$
$\texttip{\Rightarrow}{follows that} 40 = k \times 8$
$\texttip{\Rightarrow}{follows that} k = 5$
For a sphere of the same material and a radius of $5\ cm$ the weight can be calculated using:
$w = k \times 5^3$
$\texttip{\Rightarrow}{follows that} w = 5 \times 5^3$
$\texttip{\therefore}{therefore} w = 5 \times 125 = 625$
$\underline{Example\ 3:}$
Given that brightness of light is inversely proportional to the square of the distance from the source of the light.
The brightness of a fixed source of light measured at a distance of $2\ m$ from the source is $12\ lux$. (Note that lux is a unit of brightness of light).
What will be the brightness of the same source of light measured at a distance of $5\ m$?
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Since brightness is inversely proportional to distance, we can write:
$b \propto \dfrac{1}{d^2}$, where $b$ is the brightness and $d$ is the distance from the source of light
$\texttip{\Rightarrow}{follows that} b = \dfrac{1}{d^2}$
We can substitute the given brightness and distance value from the given conditions, to get:
$12 = \dfrac{k}{2^2}$
$\texttip{\Rightarrow}{follows that} k = 12 \times 4 = 48$
We can calculate the brightness at $5\ m$ by using:
$b = \dfrac{k}{5^2} = \dfrac{48}{5^2}$
$\texttip{\Rightarrow}{follows that} b = \dfrac{48}{25} = 1.92\ lux$