I introduced variation last time, a topic in mathematics that attempts to find the relationship between two quantities. I demonstrated direct variation, the type where the positive movement of the value of one commodity results in the same positive movement of another commodity. We derived the mathematical expressions that show the relationships of two quantities in direct variation. Today, we will be looking at inverse or indirect variation.
Indirect variation describes the relationship between two commodities such that as the value of one goes up, the value of the other commodity goes down. In this type, the increase of the value of one quantity leads to the decrease of the value of the other. As one goes up, the other goes down. It can be in either direction for the two quantities involved.
There are real world examples of indirection variation showing two commodities whose values go in opposite direction. Below are some specific examples:
Time and number of workers: The more people involved in doing a particular work, the less time it takes. The lesser the number of people doing a particular work, the more time it takes. For example, if it took 5 hours for 20 persons to complete a farm project, it will take 40 persons half the time it took 20 persons to do the work.
Speed and time: The faster you travel, the less time it takes to get to your definition. If it took 1 hour to get to your destination at a speed of 40km/hr, it will take less than 1 hour to reach your destination at a speed of 70km/hr.
See-saw: When two children play on a see-saw, both cannot go up at the same time. When one goes up, the other goes down and vice versa.
The above are just 3 real-world examples of inverse variation and that makes it easy for you to understand the subject a little. Now we are going to see how to derive the formula for solving inverse variations.
Assuming the time (t) it takes to do a particular work varies inversely as the number (n) of workers, the we can represent it mathematically as:
t ∝ 1/n
Introducing our constant which will be represented by letter c, we have:
t ∝ 1/n
t ∝ c/n
We will then use the formula t ∝ c/n to find the relationship between number of workers and the time it takes them to finish a particular task.
Inverse variations are solved same way as direct variation. The same two steps below:
Example 1: If A varies inversely as the cube root of B. A = 7 when B = 8. Find A when B = 64 and B when A = 4.
Solution:
A ∝ 1/(∛B)
A = k/(∛B) (Lets use k to represent the constant)
Solving for the constant k, using the initial values, we have
7=k/(∛8) (Solving the bracket first)
7=k/2 (Making k the subject)
k = 7x2 = 14
Finding A when B=64, we have
A = k/(∛B)
A=14/(∛64)(Expanding the bracket)
A=14/4
A=7/2 or 3.5
Solving for B when A=4
A = k/(∛B)
4=14/(∛B)(Making B the subject)
(∛B)x4=14
(∛B)=14/4(Removing the cube root sign by introducing cube sign on the right sid)
B=(14/4)3
B=2744/64 (reducing to lowest fraction, we have)
B=343/8
Example 2: C varies inversely as D2. C=3 when D=2. Find C when D=2 and D when C=8.
Solution:
Solving for our constant k with the initial values given
C ∝ 1/D2(Introducing our constant k)
C = k/D2
3 = k/22(Expanding the powers)
3= k/4 (Making k the subject)
k=3x4
k=12
Finding C when D=2, we have
C= 12/22(Expanding the power)
C = 12/4
C = 3
Finding D when C=8, we have
8 = 12/D2(Making D the subject)
8D2 = 12
D2 = 12/8 (Introducing square root to cancel the square sign)
D = √12/8 (Simplifying more)
D = 0.43
Inverse variation is not such a complex mathematics. It tries to find the relationship between two commodities in which one goes up as the other goes down and vice versa. Once you are able to find the constant, you could easily solve for the changing values of either quantities using the methods i have shown above.
