Direct variation is also called Direct Proportion.
It is one of the many types of proportions which we have discussed in the previous chapter.
In this post we will learn about concept of direct variation, its definition and the examples.
Concept of Direct Variation (Direct Proportion)
Two entities are directly proportional when increase in one entity will result in increase in second entity or vice-versa by the same ratio.
Also in direct variation, the increase in entity of certain times will cause the other entity to increase/decrease by same times.
For Example
Let us consider two entities; Speed and Distance.
Examine the below data points:
In the above table you can observe that:
⟹ When the speed of car is 5 km/hr, the distance travelled is 10 km.
⟹ When the speed of car is increased to 10 km/hr, the distance covered also increased to 20 km.
Note that the distance & speed are maintaining the ratio 2 : 1.
It means that when speed is increased, the distance is increased proportionally to maintain ratio 2 : 1.
Apart from maintaining the same ratio, in direct variation, if one entity is increased by certain times, the other entity will also increase by that times.
In below figure when speed is multiplied by 2, we automatically get distance data also multiplied by 2.
Similarly when speed increased by three times causes distance to increase by three times.
Hence, from the above table we conclude that the entity speed and distance are directly proportional to each other.
Conclusion
Direct variation means scaling of one entity by certain number will result in scaling of other entity by same amount
How to represent Direct Variation
Direct Variation is represented by symbol “ \mathtt{\varpropto } “
Suppose two entities x & y are proportional to each other.
Mathematically, this can be represented as:
y \mathtt{\varpropto } x
Now we convert the expression into equation by replacing \mathtt{\varpropto } with constant k.
Where k is the constant ratio maintained by entity.
Here we conclude that proportional entities maintain a common ratio k.
So if entity ” x ” increases, the variable “y” will also increase by sufficient amount so that “k” remains constant.
Cases of Direct Proportion
(a) Value of constant K > 0
It means that if x increases then value of y also increases, and if x decreases, the value of y decreases.
For k=3, the equation of proportion will be, y = 3 . x
The graph will look like this:
Note that in above graph, as x increases, the value of y also increase.
(b) Value of constant K < 0
It means when x increases, the value of y decreases, and if x decreases, the value of y increases.
Let k = -3.
The equation becomes, y = -3x
Observe the graph of equation, y = -3x
Note as the value of x increases, the value of y decrease.
Let us see some examples for conceptual understanding.
Example of Direct Proportion
Example 01
Entity x & y are proportional to each other.
Given below is the data table for value of x & y.
Find the value of y.
Solution
Since, the entities x and y are proportional, they will maintain the constant ratio.
Given below is the table with ratio y / x.
Observe that the values of x and y maintain common ratio 5.
In proportional entities;
y / x = k
Putting given values of x & y
15 / 3 = k
k = 5
Finding value of y when x = 12
\mathtt{\frac{y}{x} \ =\ k}\\\ \\ \mathtt{\frac{y}{12} \ =\ 5}\\\ \\ \mathtt{y\ =\ 12\ \times \ 5}\\\ \\ \mathtt{y\ =\ 60}
Hence for x= 12, the value of y is 60.
Example 02
Given below is the table of fuel used and distance covered by the car. Check if entities are directly proportional or not.
Solution
If the entities are proportional, the values of both entities will maintain common ratio.
Observe the below table with Fuel/Distance ratio.
Note that for all the data points, there is a constant ratio value of 6.
Hence, both the entities fuel and distance are in proportion.
Example 03
Entities x & y are direct proportional to each other.
If x = 13, then y = 52.
find the value of y, if x = 16.
Solution
Its given that x & y are proportional.
\mathtt{y\ \varpropto \ x\ }\\\ \\ \mathtt{y\ =\ k\ x}
Putting the value of x & y
\mathtt{52\ =\ k\ .\ 13}\\\ \\ \mathtt{k\ =\ \frac{52}{13}}\\\ \\ \mathtt{k\ =\ 4}
Here y & x maintain the ratio of 4 : 1.
Its given that x = 16 and we have to find value of y.
Putting the value of x and k in below equation.
\mathtt{y\ =\ k\ .\ x}\\\ \\ \mathtt{y\ =\ 4\ .\ 16}\\\ \\ \mathtt{y\ =\ 64 }
Hence for x = 16, the value of y is 64.
Example 04
Study the table below and check if the given entities are directly proportional or not.
Solution
To check if the entities are in direct variation, we need to see if they are maintaining single ratio.
Given below is the table along with ratio of all data points.
You can observe that the ratios have different values.
It means the the given entities are not direct variation to each other.
Example 05
Check if the given entities are directly proportional or not.
(a) If they are proportional, find value of x if y = -7.
Solution
If the entities are in direct variation, then the ratio y/x will have same value throughout data point.
Given below is the table with ratio y/x
Note that for each data points the ratio is same (i.e. 0.5)
Hence, the given entities are in direct variation.
(a) Now find value of x, if y = -7
Since x & y are proportional, we can write;
\mathtt{y\ \varpropto \ x\ }\\\ \\ \mathtt{y\ =\ k\ .\ x}
We found value of k = 0.5 above.
Putting the value of y and x in the equation, we get;
\mathtt{-7\ =\ 0.5\ .\ x}\\\ \\ \mathtt{x\ =\ \frac{-7}{0.5}}\\\ \\ \mathtt{x\ =\ -14}
Hence, for y = -7, the value of x is -14.